Title: Math and science / Topology
Tags: todo

Title: Metric space
Q: Metric
A metric \( d \) on a set \( X \) is a function [\( d : \text{what} \to \text{ to what?} \)] with the following three properties:
What is this called?
\( d(x, y) = 0 \iff x = y, \text{ for all } x, y \in X \)
Triangle inequality
[...]
Symmetry
\( d(x, y) = d(y, x), \text{ for all } x, y \in X \)
A metric space is a set together with a metric on it, or more formally, a pair \( (X, d) \) where \( X \) is a set and \( d \) is a metric on \( X \).
A: \( d : X \times X \to [0, \infty) \); \( d(x, y) + d(y, z) \ge d(x, z), \text{ for all } x, y, z \in X \)
Example: Some interesting metrics
1. \( d_1, d_2, d_\infty \) (instances of \( d_p \)) on \( \mathbb{R}^n \)
For all \( x = (x_1, ..., x_n) \) and \( y = (y_1, ..., y_n) \) in \( \mathbb{R}^n \) (or any subset thereof):
- The Euclidean metric, \( d_2 \) on \( \mathbb{R}^n \) is given by: \( d_2(x, y) = (\sum_{i=1}^{n} (x_i - y_i)^2)^{\frac{1}{2}} \)
- \( d_1 \) is given by: \( d_1(x, y) = \sum_{i=1}^{n}|x_i - y_i| \)
- \( d_\infty \) by: \( d_\infty(x, y) = max_{1 \le i \le n}|x_i - y_i| \)
All these metrics are particular cases of the \( d_p \) metric.
2. \( d_1, d_2, d_\infty \) (instances of \( d_p \)) on \( C[a, b] \))
All \( d_p \) metrics have parallels over function spaces. \( C[a, b] \) is the set of continuous functions \( [a,b] \to \mathbb{R} \). Three interesting metrics on \( C[a, b] \) are given by:
\[ d_1(f, g) = \int_{a}^{b} |f(t) - g(t)|dt \]
\[ d_2(f, g) = \left( \int_{a}^{b} |f(t) - g(t)|dt \right)^{\frac{1}{2}} \]
\[ d_\infty(f, g) = \sup_{a \le t \le b}|f(t) - g(t)| \]
3. Hamming metric
With \( A \) being any set, and \( x, y \in A^n \), the Hamming metric \( d \) on \( A \) is given by
\( d(x, y) = |\{i \in \{1, ..., n\} : x_i \ne y_i \}| \)
In other words, the count of how many coordinates differ.
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Order: 1.1.1

Title: Metric space. ε-balls
Q: Let \( X \) be a metric space, let \( x \in X \) and let \( \varepsilon > 0 \) be a real. The open ε-ball around \( x \) (or in more detail, the open ball around \( x \) of radius \( \varepsilon \)) is the subset of \( X \) given by
[\( B(x, \varepsilon) = \{y \in X : \text{what condition?} \} \)]
Similarly, the closed ε-ball around \( x \) is
[\( \bar{B}(x, \varepsilon) = \{y \in X : \text{what condition?} \} \)]
A: \(B(x, \varepsilon) = \{y \in X : d(x, y) < \varepsilon \} \); \(\bar{B}(x, \varepsilon) = \{y \in X : d(x, y) \le \varepsilon \} \)
BackOnlyText: The definition of open sets in metric spaces is formulated in terms of open ε-balls.
Example: Metrics such as \( d_1 \), \( d_2 \) and \( d_\infty \) for both \( \mathbb{R}^n \) and continuous function spaces \( C[a, b] \), and other metrics such as Hamming distance, each induce ε-balls within the space they are applied in. It is interesting to consider what the ε-ball in each metric space is.
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Order: 1.1.2

Title: Metric space. Open and closed sets
Q: Open and closed subsets
Let \( X \) be a metric space.
Open in \( X \)
A subset \( U \) of \( X \) is open in \( X \) iff [...].
Closed in \( X \)
A subset \( V \) of \( X \) is closed in \( X \) iff [...].
Tom Leinster describes the openness of \( U \):
Thus, \( U \) is open if every point of \( U \) has some elbow room—it can move a little bit in each direction without leaving \( U \).
Personally, I like the phrase: every element of an open set has a neighbourhood.
ε-balls
Open ε-balls are open, and closed ε-balls are closed. Consider trying to prove this. They are open on account of the definition of openness, not by their own definition alone, despite their names being suggestive.
A: for all \( u \in U \), there exists an \( \varepsilon > 0 \) such that \( B(u, \varepsilon) \subseteq U \); \( X \setminus V \) is open in \( X \)
BackOnlyText: Open, closed, neither or both
Subsets of a metric space can be any combination of open and closed.
Open subset, closed subset
There's no such thing as an open set that is not an open subset; \( U \) is open in \( X \). The importance of this is that a set \( U \) may be open in one space and not open in another. \( [0, 1) \) is not open in \( \mathbb{R} \) but is open in \( [0, 2] \).
Source: Tom Leinster's Topology notes
https://www.maths.ed.ac.uk/~tl/topology/topology_notes.pdf
Comic:
https://abstrusegoose.com/394
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Order: 1.1.3

Title: Metric space. Unions and intersections of open sets
- An arbitrary union of open subsets is open.
- A finite intersection of open subsets is open.
- Special cases: \( \emptyset \) and \( X \) are both open and closed.
BackOnlyText: Lemma. Union and intersection of open sets
Let \( X \) be a metric space.
- Let \( (U_i)_{i \in I} \) be a family (finite or not) of open subsets of \( X \). Then \( \bigcup_{i\in I} U_i \) is also open in X.
- Let \( U_1 \) and \( U_2 \) be open subsets of \( X \). Then \( U_1 \cap U_2 \) is also open in X.
- Special cases \( \emptyset \) and \( X \) are open also.
The mirror statements for closed sets are as follows.
Lemma. Union and intersection of closed subsets
Let \( X \) be a metric space.
- Let \( V_1 \) and \( V_2 \) be closed subsets of \( X \). Then \( V_1 \cup V_2 \) is also closed in X.
- Let \( (V_i)_{i \in I} \) be a family (finite or not) of closed subsets of \( X \). Then \( \bigcap_{i\in I} V_i \) is also open in X.
Special cases \( \emptyset \) and \( X \) are also closed.
Source: Tom Leinster's notes
https://www.maths.ed.ac.uk/~tl/topology/topology_notes.pdf
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Order: 1.1.4

Title: Metric space. Sequence convergence
Sequence convergence, definition
Let \( X \) be a metric space, let \( x_0 \in X \) and let \( (a_n)_{n=1}^{\infty} \) be a sequence in \( X \).
\( (a_n)_{n=1}^{\infty} \) is said to converge to \( x_0 \) iff
\( \lim_{n \to \infty} d(x_0, a_n) = 0 \)
Unfolding the definition of a limit this becomes: \( (a_n)_{n=1}^{\infty} \) converges to \( x_0 \) iff for any \( \varepsilon > 0 \) there exists an \( N \ge 1 \) such that for all \( i \ge N \), \( d(x_0, a_i) < \varepsilon \).
This next result should be internalized.
Lemma. Closed iff sequences converge inside
Let \( X \) be a metric space and \( V \subseteq X \).
Then \( V \) is closed in \( X \) \( \iff \) for all sequences \( (x_n) \) in \( V \) and all \( x \) in \( X \), if \( (x_n) \) converges to \( x \) then \( x \in X \).
The proof was not immediately obvious to me. I did, however, find a way to visualize it, which I think makes the line of reasoning clear. Part of why the proof is important, I think, is that there are only two notions available from which the result needs to be derived: the definition of open sets and the definition of sequence convergence. So, it should be simple, yet it escapes an immediately obvious proof. Thus, I feel there is a mode of thinking that has its essence somewhat distilled into the line of reasoning.
BackOnlyText: The forward part of the proof corresponds to the idea that if you choose any point not within \( V \), then it will be in an open set, and thus, there is always some ε-ball around the point that is fully contained within the enclosing open set.
The reverse part of the proof corresponds to the idea that if \( V \) is not closed, then \( X \setminus V \) is not open, so that, somewhere on the boundary between \( V \) and \( X \setminus V \), there must be a "hard" edge on the \( X \setminus V \) side of the boundary and a "soft"/open edge on the \( V \) side. Because of this, any ε-ball around this point must spill into \( V \). By choosing successively smaller ε-balls, we can make a sequence in \( V \) that converges to the point in \( X \setminus V \).
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Order: 1.1.5

Title: Metric space. Function continuity
Function continuity between metric spaces
Let \( X \) and \( Y \) be metric spaces and let \( f: X \to Y \) be a function. We say that \( f \) is continuous iff:
For all \( x_0 \in X \) and for all \( \varepsilon > 0 \) there exists a \( \delta > 0 \) such that (\( x \in B(x_0, \delta) \) \( \implies \) \( f(x) \in B(f(x_0), \varepsilon) \)).
Independence of metrics
The idea of continuity appears to depend on the notion of metric/distance as it is formulated using ε-balls. However, the following lemma reveals that this is not really so—all that is needed is the notion of open (or closed) subsets.
Equivance to function continuity
Let \( X \) and \( Y \) be metric spaces and let \( f : X \to Y \) be a function. The the following three statements are equivalent:
1. \( f \) is continuous;
2. for all open \( U \subseteq Y \), the preimage \( f^{-1}U \subseteq X \) is open;
3. for all closed \( V \subseteq Y \), the preimage \( f^{-1}V \subseteq X \) is closed.
Informally: 'the preimage of an open set is open'.
Proof at the bottom of the back side.
This result motivates the definition of a topolgical space.
BackOnlyText: The transition to Topology
With function continuity as the target concept, what is the minimum set of ideas requiring construction before function continuity can be formulated?
Without introducing the idea of a metric function, a space which we will call a topological space will be defined as being a set \( X \) paired with a collection of subsets, to be called open subsets. These subsets will be defined to satisfy the 3 properties that were previously derived for metric spaces:
1. Arbitrary unions of open subsets are open.
2. Finite intersections of open subsets are open.
3. \( \emptyset \) and \( X \) are open.
This formulation could equally have been done with closed rather than open subsets.
After this restructuring, a function will be defined to be continuous iff 'the preimage of an open set is open'.
Equivalence of Continuity, proof
Below is a proof for metric spaces of the equivalence of the three statements above regarding continuity. Below again is an attempt to visualize the line of reasoning.
The main 'trick' of the proof is to iterate through each \( y \in Y \). Consider the ε-ball around each \( y \) that must be within \( U \), and then, by the assumption of our definition of continuity applying to \( f \), we can create an ε-ball around every \( x' = f^{-1}(y) \). The the union of all these ε-balls in \( X \) must be open.
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Order: 1.1.6

Title: Topology and topological space. Definition
A topology on a set \( X \) is a collection \( \mathcal{T} \) of subsets of \( X \) with the following properties:
T1
Whenever \( (U_i)_{i \in I} \) is a family (finite or not) of subsets of \( X \) such that \( U_i \in \mathcal{T} \) for all \( i \in I \), then  \( \bigcup_{i \in I} U_i \in \mathcal{T} \).
T2
Whenever \( U_1, U_2 \in \mathcal{T} \), then \( U_1 \cap U_2 \in \mathcal{T} \).
T3
\( \emptyset \in \mathcal{T} \) and \( X \in \mathcal{T} \).
A topological space \( (X, \mathcal{T}) \) is a pair consisting of a set \( X \) together with a topology \( \mathcal{T} \) on \( X \).
T3 can be derived from T1 and T2, and is not strictly necessary as an axiom; however, most treatments of the subject introduce the definition as this triplet of statements.
In relaxed words
T1 can be phrased as 'an arbitrary union of open subsets is open'. By induction T2 can be phrased as 'a finite intersection of open subsets is open'.
Combining these two then, a topology on \( X \) can be defined as 'a collection of subsets of \( X \) that is closed under arbitrary unions and finite intersections'.
Example: Discrete and indiscrete topologies
The discrete and indiscrete topologies are two important topologies (which can be defined for any set). With \( X \) being a set, then they are defined like so:
- the collection of all subsets of \( X \) is a topology on \( X \); it is called the discrete topology.
- the collection of subsets of \( X \) consisting of \( X \) and \( \emptyset \) is a topology on \( X \); it is called the indiscrete topology.
The name 'discrete' is intended to suggest indivisibility of parts—it is not possible to arrive at a new topology by dividing elements of the discrete topology into smaller subsets (those smaller subsets must already be in the discrete topology).
Finite complement topology
Let \( X \) be a set. The collection \( (U_i)_{i \in I} \) of subsets of \( X \) such that \( X - U_i \) is finite for all \( i \in I \) forms a topology on \( X \); it is called the finite complement topology. It is worth checking that this collection of subsets does indeed meet the criteria for being a topology.
Countable complement topology
A similar construction to the finite complement topology, the countable complement topolgy is the collection of subsets of a set \( X \) such that for each \( U \) in the collection, \( X - U \) is either countable (and thus infinite) or all of \( X \). Again, it's worth checking this is indeed a topology.
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Order: 1.1.7

Title: Finer and coarser topologies
We say that one topology \( \mathcal{T} \) on \( X \) is finer than another topology \( \mathcal{T}' \) on \( X \) iff every member of \( \mathcal{T}' \) is a member of \( \mathcal{T} \). \( \mathcal{T}' \) is said to be coarser than \( \mathcal{T} \).
Stronger and weaker are alternative terminology for finer and coarser.
\( \mathcal{T} \) is strictly finer iff \( \mathcal{T}' \) is a proper subset.
Symbolically,
- \( \mathcal{T} \) is finer than \( \mathcal{T}' \) ⟺ \( \mathcal{T}' \subseteq \mathcal{T} \).
- \( \mathcal{T} \) is strictly finer than \( \mathcal{T}' \) ⟺ \( \mathcal{T}' \subset \mathcal{T} \).
BackOnlyText: Lemma. A basis for every basis.
Let \( X \) be a set. Let \( \mathcal{T} \) and \( \mathcal{T}' \) be topologies on \( X \) with bases \( \mathscr{B} \) and \( \mathscr{B}' \) respectively. Then \( \mathcal{T} \) is finer than \( \mathcal{T}' \) iff
for each \( x \in X \) and each basis element \( B' \in \mathscr{B'} \) containing \( x \), there is a basis element \( B \in \mathscr{B} \) such that \( x \in B \subseteq B' \).
It should be easy to prove this. Refer to Munkres for a proof example.
By this lemma, it can be seen how a topology induced by one metric can be equivalent to a topology induced by another.
A basis for every basis: alternative perspective
I think a better way of phasing the second part of the iff relationship is to say: every \( b' \in \mathscr{B'} \) can be expressed as a union of elements \( b \in \mathscr{B} \).
Example: For a set \( X \), the discrete topology is the finest topology while the indiscrete topology is the coarsest topology.
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Order: 1.1.8

Title: Neighbourhoods
Q: Let \( X \) be a set with topology \( \mathcal{T} \). A neighbourhood of \( x \) is [a something that something].
This is the definition according to Munkres. Tom Leinster gives a different definition that distinguishes between a 'neighbourhood' and an 'open neighbourhood'.
Neighbourhood is the closest we get to the metric space idea of open balls around a point.
The term neighbourhood packs a noun-verb pair into a noun, which I think is part of why using it makes it easier to conceptualize compared to 'open set containing \( x \)'.
A: an open subset of \( X \) containing \( x \)
BackOnlyText: The next lemma is of interest because it is a lemma expressed in the terms of topology that is in some sense an isomorphism of a definition that appeared when formulating open sets in metric spaces. When defining open sets of a topology, for our definition we took the metric space's derived result of open sets being closed under arbitrary union and finite intersection. Now we have come to a derived result of topological spaces that matches the founding definition of open sets in metric spaces. Isn't that an interesting loop.
Lemma. Open iff all elements have a neighbourhood subset
Let \( X \) be a topological space and \( U \subseteq X \). Then \( U \) is open in \( X \) iff for all \( x \in U \), there is a neighbourhood of \( x \) contained in \( U \).
For comparison, the definition of open sets for a metric space is copied below:
Metric space: Open subsets, definition
Let \( X \) be a metric space and \( U \subseteq X \). Then \( U \) is open in \( X \) iff for all \( u \in U \), there is a \( \varepsilon > 0 \) such that \( B(u, \varepsilon) \subseteq U \).
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Order: 1.1.9

Title: Closed subsets
Closed subsets, definition
Let \( X = (X, \mathcal{T}) \) be a topological space. A subset \( V \subseteq X \) is closed iff \( X \setminus V \in \mathcal{T} \).
The below properties of closed sets can be derived.
Lemma. Finite unions and arbitrary intersections
Let \( X = (X, \mathcal{T}) \) be a topological space. Then
1. Whenever \( V_1 \) and \( V_2 \) are closed subsets of \( X \), then \( V_1 \cup V_2 \) is also closed in X.
2. Whenever \( (V_i)_{i\in I} \) is a family (finite or not) of closed subsets of \( X \), then \( \bigcap_{i \in I}V_i \) is closed in \( X \).
3. \( \emptyset \) and \( X \) are closed subsets of \( X \).
This lemma follows from the application of de Morgan's laws to the definition of a topological space.
BackOnlyText: Sense of the word 'closed'
In several areas of mathematics 'closed' appears and describes some sense of an opperation on elements of a set not taking us out of thas set. For example, a vector space is 'closed' under addition and scalar multiplication. In analysis, the operation we are concerned about is the limit: a set is closed if all convergent sequences of elements of the set is another element of the set. A limit point of a set is a point that is the limit of some sequence of elements of the set. Only closed sets contain all of their own such limits.
Degrees of fundamentalness
Closed sets and the idea of closedness should be viewed as equally fundamental compared to open sets and openness. While closed sets are presented as a derivative concept when a topology is defined from open sets, the formulation could have proceeded the other way, with closed sets coming first. Kuratowski's closure axioms are an example of such a formulation.
This point is pertinent to other concepts in topology also: while it is elegant to formulate ideas of closedness, convergence, continuity, and so forth in terms of open sets, it is somewhat a distortion of the formulation process to deem one more fundamental than the other. The use of openness simply affords elegant constructions of the other concepts. There is an enlightening MathOverflow topic touching on this idea which has had mathematicians like Terrance Tao and Kevin Buzzard weigh in.
Example: Cantor set
The Cantor set is a closed subset of \( \mathbb{R} \); it acts as a reminded of the complexity or strangeness possible for closed sets.
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Order: 1.1.10

Title: Basis
Instead of specifying all possible open sets of a topology, it is convenient to be able to specify the topology in terms of a smaller set. Analogously, for metric spaces, the set of open balls could be used to describe a metric space instead of directly specifying the arbitrary open sets. For topological spaces, a basis carries out this role (plural: bases).
Basis, definition
Let \( X \) be a topological space. A basis for \( X \) is a collection \( \mathscr{B} \) of open subsets of \( X \) such that every open subset of \( X \) is a union of sets in \( \mathscr{B} \).
Two consequence of this definition are:
- Every element of \( X \) has at least one neighbourhood that is a basis element.
- the intersection of any two basis elements must be a basis element.
A bit more formally, these two consequences translate to:
Lemma.
Let \( X \) be a topological space, and let \( \mathscr{B} \) be a basis for \( X \). Then:
1. For each \( x \in X \), there is at least one basis element \( B \) containing \( x \).
2. If \( x \in B_1 \cap B_2 \) where \( B_1 \) and \( B_2 \) are basis elements, then \( x \in B_3 \) for some basis element \( B_3 \subseteq B_1 \cap B_2 \).
BackOnlyText: The choice of definition and lemma
The wording of this lemma is presented in line with Munkres's presentation. Munkres, however, presents this lemma as the definition of a basis; he then proceeds to derive the content of the definition that I've presented above. Leinster proceeds in the opposite order, as presented here. I think that the motivation of a basis is clearer from Leinster's definition, whereas the lemma presented by Leinster (definition for Munkres) is more useful to work with. I followed Munkres's wording for the lemma as I feel it encourages thinking in terms of individual elements of \( X \) which so far has been useful in proofs.
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Order: 1.1.15

Title: Subbasis
Q: Let \( X \) be a set. A subbasis for a topology on \( X \) is a collection of subsets of \( X \) whose union equals \( X \).
Let \( \mathcal{S} \) be a subbasis for a topology on \( X \). The set of all [something of something of something] is a topolgy on \( X \). This topology is said to be generated by the subbasis \( \mathcal{S} \).
It can be proved that the generated set forms a valid topology.
A: unions of finite intersections of elements of \( \mathcal{S} \)
BackOnlyText: Basis vs subbasis
A subbasis is an even more lightweight description of a topology compared to a basis. A subbasis can generate a basis for the same topology: the collection of all finite intersections of subbasis elements forms a basis.
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Order: 1.1.16

Title: Hausdorff spaces
Q: Most interesting topological spaces are Hausdorff. Hausdorffness is identified as being the second separation condition. The first condition, T1 is a sub-requirement of T2 (Housdorff), so it is useful to keep it in mind when thinking about the Hausdorff condition.
T1
A topological space \( X \) is said to be \( T_1 \) iff [...].
Now for Hausdorff.
Hausdorff
A topological space \( X \) is Hausdorff (or \( T_2 \)) iff [...].
Lemma. Every Hausdorff space is \( T_1 \).
A: every one-element subset of \( X \) is closed; for every distinct \( x, y \in X \), there exists disjoint neighbourhoods of \( x \) and \( y \)
BackOnlyText: Slightly more precise wording of the Hausdorff condition:
A topological space \( X \) is Hausdorff (or \( T_2 \)) iff for every \( x, y \in X \) such that \( x \ne y \), there exists disjoint open sets \( U, W \) of \( X \) such that \( x \in U \) and \( y \in W \).
Example: Every metrizable space is Hausdorff.
While most interesting spaces are Hausdorff, there are some non-Hausdorff spaces that are important. The Zariski topology is an example.
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Order: 1.1.11

Title: Sequence convergence
Q: Sequence convergence
Let \( X \) be a topological space, let \( (x_n) \) be a sequence in \( X \) and let \( x_0 \in X \). Then \( (x_n) \) converges to \( x_0 \) iff [for every something there is something that something].
A: for every neighbourhood \( U \) of \( x_0 \) there is an \( N \ge 1 \) such that for every \( n \ge N \), \( x_n \in U \)
BackOnlyText: In other words, every neighbourhood of the point to which a sequence converges must contain the whole tail end of the sequences starting from some arbitrary point in the sequence.
It is a good exercise to check that this definition carries the expected meaning for metric spaces.
Example: Sequences in Hausdorff spaces
A common demonstration of the importance of Hausdorff spaces:
Let \( X \) be a Hausdorff topological space. Then each sequence in \( X \) converges to at most one point.
Another good proof exercise.
Multiple convergence
An example where we get multiple convergence:
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Order: 1.1.12

Title: Continuous maps
Continuous maps
Let \( X \) and \( Y \) be topological spaces. A function \( f : X \to Y \) is continuous iff for every open subset \( U \) of \( Y \), the preimage \( f^{-1}U \) is open in \( X \).
In short, continuity means that the preimage of an open set is open.
Some results
Continuous maps preserve convergence of sequences.
Let \( f : X \to Y \) be a continuous map, and let \( (x_n) \) be a sequence in \( X \) converging to \( x \in X \); then the sequence \( (f(x_n)) \) in \( Y \) converges to \( f(x) \in Y \).
In metric spaces this lemma is an if and only if statement, whereas for topological spaces we are restricted to only the forward implication above; it is possible to construct discontinuous maps of topological spaces that, nevertheless, preserve convergence of sequences.
The composite of continuous maps is continuous.
The inverse of a continuous bijection need not be continuous. Details below.
Munkres presents three statements that are equivalent to stating that a function is continuous:
Continuity equivalences
Let \( X \) and \( Y \) be topological spaces and let \( f : X \to Y \) be a function. The the following are equivalent:
1. \( f \) is continuous.
2. For every subset \( A \) of \( X \), one has \( f(\bar{A}) \subseteq \overline{f(A)} \).
3. For every closed set \( B \) of \( Y \), the set \( f^{-1}(B) \) is closed in \( X \).
4. For each \( x \in X \) and each neighbourhood \( V \) of \( f(x) \), there is a  a neighbourhood \( U \) of \( x \) such that \( f(U) \subseteq V \).
BackOnlyText: Continuous maps preserve convergence, proof.
The sequence \( (x_n) \) in \( X \) converges to \( \lim_{n \to \infty} x_n \). Any neighbourhood \( B \subseteq Y \) of \( f(\lim_{n \to \infty} x_n) \) has an open pre-image \( A \subseteq X \) that is a neighbourhood of \( \lim_{n \to \infty} x_n \). As \( A \) is a neighbourhood of the limit of the sequence, it must fully contain some tail of the sequence (\( x_n \) for all \( n \ge N \) for some \( N > 0 \)). The image of this tail must fit within \( B \). And so, \( B \) contains a tail of \( (f(x_n))_n \). This is sufficient to conclude that \( \lim_{n \to \infty} f(x_n) \) converges to \( f(\lim_{n \to \infty} x_n) \).
The 4 equivalent statements for continuity
Some remarks about these equivalences:
- #3 implies that continuity could have been defined in terms of closed sets, inplace of opensets; more support that closedness is of equal fundamentalness to openness.
- #4 is the form in which continuity is defined for metric spaces! Just replace notions of neighbourhoods with open ε-balls.
- #2 is presented later on in Leinster's notes. In words it is saying that a sufficient and necessary requirement for a function \( f \) to be continuous is that \( f \) must map points just outside of any set \( A \) to points only just outside \( fA \).
- #2 can be unfolded (and a little bit of causality injected) to say:
no open set in \( X \) can help intersecting \( A \) when containing \( x \) ⟺ no open set in \( Y \) can help intersecting \( fA \) when containing \( f(x) \).
Warning regarding inverse
The inverse of a continuous bijection need not be continuous.
An easy way of constructing an example is to consider continuous maps to a set with the discrete topology. For example, the identity map \( i : (\mathbb{R}, \text{discrete topology}) \to (\mathbb{R}, \text{standard topology}) \) is continuous, but the inverse, \( \) \( i^{-1} : (\mathbb{R}, \text{standard topology}) \to (\mathbb{R}, \text{discrete topology}) \) is not continuous.
Naming: function, map, mapping
'Function', 'map' and 'mapping' usually mean the same thing, but in practice, people tend to talk about 'continuous maps' between topological spaces, rather than 'continuous functions'.
It is worth checking that this notion of continuity is compatible with that defined for metric spaces.
Essence of continuity #1
Continuity in terms of open sets can imagined as a test-response: choose any open set surrounding a point in the image of the continuous function and continuity can assert the existance of a "matching" open set in the domain that contains the pre-image of that point. "Matching" here means that the set in the domain is a pre-image of the set in the co-domain.
Then note that any open set around a point contains all tails of sequences approaching that point: the open set around a point \( f(x_0) \) in the co-domain contains all sequences that approach \( f(x_0) \), this open set has an open pre-image that must contain all sequences that approach \( x_0 \). There is containment of infinite sequences through the mapping.
Essence of continuity #2
Here we try to view continuity in the forward direction through a function, rather than by going backwards by going through it's image.
In a metric space, a function \( f : X \to Y \) is continuous if it maps nearby points to nearby points, which can be expressed as:
\[ \lim_{n\to \infty} f(a_n) = f( \lim_{n \to \infty} a_n) \]
Consider now the idea of closures. For a point \( x \), it is in the closure of a set \( S \subset X \) iff there is a sequence of points in \( S \) that converges to \( x \). With this in mind, we can rewrite the above condition for continuity as:
\[ f(\operatorname{cl}(S)) \subseteq \operatorname{cl}(f(S)) \]
This is the #2 equivalent form for topological continuity and might feel more familiar by how it seems closer to the idea of continuity in metric spaces.
Example: Mapping from a discrete topology
Any map from a set with a discrete topology is always continuous, as any element of such a topology is always both open and closed.
Mapping to an indiscrete topology
Any map to set with an indiscrete topology is always continuous, as the only two elements in the codomain (full set and empty set) must map to the full set and empty set of the domain. All four of which are both open and closed.
Identity map
Let \( \mathcal{T} \) and \( \mathcal{T}' \) be two different topologies on the same set \( X \). The 'identity' map \( i : (X, \mathcal{T}) \to (x, \mathcal{T}') \), defined by \( i(x) = x \), is continuous iff the codomain \( \mathcal{T}' \) is finer than the domain \( \mathcal{T} \) ( every member of \( \mathcal{T}' \) is also a member of \( \mathcal{T} \) ).
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Order: 1.1.13

Title: Compactness. Motivation
What would we have to assume about a topological space \( X \) in order to prove that every continuous map \( f : X \to \mathbb{R} \) is bounded?
Reasoning summary
Condensed version.
- What is the definition of being bounded?
- What are some cases where we know functions are bounded?
- The most basic case: the domain is finite.
- Can we generalize this?
- A little more general:
For a given \( f \), the domain can be covered by a finite collection of subsets, each of which has an image that is bounded.
- This is restricted to a given \( f \). Can we generalize?
- Every continuous \( f \) induces a neighbourhood around each \( x \in X \), and \( f \) will be bounded for each of these neigbourhoods. (Remember, the codomain is \( \mathbb{R} \)).
- So, by the definition of continuity, we have a cover where each subset is bounded.
- Sadly, this set of neighbourhoods could be infinite.
- Thus we arrive at our requirement: every open cover must have a finite subcover.
- Which means, every continuous \( f \) will induce an arbitrary open cover on \( X \) (by continuity), and we impose that this cover has a finite subcover.
BackOnlyText: Reasoning steps
The though chain in full.
1. Define bounded
For the single dimension case, the following are equivalent statements regarding a function \( f : X \to \mathbb{R} \):
1. \( f \) is bounded.
2. \( fX \) is bounded.
3. There exists an \( M \ge 0 \) such that for all \( y \in fX \), \( |y| \le M \).
4. There exists an \( M \ge 0 \) such that for all \( x \in X \), \( |f(x)| \le M \).
5. As \( \mathbb{R} \) is a metric space, we can phrase #3 with ε-balls: there is an \( B(0, M) \) for some \( M \ge 0 \) such that \( fX \subseteq B(0, M) \).
2. Functions with finite domain are bounded
If \( X \) is finite, then any function \( f : X \to \mathbb{R} \) (continuous or not) is bounded, since we can take \( M = max_{x \in X}|f(x)| \). max over a finite set of real numbers is a real number.
3. \( X \) covered by finite bounded subsets
If the domain \( X \) can be expressed as a finite union of subsets \( (A_j)_{j \in J} \) such that a function \( f : X \to \mathbb{R} \) is bounded on each individual \( A_j \), then \( f \) itself must be bounded. Why? Well, we can choose for each \( j \in J \) some \( M_j \ge 0 \) such that for all \( x \in A_j, |f(x)| \le M_j \). Then the maximum of these, \( M = \max_{j \in J} M_j \) bounds the whole of \( fX \).
Now the question becomes, given a set \( X \), how can we express it as a union of bounded sets.
4. Locally bounded near every \( x \in X \)
A continuous function need not be bounded, but it is 'locally bounded' in the following sense.
Metric space, \( X \)
Let \( x \in X \). There is some \( \delta_x > 0 \) such that \( B(x, \delta_x ) \) is a subset of  \( f^{-1}(f(x) - 1, f(x) + 1) \), the latter being an open neighbourhood of \( x \).
The same idea for a topological space, \( X \), and function \( f : X \to \mathbb{R} \)
Let \( x \in X \). The set \( U_x = f^{-1}(f(x) -1, f(x) + 1) \) is an open neighbourhood of \( x \) and satisfies \( fU_x \subseteq (f(x) - 1, f(x) + 1) \).
In both of these cases, \( f \) is bounded by \( f(x) + 1 \) when restricted to \( B(x, \delta_x) \) and \( U_x \) respectively. So each \( x \in X \) has a neighbourhood on which \( f \) is bounded.
Trying to utilize this idea to cover \( X \) with bounded subsets runs up against issues of infinite sets.
5. The maximum of an infinite set
We know that for each \( x \in X \), there is a neighbourhood \( U_x \) on which the continuous function \( f \) is bounded, with \( |f(u)| \le M_x \) for all \( u \in U_x \). The subsets \( (U_x)_{x \in X} \) cover \( X \), so why not use \( M = \max_{x \in X}M_x \) as the bounds for \( f \)? The trouble is, this maximum might not be a well-defined real number, since \( X \) is infinite.
6. Impose a finite cover requirement
Presuppose that it was possible to cover \( X \) with only finitely many of the sets \( U_x \). If \( Z \subseteq X \) was a finite set such that the neighbourhoods \( (U_x)_{x \in Z} \) covered \( X \), then we could legitimately set \( M = \max_{z \in Z} M_z \).
So, in order to be able to assert that all continuous maps \( X \to \mathbb{R} \) be continuous, we have needed to assume that there is a finite set of open subsets that cover \( X \). This is exactly what the definition of compact entails. The definition is copied below.
Cover
Let \( X \) be a topological space. A cover of \( X \) is a family \( (U_i)_{i \in I} \) of subsets of \( X \) such that \( \bigcup_{i \in I} U_i = X \). It is finite iff the indexing set \( I \) is finite, and open iff \( U_i \) is open for each \( i \in I \).
Given a cover \( (U_i)_{i \in I} \) and \( J \subseteq I \), we say that \( (U_j)_{j \in J} \) is a subcover of \( (U_i)_{i \in I} \) if it is itself a cover of \( X \).
Finally, we arrive at our definition for compact.
Compact
A topological space \( X \) is compact iff every open cover of \( X \) has a finite subcover.
We have achieved boundedness
If we take any continuous function \( X \to Y \subseteq \mathbb{R^n} \), where \( X \) is compact, \( fX \) must have a finite open cover, as \( X \) has a finite open cover. Consequently, \( fX \) must be bounded: \( fX \subseteq [-M, M]^n \), where  \( M \) is a maximum of the bounds of the finite covers of \( fX \).
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Order: 2.1.1

Title: Compactness
Compact sets can be thought of as a generalization of closed sets. See the reverse side for a justification.
There are multiple ways of formulating compactness, with some being more intuitive than others.
There are 3 formulations that are particularly important.
1. Compactness, in terms of sequences in \( \mathbb{R} \).
A set \( K \subset \mathbb{R} \) is compact iff every sequence in \( K \) has a subsequences that converges to a limit that is also in \( K \).
What is the topological generalization of this definition?
The most basic example of such a closed set in \( \mathbb{R} \) is a closed interval.
Bolzano-Weierstrass's theorem connects the first and second formulations.
2. Compactness, in terms of closed, bounded sets.
A set \( K \subset \mathbb{R}^n \) is compact iff it is closed and bounded.
The next formulation is the most general of the three, graduating to the topic of topology. It uses the idea of a cover and a subcover.
Cover
Let \( X \) be a topological space. A cover of \( X \) is a family \( (U_i)_{i \in I} \) of subsets of \( X \) such that \( \bigcup_{i \in I} U_i = X \). It is finite iff the indexing set \( I \) is finite, and open iff \( U_i \) is open for each \( i \in I \).
Given a cover \( (U_i)_{i \in I} \) and \( J \subseteq I \), we say that \( (U_j)_{j \in J} \) is a subcover of \( (U_i)_{i \in I} \) if it is itself a cover of \( X \).
3. Compactness, in terms of covers.
A topological space \( X \) is compact iff every open cover of \( X \) has a finite subcover.
What is the theorem that connects these three formulations of compactness?
BackOnlyText: Heine-Borel theorem
The Heine-Borel theorem is precicely the statement that ties these three notions of compactness together. [Hmm... actually, I think the Heine-Borel theorem might only cover the equivalence of the first two].
Heine-Borel theorem
Let \( K \) be a subset of \( \mathbb{R} \). The following three statements are equivalent:
1. \( K \) is compact.
2. \( K \) is closed and bounded.
3. Every open cover of \( K \) has a finite subcover.
Proof 1: #2 \(\implies\) #3
A MathExchange answer does a good job of motivating intuition as to why a closed interval must always have a finite subcover. Below is a screenshot of the proof (that any cover of \([0, 1]\) has a finite subcover).
Proof 1 essence
If prooving by contradiction, we must be able to zoom in on an ever narrower closed sub-interval of \([0, 1]\) that requires an infinite cover. As the intervals are closed, their intersection is not empty, but a point. This point must be in an open set of the cover, but this open set must cover at least some of the intervals that we zoomed in on.
Infinite open covers can occur when covering next to and edge. But if we need to cover the edge, then the open set that covers the edge will also cover the tail of our infinite cover.
The essence of compactness
Abbott on p97 mentions this about compact sets and closed intervals:
There may be a temptation to consider closed intervals as being a kind of standard archetype for compact sets, but this is misleading. The structure of compact sets can be much more intricate and interesting. For instance, one implication of Theorem 3.3.4 is that the Cantor set is compact. It is more useful to think of compact sets as generalizations of closed intervals. When ever a fact involving closed intervals is true, it is often the case that the same result holds when we replace “closed interval” with “compact set.” As an example, let’s experiment with the Nested Interval Property proved in the first chapter...
He then goes on to prove a version of the nested closed interval property, but with compact sets replacing closed intervals:
If \( K_1 \supseteq K_2 \supseteq K_3 \supseteq K_4 \supseteq ... \) is a nested sequence of nonempty compact sets, then the intersection \( \cap_{n=1}^{\infty} K_n \) is not empty.
Even more content...this whole section from Abbott is good to re-read.
Some phrasing that might help reading the above excerpt
The watershed case is covering a sequence of points in a set of reals (or rationals) with ever smaller open sets. Think of a sequence from the set that approaches some limit. What if the set (and thus the sequence) is bounded? What if the sequence approaches a limit within the set (i.e. think closed sets)? All points of the sequence has some wriggle-room; in other words, can be covered by an open subset. If a finite-cover is possible, then the sequence can't approach a limit outside of the set.
Other links
There are some interesting descriptions on this Math Exchange page.
Source: Abbott p97
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Order: 2.2.2

Title: Compactness. Examples
- The collection of sets \( \{(n - 1, n + 1) : n \in \mathbb{Z} \} \) form an open cover of \( \mathbb{R} \). This cover has no finite subcover. So \( \mathbb{R} \) is not compact..
- Divide the interval \( [0, 1] \) like so: \( U_0 = [0, \frac{1}{2}] \), \( U_n = (2^{-n}, 1] \) for \( n \ge 1 \), then form the cover \( (U_n)_{n \ge 0} \). This is an open cover, and has many finite subcovers (e.g. ([0, 1/2), (1/2, 1], (1/4, 1]). So we can say that \( [0, 1] \) is .We can't say anything. We would need to prove that all open covers have a finite subcover to prove compactness.
- The compact subspaces of \( \mathbb{R}^n \) are the closed bounded subsets.
- Any indiscrete space is compact.
- Any finite space is compact.
- A discrete space is compact iff it is finite.
- In a normed vector space \( V \), the closed unit ball is compact iff \( V \) is finite-dimensional.
BackOnlyText: The last example alludes to the idea that compactness is a kind of finiteness condition:
sets | finite sets
vector spaces | finite dimensional vector spaces
topological spaces | compact topological spaces
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Order: 2.3.3

Title: Compactness. Subspaces
How do compactness and subspaces interact?
1. A subspace of a compact space is not necessarily compact.
2. Every closed subspace of a compact space is compact.
3. Every compact subspace of a compact space is not necessarily closed.
4. Every compact subspace of a Hausdorff space is closed.
5. A subspace of a compact Hausdorff space is compact iff it is closed.
BackOnlyText: 1. A subspace of a compact space
A subspace of a compact space need not be compact. For example, \( [0, 1] \) is compact, yet \( (0, 1) \subseteq [0, 1] \) is not compact. To see this, consider the open cover of \( (0, 1) \) given by \( ((\varepsilon, 1)_{\varepsilon > 0}) \). This cover has no finite subcover. This example highlights that all open covers must have a subcover, and thus, to prove a space to be compact, it is not sufficient to find a single open cover having a finite subcover—we must show that it is impossible for any open cover to not have a finite subcover.
2. Every closed subspace of a compact space is compact.
Proof outline. Any cover of a closed subset \( V \) can be appended by the single open set \( X \setminus V \). This new set of subsets must cover \( X \), and must be finite as \( X \) is compact.
Proof. Let \( X \) be a compact space and \( V \) a closed subset. Let \( (U_i)_{i \in I} \) be a cover of \( V \) by open subsets of \( X \). Then \( (U_i)_{i \in I} \) together with \( X \setminus V \) is an open cover of \( X \). Since \( X \) is compact, it has some finite subcover; thus, there is some finite \( J \subseteq I \) such that
\[ \left( \bigcup_{j \in J} U_j \right) \cup (X \setminus V) = X \]
Then \( \bigcup_{j \in J} U_j \subset V \), as required.
We create an open cover of \( V \) by subsets of \( X \), not subsets of \( V \). This is for convenience, and is justified by the modified formulation of cover for subsets. See Leinster's notes and Lemma B1.7.
The converse is not necessarily true. Which is the topic of the next section.
An idea that appeals to intuition regarding closed vs. open subsets and being compact: closed sets place more stringent requirements on any collection in order for the collection to be a cover. Consequently, some covers that might not have a finite subcover don't count as covers. This conceptualization might help motivate why closed subsets of a compact space are compact, and open subsets are not necessarily.
3. Not all compact subspaces of a compact space is closed.
Trivial counter-examples make this statement false. For example, consider an indiscrete space on a set \( X \) that has two or more elements. All indiscrete spaces are compact. Now choose a proper subset \( Y \subset X \) and consider the subspace topology on \( Y \): the subspace topology on \( Y \) is indiscrete also (just \( \{Y, \emptyset \} \)), so \( Y \) is compact. But \( Y \) is not closed in \( X \), as it is neither \( \emptyset \) nor \( X \).
What other examples are there of non-closed compact subspaces of a compact space are there? Another example is the finite complement topology on \( \mathbb{R} \). Every subspace of the finite complement topology on \( \mathbb{R} \) is compact (try justify!). But the only proper subsets of \( \mathbb{R} \) that are closed in this topology are the finite sets.
With a pretty minimal extra restriction on \( X \) (being Hausdorff), we can make the desired assertion.
4. Every compact subspace of a Hausdorff space is closed.
Note first that we have relaxed the requirement that the larger/outer space be compact.
Proof. Let \( Y \) be a compact subspace of the Hausdorff space \( X \). We shall prove that \( X - Y \) is open, so that \( Y \) must be closed. Let \( x_0 \in X - Y \).
Claim: there is a neighbourhood of \( x_0 \) that is disjoint from the whole of \( Y \).
Claim proof. Since \( X \) is Hausdorff, for each \( y \in Y \), with respect to \( x_0 \) choose disjoint neighbourhoods \( U_y \) and \( U_{x_0} \): in other words \( U_y \) is a neighbourhood of \( y \) and \( U_{x_0} \) is a neighbourhood of \( x_0 \). The collection \( \{U_y : y \in Y \} \) is a covering of \( Y \) by sets open in \( X \). [This next bit is the one and only point where we utilize the compactness of \( Y \).] There must be a finite sub-cover, \( U_Y = U_{y_1}, U_{y_2}, ..., U_{y_n} \). This finite cover \( U_Y \) contains \( Y \) and is disjoint from \( x_0 \) and its neighbourhood \( Ux_0 = U_1, U_1, U_n \), formed by taking the finite intersection of neihbourhoods disjoint from each of the neigbourhoods that were unioned to form \( U_Y \). This proofs our claim.
Now we have a neighbourhood of \( x_0 \) disjoin from \( Y \). This is true for any \( x \notin Y \), so we may take the union of such open sets which will in turn be open. Let \( W \) be this open set. We have \( X - W = Y \), so \( Y \) must be closed. This completes the whole proof.
The finite complement topology was used as an example for #3. The finite complement topology is not a Hausdorff space, as any two neighbourhoods always overlap; we then fail trying to setup a neighbourhood around \( x_0 \) disjoint from \( Y \). This highlights the difference between #3 and #4).
Lemma: 'neighbourhood' of a compact subspace
If \( Y \) is a compact subspace of a Hausdorff space \( X \), and \( x_0 \) is not in \( Y \), then there exists disjoin open sets \( U_{x_0} \) and \( U_Y \) of \( X \) such that \( U_{x_0} \) contains \( x_0 \) and \( U_Y \) contains \( Y \).
This lemma is the inner claim from the proof above. Munkres makes the point this this lemma is a useful result in its own right. He even provides a nice diagram (see figure 26.1 in the text).
A subspace of a compact Hausdorff space is compact iff it is closed.
This is the combination of 2 & 4.
Source: Munkres p164
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Order: 2.5.1

Title: Connectedness, compactness and some fundamental theorems of calculus
The following three theorems in calculus, theorems about functions from and to the reals, have generalizations in topology.
Intermediate value theorem
If \( f : [a, b] \to \mathbb{R} \) is continuous, and if \( r \) is a real number between \( f(a) \) and \( f(b) \), then there exists an element \( c \in [a, b] \) such that \( f(c) = r \).
Maximum value theorem
If \( f : [a, b] \to \mathbb{R} \) is continuous, then there exists an element \( c \in [a, b] \) such that \( f(x) \le f(c) \) for every \( x \in [a, b] \).
Uniform continuity theorem
If \( f : [a, b] \to \mathbb{R} \) is continuous, then for every \( \varepsilon > 0 \) there exists a \( \delta > 0 \) such that \( |f(x_1) - f(x_2)| < \varepsilon \) for every \( x_1, x_2 \in [a, b] \) for which \( |x_1 - x_2| < \delta \).
Applications in Calculus
- The intermediate value theorem is used for constructing inverse functions, such as \( \sqrt[3]{x} \) and \( \arcsin(x) \).
- The maximum value theorem is used to prove the mean value theorem for derivatives, which in turn is used to prove the two fundamental theorems of calculus.
- The uniform continuity theorem is used for proving that every continuous function is integrable.
What is the concept in question: functions vs sets?
The three theorems can be considered to be describing facts about continuous functions; but shifting one's focus, one can view them as describing the nature of the closed interval \( [a, b] \subset \mathbb{R} \).
As topological properties
The topological property of the space \( [a, b] \) on which the intermediate value theorem depends is the topological property called connectedness.
The property which the maximum value theorem and the uniform continuity theorem depend on is called compactness.
Both of these properties are fundamental to areas beyond calculus; they are fundamental to almost any area which can be represented in topology.
Source: Munkres
p145
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Order: 4.1.1

Title: Connectedness. Definition
Connectedness
Let \( X \) be a topological space. A separation of \( X \) is a pair \( U, V \) of disjoint nonempty open subsets of \( X \) whose union is \( X \). The space \( X \) is said to be connected iff there does not exist a separation of \( X \).
This definition is from Munkres. Leinster's definition feels a little less intuitively concrete.
Connectedness, formulation 2 (or now a lemma)
A space \( X \) is connected iff the only subsets of \( X \) that are both open and closed in \( X \) are the empty set and \( X \) itself.
Again, from Munkres.
BackOnlyText: Connectedness, motivation
Focus on two sets that union to create the set in question: if two sets union to the whole but are not "touching", then the set in question is disconnected. Intuitively, a space is said to be 'connected' if it is all in one piece—it does not fall naturally into two or more pieces. 'Naturally' here carries a major component of the sense of 'connected', as any topological space can be torn into multiple pieces in some way.
Disconnected
The idea of a set being disconnected is possibly more intuitive than the idea of a set being connected. In \( \mathbb{R} \), a set \( E \subseteq \mathbb{R} \) is disconnected if it can be expressed as the union of two separated sets. Separated here means that the sets share no limit points (\( A \) and \( B \) are separated if \( \overline{A} \cap B \) and \( A \cap \overline{B} \) are empty). In other words, one can't find a sequence in \( A \) whose limit is in \( B \) and vise versa.
Yet another formulation is to say that \( E \subseteq \mathbb{R} \) is disconnected iff for some \( a, b \in E \) there is a \( c \in \mathbb{R} \) but not in \( E \) such that \( a < c < b\). Push a negation into this statement to get back to the connected version: \( E \) is connected iff for all \( a, b \in E \), whenever \( a < c < b \) for some \( c \in  \mathbb{R} \), then \( c \in E \).
Tearing
Intuitively, why can't the real line \( \mathbb{R} \) be separated naturally into two pieces? Consider the decomposition \( (-\infty, 0) \cup [0, \infty) \): in what way is this separation not natural? While it might not appeal to everyone's intuition, this separation is considered forced because a limit point of \( (-\infty, 0) \), zero, is not an element of \( (-\infty, 0) \) but an element of the other piece \( [0, \infty) \). In other words, the closure of one piece spills into or is contained or is intertwined in the other piece. Thus, naturally separating a space can be viewed from the lens of open and closed sets.
Natural separation, requirements
For a topological space \( X \), a separation of \( X \) into disjoint subsets is considered a natural separation into independent pieces \( U \) and \( V \) if no point in \( U \) is a limit point of \( V \), and vice versa. Equivalently, this means that \( Cl(U) \) is disjoin from \( V \), and vice versa, or equivalently, that both \( U \) and \( V \) are closed, or equivalently yet again, that both \( U \) and \( V \) are open.
Thus, we arrive at our definition of connectedness, by defining a space \( X \) connected if the only natural separation of \( X \) as described above is the trivial one, \( \emptyset \) and \( X \) itself.
Sufficiency of open sets
If disjoint open sets are sufficient to cover a space, then the space is not connected. Open sets don't contain their boundary limits, and two disjoint open sets must therefore not be touching. There is at least a point that exists in the sense of limits that is in neither of the open sets. In this sense, the space is disconnected.
Source: Munkres, p146
Leinster, p68
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Order: 3.1.1

Title: Connectedness Examples
- \( X = \mathbb{R} \setminus \{0\} \) is disconnected, since \( (-\infty, 0) \) and \( (0, \infty) \) are disjoint open subsets of \( X \) whose union is \( X \).
- The space of rationals numbers \( \mathbb{Q} \), topologized as a subspace of \( \mathbb{R} \), is disconnected. Consider the pair of open sets \( (-\infty, \sqrt(2)) \cap \mathbb{Q} \) and \( (\sqrt(2), \infty) \cap \mathbb{Q} \).
- A discrete space with 2 or more points is disconnected. If \( X \) is such a space, we can choose an \( x_1 \in X \) then consider the two open sets \( \{x_1\} \) and \( X \setminus \{x_1\} \).
- A non-empty indiscrete space is connected.
- An interval topologized as a subspace of \( \mathbb{R} \) is connected.
- The space \( \mathbb{R}^d \) is not necessarily connected nor disconnected—it is connected in the product topology, but not connected in the box topology.
- The letter 'O' is connected. '0' is a quotient of \( [0,1] \), the quotient map is continuous and the image of a connected space through a continuous function is connected.
- The collection of connected sets in \( \mathbb{R} \) coincide preciesly with the collection of intervals.
Source: Munkres, p149
Leinster, p69
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Order: 3.1.2

Title: Connectedness. Equivalent formulations
Q: For a nonempty topological space \( X \), the following are equivalent:
1. \( X \) is connected;
2. [something about subsets];
3. [something about maps];
4. [something similar about maps].
A: the only subsets that are both open and closed are \( \emptyset \) and \( X \); every continuous map from \( X \) to a discrete space is constant; every continuous map from \( X \) to the two-point discrete space is constant
BackOnlyText: The 3rd and 4th statement are similar, yet both useful in their own right: the 3rd statement is general and allows progress to be made in a proof when one knows that \( X \) is connected. The 4th is powerful in the reverse direction: simply proving that all continuous maps to the two-point discrete space is constant is enough to show that \( X \) is connected.
Source: Leinster, p 69
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Order: 3.1.3

Title: Connectedness. 4 lemmas
1. The edge of a connected space
Let \( X \) be a topological space. Let \( A \) and \( B \) be subspaces of \( X \) with \( A \subseteq B \subseteq \operatorname{Cl}(A) \).
If \( A \) is connected, then so is \( B \).
2. An image of a continuous function on a connected space is connected
Let \( f : X \to Y \) be a continuous map of topological spaces. If \( X \) is connected then so is \( fX \).
In particular, any quotient of a connected space is connected.
3. The product of two connected spaces is connected.
4. A space that has an overlapping covering of connected subspaces is connected.
Let \( X \) be a nonempty topological space and \( (A_i)_{i \in I} \) a family of subspaces covering \( X \). Suppose that \( A_i \) is connected for each \( i \in I \) and that \( A_i \cap A_j \neq \emptyset \) for each \( i, j \in I \), then \( X \) is connected.
This lemma says that gluing together overlapping connected spaces produces connected spaces.
BackOnlyText: 1. (\( A \text{ is connected } \implies A \subseteq B \subseteq Cl(A)) \implies B \text { is connected}\)). Proof outline.
\( B \) is within the smallest closed subset containing \( A \). How could \( B \) be divide into 2 disjoint closed/open sets with \( A \) in one and the rest in another? It can't—if it could, then \( A \) itself would be closed, and thus, it's own closure. Note: \( A \) is connected, so cannot be subdivided to create the pair of open/closed subsets of \( B \).
A good example for visualizing this lemma is to consider \( A \) to be the open unit disk, and for this lemma to represent the inability to create a separation of the closed disk with the open disk contained whollly in one of the sets of the separation.
2. A continuous image of a connected space is connected. Proof outline.
The restricted function \( f : X \to fX \) is surjective, and \( f^{-1}fX = X \). If we can split \( fX \) into two disjoint non-empty open sets, then the pre-image of those sets will be open, disjoint and non-empty in \( X \), and they will union to \( X \). But that is a contradiction, as \( X \) is connected.
3. Product of two connected spaces is connected. Proof outline.
Let \( X \) and \( Y \) be connected spaces, and consider \( X \times Y \).
Two approaches:
- Show that if \( X \times Y \) is not connected, then either \( X \) or \( Y \) cannot be connected. An ability to find a separation of \( X \times Y \) implies that we have found a separation of either \( X \) or \( Y \) or both.
- The approach taken by Leinster is to consider a continuous function \( f : X \times Y \to D \), where \( D \) is a discrete space, and show that \( f \) is a constant function. In the 'Connectedness. Equivalent formulations' card, it was shown that this is necessary and sufficient for \( X \times Y \) to be connected. The demonstration that \( f \) is constant is done by taking two arbitrary points \( t_1, t_2 \in X \times Y \) and considering a 'vertical' slice passing through \( t_1 \) and a 'horizontal' slice passing through \( t_2 \). These slices are homeomorphic to one of \( X \) and \( Y \), so \( f \) must be constant alone both. As the two slices intersect somewhere, \( f \) must be the same constant on each. As the points were arbitrary, this covers the whole space.
4. Union of overlapping connected spaces is connected. Proof.
Let \( f \) be a continuous map from \( X \) to a discrete space \( D \). For each \( i \in I \), we have the continuous map \( f|_{A_i} : A_i \to D \). \( A_i \) is connected, so \( f|_{A_i} \) has a constant value \( d_i \). But for each \( i, j \in I \), we have \( A_i \cap A_j \neq \emptyset \), so \( d_i = d_j \). Hence \( d_i \) is independent of \( i \in I \), so \( f \) is constant, as required.
Example: the letter O is a quotient of \( [0, 1] \), which is connected, so \( O \) is connected. Gluing O to another line segment gives P, which must now also be connected. Gluing on another line segment to give A shows us that A too is connected.
Source: Leinster, p70-71
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Order: 3.1.4

Title: Homeomorphisms. Homeomorphic spaces.
Q: Homeomorphisms and homeomorphic spaces.
Let \( X \) and \( Y \) be topological spaces.
Homeomorphism
A homeomorphism from \( X \) to \( Y \) is a [...].
Homeomorphic spaces
The spaces \( X \) and \( Y \) are homeomorphic iff [...].
\( X \cong Y \) is the notation for \( X \) to be homeomorphic to \( Y \). 'Topologically equivalent' is an alternative term for 'homeomorphic'.
When are two spaces homeomorphic?
To show that two spaces are homeomorphic, find a homeomorphism between them.
Roughly, two spaces are homeomorphic iff one can be deformed into the other by bending and reshaping, but without tearing or gluing.
Equivalence relation
Being homeomorphic is an equivalence relation on the class of all topological spaces.
Reflexive
Let \( X \) be a topological space. Then the identity map on \( X \) is a homeomorphism.
Symmetric
Let \( f : X \to Y \) be a homeomorphism. Then \( f^{-1} : Y \to X \) is a homeomorphism.
Transitive
Let \( f : X \to Y \) and \( g : Y \to Z \) be homeomorphisms. Then \( g \circ f : X \to Z \) is a homeomorphism.
Thus, \( X \cong X \); \( X \cong Y \iff Y \cong X \); \( X \cong Y \land Y \cong Z \implies X \cong Z \).
A: continuous bijection whose inverse is also continuous; there exists a homeomorphism between them
BackOnlyText: Knots homeomorphic to the circle
Some cases of homeomorphisms seem to defy the restriction to bending and reshaping: all knots are homeomorphic to the circle. To see this, take two knots of 1m string; choose a point on each as a starting point; then, trace around each knot at the same rate until you reach the start again. This defines a homeomorphism. While we can't bend a knot into a circle in \( \mathbb{R}^3 \), any knot can be reshaped into a circle if both are first placed in \( \mathbb{R}^4 \).
Bijection with non-continuous inverse
The inverse of a continuous bijection need not be continuous.
An easy way of constructing an example is to consider continuous maps to a set with the discrete topology. For example, the identity map \( i : (\mathbb{R}, \text{discrete topology}) \to (\mathbb{R}, \text{standard topology}) \) is continuous, but the inverse, \( \) \( i^{-1} : (\mathbb{R}, \text{standard topology}) \to (\mathbb{R}, \text{discrete topology}) \) is not continuous.
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Order: 1.1.14

Title: Path connectedness. Definition
Q: Viewing topology from the lens of Euclidean space suggests a variant of the notion of connectedness: path-connectedness.
The Euclidean lens
A topological space is said to be an n-dimensional manifold if it is Hausdorff and has an open cover by subsets each homeomorphic to an open ball in \( \mathbb{R}^n \). Typical examples of 2-dimensional manifolds (surfaces) are the sphere, the torus and the Klein bottle. Manifolds are enormously important.
Paths
Let \( X \) be a topological space.
A path in \( X \) is a [something]. If \( \gamma(0) = x \) and \( \gamma(1) = y \), then we say that [something] is a path from \( x \) to \( y \).
Path-connectedness
A topological space \( X \) is path-connected if it is non-empty and [...].
The relevance to standard connectedness is quickly apparent:
Every path-connected space is connected.
Proof on reverse side. Note that the converse is false, and thus, path-connectedness is a stronger condition than vanilla conectedness.
A: continuous map \( \gamma : [0, 1] \to X \); \( \gamma \); for all \( x, y \in X \), there exists a path from \( x \) to \( y \) in \( X \)
BackOnlyText: Every path-connected space is connected. Proof.
We know that all continuous maps from a path-connected space \( X \) to a discrete space \( D \) are constant iff \( X \) is connected. Let \( f \) be an arbitrary continuous map from \( X \) to \( D \). Claim: \( f \) is constant. Let \( x, y \in X \). There is a path \( \gamma : [0, 1] \to X \) from \( x \) to \( y \), and \( f \circ \gamma \) is then a continuous map \( [0, 1] \to D \). But \( [0, 1] \) is connected, so \( f \circ \gamma \) is constant, and in particular, \( f(x) = f(\gamma(0)) = f(\gamma(1)) = f(y) \), as required.
Proving connectedness
It can be quite hard to prove that a space is connected by relying on the definition of connectedness. For example, how do you show that a disk in \( \mathbb{R}^2 \) doesn't have a separation? The implication from path-connectedness to connectedness can be very useful in such situations.
Example: Convex sets
A subset \( X \) of \( \mathbb{R}^n \) is convex iff for all \( x, y \in X \) and \( t \in [0, 1] \), we have \( (1-t)x + ty \in X \). For instance, the convex subsets of \( \mathbb{R} \) are precisely the intervals. Every convex subset of \( \mathbb{R}^n \) is path-connected (with \( x, y \in X \), \( t \mapsto (1 - t)x + ty \) is a path from \( x \) to \( y \)). Hence, every convex subset of \( \mathbb{R}^n \) is connected.
Geometrically, convex sets are those where every point in the set is connected by a straight line segment.
Source: Leinster p76
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Order: 3.3.1

Title: Path-connectedness. 4 statements
Path-connectedness implies connectedness, the converse is not generally true through.
The topologist's sine curve is an example of a space that is connected but not path-connected. Leinster covers the topologist's sine curve in some detail.
Below, we introduce an iff statement that does hold. It has the form: something ∧ connected ⟺ path-connected. After this, 3 conditions that each imply path-connectedness are presented.
1. Path-connected ⟺ connected and all points have a path-connected neighbourhood
Let \( X \) be a topological space. \( X \) is path-connected if and only if \( X \) is connected and every point of \( X \) has at least one path-connected neighbourhood.
2. Corollary. Every connected open subset of \( \mathbb{R}^d \) is path-connected.
3. Path-connectedness and continuous functions
Let \( f : X \to Y \) be a continuous map of topological spaces. If \( X \) is path-connected, then so is \( fX \).
4. Path-connectedness and products
The product of two path-connected spaces is path-connected.
BackOnlyText: Proofs
Path-connected ⟺ connected and all points have a path-connected neighbourhood
Let \( X \) be a topological space.
Forward implication
Assume that \( X \) is path-connected. As shown in a previous card, \( X \) is connected as it is path-connected. In addition, we have a path-connected neighbourhood for every point as the set \( X \) is a neighbourhood for every point and \( X \) is path-connected.
Reverse implication
Assume \( X \) is connected and all points have a path-connected neighbourhood. Let \( x \in X \) and let \( U \) be the set of all elements having a path to \( x \):
\[ U = \{ y \in X : \text{ there exists a path from } x \text{ to } y \text{ in } X \} \]
Claim: both \( U \) and \( X \setminus U \) are open in \( X \). Show this in two parts.
\( U \) is open
Let \( y \in U \). By assumption, there is an path-connected neighbourhood \( W \) of \( y \). Let \( w \in W \). There is a path from \( x \) to \( w \) as we can join a path from \( x \) to \( y \) and a path from \( y \) to \( w \). Thus, \( W \subset U \). As every element of \( U \) has a neighbourhood contained in \( U \), \( U \) must be open (A2.9 in Leinster's notes).
\( X \setminus U \) is open
Let \( y \in X \setminus U \). By assumption, there is a path-connected neighbourhood \( W \) of \( y \). \( W \cap U = \emptyset \), as if there was an element \( t \in W \cap U \) we could find a path from \( x \) to \( y \) by joining the path from \( x \) to \( t \) and the path from \( t \) to \( y \); but such a path would contradict \( y \notin U \). We can now state that \( W \in X \setminus U \), as \( W \subseteq X \) and \( W \cap U = \emptyset \). As every element of \( X \setminus U \) has a neighbourhood contained in \( X \setminus U \), \( X \setminus U \) must be open (again by A2.9).
So, we have both \( U \) and \( X \setminus U \) being open. However, \( X \) is connected, so one of the sets must be empty and the other must equal \( X \). \( x \) itself is in \( U \), so \( U = X \). \( U \) is path-connected by definition, so \( X \) is path-connected.
Source: Leinster, p79
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Order: 3.3.2

Title: Topological properties
Text: TODO
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Order: 1.6.1

Title: Product spaces
Text: TODO
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Order: 1.10.1

Title: Quotient spaces
Text: TODO
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Order: 1.11.1

Title: Compactness. Products
Text: TODO
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Order: 2.4.1

Title: Compactness. Euclidean space
Text: TODO
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Order: 2.2.1

Title: Compactness. Quotients
Text: TODO
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Order: 2.6.1

Title: Compactness. Metric spaces
Text: TODO
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Order: 2.7.1

Title: Connectedness. The reals
Text: TODO
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Order: 3.2.1

Title: Connectedness and path-connectedness
Text: TODO
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Order: 3.4.1