Title: Math and science / Analysis / Tao / 05. The real numbers
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Title: Sequences
Q: Let \( m \) be an integer. A sequence \( (a_n)_{n=m}^{\infty} \) of rational numbers is any function from the set [...] to \( \mathbb{Q} \).
A: \( \{ n \in \mathbb{Z} : n \ge m \} \)
BackOnlyText: In other words, a sequence is a mapping that assigns to each integer greater than or equal to \( m \) a rational number \( a_n \).
From sequences to reals
sequence → ε-steady sequence → eventually ε-steady sequence → Cauchy sequence → ε-close sequences → eventually ε-close sequences → equivalent sequences → real numbers.
Example: \( (n^2)_{n=0}^{\infty} \) is the collection 0, 1, 4, 9..., where the function \( f: \mathbb{Z} \rightarrow \mathbb{Q} \) is the set of tuples \( \{ (0, 0), (1, 1), (2,4), (3, 9)... \} \) where for every first element of the tuple, \( n \), the second element is \( n^2 \).
Source: Tao, Analysis I

Title: ε-steadiness
Q: Let \( \varepsilon > 0 \) be a rational. A sequence \( (a_n)_{n=m}^{\infty} \) is ε-steady iff each pair \( a_j, a_k \) of sequence elements are [...] for every [...].
A: ε-close; \( j,k \ge m \)
BackOnlyText: In other words, the sequence \( a_0, a_1, a_2, ... \) is ε-steady iff each \( |a_j - a_k| \le \varepsilon \) for all \( j,k \).
Example: The sequence \( 1, 0, 1, 0, 1,... \) is 1-steady, but not 0.5-steady. The sequence \( 0.1, 0.01, 0.001, 0.0001, ... \) is 0.1 steady, but not 0.01 steady. The sequence \( 1,2,3,4,5,... \) is not steady for any ε, and the sequence \( 2, 2, 2,... \) is steady for all \( \varepsilon > 0 \).
Source: Tao, Analysis I
Context: From sequences to reals
sequence → ε-steady sequence → eventually ε-steady sequence → Cauchy sequence → ε-close sequences → eventually ε-close sequences → equivalent sequences → real numbers.

Title: Bounded sequences
Q: Let \( M >0 \) be a rational.
A finite sequence (of rationals, but equally applies to reals) \( a_1, a_2, a_3, ..., a_n \) is bounded by M iff [...] for all \( 1 \ge i \ge n  \).
An infinite sequence \( (a_n)_{n=1}^{\infty} \) is bounded by M iff [...] for all \( i \ge 1 \).
A sequence is said to be bounded iff it is bounded by some \( M \ge 0 \).
A: \( |a_i| \le M \); \( |a_i| \le M \)
BackOnlyText: Note how the definition is using only a sequence beginning at index 1 (the sequences of form \( (a_n)_{n=1}^{\infty} \) as opposed to the more general form \( (a_n)_{n=m}^{\infty} \) for some integer m). Tao mentions earlier in the text that the beginning index is irrelevant.
Two propositions/lemmas that follow:
1. All finite sequences are bounded.
2. All Cauchy sequences are bounded. (proof: exercise 5.1.1).
Source: Tao, Analysis I

Title: ε-close sequences
Q: Let \( (a_n)_{n=0}^{\infty} \) and \( (b_n)_{n=0}^{\infty} \) be two sequences and let \( \varepsilon > 0 \) be a rational. \( (a_n)_{n=0}^{\infty} \) is said to be ε-close to \( (b_n)_{n=0}^{\infty} \) iff [...] for all \( k \ge 0 \).
A: \( a_k \) is ε-close to \( b_k \)
BackOnlyText: In other words, the sequence \( a_0, a_1, a_2, ... \) is ε-close to the sequence \( b_0, b_1, b_2, ... \) iff \( |a_k - b_k| \le \varepsilon \) for all \( k = 0, 1, 2, ... \).
From sequences to reals
sequence → ε-steady sequence → eventually ε-steady sequence → Cauchy sequence → ε-close sequences → eventually ε-close sequences → equivalent sequences → real numbers.
Example: The two sequences:
\[ 1, -1, 1, -1, 1, ... \]
and
\[ 1.1, -1.1, 1.1, -1.1, 1.1, ... \]
are 0.1-close to each other. Note how neither of the sequences are 0.1-steady.
Source: Tao, Analysis I

Title: Eventually ε-close sequences
Q: Let \( (a_n)_{n=0}^{\infty} \) and \( (b_n)_{n=0}^{\infty} \) be two sequences of rational numbers and let \( \varepsilon >0 \) be a rational. The sequences are said to be eventually ε-close iff there exists an integer \( N \ge 0 \) such that [...] are ε-close.
A: the sequences \( (a_n)_{n=N}^{\infty} \) and \( (b_n)_{n=N}^{\infty} \)
BackOnlyText: In other words \( a_0, a_1, a_2, ... \) is eventually ε-close to \( b_0, b_1, b_2, ... \) if there exists an \( N \ge 0 \) such that \( |a_j - b_j| \le \varepsilon \) for all \( j \ge N \).
From sequences to reals
sequence → ε-steady sequence → eventually ε-steady sequence → Cauchy sequence → ε-close sequences → eventually ε-close sequences → equivalent sequences → real numbers.
Example: The two sequences
\[ 1.1, 1.01, 1.001, 1.0001, ... \]
and
\[ 0.9, 0.99, 0.999, 0.9999, ... \]
are not 0.1-close but are eventually 0.1-close. They are also eventually 0.01-close.
Source: Tao, Analysis I

Title: Equivalent sequences
Q: Two sequences \( (a_n)_{n=0}^{\infty} \) and \( (b_n)_{n=0}^{\infty} \) are equivalent iff for each rational \( \varepsilon \gt 0 \), the sequences [...] are [...].
A: \( (a_n)_{n=0}^{\infty} \) and \( (b_n)_{n=0}^{\infty} \); eventually ε-close
BackOnlyText: In other words the sequences \( a_0, a_1, a_2, ... \) and \( b_0, b_1, b_2, ... \) are equivalent iff for every rational \( \varepsilon \ge 0 \) there exists an \( N \ge 0 \) such that \( |a_k - b_k| \le \varepsilon \) for all \( k \ge N \).
From sequences to reals
sequence → ε-steady sequence → eventually ε-steady sequence → Cauchy sequence → ε-close sequences → eventually ε-close sequences → equivalent sequences → real numbers.
Example: The two sequences
\[ 1.1, 1.01, 1.001, 1.0001, ... \]
and
\[ 0.9, 0.99, 0.999, 0.9999, ... \]
are equivalent.
Source: Tao, Analysis I

Title: Real numbers, the construction from Cauchy Sequences
Q: A real number is defined to be a new type of object, written as \( LIM_{n \rightarrow \infty} a_n \). This object has a (1 to many) correspondence to a Cauchy sequence \( (a_n)_{n=1}^{\infty} \). The Cauchy sequence is used to define an equivalence relation between real numbers: two real numbers \( LIM_{n \rightarrow \infty} a_n \) and \( LIM_{n \rightarrow \infty} b_n \) are said to be equal iff the corresponding sequences \( (a_n)_{n=1}^{\infty} \) and \( (b_n)_{n=1}^{\infty} \) are [...].
A: equivalent Cauchy sequences
BackOnlyText: The set of real numbers is denoted \( \mathbb{R} \).
\( LIM_{n \rightarrow \infty}a_n \) is referred to as a formal limit. It is not the standard limit, but an object taking the form of a limit, and existing by definition.
The next step (for another card) is to ensure via proof that the equivalence relation that has just been defined obeys the axioms of equality (all 4 of them).
From sequences to reals
sequence → ε-steady sequence → eventually ε-steady sequence → Cauchy sequence → ε-close sequences → eventually ε-close sequences → equivalent sequences → real numbers.
Example: Let \( a_0, a_1, a_2, ... \) be the sequence:
\[ 1.4, 1.41, 1.414, 1.4142, 1.41421, ... \]
and let \( b_0, b_1, b_2, ... \) be the sequence:
\[ 1.5, 1.42, 1.415, 1.4143, 1.41422, ... \]
Then \( LIM_{n \rightarrow \infty}a_n \) is a real number and is the same real number as \( LIM_{n \rightarrow \infty}b_n \), as their corresponding Cauchy sequences are equivalent Cauchy sequences.
Source: Tao Analysis, I

Title: Sequences bounded away from zero
Q: A sequence \( (a_n)_{n=1}^{\infty} \) is said to be bounded away from zero iff there exists a rational \( c > 0 \) such that [...] for all integers \( k \ge 1 \).
A sequence  \( (a_n)_{n=1}^{\infty} \) is said to be positively bounded away from zero iff there exists a rational \( c > 0 \) such that [...] for all integers \( k \ge 1 \). (In other words, the sequence is comprised entirely of positive rationals.)
A sequence  \( (a_n)_{n=1}^{\infty} \) is said to be negatively bounded away from zero iff there exists a rational \( c > 0 \) such that [...] for all integers \( k \ge 1 \). (In other words, the sequence is comprised entirely of negative rationals.)
A: \( |a_k| \ge c \); \( a_k \ge c \); \( a_k \le -c \)
BackOnlyText: These properties are used, among other things, to define reciprocation for the reals, ordering of reals and by extension the Archimedean property.
Source: Tao, Analysis I

Title: Reciprocals of real numbers
Q: In order to define reciprocation, it is required to insure that division [...] is avoided. Eventually, we define the reciprocal of a real by taking a Cauchy sequence that represents the real, then taking the reciprocal of each elements to form a new Cauchy sequence. To do this requires the following two lemmas.
Lemma 5.3.14
Let \( x \) be a non-zero real number. Then \( x = LIM_{n\rightarrow \infty}a_n \) for some Cauchy sequence \( (a_n)_{n=1}^{\infty} \) which is bounded away from zero.
In other words, out of the many Cauchy sequences that can represent the real \( x \), there must be at least one that is bounded away from zero.
The proof for Lemma 5.3.14 is a good exercise and is listed out in the book.
Lemma 5.3.15
Suppose that \( (a_n)_{n=1}^{\infty} \) is a Cauchy sequence which is bounded away from zero. Then the sequence \( (a_n^{-1})_{n=1}^{\infty} \) is also a Cauchy sequence.
Finally we get to the definition:
Reciprocation for reals
Let \( x \) be a non-zero real number. Let \( (a_n)_{n=1}^{\infty} \) be a Cauchy sequence bounded away from zero such that \( x = LIM_{n\rightarrow \infty} a_n \), which exists by Lemma [...]. Then we define the reciprocal \( x^{-1} \) as:
\[ x^{-1} := LIM_{n\rightarrow \infty}a_n^{-1} \]
which we know to be a Cauchy sequence from Lemma [...].
A: by zero; 5.3.14; 5.3.15
Source: Tao, Analysis I

Title: Upper bound
Q: Let \( E \) be a subset of \( \mathbb{R} \), and let \( M \) be a real number. We say that \( M \) is an upper bound for \( E \) if [...] for every element \( x \) in \( E \).
A: \( M \ge x \)
BackOnlyText: Upper bound def → least upper bound def→ uniqueness of least upper bound → existence of least upper bound → supremum def
Example: The interval \( E:= \{ x \in R : 0 \le x \le 1 \} \) has 1 as an upper bound. All numbers greater than 1 are also upper bounds.
The set of positive reals, \( \mathbb{R}^+ \), has no upper bound.
The empty set, \( \emptyset \), has every real as an upper bound. This is true as any real is greater than all elements of the empty set (vacously true, but still true).
Source: Tao, Analysis I

Title: Least upper bound
Q: Let \( E \) be a subset of \( \mathbb{R} \) and let \( M \) be a real number. We say that \( M \) is a least upper bound for E iff:
1. \( M \) is an [...] for \( E \).
2. Any other upper bound for \( E \) is [...].
A: upper bound; greater or equal to \( M \)
BackOnlyText: Upper bound def → least upper bound def→ uniqueness of least upper bound → existence of least upper bound → supremum def
Example: The interval \( E:= \{ x \in R : 0 \le x \le 1 \} \) has 1 as a least upper bound.
The empty set does not have any least upper bound (why?).
Source: Tao, Analysis I

Title: Uniqueness of least upper bound, proposition
Q: Let \( E \) be a subset of \( \mathbb{R} \). Then \( E \) can have [...].
A: at most one least upper bound
BackOnlyText: This is a simple proposition.
Proof
Let \( E \) be a subset of \( \mathbb{R} \), and let \( M \) and \( M' \) be two upper bounds of \( E \). If both upper bounds are least upper bounds then both \( M \le M' \) and \( M' \le M \), by the definition of least upper bounds. This implies that \( M = M' \) and that there is only a single least upper bound.
Upper bound def → least upper bound def→ uniqueness of least upper bound → existence of least upper bound → supremum def
Source: Tao, Analysis I

Title: Existence of least upper bound, theorem
Q: Let \( E \) be a subset of \( \mathbb{R} \). If \( E \) has some upper bound, then it must have [...].
A: a least upper bound
BackOnlyText: While it may seem banal at first, the result is quite amazing. To appreciate this, consider why a set of rationals does not necessarily have a rational least upper bound.
Proof outline
The proof of this is somewhat involved (but not difficult). The proof (p118) is a great example of the utilization of the properties of Cauchy sequences and equivalent sequences.
It starts by constructing two Cauchy sequences. One sequence, \( (a_n)_{n=1}^{\infty} \), consists of only upper bounds, and the other, \( (b_n)_{n=1}^{\infty} \) only of not upper bounds. There is a relationship between elements in the sequences: \( a_k \) and \( b_k \) differ by a rational \( \frac{1}{n} \). As \( n \) increases, it is shown that both sequences are equivalent and define the same real, \( S \). Then, considering any other upper bound \( M \), it must be greater or equal to all elements of \( (b_n)_{n=1}^{\infty} \), otherwise it would not be an upper bound. Consequently, it must be greater to or equal to \( S \). This makes \( S \) less than or equal to all other upper bounds.
Example: Upper bound def → least upper bound def→ uniqueness of least upper bound → existence of least upper bound → supremum def
Source: Tao, Analysis I

Title: Supremum
Q: Let \( E \) be a subset of \( \mathbb{R} \). If \( E \) is not empty and has some upper bound, we define \( sup(E) \) to be [...].
We introduce two new symbols, \( +\infty, -\infty \), to deal with two special cases. If \( E \) is non-empty and [...], we set \( sup(E) := +\infty \); if \( E \) [...], we set \( sup(E) := -\infty \).
We refer to \( sup(E) \) as the supremum of E, and denote it as sup E.
A: the least upper bound of \( E \); has no upper bound; is empty
BackOnlyText: This definition simply creates syntax to refer to the least upper bound, which was shown to exist in the 'existence of least upper bound' theorem. To insure that the newly defined supremum always exists, \( +\infty \) and \( -\infty \) are introduced to handle the two edge cases where a least upper bound doesn't exist.
Thus the supremum is defined as the least upper bound of a set of reals, if it exists, or one of negative or positive infinity, which exist at the moment as symbobls only.
The least upper bound and supremum are a basic property of the reals that does not exist for the rationals. With them, the \( \sqrt{2} \) can be shown to exist as a real.
Upper bound def →least upper bound def→ uniqueness of least upper bound → existence of least upper bound → supremum def
Source: Tao, Analysis I

Title: Cauchy sequences
Q: A sequence \( (a_n)_{n=m}^{\infty} \) of rational numbers is a Cauchy Sequence iff the sequence is [...] for every [...] \( \varepsilon > 0 \).
A: eventually ε-steady; rational
BackOnlyText: In other words, a sequence \( a_0, a_1, a_2, ... \) of rational numbers is a Cauchy Sequence iff for every rational \( \varepsilon > 0 \) there exists an integer N such that \( d(a_j - a_k) \le \varepsilon \) for every \( j, k \ge N \). Where \( d(a,b) = |a - b| \).
From sequences to reals
sequence → ε-steady sequence → eventually ε-steady sequence → Cauchy sequence → ε-close sequences → eventually ε-close sequences → equivalent sequences → real numbers.
Example: Proposition
The sequence \( a_1, a_2, a_3, ... \) defined by \( a_n := \frac{1}{n} \) (i.e., the sequence \( 1, \frac{1}{2}, \frac{1}{3}, ... \)) is a Cauchy sequence.
Proof
We have to show that for every \( \varepsilon > 0 \) there exists an integer \( N > 0 \) such that the sequence \( (a)_{N}^{\infty} \) is ε-steady, which would mean that \( d(a_j, a_k) \le \varepsilon \) for all \( j, k \ge N \):
\[ | \frac{1}{j} - \frac{1}{k}| \le \varepsilon \text{ for every } j, k \le N \]
As \( j, k \ge N \), we know that \( 0 \le \frac{1}{j}, \frac{1}{k} \le \frac{1}{N} \), so \( |\frac{1}{j} - \frac{1}{k}| \le \frac{1}{N} \). Thus, to force \( |\frac{1}{j} - \frac{1}{k}| \) to be less than or equal to ε, it is sufficient to choose an N such that \( \frac{1}{N} \) is less than ε, or in other words, an N such that \( N \gt \frac{1}{\varepsilon} \), and this can be done due to Prop. 4.4.1 (interspersing of integers by rationals).
Source: Tao, Analysis I

Title: Eventual ε-steadiness
Q: Let \( \varepsilon > 0 \) be a rational. A sequence \( (a)_{n=m}^{\infty} \) is eventually ε-steady iff the sequence [...] is ε-steady for some integer [...].
A: \( a_N, a_{N+1}, a_{N+2}, ... \); \( N \ge m \)
BackOnlyText: In other words, a sequence \( (a)_{n=m}^{\infty} \) is eventually ε-steady iff there exists an integer \( N \ge m \)  such that \( |a_j - a_k| \le \varepsilon \) for all \( j, k \ge N \).
From sequences to reals
sequence → ε-steady sequence → eventually ε-steady sequence → Cauchy sequence → ε-close sequences → eventually ε-close sequences → equivalent sequences → real numbers.
Example: The sequence \( 1, 0.1, 0.001, ... \) is not 0.1-steady, but is eventually 0.1-steady. The sequence \( 10, 0, 0, 0, ... \) is not ε-steady for any ε less than 10, but it is eventually ε-steady for every \( \varepsilon > 0 \).
Source: Tao, Analysis I

Title: \( x^2 = 2 \). There exists a positive real whose square is 2. Proposition.
This proposition's proof is enabled by the existance of the least upper bound for sets of reals.
Proof
Let \( E \) be the set \( \{y \in R: y \ge 0 \text{ and } y^2 < 2 \} \). \( E \) has an upper bound of 2, as if \( y > 2 \) then \( y^2 > 4 > 2 \). As \( E \) is a non-empty set of reals and has an upper bound, it must have a least upper bound. Let \( x = \sup E \) be the least upper bound of \( E \).
Exactly one of \( x^2 < 2, x^2 = 2 \text{ or } x^2 > 2 \) must be true. We proceed to show that \( x^2 < 2 \) and \( x^2 > 2 \) lead to contradictions and must be false.
\( x^2 < 2 \) must be false.
Assume \( x^2 < 2 \) is true. (This means that \( x \in E \), but we must go further to impose that \( x \) is not an upper bound of \( E \)). Consider \( (x + \varepsilon)^2 \), where \( 0 < \varepsilon < 1 \) is a real:
\[\begin{align}
(x + \varepsilon)^2 &= x^2 + 2x\varepsilon + \varepsilon^2 \\
                              &\le x^2 + 4\varepsilon + \varepsilon^2 && \text{as } x \le 2 \\
                               &\le x^2 + 5\varepsilon && \text{as } \varepsilon < 1 \\
\end{align} \]
We can choose an \( \varepsilon \) such that \( x^2 + 5\varepsilon < 2 \). From above, \( (x + \varepsilon)^2  \le x^2 + 5\varepsilon \), so it must be the case that \( (x + \varepsilon)^2 < 2 \) is also true. This implies that \( x + \varepsilon \in E \). As \( x + \varepsilon > x \), \( x \) cannot be an upper bound of \( E \), which is a contradiction. Thus, \( x^2 < 2 \) is false.
\( x^2 > 2 \) must be false.
A similar argument to above. Assume \( x^2 > 2 \). Consider \( (x - \varepsilon)^2 \), where \( 0 < \varepsilon < 1 \) is a real:
\[\begin{align}
(x - \varepsilon)^2 &= x^2 - 2x\varepsilon + \varepsilon^2 \\
                              &> x^2 - 4\varepsilon + \varepsilon^2 && \text{as } x < 2 \\
                              &> x^2 - 4\varepsilon && \text{as } \varepsilon > 0 \\
\end{align}
\]
Since \( x^2 > 2 \), we can choose an \( \varepsilon \) such that \( x^2 - 4\varepsilon > 2 \) (why? This reasoning tripped me up when I first attempted the proof). In this case, \( (x-\varepsilon)^2 > 2 \) is also true.
We want \( x-\varepsilon \) to be an upper bound, and to show this, it must be true that \(  x - \varepsilon \gt y \) for all \( y \in E \). We show this by contradiction: if \( x-\varepsilon < y \) for some \( y \in E \), then \( (x-\varepsilon)^2 < y^2 < 2 \) , a contradiction. Thus \( x - \varepsilon \) is an upper bound for E.
As \( x - \varepsilon \) is an upper bound for E and is less than \( x \), it cannot be true that \( x \) is a least upper bound. Thus, \( x^2 > 2 \) mus be false.
As both \( x^2 < 2 \) and \( x^2 > 2 \) are false, it must be the case that \( x^2 = 2 \) is true.

Title: \( n^{th} \) root of a real
Q: The definition of the n^th root of a real utilizes the supremum, hence why the \( n^{th} \) doesn't exist for rationals.
Let \( 0 \ge x \) be a non-negative real. Let E be the set [\( E = \{ ? \} \)]. We define the \( n^{th} \) root of \( x \), denoted as \( x^{\frac{1}{n}} \), to be \( x^{\frac{1}{n}} := \sup E \). Thus,
[\[ x^{\frac{1}{n} } := \sup\{y \in \mathbb{R} : ? \]]
A: \( E = \{y \in \mathbb{R} : y \ge 0 \text{ and } y^n < x \} \); \[ x^{\frac{1}{n} } := \sup\{y \in \mathbb{R} : y \ge 0 \text{ and } y^n < x \} \]
BackOnlyText: The existence of the n^th root, proposition
Let \( x \ge 0 \) be a positive real and let \( n \ge 1 \) be a positive integer. Then \( x^{\frac{1}{n}} \) is a real number (it exists).
An easy proof
Outline: we must show that the set \( \{y \in \mathbb{R} : y \ge 0 \text{ and } y^n < x \} \) has an upper bound; if it has an upper bound, it follows that it has a least upper bound and this least upper bound is the n^th root of \( x \), by definition.
Let E be the set \( \{y \in \mathbb{R} : y \ge 0 \text{ and } y^n < x \} \). 0 is an element of E ( \( 0 \ge 0 \text{ and } 0^k = 0 \) for any natural number \( k \)). Thus E is non-empty. E also has an upper bound. To show this consider the two possible cases, \( x < 1 \) and \( x \ge 1 \).
If \( x < 1 \), then 1 is an upper bound of E. If it isn't then \( 1 < y \) for some \( y \in E \). \( y > 1 \) implies that \( y^n > 1 \) so that \( y \) doesn't meet the criteria for being an element of E, a contradiction.
If \( x \ge 1 \), then \( x \) is an upper bound of E. If it isn't then \( y > x \) for some \( y \in E \). This implies that \( y^n > x \), so again, \( y \) cannot be an element of E, a contradiction.
So E must have an upper bound. As it is a non-empty set of reals with an upper bound, it must have a least upper bound. Thus \( x^{\frac{1}{n}} \) is a real number.
Other fields (not shown on cards):
Order: 122

Title: The irrationality of \( \sqrt(2) \)
Q: Consider a rational number fully reduced, \( \frac{p}{q}\) such that \(\frac{p^2}{q^2} = 2\). Investigate whether \(p\) or \(q\) are odd or even numbers leads to a contradiction which shows that \(2\) cannot be rational.
- \(p\) and \(q\) can't both be even, as [...].
- \(p\) and \(q\) must both be odd, as [...].
- \(p\) is even as [...].
These points lead to a contradiction.
A: otherwise they could be simplified; an odd \(\times\) odd is odd, so if \(p\) is odd, then \(q\) must be odd also, as \(p^2\) (odd) will divide by \(q^2\) to produce 2 only if \(q^2\) is also odd. Similarly, if \)q\) is odd, so too must \(p\) be; \(p^2\) is even. This can be seen by rearranging the original expression to get \(p^2 = 2q^2\)

Title: Cut property for real numbers
Q: Cut property for real numbers
If \( A \) and \( B \) are nonempty, disjoint sets with \( A \cup B = \mathbb{R} \) and \( a < b \) for all \( a \in A \) and \( a < b \), then there exists a [...] such [...].
This a Dekekind cut and Dedekind completeness.
The cut property expresses the idea that sets of reals always have a boundary which is itself a real.
A: \( c \in \mathbb{R} \); that \( x \le c \) whenever \( x \in A \) and \( x \ge c \) whenever \( x \in B \)
BackOnlyText: Cut property is equivalence to the idea of supremum
The cool thing about the cut property is that it that it is equivalent to 'Cauchy completeness', which is what Tao covered (every Cauchy sequence of real numbers converges to a real number). In addition, both of these are logically equivalent to the existence of supremum & infimum for sets of reals. The cut property feels a bit more tangible compared to the supremum & infimum, and it can be good to remember that the property encapsulates the same power.
In "Understanding Analysis", Stephen Abbott introduces the existence of supremums and infimums as an axiom which imbues the reals with completeness. He points out that the cut property instead could have been given as an axiom, and the existence of the supremum and infimum would then be implied.
Source: Abbott, p19

Title: Nested intervals property for reals
Q: Nested interval property
A sequence of nested closed intervals of reals has a [something something].
In other words, if one considers a sequence of nested closed intervals \( (I_n)_{n=0}^{\infty} \) such that for each \( n \in \mathbb{N} \), \( I_n = [a_n, b_n] \) for some \( a_n, b_n \in \mathbb{R} \) and \( I_{n+1} \subseteq I_n \), then It holds that [\( \cap_{n=1}^{\infty} I_n \;\; ? \quad  ? \;\;  \)].
Proof on the reverse.
A: nonempty intersection; \( \cap_{n=1}^{\infty} I_n \neq \emptyset \)
BackOnlyText: Proof
The proof uses the property that sets of reals have a supremum; this is used to find an \( x \in \mathbb{R} \) such that \( x \in \cap_{n=0}^{\infty} I_n \).
Proof. Let \( A = \{a_n : n \in \mathbb{N} \} \) be the set of left-endpoints for the intervals. Note that for every \( i \in \mathbb{N} \), \( b_i \) is an upper bound for \( A \). Let \( x = \sup A \).
- \( \forall i \in \mathbb{N}, x \le b_i \), as \( x \) is the least upper bound for \( A \), and every \( b_i \) is an upper bound.
- \( \forall i \in \mathbb{N}, x \ge a_i \), as \( x \) is an upper bound for \( A \) and all \( a_i \in A \).
So, we have \( \forall i \in \mathbb{N}, a_n \le x \le b_n \), which means that \( \forall i \in \mathbb{N}, x \in I_n \), and furthermore, \( x \in \cap_{n=0}^{\infty} I_n \) meaning that this intersection is not empty.
Visualization of the interval endpoints
Source: Abbott, p20

Title: Archimedean property
The Archimedean property is a property of the reals. Roughly, it says: there are no infinitely small or infinitely large elements, when compared to the rationals. It also says that the naturals are not bounded by any real.
Archimedean property
1. Given any number \( x \in \mathbb{R} \), there exists an \( n \in \mathbb{N} \) satisfying \( n > x \).
2. Given any real number \( x > 0 \), there exists an \( n \in \mathbb{N} \) satisfying \( \frac{1}{n} < x \).
This is Abbott's description of the Archimedean property.
Tao presents the more common form which goes like:
For any two positive reals \( x \) and \( y \) there exists a natural number \( n \) such that \( ny > x \).
Density of \( \mathbb{Q} \) in \( \mathbb{R} \)
The Archimedean property leads to an important result:
For any reals \( a, b \in \mathbb{R} \) such that \( a < b \) there exists a rational \( q \in \mathbb{Q} \) such that \( a < q < b \).
BackOnlyText: Proof
The below proof assumes the truth of two statements:
1. Interspersing of integers by rationals: let \( x \) be a rational number, then there exits an integer \( n \) such that \( n \ge x \ge n+1 \).
2. Cauchy sequences (sequences of rationals) are bounded (by rationals). Note that a sequence \( (a_n)_{n=1}^{\infty} \) is bounded by \( M \in \mathbb{Q} \) iff \( \forall i \in \mathbb{N}, |a_i| \le M \).
Proof.
- First prove 1. Let \( x \in \mathbb{R} \). \( x \) is a Cauchy sequence of rationals, and is thus bounded by rationals. Any rational is also bounded by integers, so we have the desired result by transitivity.
- For part 2, rewrite as \( n > \frac{1}{y} \) and apply part 1.
Density of \(\mathbb{Q} \) in \( \mathbb{R} \), proof.
Source: Tao, p114, p115
Abbott, p21