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Math and science::Topology

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.

Finally, we arrive at our definition for compact.

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 \( fX \).