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.
Reasoning steps
The though chain in full. Consider functions such as \( f \) from \( X \to Y \) where \( Y \subset \mathbb{R}^{n} \). Focus on the simple case when \( \mathbb{R}^n = \mathbb{R} \).
1. Define bounded
For the single dimension case, the following are equivalent statements regarding a function \( f : X \to \mathbb{R} \):
- \( f \) is bounded.
- \( fX \) is bounded.
- There exists an \( M \ge 0 \) such that for all \( y \in fX \), \( |y| \le M \).
- There exists an \( M \ge 0 \) such that for all \( x \in X \), \( |f(x)| \le M \).
- 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 \( f : X \to Y \) (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 \( f : X \to Y \subset \mathbb{R} \)
- 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. Every function induces a cover
Form a bounded open ball around every \( y \in Y \subset \mathbb{R} \), for example \( B(y_0 -1, y_0 + 1) \). Each of these balls induces an open ball in the domain \( X \), for example, \( (f^{-1}(y_0 -1), f^{-1}(y_0 + 1) ) \). Collecting every such ball in \( X \) forms a cover of \( X \). The cover will be infinite if \( fX \) is infinite.
7. Impose a finite cover requirement
Enter the defining restriction: if we simply assert that \( X \) is such that every possible cover has a finite subcover, then any continuous function from \( X \) to \( Y \) must induce a finite cover of \( X \), and this allows us to use the idea above of using max over these subsets. If \( Z \subseteq X \) was a finite set such that the neighbourhoods \( (U_x)_{x \in Z} \) covered \( X \), then we could legitimately set \( M = \max_{z \in Z} M_z \).
The requirement of every cover having a finite subcover 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 \( fX \).