There are a number of theorems in various settings that link continuity of a function \(f\colon X \to Y\) to closedness of that function’s graph – i.e., the set of all points \((x,y)\in X \times Y\) such that \(y = f(x)\).
For example, we learn in functional analysis that if \(f \colon X \to Y\) is a linear map between Banach spaces, then \(f\) is continuous if and only if its graph is a closed subspace of \(X \times Y\).
Normally, when we learn this fact, we also learn a particular way of looking at it. We know that a function \(f \colon X \to Y\) is continuous if and only if whenever \(x_n \to x\) is a convergent sequence in \(X\), we have \(f(x_n) \to f(x)\) in \(Y\).
Meanwhile, to say that the graph of \(f\) is closed means that it contains all its limit points in \(X \times Y\): i.e., the graph of \(f\) is closed if and only if whenever \(x_n \to x\) is a convergent sequence in \(X\) such that \(f(x_n) \to y\) for some \(y\), we have \(y = f(x)\).
It’s clear then, that continuity of \(f\) is a stronger statement than closedness of its graph: to prove that \(f\) has a closed graph, we can proceed exactly as we would for proving continuity of \(f\), except that we only need to consider those convergent sequences \(x_n \to x\) such that \(f(x_n)\) converges to some limit \(y\).
Let’s turn this into a formal proof: we’re given a continuous function \(f \colon X \to Y\) between (say) normed spaces \(X\) and \(Y\), and we want to prove that the graph of \(f\) is closed. Let \(x_n \to x\) be a convergent sequence in \(X\), and suppose that \(f(x_n) \to y\) in \(Y\). Then, by continuity of \(f\), we know that \(f(x_n) \to f(x)\). Therefore, \(y = f(x)\) by uniqueness of limits.
This last part – uniqueness of limits in \(Y\) – plays an essential role, which we see when we try to generalize this implication to arbitrary topological spaces. In more general topological spaces, sequences can converge to more than one limit – and, indeed, it is possible for a continuous function to have a graph that is not closed.
However, if \(Y\) is a Hausdorff space, then limits of sequences in \(Y\) are unique, and we get our first topological closed graph theorem.
Exercise 1. Let \(X,Y\) be topological spaces, where \(Y\) is a Hausdorff space. Prove that if \(f \colon X \to Y\) is a continuous function, then the graph of \(f\) is closed.
More exciting is that we can use this as a characterization of Hausdorff spaces: i.e., if \(Y\) is a topological space such that the graph of any continuous function into \(Y\) is closed, then \(Y\) must be a Hausdorff space. In fact, it suffices to consider the identity function \(Y \to Y\).
Exercise 2. Convince yourself that the statement ‘\(Y\) is a Hausdorff space’ is exactly the same thing as saying that the graph of the identity function \(Y \to Y\) is closed in \(Y \times Y\).
That’s nice, but what about the other direction? In functional analysis, we know that the converse holds when \(X\) and \(Y\) are Banach spaces, but the topological version is a bit different.
Exercise 3. Let \(X,Y\) be topological spaces, where \(Y\) is compact. Let \(f\colon X \to Y\) be a function such that the graph of \(f\) is closed in \(X \times Y\). Prove that \(f\) is continuous.
(This is very different from the theorem in analysis: non-zero dimensional Banach spaces are never compact!)
It would be very nice if this theorem gave a characterization of compact topological spaces in the same way that the other direction gave a characterization of Hausdorff spaces: we often see ‘compact’ and ‘Hausdorff’ together as though they were brother concepts, even though the definitions are very different, so the prospect of giving new, more explicitly dual definitions for them is attractive.
Unfortunately, there are spaces \(Y\) that satisfy the conclusion of Exercise 3 (for arbitrary \(X\)) without being compact. As an example, give the set \(\mathbb N\) of natural numbers a topology by declaring the open sets to be the downward closed sets (i.e., those sets \(U\) such that if \(n \in U\) and \(m \le n\) then \(m \in U\)).
Exercise 4. Prove that if \(X\) is a non-empty topological space and \(f \colon X \to \mathbb N\) is a function, then the graph of \(f\) is not closed.
Thus, the conclusion of Exercise 3 is vacuously satisfied for this space \(\mathbb N\): the only function into \(\mathbb N\) that has a closed graph is the empty function, which is obviously continuous. Nevertheless, \(\mathbb N\) with this topology is certainly not compact.
The problem here is that some particularly nasty topological spaces throw up barriers to graphs being closed that are unrelated to whether or not the underlying functions are continuous. If we examine the example above a bit more closely, it becomes clear that the fundamental property of a graph of a function \(X \to Y\) – namely, that it intersects each set \(\{x\} \times Y\) in a single point – is not in general compatible with being a closed set.
To try and get around this, we’ll relax this property by generalizing from functions to relations: i.e., functions that can take multiple values – or no value at all. We can identify a relation between spaces \(X\) and \(Y\) with its graph, which is an arbitrary subset of \(X \times Y\).
Let’s start by trying to extend the definition of continuity to relations. We know that a function \(f \colon X \to Y\) is continuous if and only if \(f^{-1}(U)\) is open in \(X\) whenever \(U\) is an open set in \(Y\).
Definition 5. Let \(R \subset X \times Y\) be a relation. Given a subset \(Z\) of \(Y\), we write \(R^{-1}(Z)\) for the set of all \(x\in X\) such that \((x, z)\in R\) for some \(z\in Z\). We say that \(R\) is continuous if \(R^{-1}(U)\) is open in \(X\) for any open subset \(U\) of \(Y\).
We can readily extend our earlier result to continuous relations.
Exercise 6. Let \(X,Y\) be topological spaces, where \(Y\) is compact. Let \(R\subset X \times Y\) be a closed set. Prove that \(R\) is a continuous relation between \(X\) and \(Y\).
What is more exciting is that we can use this to give a characterization of compactness. We will use an adapted version of a lovely argument due to Martín Escardó (see the brilliantly named Intersections of Compactly Many Open Sets are Open [1]).
Theorem 7. Let \(Y\) be a topological space, and suppose that whenever \(X\) is a topological space and \(R\subset X \times Y\) is a closed space, then \(R\) is a continuous relation between \(X\) and \(Y\). Then \(Y\) is compact.
Proof. Let \((U_\alpha)_{\alpha\in A}\) be an open cover of \(Y\). Define a topological space \(X\) as follows.
- The points of \(X\) are the open sets \(U\) of \(Y\).
- A set \(\mathcal U\subset X\) is open if:
- it is upwards closed – i.e., if \(U \in \mathcal U\) and \(U \subset V\), then \(V \in \mathcal U\); and if
- there is some \(U \in \mathcal U\) such that \(U\) may be written as the union of finitely many of the \(U_\alpha\).
To show that \((U_\alpha)\) admits a finite subcover, it will suffice to prove that the set \(\{Y\}\) is open in \(X\). Since \((U_\alpha)\) was arbitrary, this will suffice to prove that \(Y\) is compact.
First, however, we need to show that this is indeed a valid topology on \(X\). It is easy to see that it is closed under unions. Moreover, if we have two such sets \(\mathcal U\) and \(\mathcal V\), then the intersection \(\mathcal U \cap \mathcal V\) is certainly upwards closed; moreover, we have some \(U \in \mathcal U\) and \(V \in \mathcal V\) such that \(U\) and \(V\) are the union of finitely many of the \(U_\alpha\): then \(U \cup V\) is also the union of finitely many of the \(U_\alpha\), and is contained in both \(\mathcal U\) and \(\mathcal V\) by upwards closedness.
[As an aside, notice how our new definition of compactness doesn’t mention finiteness at all: the argument above shows that this is because it has, in some sense, been ‘cancelled out’ by the fact that open sets in a topological space must be closed under finite intersections. If you’ve ever used compactness in a proof in order to show that a particular collection of open sets can be taken to be finite and therefore to have an open intersection (for example, in Exercise 3), then there’s a good chance that you’d have been better served by using this new definition.]
We will now consider the relation \(\not\ni\) between \(X\) and \(Y\) (i.e., \(U \not\ni x\) if it is not the case that \(x \in U\)). We claim that the graph of this relation is closed in \(X \times Y\).
Indeed, suppose that \(x\in V\) – so \((V, x)\) is in the complement of \(\not\ni\). We seek an open neighbourhood of \((V, x)\) that does not intersect with \(\not\ni\). Indeed, since \((U_\alpha)\) is an open cover of \(Y\), there is some \(U_\beta\) such that \(x\in U_\beta\). Then let \(\mathcal V\) be the set of all open sets in \(Y\) that contain \(U_\beta \cap V\).
\(\mathcal V\) is certainly upwards closed: moreover, it contains \(U_\beta\), so it is open in \(X\). Then the set $$\mathcal V \times (U_\beta \cap V)$$ is an open neighbourhood of \((V, x)\) in \(X \times Y\), and it does not intersect the graph of \(\not\ni\), since every set in \(\mathcal V\) contains every element of \(U_\beta \cap V\) by definition. This completes the proof that \(\not\ni\) is closed in \(X \times Y\).
By hypothesis, we can deduce that \(\not\ni\) is a continuous relation. In particular, this means that \(\not\ni^{-1}(Y)\) is closed in \(Y\). But this is precisely the set of all open subsets \(U \subset Y\) such that \(x \not \in U\) for some \(x\) – i.e., the set of all proper open subsets of \(Y\). So \(\{Y\}\) is the complement of this closed set and is therefore open as desired. \(\Box\)
Unfortunately, it’s not necessarily true that the graph of a continuous relation into a Hausdorff space is always closed, so we don’t quite get that explicitly dual pair of definitions for ‘compact space’ and ‘Hausdorff space’. I think we’ve come pretty close though!
@MISC{MartinCompact,
author = {Martín Escardó},
title = {Intersections of compactly many open sets are open},
year = {2009}
}