Continuity, inverse images, and preserving closure

Here we address the standard characterization of a continuous function between subset spaces: The inverse image of an open (closed, etc.) set is open (closed, etc.). Why do we use inverse images? Certainly the inverse-image criterion has a lot of support. Inverse images preserve unions, intersections, and complements while images preserve only unions. Also, the inverse-image criterion captures metric-space continuity exactly. Still, the first of these is circumstantial and the second is a posteriori. Meanwhile, images preserve compactness, connectedness, and unions, so there is evidence that using images could be fruitful. How might we decide between these options a priori?

First, let’s discuss the interpretation of open and closed sets. It is common to interpret open sets as measuring closeness: “Close” points share more open sets than “far” points; in particular, in a TVS, points in an open neighborhood of \(\vec0\) are “small” (being “close to zero”). Of course, the whole space is an open neighborhood of \(\vec0\), so we find this interpretation unsatisfactory.

Here is an alternative. Taking a cue directly from the word, we interpret closed sets as closed under some unspecified operation. Likewise, open sets prevent their contents from being absorbed by that operation from without. The obvious “operation” here is the limit of a sequence but, more generally, one can consider any (finitary, \((n+1)\)-ary) relation \(R\) where, for a subset \(C\subseteq X\), we say \(C\) is closed under \(R\) in \(X\) (or \(R\)-closed in \(X\)) if, whenever \(x_i\in C\) and \(\vec{x}Ry\), then \(y\in C\). If you wish, you may think of \(R\) as the relational version of a function equation \(g(\vec{x})=y\). Then being \(R\)-closed is the same as being closed under \(g\).

At the same time, for \((n+1)\)-ary relations \(Q\) and \(R\) on sets \(U\) and \(X\) respectively, we already have the notion of a homomorphism \(f\) from \(\langle U,Q\rangle\) to \(\langle X,R\rangle\) : \(\vec{u}Qv\Rightarrow f(\vec{u})R f(v)\) (with the obvious meaning for \(f(\vec{u})\)). Is this equivalent to preservation of closure by inverse images? Yes, provided the relation \(R\) is reflexive and transitive, which, as we will see, is a reasonable condition. One direction of the equivalence does not require these properties:

Lemma 1. If a function f is a homomorphism between \((n+1)\)-ary relations, then the inverse images of closed sets are closed. That is, if \(f:\langle U,Q\rangle\hat{\rightarrow}\langle X,R\rangle\) and \(C\subseteq X\) is \(R\)-closed in \(X\), then \(f^<C\) is \(Q\)-closed in \(U\).

Proof. Suppose \(f:\langle U,Q\rangle\hat{\rightarrow}\langle X,R\rangle\) and \(C\subseteq X\) is \(R\)-closed in \(X\). We will show that the inverse image of \(C\) is \(Q\)-closed in \(U\).

Suppose now that \(u_i\in f^<C\) and \(\vec{u}Rv\). Then (by definition of inverse image and the hypothesis of homomorphism), we have \(f(u_i)\in C\) and \(f(\vec{u})S f(v)\). Then \(f(v)\in C\) (because \(C\) is \(R\)-closed by hypothesis), whence \(v\in f^<C\). This shows that \( f^<C\) is \(Q\)-closed in \(U\). Thus, inverse images of closed sets are closed, as promised.

The reverse implication is too much to ask without additional conditions, alas. Consider the structures

\(\langle\Bbb N,=\rangle\) i.e. \(Q\) is eQuality (the diagonal of \({\Bbb N}^2\)) and
\(\langle\Bbb N,R\rangle\) where \(R\) comprises all ordered pairs except those the diagonal.

Then every subset of \(\Bbb N\) is \(Q\)-closed while the only \(R\)-closed subsets are Ø and \(\Bbb N\) itself. Thus, the identity map from \(\langle\Bbb N,=\rangle\) to \(\langle\Bbb N,R\rangle\) has the inverse-image property but does not preserve the relation.

We can accommodate this setback by requiring that the relations be reflexive and transitive. This is not burdensome because (1) it is reasonable to consider transitive relations in the context of closure (the closed sets are simply generated in one step instead of successive steps) and (2) a relation’s closed sets are the same as those of its transitive closure.

To keep the exposition manageable, we will use only binary relations; it is straightforward (but really tedious) to generalize the exposition to multary relations . (An \((n+1)\)-ary relation \(R\) is transitive iff \(\vec{x}Ry_i\) and \(\vec{y}Rz\) entail \(\vec{x}Rz\). It is reflexive iff \(\vec{x}Rx_i\) for each \(i\).)

Let’s verify assertion (2) above. It has no bearing on the ultimate result but, again, it is evidence that requiring transitivity is reasonable (which is necessary for me to consider the inverse-image criterion "right").

Proposition. Suppose \(R\) is the transitive closure of \(R_0\) in \(X\). Then, for any subset \(C\) of \(X\), \(C\) is \(R_0\)-closed in \(X\) iff \(C\) is \(R\)-closed in \(X\). Thus, \(R_0\) and \(R\) have exactly the same closed sets, whence we deduce that the \(R_0\)-closure of \(A\) = the \(R\)-closure of \(A\).

Proof. Given \(X\), \(R_0\), \(R\), and \(C\) as above, we will show that \(C\) is \(R_0\)-closed in \(X\) iff \(C\) is \(R\)-closed in \(X\).

\((\Rightarrow)\) Suppose \(C\) is \(R_0\)-closed in \(X\). To show that \(C\) is \(R\)-closed in \(X\), we suppose further that \(xRy\) where \(x\in C\). By definition of transitive closure, there is a finite sequence \(xR_0z_1, z_1R_0z_2, z_2R_0z_3,\ldots,z_nR_0y\). An easy induction shows that all \(z_i\)s are in \(C\) (by hypothesis that \(C\) is \(R_0\)-closed). Since \(z_nR_0y\) and \(z_n\in C\), we get \(y\in C\). Thus, \(C\) is \(R\)-closed in \(X\).

\((\Leftarrow)\) Suppose \(C\) is \(R\)-closed in \(X\). Suppose further that \(xR_0y\) where \(x\in C\). \(R\) is an extension of \(R_0\), so \(xRy\). \(C\) is \(R\)-closed by hypothesis, so \(y\in C\). Thus, \(C\) is \(R_0\)-closed in \(X\).

Now suppose \(A\) is a subset of \(X\). The \(R_0\)-closure of \(A\) in \(X\) is the intersection of all \(R_0\)-closed subsets of \(X\) that include \(A\)—but these are the same as the \(R\)-closed subsets of \(X\) that include \(A\). Thus, the \(R_0\)-closure of \(A\) = the \(R\)-closure of \(A\), as promised.

A few more technical lemmata:

Lemma 2. For any subset \(A\subseteq X\), the following two sets are equal and they are both the \(R\)-closure of \(A\) in \(X\).

\(K^+:=\bigcap\{C\supseteq A\ \vert\ C\mbox{ is }R\mbox{-closed}\}\)
\(K^-:=\bigcup C_n\) where \(C_n=\begin{cases} A & (n=0) \\ C_{n-1}\cup\{y\ \vert\ \exists x\in C_{n-1}: xRy\} & (n>0)\end{cases}\)

Sketch of Proof. First, observe that both \(K^+\) and \(K^-\) are \(R\)-closed by easy definition-chasing. Now for the double-inclusion:

- \((K^-\subseteq K^+)\) An easy induction on \(n\) establishes that \(C_n\subseteq K^+\) for all \(n\), whence their union \(\subseteq K^+\).
- \((K^+\subseteq K^-)\) \(K^-\) is \(R\)-closed and \(\supseteq A\), hence it is in the family whose intersection is \(K^+\).

Thus, \(K^+=K^-\), as promised.

Lemma 3. In the notation for \(K^-\) above, \(z\in C_{n+1}\) iff \(z\in A\) or \(\exists x_0,x_1,\ldots,x_n\in C_n: x_0Rx_1R\cdots Rx_nRz\).

Sketch of Proof. Induction on \(n\).

Lemma 4. In the same notation, if \(R\) is reflexive and transitive, then \(C_1\) is the \(R\)-closure of \(A\). That is, \(y\) is in the \(R\)-closure of \(A\) iff there is some \(a\in A\) for which \(aRy\), i.e., the "tower" has only one storey.

Sketch of Proof. More induction.

Lemma 5. Suppose \(Q\), \(R\) are (binary) relations on \(U\), \(X\) respectively, with \(R\) reflexive and transitive, and \(f\) is a function for which inverse images of sets \(R\)-closed in \(X\) are \(Q\)-closed in \(U\). Then \(f\) is a homomorphism from \(\langle U,Q\rangle\) to \(\langle X, R\rangle\).

Proof: Suppose that inverse images of \(R\)-closed subsets of \(X\) are \(Q\)-closed in \(X\). We will show that \(f\) is a homomorphism. To that end, suppose that \(uQv\) in \(U\). We will show that \(f(u) R f(v)\) in \(X\).

Let \(C\) be the \(R\)-closure of \(\{f(u)\}\) in \(X\) (i.e., the union of the tower of nested subsets of \(X\) where each set is one step closer to the complete closure). Obviously \(C\) is \(R\)-closed in \(X\). Since \(f(u)\in\{f(u)\}\subseteq C\), we have \(u\in f^<C\), which is \(Q\)-closed (by hypothesis on \(f\)), whence \(v\in f^<C\) and so \(f(y)\in C\). Since \(R\) is reflexive and transitive, we have \(f(u)Rf(v)\) in \(X\) by Lemma 4, thus proving that \(f\) is a homomorphism from \(\langle U,Q\rangle\) to \(\langle X, R\rangle\), as promised.

Theorem. If \(f:\langle U,Q\rangle\to\langle X,R\rangle\), then equivalent are

The inverse image under \(f\) of a set \(R\)-closed in \(X\) is \(Q\)-closed in \(U\).
\(f\) is a homomorphism of relations.

Proof. Lemmata 1 and 5.