Ultrafilters on a set
\small \rule[-1.5]{0.1}{0.1} X are a bit like generalised points of that set. Of course, principal ultrafilters correspond to points of the set in an obvious way. If \small \rule[-1.5]{0.1}{0.1} X lives in some model of set theory, and we take an ultrapower of that model by some ultrafilter, then in the ultrapower \small \rule[-1.5]{0.1}{0.1} X gains a generic point, with respect to which the ultrafilter is principal.Sometimes we might want to associate actual points of the set to these ultrafilters: We are interested in relations
\small \rule[-1.5]{0.1}{0.1} R between \small \rule[-1.5]{0.1}{0.1} X and the set \small \rule[-1.5]{0.1}{0.1} \beta(X) of ultrafilters on \small \rule[-1.5]{0.1}{0.1} X. Such a relation is a set of ordered pairs \small \rule[-1.5]{0.1}{0.1} (x, {\cal U}) with \small \rule[-1.5]{0.1}{0.1} {\cal U} an ultrafilter on \small \rule[-1.5]{0.1}{0.1} X, and \small \rule[-1.5]{0.1}{0.1} x a point of \small \rule[-1.5]{0.1}{0.1} X. \small \rule[-1.5]{0.1}{0.1} {\cal U} may be thought of as specifying a set of subsets of \small \rule[-1.5]{0.1}{0.1} X to which some imaginary point belongs. Unless \small \rule[-1.5]{0.1}{0.1} {\cal U} is the principal ultrafilter at \small \rule[-1.5]{0.1}{0.1} x, there will be some sets containing \small \rule[-1.5]{0.1}{0.1} x but not in \small \rule[-1.5]{0.1}{0.1} {\cal U}: Since such sets are witnesses of the fact that \small \rule[-1.5]{0.1}{0.1} {\cal U} isn't \small \rule[-1.5]{0.1}{0.1} x, and so I'll call them inconsistent with the pairing \small \rule[-1.5]{0.1}{0.1} (x, {\cal U}). All other sets are consistent with the pairing.This consistency relation induces a Galois connection from the set of relations
\small \rule[-1.5]{0.1}{0.1} R of the type described above to the power set of the power set of \small \rule[-1.5]{0.1}{0.1} X. It is here, on this bleak mountaintop of abstraction, that there is a surprise. The sets of subsets of \small \rule[-1.5]{0.1}{0.1} X which are closed with respect to this connection are precisely the topologies on \small \rule[-1.5]{0.1}{0.1} X.Proof: Let
\small \rule[-1.5]{0.1}{0.1} {\cal T} be a set of subsets of \small \rule[-1.5]{0.1}{0.1} X closed with respect to the connection. Then there is a relation \small \rule[-1.5]{0.1}{0.1} R which is taken to \small \rule[-1.5]{0.1}{0.1} {\cal T} by the connection. That is, \small \rule[-1.5]{0.1}{0.1} {\cal T} is the set of subsets of \small \rule[-1.5]{0.1}{0.1} X consistent with \small \rule[-1.5]{0.1}{0.1} R; \small \rule[-1.5]{0.1}{0.1} {\cal T} is the set of sets \small \rule[-1.5]{0.1}{0.1} O such that, for all \small \rule[-1.5]{0.1}{0.1} x and \small \rule[-1.5]{0.1}{0.1} {\cal U} with \small \rule[-1.5]{0.1}{0.1} xR{\cal U}, if \small \rule[-1.5]{0.1}{0.1} x \in O then \small \rule[-1.5]{0.1}{0.1} O \in {\cal U}. In particular, as the empty set \small \rule[-1.5]{0.1}{0.1} \emptyset contains no points, it is in \small \rule[-1.5]{0.1}{0.1} {\cal T}. As \small \rule[-1.5]{0.1}{0.1} X is in every ultrafilter, \small \rule[-1.5]{0.1}{0.1} X \in {\cal T}. If \small \rule[-1.5]{0.1}{0.1} A and \small \rule[-1.5]{0.1}{0.1} B are in \small \rule[-1.5]{0.1}{0.1} {\cal T}, \small \rule[-1.5]{0.1}{0.1} xR{\cal U}, and \small \rule[-1.5]{0.1}{0.1} x \in A \cap B, then \small \rule[-1.5]{0.1}{0.1} x is in both, so both are in \small \rule[-1.5]{0.1}{0.1} {\cal U}. But then \small \rule[-1.5]{0.1}{0.1} A \cap B \in {\cal U}, as \small \rule[-1.5]{0.1}{0.1} {\cal U} is an ultrafilter. That is, \small \rule[-1.5]{0.1}{0.1} A \cap B \in {\cal T}. If each set in a family \small \rule[-1.5]{0.1}{0.1} ({\cal U}_i)_{i \in I} is in \small \rule[-1.5]{0.1}{0.1} {\cal T}, and \small \rule[-1.5]{0.1}{0.1} x \in \bigcup_{i \in I}{\cal U}_i, then \small \rule[-1.5]{0.1}{0.1} x is in one of them, so one of them (and hence their union) is in \small \rule[-1.5]{0.1}{0.1} {\cal U}. That is, \small \rule[-1.5]{0.1}{0.1} {\cal T} is closed under arbitrary unions. Putting it all together, \small \rule[-1.5]{0.1}{0.1} {\cal T} is a topology on \small \rule[-1.5]{0.1}{0.1} X.Suppose that
\small \rule[-1.5]{0.1}{0.1} {\cal T} is a topology on \small \rule[-1.5]{0.1}{0.1} X, and let \small \rule[-1.5]{0.1}{0.1} R be the relation \small \rule[-1.5]{0.1}{0.1} {\cal T} is taken to by the Galois connection, and suppose that the connection takes \small \rule[-1.5]{0.1}{0.1} R to \small \rule[-1.5]{0.1}{0.1} {\cal T}'. It is enough to show that \small \rule[-1.5]{0.1}{0.1} {\cal T}' = {\cal T}. Evidently \small \rule[-1.5]{0.1}{0.1} {\cal T} \subseteq {\cal T}', so it's enough to show that, for any set \small \rule[-1.5]{0.1}{0.1} C \not\in {\cal T}, we have \small \rule[-1.5]{0.1}{0.1} C \not\in {\cal T}'. Let \small \rule[-1.5]{0.1}{0.1} C be such a set, and let \small \rule[-1.5]{0.1}{0.1} O be the interior of \small \rule[-1.5]{0.1}{0.1} C. As \small \rule[-1.5]{0.1}{0.1} C isn't open, there is some \small \rule[-1.5]{0.1}{0.1} x \in C \setminus O. Let \small \rule[-1.5]{0.1}{0.1} {\cal F} be the set of all open neighbourhoods of \small \rule[-1.5]{0.1}{0.1} x, together with the complement of \small \rule[-1.5]{0.1}{0.1} C. Any finite intersection of sets in \small \rule[-1.5]{0.1}{0.1} {\cal F} is nonempty, so \small \rule[-1.5]{0.1}{0.1} {\cal F} can be extended to an ultrafilter \small \rule[-1.5]{0.1}{0.1} {\cal U}. Any neighbourhood of \small \rule[-1.5]{0.1}{0.1} x is in \small \rule[-1.5]{0.1}{0.1} {\cal U}, so \small \rule[-1.5]{0.1}{0.1} xR{\cal U}. But \small \rule[-1.5]{0.1}{0.1} C \not\in {\cal U}, so \small \rule[-1.5]{0.1}{0.1} C isn't in \small \rule[-1.5]{0.1}{0.1} {\cal T}', as required.This result is remarkable enough, but there's more. It turns out that compactness and Hausdorffness correspond closely with similar properties of relations. Say a relation
\small \rule[-1.5]{0.1}{0.1} R is surjective if, for every \small \rule[-1.5]{0.1}{0.1} {\cal U} there is at least one \small \rule[-1.5]{0.1}{0.1} x with \small \rule[-1.5]{0.1}{0.1} xR{\cal U}. Say \small \rule[-1.5]{0.1}{0.1} R is injective if, for any \small \rule[-1.5]{0.1}{0.1} {\cal U}, there is at most one \small \rule[-1.5]{0.1}{0.1} x with \small \rule[-1.5]{0.1}{0.1} xR{\cal U}. If \small \rule[-1.5]{0.1}{0.1} R is a function, these definitions are exactly the usual definitions of injectivity and surjectivity.Claim
\small \rule[-1.5]{0.1}{0.1} 1: Let \small \rule[-1.5]{0.1}{0.1} R be a relation, as above, and let \small \rule[-1.5]{0.1}{0.1} {\cal T} be the topology that \small \rule[-1.5]{0.1}{0.1} R is taken to by the Galois connection. Then \small \rule[-1.5]{0.1}{0.1} {\cal T} is compact if \small \rule[-1.5]{0.1}{0.1} R is surjective.Proof: By contradiction. Pick any open cover of
\small \rule[-1.5]{0.1}{0.1} X with no finite subcover, and let \small \rule[-1.5]{0.1}{0.1} {\cal F} be the set of complements of the sets in the cover. Then any finite intersection of sets in \small \rule[-1.5]{0.1}{0.1} {\cal F} is nonempty, so \small \rule[-1.5]{0.1}{0.1} {\cal F} may be extended to an ultrafilter \small \rule[-1.5]{0.1}{0.1} {\cal U} on \small \rule[-1.5]{0.1}{0.1} X. By surjectivity, there is some point \small \rule[-1.5]{0.1}{0.1} x with \small \rule[-1.5]{0.1}{0.1} xR{\cal U}. \small \rule[-1.5]{0.1}{0.1} x must lie in some set \small \rule[-1.5]{0.1}{0.1} O of the original cover. But \small \rule[-1.5]{0.1}{0.1} O can't be in \small \rule[-1.5]{0.1}{0.1} {\cal U} (its complement is), contradicting the definition of \small \rule[-1.5]{0.1}{0.1} {\cal T}.Claim
\small \rule[-1.5]{0.1}{0.1} 2: Let \small \rule[-1.5]{0.1}{0.1} {\cal T} be any topology on \small \rule[-1.5]{0.1}{0.1} X, and let \small \rule[-1.5]{0.1}{0.1} R be the relation that \small \rule[-1.5]{0.1}{0.1} {\cal T} is taken to by the Galois connection. Then \small \rule[-1.5]{0.1}{0.1} {\cal T} is compact iff \small \rule[-1.5]{0.1}{0.1} R is surjective.Proof: The 'if' follows from Claim
\small \rule[-1.5]{0.1}{0.1} 1 and the fact that \small \rule[-1.5]{0.1}{0.1} {\cal T} is closed with respect to the Galois connection. To prove the 'only if', suppose that \small \rule[-1.5]{0.1}{0.1} {\cal T} is compact, and let \small \rule[-1.5]{0.1}{0.1} {\cal U} be any ultrafilter on \small \rule[-1.5]{0.1}{0.1} X. Suppose for a contradiction that every \small \rule[-1.5]{0.1}{0.1} x \in X has an open neighbourhood not in \small \rule[-1.5]{0.1}{0.1} {\cal U. These neighbourhoods form an open cover, which therefore has a finite subcover. The complements of the sets in this subcover are in \small \rule[-1.5]{0.1}{0.1} {\cal U}, and their intersection is empty, contradicting the fact that \small \rule[-1.5]{0.1}{0.1} {\cal U} is an ultrafilter. So there is an \small \rule[-1.5]{0.1}{0.1} x such that every open neighbourhood of \small \rule[-1.5]{0.1}{0.1} x is in \small \rule[-1.5]{0.1}{0.1} {\cal U}, so that \small \rule[-1.5]{0.1}{0.1} xR{\cal U}. As \small \rule[-1.5]{0.1}{0.1} {\cal U} was arbitrary, \small \rule[-1.5]{0.1}{0.1} R is surjective.Claim
\small \rule[-1.5]{0.1}{0.1} 3: Let \small \rule[-1.5]{0.1}{0.1} R be a relation, as above, and let \small \rule[-1.5]{0.1}{0.1} {\cal T} be the topology that \small \rule[-1.5]{0.1}{0.1} R is taken to by the Galois connection. Then \small \rule[-1.5]{0.1}{0.1} R is injective if \small \rule[-1.5]{0.1}{0.1} {\cal T} is Hausdorff.Proof: By contradiction. Let
\small \rule[-1.5]{0.1}{0.1} {\cal U} be an ultrafilter, and let \small \rule[-1.5]{0.1}{0.1} x \neq y \in X be such that \small \rule[-1.5]{0.1}{0.1} xR{\cal U} and \small \rule[-1.5]{0.1}{0.1} yR{\cal U}. Then we can find disjoint open sets \small \rule[-1.5]{0.1}{0.1} O and \small \rule[-1.5]{0.1}{0.1} P with \small \rule[-1.5]{0.1}{0.1} x \in O and \small \rule[-1.5]{0.1}{0.1} y \in P. Then \small \rule[-1.5]{0.1}{0.1} xR{\cal U} implies that \small \rule[-1.5]{0.1}{0.1} O \in {\cal U}, and \small \rule[-1.5]{0.1}{0.1} yR{\cal U} implies that \small \rule[-1.5]{0.1}{0.1} P \in {\cal U}. But \small \rule[-1.5]{0.1}{0.1} O \cap P = \emptyset \not\in {\cal U}, contradicting the fact that \small \rule[-1.5]{0.1}{0.1} {\cal U} is an ultrafilter.Claim
\small \rule[-1.5]{0.1}{0.1} 4: Let \small \rule[-1.5]{0.1}{0.1} {\cal T} be any topology on \small \rule[-1.5]{0.1}{0.1} X, and let \small \rule[-1.5]{0.1}{0.1} R be the relation that \small \rule[-1.5]{0.1}{0.1} {\cal T} is taken to by the Galois connection. Then \small \rule[-1.5]{0.1}{0.1} R is injective iff \small \rule[-1.5]{0.1}{0.1} {\cal T} is Hausdorff.Proof: The 'if' part follows from Claim
\small \rule[-1.5]{0.1}{0.1} 3 and the fact that \small \rule[-1.5]{0.1}{0.1} {\cal T} is closed with respect to the Galois connection. To prove the 'only if', suppose \small \rule[-1.5]{0.1}{0.1} {\cal T} isn't Hausdorff, and let \small \rule[-1.5]{0.1}{0.1} x and \small \rule[-1.5]{0.1}{0.1} y in \small \rule[-1.5]{0.1}{0.1} {\cal T} be distinct but not separated by any pair of open sets. Let \small \rule[-1.5]{0.1}{0.1} {\cal F} be the set of open sets containing either \small \rule[-1.5]{0.1}{0.1} x or \small \rule[-1.5]{0.1}{0.1} y. Any finite intersection of sets in \small \rule[-1.5]{0.1}{0.1} {\cal F} is an intersection of an open set containing \small \rule[-1.5]{0.1}{0.1} x with one containing \small \rule[-1.5]{0.1}{0.1} y, so is nonempty. Hence \small \rule[-1.5]{0.1}{0.1} {\cal F} can be extended to some ultrafilter \small \rule[-1.5]{0.1}{0.1} {\cal U}. Then \small \rule[-1.5]{0.1}{0.1} xR{\cal U} and \small \rule[-1.5]{0.1}{0.1} yR{\cal U}, so \small \rule[-1.5]{0.1}{0.1} R isn't injective.The converses to claims
\small \rule[-1.5]{0.1}{0.1} 1 and \small \rule[-1.5]{0.1}{0.1} 3 are false. This remarkable pattern is a shadow of a pair of adjoint functors, which I hope to say a little more about soon.
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