Given a set $X$, let $\beta X$ denote the set of ultrafilters. The following theorems are known:
- $X$ canonically embeds into $\beta X$ (by taking principal ultrafilters);
- If $X$ is finite, then there are no non-principal ultrafilters, so $\beta X = X$.
- If $X$ is infinite, then (assuming choice) we have $|\beta X| = 2^{2^{|X|}}$.
These are reminiscent of similar claims that can be made about vector spaces and double duals:
- $V$ canonically embeds into $V^{\star \star}$;
- If $V$ is finite-dimensional, then we have $V = V^{\star \star}$;
- If $V$ is infinite-dimensional, then (assuming choice) we have $\dim(V^{\star \star}) = 2^{2^{\dim(V)}}$.
This suggests that the operation of taking the collection of ultrafilters on a set can be viewed as a double iterate of some 'duality' of sets. Can this be made precise: that is to say, is there some notion of a 'dual' of a set $X$, $\delta X$, such that the following are true?
- The double dual $\delta \delta X$ is (canonically isomorphic to) the set $\beta X$ of ultrafilters on $X$;
- If $X$ is finite, then $|\delta X| = |X|$ (but not canonically so);
- If $X$ is infinite, then (assuming choice) $|\delta X| = 2^{|X|}$.
Apart from the tempting analogy between $\beta X$ and $V^{\star \star}$, further evidence for this conjecture is that $\beta$ can be given the structure of a monad (the 'ultrafilter monad'), and monads can be obtained from a pair of adjunctions.
Best Answer
This is a quite standard idea in functional analysis. Let $X$ be any set and let $c_0(X)$ be the space of all functions from $X$ to $\mathbb{C}$ which go to zero at infinity. Then the algebra homomorphisms from $c_0(X)$ to $\mathbb{C}$ are precisely the point evaluations at elements of $X$, i.e., the spectrum of $c_0(X)$ is naturally identified with $X$.
Going to the second dual we get $l^\infty(X)$, the space of all bounded functions from $X$ to $\mathbb{C}$, whose spectrum is naturally identified with $\beta X$.
[deleted an additional comment which wasn't accurate]