# On ultraproducts of topological spaces

Intuitively, I understand the construction of the hyperreals by ultraproducts as a process of turning the limit operation into an algebraic object. More precisely, to check the existence of the limit $\lim_{x\to a} f(x)$, is equivalent to calculate $\overline f(\tilde a)$ for every $\tilde a$ such that $\tilde a\neq a$ and, $a-u<\tilde a<a+u,\ \forall u>0$, and check if its projection on $\mathbb R$ is the same for every $\tilde a$ of that form; and if it does, then the common value is the limit.

I'd like to work with this kind of tecnique, but for general topological spaces. However, it is not that simple when the topological space has points of different character. Like, a point that admits countable local basis and other point that does not admit it.

When we construct the hyperreals we take a non-principal ultrafilter $\mathcal U\subset\mathcal P(\mathbb N)$ containing the cofinite subsets of $\mathbb N$. The convergence with respect to this ultrafilter is equivalent to the convergence of a sequence in the classical sense. Since every point in $\mathbb R$ has a local basis $\mathcal B$ such that $(\mathcal B,\supset) \equiv(\mathbb N,\leq)$, the ultrapower of $\mathbb R$ by $\mathcal U\subset\mathcal P(\mathbb N)$ "translates" the convergence to every point $a\in\mathbb R$ via the elements $\tilde a$ such that $\tilde a\neq a$ and, $a-u<\tilde a<a+u,\ \forall u>0$, as I said in the first paragraph. This kind of idea should work analogously in any first countable topological space.

But what about topological spaces that are not first countable? I know, a priori, that if every point $x\in X$ has a local basis $\mathcal B$ such that $(\mathcal B,\supset) \equiv(I,\leq)$, for a common directed set $(I,\leq)$, then, we may pick an ultrafilter $\mathcal U\subset \mathcal P(I)$ containing the subsets of the form $S_i=\{j\in I:i\leq j\}$ and construct the ultrapower of $X$ by $\mathcal U$. This construct works in the way I'd like to. However, it relies on the hypothesis that

every point $x\in X$ has a local basis $\mathcal B$ such that $(\mathcal B,\supset) \equiv(I,\leq)$, for a common directed set $(I,\leq)$.

I'd like to know if there is a construction like that for any topological space $X$, or at least for a good class of them, say compact or locally compact and Hausdorff, not requiring this hypothesis.

• Could you make it more clear exactly what you are asking? Your way of describing the ultrapower construction is not exactly how I think about it, and so I am not sure exactly what you mean when you say that you want to know "if there is a construction like that". In general, one can take the ultrapower of any given mathematical structure, even the entire set-theoretic universe, and not just by ultrafilters on $\mathbb{N}$, but by any ultrafilter on any set. This ultrapowers will have certain properties and relate to the original structures in certain ways. So what is it that you want to know? – Joel David Hamkins May 23 '17 at 15:16
• Andre. Have you looked at the work on Paul Bankston? Paul Bankston has spent his entire career looking at ultraproducts of topological spaces. – Joseph Van Name May 23 '17 at 15:47
• @JoelDavidHamkins, the construction of ultraproduct by filters on $\mathbb N$ "codifies" the notion of convergence into some algebraic objects (the ones called "infinitesimals"), like the sequence $\tilde a= (1,1/2,1/3,...,1/n,...)$ which is different of $0$ but greater than any other real number greater than $0$. When I calculate $\overline f(\tilde a)$, it gives the hyperreal $(f(1),f(1/2),...,f(1/n),...)$. If $\lim f(1/n)=b$, then $\overline f(\tilde a)= (b,b,b,...,b,...) + \alpha$, for some $\alpha$ which different of $0$ and contained in $(-a,a)$ for every real number $a>0$. – André Porto May 23 '17 at 22:29
• @JoelDavidHamkins, I'd like to do this same approach to calculate limits in an arbitrary topological spaces. For that, I don't really need to define a topology in the ultraproduct, I'd like just to translate limits via "infinitesimals" just like we are able to do in the hyperreals, as I ilustrated above. – André Porto May 23 '17 at 22:33
• The limit of a sequence in the reals is invariant under passing to a subsequence (and re-indexing), but the ultrapower representation is not. Also, limits of a sequence are invariant under arbitrary permutations of the sequence, but this is not true for the ultrapower representation. So I wonder how that affect your underlying metaphor. – Joel David Hamkins May 23 '17 at 22:47

See the paper [1] for an introduction of the ultraproduct construction of topological spaces. Suppose that $(X_{i},\mathcal{T}_{i})$ is a topological space for each $i\in I$ and $M$ is an ultrafilter on $I$. Then $\prod_{i\in I}X_{i}/M$ may be given the topology generated by $\prod_{i\in I}U_{i}/M$ where $U_{i}\in\mathcal{T}_{i}$ for $i\in I$ which is called the ultraproduct of the topological spaces $(X_{i},\mathcal{T}_{i})$. The ultraproduct construction preserves separation axioms including Hausdorff axioms, regular spaces, and completely regular spaces. The ultraproduct of regular spaces by a non-$\sigma$-complete ultrafilter is always a $P$-space (A $P$-space is a regular space where every $G_{\delta}$-set is open). Most topological spaces that mathematicians encounter in nature are not $P$-spaces, so I am unsure if you would consider these spaces as being natural (I consider the ultraproduct construction of new topological spaces to be natural simply because it preserves the separation axioms. Furthermore, since ultrafilters are ). If you want a construction that preserves compactness, then you should look at the ultracoproduct construction instead of the ultraproduct. If you want a construction that preserves metric spaces (or more generally uniform spaces (uniform spaces have local bases which can be index by the same ordering as you want)), then you should probably look at the notion of an ultralimit of metric spaces.