7
$\begingroup$

I have recently been reading about the non-standard characterisation of topological spaces, by saying which points of ${^*X}$ are infinitesimally close to which standard points. The theory looks a lot like that of ultrafilter convergence. Let me clarify:

  • A point $x$ lies in the closure of $A$ iff there is an ultrafilter $\mathcal{U}$ on $A$ with $\mathcal{U}\to x$
  • A point $x$ lies in the closure of $A$ iff there is a non-standard $y \in {^*A}$ with $y\simeq x$

Here, ultrafilters play the same role as non-standard points, and convergence plays the same role as the $\simeq$ relation. Another example:

  • X is Hausdorff iff for every ultrafilter $\mathcal{U}$ we have $(\mathcal{U}\to x \mbox{ and } \mathcal{U}\to y) \implies x=y$
  • X is Hausdorff iff for every non-standard $z$ we have $(z\simeq x \mbox{ and } z\simeq y) \implies x=y$

Again we just have to change ultrafilters to non-standard points and convergence to $\simeq$ to get the non-standard version of the property. This analogy also works for compactness and products, yielding the same easy proof of Tychonov.

We have the following correspondence (let $A\subseteq X, x\in X$): $$\begin{aligned} y\in {^*X} &\leftrightarrow \mathcal{U} \mbox{ ultrafilter on } X\\ y \simeq x &\leftrightarrow \mathcal{U} \to x\\ y\in {^*A} &\leftrightarrow \mathcal{U} \mbox{ ultrafilter on } A \mbox{ (i.e. containing $A$)}\\ ^*A &\leftrightarrow \{\mathcal{U}\mbox{ ultrafilter } \mid A \in \mathcal{U} \}\\ ^*x &\leftrightarrow \mbox{ the principal filter on $x$} \end{aligned} $$

My first question is, how far does this analogy go? Is the ultrafilter characterisation a special case of non-standard topology? Can we then also use it to describe the star of functions, relations, ... , i.e. does the above correspondence induce a monomorphism?

My second question is, if we apply this analogy to the space $({^*X}, \mathcal{T}_S)$ carrying the standard topology (which is compact if the extention is saturated), then we get the space of all ultrafilters on $X$ with basis open sets $\{\mathcal{U} \mid G \in \mathcal{U} \}$ for all opens sets $G \subseteq X$. Is this space compact (like the analoguous space in the non-standard setting)? Does this construction have a name, in the ultrafilter setting? According to this analogy, its Hausdorff reflection should be the Stone-Cech compactification $\beta X$. This does agree with the ultrafilterconstruction of $\beta X$ for discrete $X$

Remark: It is very difficult to google this question because ultrafilters (on a different set) are used to construct non-standard extentions, so any search containing ultrafilters and non-standard analysis leads me to that construction and not to the use of ultrafilter convergence to describe topology. A link to a book or paper on this topic, or even a title would be a satisfactory answer.

$\endgroup$
  • $\begingroup$ Have you heard of the Wallmann Extension? $\endgroup$ – DanielWainfleet May 15 '18 at 13:26
  • $\begingroup$ @DanielWainfleet that sounds interesting. Can you comment more fully? $\endgroup$ – Mikhail Katz May 16 '18 at 8:47
  • $\begingroup$ @MikhailKatz . See General Topology by R. Engelking. For a completely regular space $X$ there's is a dense homeomorphic embedding $w:X\to wX$ where $wX$ is compact and any continuous $f:X\to Y$ (with $Y$ compact Hausdorff) extends to a continuous $f^*:wX\to Y.$... If $X$ is normal, $wX$ is equivalent to $\beta X.$ If $X$ is not normal, $wX$ is not a $T_1$ space....Let $C$ be the set of closed subsets of $X.$ Let $F \subset C$ such that (i) $\phi \not \in F \ne \phi$, (ii) $A,B\in F\implies A\cap B\in F$, (iii) $(A\subset B\in C \land A\in F) \implies B\in F$, (continued) $\endgroup$ – DanielWainfleet May 17 '18 at 14:33
  • $\begingroup$ ...(continued)...and (iv) $F$ is subset-maximal among subsets of $C$ with properties (i),(ii),(iii). Then $wX$ is the set of all such $F$.... For $p\in X$ let $w(p)=\{c\in C:p\in C\}$.... Let $O$ be the set of open subsets of $X$. A base for the topology on $wX$ is $\{U^*: U\in O\}$ where $U^*=\{F\in wX: \exists c\in F\;(c\subset U\}$....Discrete spaces are normal. If $X$ is discrete then $wX$ is the set of all ultrafilters on $X$ and for $p\in X,\;$ $w(p)$ is the principal ultralfilter $\{A\subset X:p\in A\},$ and for $U\subset X,\;$ $U^*=\{F\in wX:U\in F\$. $\endgroup$ – DanielWainfleet May 17 '18 at 14:48
  • $\begingroup$ Engelking means "compact Hausdorff" when he writes "compact". He calls a space with the "compactness property" but not necessarily Hausdorff a "pseudo-compact" space..... Proving that $wX$ has the compactness property is tricky but not difficult/ $\endgroup$ – DanielWainfleet May 17 '18 at 14:54
3
$\begingroup$

The existence of such a relation is not surprising because a nonstandard number $x\in{}^{\ast}\mathbb R$ can be used to specify an ultrafilter. If $x\in{}^{\ast}\mathbb N\setminus \mathbb N$ then the corresponding ultrafilter is $\{A\in\mathcal{P}(\mathbb N)\colon {}^{\ast}\!A\ni x\}$. If $x\not\in{}^\ast\mathbb N$ argue similarly. For instance if $x$ is infinite, use instead $\lfloor x \rfloor\in{}^\ast\mathbb N$. If $x$ is finite use $\frac{1}{x-\mathbf{st}(x)}$, etc.

Benci and di Nasso have studied conditions that allow one to assume that the correspondence surjective, injective, etc. Some of the axioms involved turn out to be independent of ZFC. One of the relevant articles is

Di Nasso, Mauro. Hypernatural numbers as ultrafilters. Nonstandard analysis for the working mathematician, 443-474, Springer, Dordrecht, 2015

A few years ago I presented a hyperreal-based proof of Ramsey theorem following Goldblatt. Our local ultrafilter expert attended all the lectures studiously, and at the conclusion of the series exclaimed: "But this is the same as the ultrafilter proof!" I am not sure whether he was right or not, but there is clearly a close connection :-)

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.