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Consider the pairs $(A_i,a_i)$ where $A_i$ is a set and $a_i: U(A_i)\rightarrow A_i$ is a relation where $U(A_i)$ is denotes the set of ultrafilters on $A_i$.

Now consider the topological spaces $(A_i,\tau_i)$ where $\tau_i$ is the topology generated as follow:

$U\in \tau_i \iff$ for every $x\in U$ whenever $(\mathcal{F},x)\in a_i $ we have $U\in \mathcal{F}$ (here $\mathcal{F}$ is an ultrafilter on $A_i$). $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (*)$

Now consider the pair $(\Pi_{i\in I}A_i, a_p)$ where $a_p: U(\Pi_{i\in I}A_i)\rightarrow \Pi_{i\in I}A_i$ is a relation between ultrafilters on the product and the product itself (as before).

Suppose that the relation $a_p$ is described as follows:

$$\big(\mathcal{F},(x_i)_{i\in I}\big)\in a_p \iff ({\pi_i}_*(\mathcal{F}),x_i) \in a_i \text{ for every } i\in I$$

Here ${\pi_i}_*(\mathcal{F}) = \{X\subseteq A_i: \pi_i^{-1}(X)\in\mathcal{F}\} $ as usual where $\mathcal{F}$ is an ultrafilter on the product.

Question: is the topology generated by $a_p$ on $\Pi_{i\in I}A_i$ (by using the characterization of open sets in $(*)$ in the orange block) the product topology?

Background: the reason I suspect this is that if $\tau$ is the product topology on $\Pi_{i\in I}A_i$, because ultrafilter convergence in the topological sense satisfies the same relation as $a_p$.

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  • $\begingroup$ In the definition of $a_p$ on the right side shouldn't it be: $({\pi_i}_*(\mathcal{F}), x_i) \in a_i$ ? Also $i$ is not quantified. $\endgroup$ – Adayah Jul 29 '17 at 19:38
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No. For instance, let $a_i$ be the empty relation for all $i$. Then $\tau_i$ is the discrete topology, and $a_p$ is also the empty relation (assuming $I$ is nonempty) so it induces the discrete topology as well. But the product topology on $\prod A_i$ is not the discrete topology as long as each $A_i$ is nonempty and infinitely many of the $A_i$ have more than one point.


For an example where principal ultrafilters converge to the corresponding points, let $A$ be an infinite set and choose two distinct elements $s,t\in A$ and a nonprincipal ultrafilter $\mathcal{G}$ on $A$. Define a relation $a:U(A)\to A$ by saying $(\mathcal{F},x)\in a$ iff $\mathcal{F}$ is the principal ultrafilter at $x$, or $\mathcal{F}\neq\mathcal{G}$ is nonprincipal and $x=s$, or $\mathcal{F}=\mathcal{G}$ and $x=t$. Note that if $U$ is any open set containing $s$ in the induced topology, then $U$ is cofinite (since if $U$ is coinfinite, you can choose a nonprincipal ultrafilter $\mathcal{F}\neq\mathcal{G}$ which contains $A\setminus U$ and $(\mathcal{F},s)\in a$). In particular, it follows that with respect to the induced topology, the ultrafilter $\mathcal{G}$ converges to both $s$ and $t$, so the induced topology is not Hausdorff.

Now consider the product relation $a_p$ on $A\times A$. I claim that the set $\Delta=\{(x,x):x\in A\}$ is closed in the topology induced by $a_p$. Indeed, suppose $\mathcal{F}$ is an ultrafilter on $A\times A$ such that $\Delta\in\mathcal{F}$. For any $X\subseteq A$, $(X\times A)\cap \Delta=(A\times X)\cap\Delta$, and it follows that ${\pi_0}_*(\mathcal{F})={\pi_1}_*(\mathcal{F})$. Thus if $\mathcal{F}$ converges to a point $(x,y)$ according to $a_p$, then the ultrafilter ${\pi_0}_*(\mathcal{F})={\pi_1}_*(\mathcal{F})$ converges to both $x$ and $y$ according to $a$. Since no ultrafilter converges to two different points according to $a$, this means $x=y$, so $(x,y)\in\Delta$. Thus $\Delta$ is closed in the topology induced by $a_p$. However, $\Delta$ is not closed in the product topology since the topology on $A$ is not Hausdorff.

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  • $\begingroup$ Can the second example be made to work for this question, perhaps? It's for convergence spaces for filters, not only ultrafilters, don't know if that matters much.. $\endgroup$ – Henno Brandsma Jul 29 '17 at 20:41
  • $\begingroup$ Would it change the example much we also let the principal filters at $s$ and $t$ converge to $s$ resp. $t$? This is often in axiom in ultrafilter-convergence spaces. $\endgroup$ – Henno Brandsma Jul 29 '17 at 20:43
  • $\begingroup$ (1) I'm not quite sure what you mean by "work". If you mean it gives a convergence space in which no proper filter has more than one limit but the induced topology is not Hausdorff, then yes, it works. (2) That is included in the example. $(\mathcal{F},x)\in a$ if $\mathcal{F}$ is the principal ultrafilter at $x$, including the cases $x=s$ and $x=t$. $\endgroup$ – Eric Wofsey Jul 29 '17 at 20:47
  • $\begingroup$ OK, I see. But what would the convergence be in terms of a relation between filters and points? $\endgroup$ – Henno Brandsma Jul 29 '17 at 20:54
  • $\begingroup$ A filter converges to $x$ iff every ultrafilter containing it converges to $x$. $\endgroup$ – Eric Wofsey Jul 29 '17 at 20:54

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