# Is this induced topology the product topology?

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 $a_p$.

• 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. – Adayah Jul 29 '17 at 19:38

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.
• 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. – Henno Brandsma Jul 29 '17 at 20:43
• (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 $x=t$. – Eric Wofsey Jul 29 '17 at 20:47
• A filter converges to $x$ iff every ultrafilter containing it converges to $x$. – Eric Wofsey Jul 29 '17 at 20:54