How do I show this statement about convergence in filters?

Question

I have the following problem:

Let $\{(M_i,\tau_i)\}_{i\in I}$ be nonempty topological spaces where $I$ is arbitrary but non empty. Let $M=\prod_{i\in I} M_i$. Let $F$ be a filter on $M$ and denote by

$$F_i=(pr_i)_*F=\{B\subseteq M_i: \exists\,\,A\in F\,\,s.t.\,\,pr_i(A)\subseteq B\}$$ the corresponding image filter on each componente $M_i$. Show that $F$ converges to $p\in M$ iff $F_i$ converges to $p_i=pr_i(p)$ for all $i\in I$.

Then $\Rightarrow$ implication worked perfectly by using some theorem from the lecture. But I somehow struggled a bit with the other implication. In the solution they did something with subbasis ect but I didn't understand that. Therefore I wandet do do it different. My Idea was the following:

$\Leftarrow$ Let us assume tat $F_i$ converges to $p_i$. We need to show that $F$ converges to $p$, i.e. we need to show that $\mathfrak{U}(p)\subset F$ where $\mathfrak{U}(p)$ is the neighbourhood filter of $p$. Now let us take any basis open set $B$ in M. $$B=\bigcap_{i\in I} pr_i^{-1}(O_i)$$where $I$ is finite and $p_i\in O_i$ are open sets in $M_i$. We see that it is enought to show that $pr_i^{-1}(O_i)\in F$. To do so let us use that $F_i\rightarrow p_i\Leftrightarrow \mathfrak{U}(p_i)\subset F_i$. But then $O_i\in (pr_i)_*F$ which implies that there is some $A_i\in F$ such that $pr_i(A_i)\subset O_i$. This implies that $A_i\subset pr_i^{-1}(O_i)\in F$. Then since the intersection is finite also $B\in F$.

Now what I don't see is why then $\mathfrak{U}(p)\subset F$.

I'm really not sure if this workes therefore it would be nice if someone could take a look at it. Thanks a lot

Answer 1

Let $O$ be a subbasic element in the product i.e. $O=\pi_i^{-1}[U]$ for some $i \in I$ and some open $U \subseteq X_i$ (I use $\pi_i$ for the projections), so that $p \in O$ i.e. $p_i \in U$.

As by assumption $(\pi_i)_\ast(\mathcal{F}) \to p_i$ we know that $U \in (\pi_i)_\ast(\mathcal{F})$ so for some $F_i \in \mathcal{F}$, $\pi_i[F_i] \subseteq U$, which implies $F_i \subseteq O$ and hence $O \in \mathcal{F}$ as $\mathcal{F}$ is a filter.

So all subbasic elements in the product that contain $p$ are in the filter, and so all basic elements (which are the finite intersections of such subbasic sets) also are, and hence $\mathcal{F}\to p$, as required.

Answer 2

You have to be a bit careful with the details here. When you write $U=\prod_{i \in I} O_i$, it is not true that every open set of $M$ is of this form. Rather, you can show that given a neighborhood $p \in U$, there are some open sets $p_i \in O_i \subseteq M_i$, all but finitely many equal to $M_i$, such that $U \supseteq \prod_{i \in I} O_i$. (This is essentially the claim that the $\mathrm{pr}_i^{-1}(O_i)$ form a subbasis of the product topology on $M$.) Now, the convergence tells you that $O_i \in F_i$, which implies that some set $B_i \subseteq M$ has $\mathrm{pr}(B_i)=O_i$. Here you can asusme that $B_i=\mathrm{pr}_i^{-1}(O_i)$ by the monotonicity of the filter, and then this intersection (which is finite!) tells you that $U \in F$.