How do I show that the filter is convergent in a cartesian product?

Question

I have the following problem:

Let $\{(M_i,T_i)\}_{i\in I}$ be non-empty top. spaces where $I$ is an arbitrary non empty index set. Moreover let $F$ be a filter on $M=\prod_{i\in I}M_i$ and denote $F_i=(pr_i)_*F$ the corresponding image filter on each component $M_i$. Show that $F$ converges to $p\in M$ iff $F_i$ converges to $p_i=pr_i(p)$ forall $i\in I$

I wanted to do it as follows:

$\Rightarrow$ Let us assume that $F$ converges to $p\in M$. Let us remark that $M$ is endowed with the product topology. But we just know that in this topology all projections $pr_i$ are continuous forall $i\in I$ and at each $m\in M$, so naturally they are also continuous at $p\in M$. By a corollary from the lecture this is equevalent to say that $F$ has a convergent image filter $F_i$ converging to $pr_i(p)$ for all $i\in I$

I hope this works till here.

$\Leftarrow$ Let us assume that $F_i=(pr_i)_*F=\{B\subset M_i|\exists A\in F\,\,\,\text{with}\,\,\,pr_i(A)\subset B\}$ converges to $p_i$ for all $i\in I$

Now in this direction I don't know how to procede because we only have the definition of a converging filter, so that it is finer that the neighbourhood filter and we also know that a filter converges to $p$ if its associated net converges to $p$. But I don't see how to use this informations. Could someone give me a hint?

Thanks a lot.

Answer 1

Let $U$ be a neighborhood of $p$. Then $U$ contains some open set $A$ such that $p\in A$. And, by the definition of the product topology, there are open sets $A_i$ of $M_i$ such that:

• $(\forall i\in I):p_i\in A_i$;
• for some finite subset $\varphi$ of $I$, $A_i=M_i$ if $i\in I\setminus\varphi$;
• $A\supset\prod_{i\in I}A_i$

If $\varphi=\{i_1,i_2,\ldots,i_n\}$ then, for each $k\in\{1,2,\ldots,n\}$, there is some $F_k\in F$ such that $A_{i_k}\supset\operatorname{pr}_{i_k}(F_k)$. Therefore, each $A_{i_k}$ contains $\operatorname{pr}_{i_k}\left(\bigcap_{j=1}^nF_j\right)$, and so$$U\supset A\supset\prod_{i\in I}A_i\supset\bigcap_{j=1}^nF_j,$$and therefore $U\in F$.