I have the following problem:
Let $\{(M_i, \mathfrak{M}_i)\}$ be nonempty topopological spaces. And let $M=\prod_{i\in I} M_i$. Let $\mathfrak{F}$ be a filter on $M$ and so $\mathfrak{F}_i=(\pi_i)_* \mathfrak{F}$ it's image filter on each $\mathfrak{M}_i$ ($\pi_i$ is the projection). I need to show that $\mathfrak{F}$ converges to $p\in M$ iff $\mathfrak{F}_i$ converges to $p_i=\pi_i(p)$.
My Idea was the following:
$\Rightarrow$ Let us assume that $\mathfrak{F}$ converges to $p\in M$. Let us remark that $M$ is endowed with the product topology. Then we know by it's definition that all the projections $$\pi_i:M\rightarrow M_i$$are continuous for all $i\in I$. Now we have a continuous function $\pi_i$ and a convergent filter $\mathfrak{F}$ and this implies by a corollary of the lecture that $(\pi_i)_*\mathfrak{F}$ converges to $\pi_i(p)$ for all $i\in I$.
Is this correct so far? So in the other direction I had some problems, I wanted to prove this by contradiction and that's what I get so far:
$\Leftarrow$ Let us assume that $(\pi_i)_*\mathfrak{F}$ converges to $p_i$ and that $\mathfrak{F}$ does not converge to $p$. By definition since $\mathfrak{F}$ does not converge to $p$ we know that $\mathfrak{F}$ is coarser that the the neighbourhood filter $\mathfrak{U}(p)$ of $p$, i.e. there exists a neighbourhood $U\in \mathfrak{U}(p)$ such that $U\notin \mathfrak{F}$. On the other hand since $(\pi_i)_*\mathfrak{F}$ converges to $p_i$ we know that the following inclusion holds: $$\mathfrak{U}(p_i)\subset (\pi_i)_*\mathfrak{F}=\{B\subset M_i: \exists A\in \mathfrak{F}\,\,\text{with}\,\,\pi_i(A)\subset B\}$$...
Now I wanted to work with this explicit $U$ to produce a contradiction but I somehow don't see how. The only thing I remarked in addition is that for $V\in \mathfrak{U}(p_i)\Rightarrow \pi_i^{-1}(V)\in \mathfrak{U}(p)$ since $\pi_i$ are continuous.
Could someone maybe tell me if $\Rightarrow$ direction works and give me a hint for the other one how to create a contradiction?
Thanks a lot
The forward direction is fine, using the continuity of the projections.
For the reverse, we have a point $p = (p_i)_{i \in I}$ from the component spaces and we want to show that $\mathfrak{F}$ converges to $p$. We don't have to use contradiction, but use the standard base of the product topology: a basic open subset $B$ of $p$ is of the form $B=\bigcap_{i \in F} \pi_i^{-1}[O_i]$ where $O_i \subseteq M_i$ is open and contains $p_i$ and $F \subseteq I$ is finite.
Now show that all $\pi_i^{-1}[O_i] \in \mathfrak{F}$ from the fact that the "component filters" $(\pi_i)_\ast(\mathfrak{F})$ converge to $p_i$ for all $i$. Then the filter axioms tell us that $B \in \mathfrak{F}$ and as this holds for all basic open subsets of $p$, we have that $\mathfrak{U}(p) \subseteq \mathfrak{F}$ and so $\mathfrak{F} \to p$.