Let $A$ be an index set, $X$ a topological space. Define $X^A$ to be the product $\displaystyle\prod_{\alpha \in A}X_\alpha$ where $X_\alpha = X, \forall \alpha \in A$. We can think the elements of $X^A$ as functions $f$ so that $f: A \to X$, $f(\alpha)=\pi_\alpha(f)$.
Now I have the following theorem:
Theorem.$f_n,f \in X^A$ and $f_n \to f$ in the product topology $\iff$ $f_n,f: A \to X$ and $f_n\to f$ pointwise.
I could not start, how should I think the pointwise convergence here?
Best Answer
$(\Rightarrow)$ Suppose that $f_n,f\in X^A$ are such that $f_n\to f$ in the product topology. This means that, for any finite set of points $\{a_1,a_2,\dots,a_k\}\subseteq A$ and and any choice of open neighborhoods $U_i\subseteq X$ of $f{\left(a_i\right)}$, $1\leq i\leq k$, there exists an $N\in\mathbb{N}$ such that if $n\geq N$, then $f_n{\left(a_i\right)}\in U_i$ for all $1\leq i\leq k$. Hence in particular, for each singleton $\{a\}\subseteq A$ and each choice of open neighborhood $U$ of $f(a)$, there is an $N\in\mathbb{N}$ such that if $n\geq N$, then $f_n(a)\in U$. Therefore $f_n\to f$ pointwise.
$(\Leftarrow)$ Suppose that $f_n\to f$ pointwise. This means that, for any $a\in A$ and any open neighborhood $U$ of $f(a)$, there is an $N\in\mathbb{N}$ such that if $n\geq N$, then $f_n(a)\in U$. Let $\left\{a_1,a_2,\dots,a_k\right\}\subseteq A$ be an arbitrary finite subset, and for each $1\leq i\leq k$, let $U_i\subseteq X$ be an open neighborhood of $f{\left(a_i\right)}$. By pointwise convergence of $f_n$, for each $1\leq i\leq k$, there is an $N_i\in\mathbb{N}$ such that if $n\geq N_i$, then $f_n{\left(a_i\right)}\in U_i$. Now set $N:=\max{\left\{N_1,N_2,\dots,N_k\right\}}$. Thus if $n\geq N$, then $f_n{\left(a_i\right)}\in U_i$ for all $1\leq i\leq k$. Therefore $f_n\to f$ in the product topology. $\blacksquare$