Convergence in the product topology iff mappings converge

Solution 1:

Your argument is worded just a little clumsily, but the basic idea is just fine. Here’s one way to clean it up a bit.

Let $A$ be the index set, and suppose that $\langle\pi_\alpha(x_n):n\in\Bbb N\rangle\to\pi_\alpha(x)$ for each $\alpha\in A$; we want to show that $\langle x_n:n\in\Bbb N\rangle\to x$. To this end let $U$ be any open nbhd of $x$ in $X=\prod_{\alpha\in A}X_\alpha$. Then there is a finite $F\subseteq A$ and open sets $V_\alpha\subseteq X_\alpha$ for $\alpha\in A$ such that $V_\alpha=X_\alpha$ for all $\alpha\in A\setminus F$, and $$x\in\prod_{\alpha\in A}V_\alpha\subseteq U\;.$$

For each $\alpha\in F$ there is an $m_\alpha\in\Bbb N$ such that $\pi_\alpha(x_n)\in V_\alpha$ whenever $n\ge m_\alpha$; $F$ is finite, so let $m=\max\{m_\alpha:\alpha\in F\}$. Then for any $n\ge m$ we have $\pi_\alpha(x_n)\in V_\alpha$ for all $\alpha\in A$ and hence $$x_n\in\prod_{\alpha\in A}V_\alpha\subseteq U\;.$$ $U$ was an arbitrary open nbhd of $x$, so $\langle x_n:n\in\Bbb N\rangle\to x$.