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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?

Clemens Bartholdy
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Ninja
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  • $f_n\to f$ pointwise means that, for any $a\in A$ and any neighborhood $U$ of $f(a)$, there is an $N\in\mathbb{N}$ such that, if $n\geq N$, then $f_n(a)\in U$. Colloquially, we can say that "$f_n(a)$ is eventually close to $f(a)$" in some sense. – anonymous Mar 29 '17 at 14:37

1 Answers1

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$(\Rightarrow)$ 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. $\blacksquare$

anonymous
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    My first question is, why do we consider a finite set like ${a_1,\dots,a_n}$? – Ninja Mar 29 '17 at 17:24
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    Because a basis for the product topology on $\prod_{\alpha\in A}{X_\alpha}$ consists of sets of the form $\prod_{\alpha\in A}{U_\alpha}$, where $U_\alpha\subseteq X_\alpha$ are open sets and only finitely many $U_\alpha$s are not equal to the corresponding $X_\alpha$. Considering only basic open sets of this form simplifies the problem. – anonymous Mar 29 '17 at 17:34
  • Okay, thank you, let me think more. – Ninja Mar 29 '17 at 17:37
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    As an aside: almost the same proof holds for nets, using the directnedness of the index set. The forward proof is just saying that all projections are continuous, so preserve convergence. Going back we just need that finitely determined open sets are a base and the index set is directed, basically. – Henno Brandsma Mar 31 '17 at 04:25