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Let $x_1,x_2,\ldots$ be a sequence of points of the product space $\prod X_\alpha$. Show that this sequence converges to a point $x$ if and only if the sequence $\pi_\alpha(x_1),\pi_\alpha(x_2),\ldots$ converges to $\pi_\alpha(x)$ for each $\alpha$.

$\Rightarrow$ Suppose that $x_n \to x$. The $\pi_\alpha(x_n) \to \pi_\alpha(x)$ since each of the $\pi_\alpha$ are continuous.

$ \Leftarrow $ Suppose that $\pi_\alpha(x_n) \to \pi_\alpha(x)$ for each $\alpha$. Let $ x \in \prod_\alpha U_\alpha$. Then for some natural number $N$, for $n \geq N$ we have $x_n \in \prod_\alpha U_\alpha$. Now we have $U_\alpha \neq X_\alpha$ for only finitely many $\alpha$, say $\alpha_1,\ldots,\alpha_k$. For each $j \in 1,\ldots,k$ let $N_j$ be such that $\pi_{\alpha_k}(x_n) \in U_{\alpha_j}$ for all $n \geq N_j$. Let $\overline{N}=\max\{N_1,\ldots,N_k\}$. Then $x_n \in \prod_\alpha U_\alpha$ for $n \geq \overline{N}$.

I'm stuck on the $\Leftarrow$. I tried to argue through...but I'm having problems making my argument eloquent (and I am not confident that it is correct).

Zev Chonoles
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emka
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1 Answers1

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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 $\langle x_n:n\in\Bbb N\rangle\to x$.

Brian M. Scott
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  • Thanks for helping me clean up the converse. – emka Apr 08 '13 at 04:21
  • @abet: You’re welcome. You really did understand what was going on; you just need more practice writing. – Brian M. Scott Apr 08 '13 at 04:23
  • Hi Brian...I dont fully understand why this is the case: "For each $\alpha\in F$ there is an $m_\alpha\in\Bbb N$ such that $\pi_\alpha(x_n)\in V_\alpha$ whenever $n\ge m_\alpha$" Thanks if you have time to explain that. Thanks in any case tho, i learned from your answer. – JKnecht Mar 10 '16 at 16:46
  • @JKnecht: Pick any $\alpha\in A$. We’ve assumed that the sequence $\langle\pi_\alpha(x_n):n\in\Bbb N\rangle$ converges to $\pi_\alpha(x)$, and we know that $V_\alpha$ is an open nbhd of $\pi_\alpha(x)$. By the definition of convergence this means that there is some $m_\alpha\in\Bbb N$ sich that $\pi_\alpha(x_n)\in V_\alpha$ whenever $n\ge m_\alpha$. It may help to get rid of some of the notation temporarily. Basically we have a sequence $\langle y_n:n\in\Bbb N\rangle$ converging to a point $y$, which means that if $V$ is any open nbhd of $y$, there is some $m\in\Bbb N$ such that $y_n\in V$ ... – Brian M. Scott Mar 10 '16 at 20:33
  • ... whenever $n\ge m$. Here $y_n=\pi_\alpha(x_n)$, $y=\pi_\alpha(x)$, and the $V$ in which we’re interested is $V_\alpha$. – Brian M. Scott Mar 10 '16 at 20:33
  • +1 Thanks. What confused me was how we can be sure that the product of $V_\alpha$ we sort of randomly say $x$ belongs to automatically have to be the same $V_\alpha$ we have for $\pi_\alpha(x_n)$. I dont see how this is proven. But if i let it sink in for a while i will probably understand it.

    $$x\in\prod_{\alpha\in A}V_\alpha\subseteq U;.$$

    $$\pi_\alpha(x_n)\in V_\alpha$$ whenever $n\ge m_\alpha$

     – JKnecht Mar 11 '16 at 14:59
  • @JKnecht: You’re welcome. The $V_\alpha$s are jointly determined by $x$ and $U$: $U$ is an open nbhd of $x$ in $X$, so there must be a basic open nbhd of $x$ contained in it. Such a set by definition has the form $\prod_{\alpha\in A}V_\alpha$, where each $V_\alpha$ is an open nbhd of $\pi_\alpha(x)$ in $X_\alpha$, and $V_\alpha=X_\alpha$ for all but finitely many of the coordinates $\alpha$. – Brian M. Scott Mar 11 '16 at 22:14
  • ahh, now i get it. Thanks again! There are SO MANY definitions and theorems to keep track of :) But i am slowly getting there...:) – JKnecht Mar 11 '16 at 22:32