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I asked this question yesterday. I thought I had the forward direction figured out but now I've lost confidence.

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 $\alpha$.

My attempt at a proof.

Suppose $x_n \to x$. Let $A$ be an index set and let $\alpha_0 \in A$. Let $U_{\alpha_0}$ be a neighborhood such that $\pi_{\alpha_0(x)}\in U_{\alpha_0}$. Then $\pi_{\alpha_0}: \Pi_{\alpha \in A}X_\alpha \to X_{\alpha_0}$. Then $\pi_{\alpha_0}^{-1}(U_{\alpha_0})=U_{\alpha_0}\times\Pi_{\alpha \in A \setminus {\alpha_0}}X_\alpha$.

Let $N \in \mathbb{N}$. Then for $n > N$ we have that $x_n=x_{n,\alpha_0} \in U_{\alpha_0}\times\Pi_{\alpha \in A \setminus \alpha_0}X_\alpha=\pi_{\alpha_0}^{-1}(U_{\alpha_0})$.

Thus $\pi_{\alpha_0}(x_n)\in U_{\alpha_0}$ for $n > N$.

Community
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emka
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1 Answers1

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It’s basically fine, but it can be simplified a bit. Fix $\alpha\in A$, and let $U$ be any open nbhd of $\pi_\alpha(x)$ in $X_\alpha$. The projection maps are continuous, so $\pi_\alpha^{-1}[U]$ is an open nbhd of $x$ in $X=\prod_{\beta\in A}X_\beta$. By hypothesis, therefore, there is an $m\in\Bbb N$ such that $x_n\in\pi_\alpha^{-1}[U]$ whenever $n\ge m$. But if $x_n\in\pi_\alpha^{-1}[U]$, then $\pi_\alpha(x_n)\in\pi_\alpha\big[\pi_\alpha^{-1}[U]\big]=U$, so $\pi_\alpha(x_n)\in U$ whenever $n\ge m$. Since $U$ was an arbitrary open nbhd of $\pi_\alpha(x)$, it follows that $\langle\pi_\alpha(x_n):n\in\Bbb N\rangle\to\pi_\alpha(x)$.

Brian M. Scott
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