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Let $Y$ is a topological space and for $\alpha\in I$, $X_\alpha\subset Y$. Then $\bigcap\limits_{\alpha\in I}X_\alpha$ is homeomorphic to a closed subspace of $\prod\limits_{\alpha\in I}X_\alpha$.

My main problem in proving this proposition is closedness. Thanks for any advice.

TXC
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    Do we have a separation axiom, like $T_1$ assumed on $Y$? – Berci May 09 '13 at 07:30
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    @Berci: My main goal of this question is show this proposition on metric space $Y$. But i think we can add hausdorffness in space $Y$, since in most papers all spaces considered are assumed to be hausdorff. – TXC May 09 '13 at 07:49
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    You can suppose $|I| = 2$, the proof is almost the same. – Damien L May 09 '13 at 07:59

1 Answers1

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Assume that $Y$ is Hausdorff. Then consider the diagonal $$\Delta_Y:=\{(..,y,y,y,..)_\alpha\,\mid\, y\in Y\}\ \subseteq Y^I$$ It is closed, because whenever $x=(x_\alpha)\notin \Delta_Y$, we have $x_\alpha\ne x_\beta$, so there are disjoint open sets $U$ and $V$ around $x_\alpha$ and $x_\beta$, so $$x\in Y\times Y\times \dots\times U\times\dots\times V\times\dots\times Y\times Y\times\dots$$ which is open and disjoint from $\Delta_Y$.

So, we have that $\Delta_Y\,\cap\,\prod_\alpha X_\alpha$ is closed in $\prod_\alpha X_\alpha$ w.r.t. to the subspace topology. And, finally $$\Delta_Y\,\cap\, \prod_\alpha X_\alpha\ =\ \Delta_{\left(\bigcap_\alpha X_\alpha \right)} \ \cong\ \bigcap_\alpha X_\alpha\,.$$

Berci
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    Aren't you assuming the sets $X_\alpha$ are closed? I was wondering about that in my attempt. – Henno Brandsma May 09 '13 at 11:45
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    Yes. I thought it was assumed. But not needed. Just updated my answer. – Berci May 09 '13 at 13:16
  • @Berci: I can't construct a homeomorphism from $\Delta_{\left(\bigcap_\alpha X_\alpha \right)}$ to $\bigcap_\alpha X_\alpha$. please give me more hints. – TXC May 10 '13 at 16:44
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    @TXC: $\langle x_\alpha:\alpha\in I\rangle\in\Delta_Y\cap\prod_\alpha X_\alpha$ iff there is an $x\in\bigcap_\alpha X_\alpha$ such that $x_\alpha=x$ for all $\alpha\in I$. Take your homeomorphism $h$ to be $h(\langle x_\alpha:\alpha\in I\rangle=x$. – Brian M. Scott May 10 '13 at 17:27