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Let I be a set (possibly infinite), and for each $\alpha$ $\in$ I let $X_\alpha$ be a closed subset of $\mathbb{R}$. Show that the intersection of $ \bigcap_{\alpha \in I} X_\alpha$ is also closed

My attempt

Since we know that $X_\alpha$ is closed, then there is a convergent sequence consisting of elements in $X_\alpha$, which converges to an adherent point in $X_\alpha$.

Is it right to assume that the intersection of these sets means that every convergent sequence consisting of elements of $X_\alpha$ converges to a point in the intersection? Then I think we can say because of that, the intersection of all $X_\alpha$ is also closed.

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    $\begingroup$ You could also prove that if each $U_i$ is open, then $\bigcup_{i\in I}U_i$ is open. Depending on your definition of openness, this could be obvious. $\endgroup$ – neth May 4 '16 at 2:12
  • $\begingroup$ observe that $\cap_{\alpha}X_{\alpha}\subset X_{\beta}$ for each $\beta$ $\endgroup$ – janmarqz May 4 '16 at 2:13
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Your assumption is right, here's short proof using it:

Let $X=\bigcap_{\alpha \in I} X_\alpha$ and $(x_n)_{n \in \mathbb{N}}$ an arbitrary sequence in $X$, as janmarqz points out, $(x_n)_{n \in \mathbb{N} }\subset X_\alpha$ for all $\alpha \in I$. That means, if $(x_n)_{n \in \mathbb{N} }$ has a limit $x$, then $x \in X_\alpha$ for each $\alpha \in I$ because each $X_\alpha$ is a closed set, that implies by definition of $X$, that $x \in X$ which proves your statement.

I hope it helps.

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