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Let $\Lambda$ be an indexing set and suppose that for each $\lambda \in \Lambda$ the set $E_{\lambda}\subseteq \mathbb{R}$ is closed. Using the definition of a closed set prove that the set $E$ defined by $$E= \cap_{\lambda \in \Lambda}E_{\lambda}$$ is closed as well.

My workings: The definition of a closed set is: A set $G$ is closed if it contains all its limit points. I wanted to argue by contradiction and suppose there exists a limit point $x_0$ not contained in $E$. If it's a limit point of $E$ then it must be the limit point of at least one $E_{\lambda}$ and therefore contained in that $E_{\lambda}$. Now intuitively I thought that intersecting a closed set $F$ with a closed $G$ means you keep the limit points of $F$ if $G$ is 'bigger' (or vice versa) meaning that the limit points stay in the set. Or possibly the left limit point of $F$ is kept and the right limit point of $G$ or vice versa. However I do not know how to make this rigorous especcially in the case of an infinite (possbly uncountable) intersection. Any help would be appreciated. Thanks!

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up vote 1 down vote accepted

Start: Let $x$ be a limit point of $E$. What does it mean? Each neighborhood of $x$ intersects $E$, that is, intersects all $E_\lambda$.

So, $x$ is a limit point of each $E_\lambda$, hence $x\in E_\lambda$ for each $\lambda$ because they were closed. And we're ready.

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So therefore $x$ is internal to some of the $E_{\lambda}$ and a limit point of the rest and therefore $x\in E_{\lambda}$ for each $\lambda \in \Lambda$ and thus it is contained in $E= \cap_{\lambda \in \Lambda}E_{\lambda}$. Would that be a correct finish to the argument? – Slugger Feb 8 '13 at 11:42
I have finished it. – Berci Feb 8 '13 at 12:23

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