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!
