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I have been trying to come up with a proof, but I don't think I am making leaps in logic and assuming things that I'm possibly trying to prove, if anyone could help correct/(or completely discard) what I have done so far and help me towards a rigerous proof I would greatly appreciate it. This is what I have come up with so far.

Let $U_1,U_2,...,U_n$ be subsets of $\mathbb{R^{n}}$ and closed, define

$U:=\bigcap _{i=1}^nU_i$

And the boundary of $U$ as $B:=CL(U)\cap CL(\mathbb{R^{n}}-U)$

WLG take $U_1,U_2,...,U_m$ where $m\leq n$ such that for some $W_1,W_2,...,W_m$ where $W_i\subset U_i$

$B:=\bigcup _{i=1}^mW_i$

Take $x\in B$ $\Rightarrow x\in \bigcup _{i=1}^mW_i$ $\Rightarrow x\in U_i$ for some $i=1,2,...,m$ $\Rightarrow x\in \bigcap _{i=1}^nU_i \Rightarrow x\in U $

Therefore $B\subset U$ therefore $U$ is closed

Pretty sure this is wrong, but I thought it would be good to show you my train of thought

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So you want to show that the union? (title says intersection) of a finite collection of closed sets is closed? What is your definition of closed? – Henno Brandsma Sep 21 '11 at 14:23
In a topological space the intersection of any family of closed sets is closed, even infinite families. – lhf Sep 21 '11 at 14:25
@HennoBrandsma: Sorry, just corrected it.. intersection is correct. My definition of closed a closed set is one that contains all it's closure points, e.g. i'm trying to show that the boundary of $U$ is a subset of $U$ – Freeman Sep 21 '11 at 14:25
up vote 3 down vote accepted

The intersection of any number of closed subsets if always closed.

Let $X$ be a topological space. Suppose $C_i$ are closed sets ($i\in I$ any index set). Set $U_i=\{x\in X\mid x\notin C_i\}$ the complement of $C_i$ is an open set.

The union of any number of open sets is open, therefore $U=\bigcup_{i\in I} U_i$ is open.

We have that $\{x\in X\mid x\notin U\}$ is closed. This is exactly $\bigcap_{i\in I}C_i$, since $x\notin U$ if and only if for all $i\in I$, $x\notin U_i$, if and only if for all $i\in I$, $x\in C_i$, if and only if $x\in\bigcap_{i\in I}C_i$.

Thus the intersection of closed sets is closed.

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Brilliant, that's much more elegant than what I was trying to do. – Freeman Sep 21 '11 at 14:39

Suppose the definition of a closed set is being taken as "a set which contains all its accumulation points". Then examine an accumulation point of the intersection. All of the points in the intersection are contained in each one of the original collection of closed sets. Looking at the closed sets one by one, because they are closed they all contain the accumulation point we have identified. Because they all contain that point, it is also contained in the intersection, and the intersection is therefore closed.

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