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I need to prove that the intersection of any number of closed sets is a closed set itself. There are some proofs that come up upon doing some research but I want to do it with a very specific definition. Here it is:

A set A is closed if every point that is arbitrarily close to A is a member of A.

(Equivalently , if y --> x with y being an element of A, then x is an element of A)

I've been trying to figure out how to use this definition to prove that the intersection of any number of sets is closed as well.

myriagon
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  • https://proofwiki.org/wiki/Intersection_of_Closed_Sets_is_Closed – Jim Jan 25 '17 at 21:44
  • Not really what I'm looking for here but thanks. Doesn't seem to apply this specific version of the definition of a closed set in any way. – myriagon Jan 25 '17 at 21:52
  • A couple of comments. First, please use $\LaTeX$. Second, I think you mean that $A$ is closed if $y_n\to x$ (with ${y_n}$ a sequence or more generally a net) with $y_n\in A$ for all $n$ implies $x\in A$. Either that or $A$ is closed if for any $x$ and any neighborhood $U$ of $x$, $U\cap A\neq\varnothing$ implies $x\in A$. – Aweygan Jan 25 '17 at 21:53
  • what is a point that is arbitarily close? Please define! Are we in a metric setting? – Henno Brandsma Jan 26 '17 at 04:48

2 Answers2

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Assuming (As this is not stated) that $x$ is "infinitely close" to a set $A$ means that every open set $O$ that contains $x$ intersects $A$ (or with open balls of radius $>0$ in a metric setting) (which is usually called that $x$ is an adherent point of $A$), the proof could go:

Suppose $A_i, i \in I$ are all closed in this sense. Define $A = \cap_i A_i$. Let $x$ be "close to" $A$. We want to show that $x \in A$. So suppose for a contradiction that $x \notin A$. Then there is some $i$ such that $x \notin A_i$. As $A_i$ is closed, this would mean that $x$ cannot be "close to" $A_i$, so there exists some open set (or ball depending on your exact definition) $O$ that contains $x$ and does not intersect $A_i$. But if it does not intersect $A_i$ it certainly does not intersect $A$. But that would make $x$ not close to $A$, a contradiction. So $x \in A$ must be the case. So $A$ is closed.

Henno Brandsma
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If $\{E_\alpha\}$ is a collection of closed set, then $\bigcap_{\alpha} E_\alpha $ is a closed set.

Proof:

$p_\alpha$ is a limit point of $E_\alpha$ implies that $p_\alpha \in E_\alpha$, since $E_\alpha$ is a closed set.

For every $\epsilon > 0$ there is a point $q_\alpha\in E_\alpha$ such that $\operatorname{d}(p_\alpha,q_\alpha) < \epsilon$, by the definition of limit points. Also if $p$ is a limit point of $\bigcap_\alpha E_\alpha$, there is a point $q\in \bigcap_\alpha E_\alpha$ such that $\operatorname{d}(p,q) < \epsilon$. Hence $$ \operatorname{d}(p_\alpha,q_\alpha) \leq \operatorname{d}(p,p_\alpha)+\operatorname{d}(p,q)+\operatorname{d}(q,q_\alpha)\leq\epsilon +\operatorname{d}(p,p_\alpha)+\operatorname{d}(q,q_\alpha) $$ Also $\operatorname{d}(p_\alpha,q_\alpha) < \epsilon$, then $\operatorname{d}(p,p_\alpha) = \operatorname{d}(q,q_\alpha) = 0$. It concludes that $p\in E_\alpha$ for any $\alpha$, which completes the proof.

Shuqi XIAO
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