Ashley H.

asked • 06/06/18# How do I prove that a set of closed sets is itself closed.

Hi, I have been asked the following but am struggling.

Let {Ω

_{a}: a ∈ I} be an arbitrary family of closed sets Ω_{a}⊆ R^{d }with an index set I. Prove that ∩ Ωa

a ∈ I

is a closed set.

Firstly, I am not sure exactly what this set is, so if anyone knows what it means then that would be great. I am particularly confused by the ∩ symbol. Secondly, can anyone show me how to prove that it is closed? Thanks.

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## 1 Expert Answer

Maurizio T. answered • 06/06/18

Tutor

5.0
(759)
Stat PhD with experience in both theoretical and applied Econometrics

Recall that a set is closed if and only if (iff) its complement is open.

So to prove that ∩Ω

_{a}is closed is equivalent to proving that (∩Ω_{a})^{c}= ∪Ω_{a}^{c }where Ω_{a}^{c}is open as, by assumption the Ω_{a}are closed. Now, there is a theorem (easy) that says that the union of any number of open sets (countable or uncountable!) is again an open set. If you have never seen this kind of proofs, then why are you doing this kind of problems. Proving it is not complicated. Let Ψ = ∪Ω_{a}^{c }and consider x ∈Ψ be an arbitrary point; then x ∈ Ω_{a}^{c}for some a ∈ I. Since this set is open, it contains an open ball around x that lies in Ω_{a}^{c}. Clearly, this open ball lies in Ψ. The choice of a in Ψ is arbitrary and then, by definition, Ψ is open which proves your statement.Ashley H.

Thanks for this. The next part of the question asks me to set d=2 and prove that ∪ Ω

_{a}is not necessarily closed using a counterexample. How would this work given the above? Thanks. a∈I
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06/07/18

Maurizio T.

tutor

A finite union of closed sets is closed! No doubts there. However, an infinite union of closed sets can fail to be closed. Not sure what you mean by d=2. A well known type of example for this using the real line, R, is the following: let Ω

_{n}= [1/n, 1]. Then ∪_{n}Ω_{n}= (0, 1] $ which is not a closed set. There are many of these counterexamples.
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06/07/18

Ashley H.

As you can see in the original question, we have ?

_{a}⊆ R^{d}, so setting d=2 we are talking specifically about the space R^{2}, rather than using some arbitrary d.
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06/10/18

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Maurizio T.

^{c}denotes the complement of a set.06/06/18