How do I prove that a set of closed sets is itself closed.

Question

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.

 

Answer 1

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.