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 {Ωa : a ∈ I} be an arbitrary family of closed sets Ωa ⊆ Rwith 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.
 

Maurizio T.

tutor
How can someone not know the meaning of ∩ and ∪ (intersection and union) and be asked to do prove statements like this? That is bizarre. This is a honest master's level question in analysis  The operator c denotes the complement of a set. 
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06/06/18

Ashley H.

The fact that you have said this is quite intriguing to me, as this is a question that I have been asked as part of my Mathematics course in first year (freshman year!). If it is masters level then i have no idea why it is being asked in all honesty
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06/07/18

Maurizio T.

tutor
There is nothing difficult, really. But a student must know the notions of open and close sets which usually are introduced in general topology.  Calculus classes di not deal with this, but Principles of Real Analysis type classes do.
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06/07/18

1 Expert Answer

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Ashley H.

Thanks for this. The next part of the question asks me to set d=2 and prove that   ∪ Ωa 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 Ωn = [1/n, 1]. Then ∪n Ωn = (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 ⊆ Rd, so setting d=2 we are talking specifically about the space R2, rather than using some arbitrary d.
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06/10/18

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