Closed set as infinite intersection of open sets

In summary, in a general metric space, it is possible to express every closed set as the intersection of an infinite collection of open sets. This is because in a metric space, a closed set can be written as a countable intersection of open sets. This is not always true in an arbitrary topological space. The reason for this is that a continuous function can be used to create a countable intersection of open sets that equals the closed set. The notation used for the open sets is typically ]-1/n,1/n[, although (-1/n,1/n) is also acceptable.
  • #1
lonewolf5999
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This is not a homework problem, just something I was thinking about. In a general metric space, is it true that every closed set can be expressed as the intersection of an infinite collection of open sets?

I don't really know where to begin. Since the finite intersection of open sets is open, and the infinite intersection of open sets may or may not be open or closed, this suggests to me that the statement may not be true; however, I'm not sure how I would use this fact to construct a counter-example. Any help is much appreciated!
 
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  • #2
The answer is yes. Every closed set can be written as a countable intersection of open sets in a metric space. (this is not true anymore in an arbitrary topological space)

The reason is this: take a closed set F. We have a continuous function [itex]f(x)=d(x,F)[/itex]. Notice that [itex]f^{-1}(0)=F[/itex].
Of course, we have that [itex]\{0\}=\bigcap_{n>0} ]-1/n,1/n[[/itex], thus

[tex]F=\bigcap_{n>0} f^{-1}(]-1/n,1/n[)[/tex]
 
  • #3
Thanks for the reply. I hope you can clarify a couple of points in your answer, however. When you say "We have a continuous function [itex]f(x)=d(x,F)[/itex]," am I correct in saying that you mean d(x,F) = inf{d(x,f): f lying in F}?

Also, what do you mean by [itex]]-1/n,1/n[[/itex] ? Is it something like an open neighbourhood of radius 1/n around the zero element, and if so, is there any reason why you write it that way as opposed to (-1/n, 1/n)?
 
  • #4
lonewolf5999 said:
Thanks for the reply. I hope you can clarify a couple of points in your answer, however. When you say "We have a continuous function [itex]f(x)=d(x,F)[/itex]," am I correct in saying that you mean d(x,F) = inf{d(x,f): f lying in F}?

Yes.

Also, what do you mean by [itex]]-1/n,1/n[[/itex] ? Is it something like an open neighbourhood of radius 1/n around the zero element, and if so, is there any reason why you write it that way as opposed to (-1/n, 1/n)?

I meant (-1/n,1/n). But I always write ]-1/n,1/n[. It's an accepted notation.
 
  • #5
Ok, thanks very much for the help and the quick reply!
 

What is a closed set?

A closed set is a set that contains all of its limit points. This means that for any sequence of points within the set that converges to a point outside of the set, the limit point is also included in the set.

What is an open set?

An open set is a set that does not contain its boundary points. This means that for any point within the set, there exists a small enough neighborhood around that point that is also contained within the set.

Why is a closed set equal to the infinite intersection of open sets?

This is a fundamental property of topology, where a closed set can be defined as the complement of an open set. Therefore, by taking the infinite intersection of open sets, we are essentially finding the complement of a set, which results in a closed set.

Can a closed set be represented as a finite intersection of open sets?

Yes, it is possible for a closed set to be represented as a finite intersection of open sets. However, in the case of an infinite intersection, the resulting set may not be closed.

How is the concept of closed set as infinite intersection of open sets used in mathematical analysis?

This concept is used in mathematical analysis to prove theorems and solve problems related to continuity, compactness, and convergence of sequences or functions. It also helps to define important concepts such as compact sets and Baire spaces.

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