Set closure and interior points

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  • #1
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If [tex]A \subset X[/tex] where [tex]X[/tex] has a topology, is it generally true that the interior of [tex]A[/tex] is equal to the interior of the closure of [tex]A[/tex]? This seems very reasonable to me, but probably only because I'm visualizing [tex]A[/tex] as a disc in the real plane. If it isn't true, what would be a counterexample?

thanks
 

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  • #2
HallsofIvy
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Take X to be the set of real numbers with the usual topology.

Let A be the set of all rational numbers between 0 and 1 (inclusive).

What is the interior of A? What is the closure of A?

What is the interior of the closure of A?
 
  • #3
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HallsofIvy said:
Take X to be the set of real numbers with the usual topology.

Let A be the set of all rational numbers between 0 and 1 (inclusive).

What is the interior of A? What is the closure of A?

What is the interior of the closure of A?
Okay this is a good example! Let me see if I have it straight. The interior of A is the largest open set contained in A, which would be the empty set in this case. I say this because the set has no interior points, or points for which a neighborhood can be found containing only points in A. The closure of A is the smallest closed set containing A, which is the interval [0, 1]. I say this because for every point in [0, 1], every neighborhood of that point contains points in A. But the interior of [0, 1] is (0, 1), which is not the empty set.

Did I understand correctly? I am a little foggy on the properties of real numbers, so I can't really back up my claims about the density of rationals at the moment.

Thanks for your help!
 
  • #4
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yes....you've got it right...
 

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