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Connectedness math problem

  • Thread starter kingwinner
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  • #1
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"Let S be a subset of R^n.
If S is connected, then the interior of S is connected."

Is this true or not?

I can't think of a counterexample, but I don't know how to prove it either...
 

Answers and Replies

  • #2
Hurkyl
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"Let S be a subset of R^n.
If S is connected, then the interior of S is connected."

Is this true or not?

I can't think of a counterexample, but I don't know how to prove it either...
It might help to think "backwards" -- rather than thinking of starting with a set S, and then working with its interior, why not start with an open set, and then consider its closure (possibly excluding part of the boundary)?
 
  • #3
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But that "backward" one is not equivalent to the original one.

How can I find a counterexample?
 
  • #4
Hurkyl
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True, the backward case is just a subset of the possibilities -- but I assert that it covers enough of the possibilities that it should suggest a proof or yield a counterexample.
 
  • #5
MathematicalPhysicist
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I think you can take something like
an interval (a,b] or [a,b) (the same should work also for R^n).
obviously if [a,b) is connected, then (a,b) is'nt necessarily.
 
  • #6
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Is the claim true or false?
I can't figure it out and I don't know which direction to push my proof towards...
 
  • #7
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It seems true with simple examples, but we can't generalize from specifics. How can I start the proof in the general situation?
 
  • #8
HallsofIvy
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I think you can take something like
an interval (a,b] or [a,b) (the same should work also for R^n).
obviously if [a,b) is connected, then (a,b) is'nt necessarily.
?? All of [a,b], [a,b), (a,b], (a,b) are connected!
 

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