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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...

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- Thread starter kingwinner
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- #1

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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...

- #2

Hurkyl

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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?

How can I find a counterexample?

- #4

Hurkyl

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- #5

MathematicalPhysicist

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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.

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I can't figure it out and I don't know which direction to push my proof towards...

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- #8

HallsofIvy

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?? All of [a,b], [a,b), (a,b], (a,b)

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.

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