Proving the Connectedness Math Problem in R^n: Is it True?

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Homework Help Overview

The discussion revolves around the connectedness of subsets in R^n, specifically examining the claim that if a set S is connected, then its interior is also connected. Participants are exploring the validity of this statement and seeking to understand its implications.

Discussion Character

  • Conceptual clarification, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants are attempting to find counterexamples to the claim and are considering different approaches, such as examining open sets and their closures. Some are reflecting on specific examples like intervals in R.

Discussion Status

The discussion is ongoing, with various perspectives being shared. Some participants suggest that the backward approach may provide insights, while others express uncertainty about how to construct a proof or counterexample. There is a recognition that simple examples may not suffice for generalization.

Contextual Notes

Participants note the challenge of generalizing from specific examples and the potential limitations of the original claim. There is an emphasis on the need for clarity regarding the definitions and properties of connectedness in the context of subsets of R^n.

kingwinner
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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...
 
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kingwinner said:
"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)?
 
But that "backward" one is not equivalent to the original one.

How can I find a counterexample?
 
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.
 
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.
 
Is the claim true or false?
I can't figure it out and I don't know which direction to push my proof towards...
 
It seems true with simple examples, but we can't generalize from specifics. How can I start the proof in the general situation?
 
loop quantum gravity said:
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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