Is a connected subset of ? I'm sure it is, although I can't find a proof. (Note the "overline" denotes the closure).

Let A,B be two connected subsets of a topological space X such that . Prove that is connected.

Printable View

- Feb 24th 2009, 11:46 AMAmanda1990Prove that this is a connected subset of R^2
Is a connected subset of ? I'm sure it is, although I can't find a proof. (Note the "overline" denotes the closure).

Let A,B be two connected subsets of a topological space X such that . Prove that is connected. - Feb 24th 2009, 12:03 PMPlato
Hint: Is it true that the point in one of those sets and a limit point of the other?

- Feb 24th 2009, 12:05 PMAmanda1990
Certainly I'd agree that is in and is a limit point of . This seems to suggest that there may be some "standard result" that I'm unaware of because I don't see how this helps...

- Feb 24th 2009, 12:10 PMAmanda1990
Oh! Does this mean that the set is path connected? Intuitively it seems like you can "pass through (0,0)" but it's a bit tricky to prove this formally.

- Feb 24th 2009, 12:13 PMPlato
- Feb 24th 2009, 12:43 PMAmanda1990
No, I haven't come across that term before. Basically, my understanding of a connected set X is one that satisfies any one of the following:

1) If {U,V} is an open cover of X such that then either or .

2) The only subsets of X which are both open and closed in X are and X.

3) There exists no open cover {U,V} of X composed of non-empty disjoint sets.

4) For any open (resp. closed) cover {U,V} of X composed of non-empty sets we have - Feb 24th 2009, 12:50 PMPlato
- Feb 24th 2009, 12:54 PMAmanda1990
But surely in this case and so the conditions in the second part (which I'm

struggling with anyway, but we'll assume I've proved it) don't seem to hold. So we can't use it to prove the first part. (?)

EDIT: Ah, I see...I was thinking with A and B "the wrong way round". I see now how this second part helps. - Feb 24th 2009, 01:05 PMPlato
- Feb 24th 2009, 01:33 PMAmanda1990
I don't understand where we're going with the hint...as in I don't know what this means about open sets containing this point and don't see the relevance. What I have found is that implies that is connected, but clearly this isn't quite what we are after.