# Connectedness again

1. Oct 13, 2010

1. The problem statement, all variables and given/known data

This one is giving me serious trouble.

Let Y be a subset of X. Let both X and Y be connected. SHow that if A and B form a separation of X\Y, then Y U A and Y U B are connected.

3. The attempt at a solution

I know all the basic definitions and theorems from the chapter about connectedness preceding this exercise section, but any way I try it, I don't seem to get anywhere.

Any ideas?

2. Oct 13, 2010

### ystael

When you say "$$A$$ and $$B$$ form a separation of $$X \setminus Y$$", what does this mean, exactly? My guess would be what I usually call "$$\{A, B\}$$ disconnect $$X \setminus Y$$", that is, $$A$$ and $$B$$ are disjoint (relatively) open subsets of $$X \setminus Y$$, whose union is $$X \setminus Y$$. Is this correct?

3. Oct 13, 2010

Well, partially. According to Munkres, there's a subtle difference:

If X is a topological space, a separation of X is a pair of non empty, disjoint and open subsets U and V of X whose union is X. (This is a definition.)

If Y is a subspace of X, a separation of Y is a pair of non empty disjoint sets U, V whose union if Y. (This is a Lemma)

Note that in the Lemma it is not required for the separation sets to be open.

So, I assume if we're talking about X\Y, which is a subspace of X, the sets in the separation are not required to be open.

4. Oct 13, 2010

### ystael

That won't work. You need some kind of relative openness hypothesis, otherwise the result is false. Take $$X = \mathbb{R}$$, $$Y = [0, 1]$$, $$A = (-\infty, -1] \cup (1, 2)$$, $$B = (-1, 0) \cup [2, \infty)$$. Then $$A$$ and $$B$$ form a separation (according to the conditions you gave) of $$X \setminus Y = (-\infty, 0) \cup (1, \infty)$$, but $$Y \cup A = (-\infty, -1] \cup [0, 2)$$ and $$Y \cup B = (-1, 1] \cup [2, \infty)$$ are both disconnected.

Also, with the definition of "separation" you cite for subspaces, you can give a separation for a connected subspace, such as $$X = \mathbb{R}, Y = [0, 1], U = [0, \textstyle\frac12), V = [\textstyle\frac12, 1]$$. That doesn't make sense.

5. Oct 14, 2010

Interesting, it's exactly what says in the book. So, in the "separation lemma" for subspaces, the sets should be open too?

6. Oct 14, 2010

### ystael

At least relatively open in the subspace, otherwise they are useless for establishing disconnectedness. I don't have a copy of the new edition of Munkres, so I can't guess what he might be thinking there.

7. Oct 14, 2010

I found the solution to this exercise (Ex 23.12)

http://www.math.ku.dk/~moller/e02/3gt/opg/S23.pdf

Consider this problem solved, I'll go through it by myself, although I don't like the looks of it, for some reason.

I posted it if you happen to be interested to look at it regardless.

8. Oct 14, 2010

### ystael

OK, I looked it up. The problem is that you missed an important hypothesis in the lemma 23.1, which changes things entirely.

Lemma 23.1 of Munkres says that a separation of $$Y \subset X$$ is a disjoint pair of nonempty sets $$A, B$$ whose union is $$Y$$, which satisfy the condition that neither of $$A, B$$ contains a limit point of the other. This latter condition is exactly what you need for $$A$$ and $$B$$ to be relatively open (in fact, clopen) in $$Y$$. The examples I gave above do not satisfy the latter condition.

9. Oct 15, 2010