Connectedness: Show Y U A & Y U B are Connected

[radou]

Homework Statement

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.

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?

 

[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?

 

[radou]

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.

 

[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.

 

[radou]

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.

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

 

[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.

 

[radou]

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.

 

[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.

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.

 

[radou]

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.

You're right, for some reason I forgot to mention this.