Separation axioms vs subspaces

[Hurkyl]

How do separation axioms carry over to subspaces?

Some are clear -- it's easy to see that if any two points of a space X are separated by neighborhoods, then the same must be true of any subset S of X.

But what about the nicer ones? Is it true that if S is a subset of a normal space, that S is itself normal?

This one is less obvious... one example that worries me is this:

Consider the union S of two open discs in R^2. (that aren't disjoint) Consider the two closed sets formed by restricting the boundaries of the two discs to S. We can't directly appeal to the normality of R^2, because the closure of these sets aren't disjoint in R^2.

It is still easy to see S is normal, because it's homeomorphic to R^2, but that doesn't help me in the general case of a subset of an aribtrary normal space.

 

[NateTG]

It should be pretty easy to see that subspaces on closed subsets of normal spaces are normal. Or, for that matter that on any normal space [itex]S[/itex], the subspace formed on [tex]S_1[/tex] is normal if, for any two closed sets [tex]U,V[/tex] in [itex]S[/itex] there exists an open set [itex]O[/itex] so that [itex]O\supset U \cap V[/itex] and [itex]O \cap S_1 = \emptyset[/itex]

It's pretty straightforward to come up with examples of subspaces of normal spaces that aren't normal. For example, consider the space
[tex]\{0,1,o,c\}[/tex]
Where the open sets are
[tex]\emptyset,\{o\},\{0,o\},\{1,o\},\{0,1,o\},\{0,1,o,c\}[/tex]

Which is normal, since the only closed set that does not contain [itex]c[/itex] is [itex] \emptyset[/itex].

However, the subspace [itex]\{0,1,o\}[/itex] is not normal since [itex]{0}[/itex] and [itex]{1}[/itex] are both closed, but any open set containing either also contains [itex]o[/itex].

 

[Hurkyl]

Bleh. Well, what if the space is nicer? Like a normal T2 space?

 

[NateTG]

Hmmm. I have a hard enough time thinking of a non-normal T2 space.

 

[Hurkyl]

Well, http://en.wikipedia.org/wiki/Normal_space suggests that if you take the uncountable product of noncompact Hausdorff spaces, it won't be normal.

Oh, I guess that suggests an example -- take an uncountable product of compact Hausdorff spaces. That should be normal, right? Then, take as a subset the uncountable product of an open subset.


*sigh* How disappointing. Topology is hard! Maybe I should stick to the world of metric spaces, I understand those better!

 

[NateTG]

Oh, I guess that suggests an example -- take an uncountable product of compact Hausdorff spaces. That should be normal, right? Then, take as a subset the uncountable product of an open subset.

I was wondering if the long line was going to figure into this.

Topology is nice. It's just that negative examples can be really hard to think of.

 

[Hurkyl]

Ack, open sets that are compact? Well, doesn't the quote from the wikipedia article imply the open subset wouldn't be normal?


Incidentally, the property I want to be true is this:

If I have closed sets [itex]A \cup B = C[/itex], then there exists closed sets A' and B' such that [itex]A' \cup B' = X[/itex], [itex] A' \cap C = A[/itex], and [itex]B' \cap C = B[/itex].

Actually, I don't even need that strong -- I think that it's good enough that if C is a reducible closed set, then there is a way to write X as the union of two closed sets, neither containing C.


I think that this is provable if the complement of C is a normal space, but I guess I can't guarantee that, even if X is a "nice" space.