# Separation axioms vs subspaces

1. Jun 8, 2005

### Hurkyl

Staff Emeritus
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.

2. Jun 8, 2005

### 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 $S$, the subspace formed on $$S_1$$ is normal if, for any two closed sets $$U,V$$ in $S$ there exists an open set $O$ so that $O\supset U \cap V$ and $O \cap S_1 = \emptyset$

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

Which is normal, since the only closed set that does not contain $c$ is $\emptyset$.

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

3. Jun 8, 2005

### Hurkyl

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

4. Jun 8, 2005

### NateTG

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

5. Jun 8, 2005

### Hurkyl

Staff Emeritus
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!

6. Jun 8, 2005

### NateTG

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.

Last edited: Jun 8, 2005
7. Jun 8, 2005

### Hurkyl

Staff Emeritus
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 $A \cup B = C$, then there exists closed sets A' and B' such that $A' \cup B' = X$, $A' \cap C = A$, and $B' \cap C = B$.

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.