Connected normal space problem

[radou]

Homework Statement

Show that a connected normal space having more than one point is uncountable.

The Attempt at a Solution

First of all, if X has more than one point, take the points a and b in X. Let U be an open set containing a. Take a basis element B around a contained in U. Then Cl(B) and X\U are disjoint closed sets containing a and b, respectively.

Since X is normal, by the Urysohn lemma there exists a continuous function f : X --> [0, 1] such that f(Cl(B)) = 0 and f(X\U) = 1.

Now, assume that X is countable, and let f(X) be the image set of f. f(X) is a countable subset of [0, 1], and since X is connected, by the intermediate value theorem, for any r between 0 and 1 there exists some point c of X such that f(c) = r. Let r be some point of [0, 1]\f(X). Then there must exist some c in X which maps to r. Since [0, 1] is uncountable, we arrive at a contradiction with the assumption that X is countable.

I hope this works, thanks in advance.

 

[micromass]

I agree with the general proof. But I'm a bit skeptic of this part:

First of all, if X has more than one point, take the points a and b in X. Let U be an open set containing a. Take a basis element B around a contained in U. Then Cl(B) and X\U are disjoint closed sets containing a and b, respectively.

Some remarks:
- What do you mean with "Take a basis element B". What basis are you talking about?
- How do you know for certain that X\U contains b? What happens if you took U such that a and b are in U?

You will have to use somewhere that the one-point sets are closed...

 

[radou]

Ahhh, a good remark indeed!

I completely forgot that one-point sets closed are a part of the definition of normality! I was skeptic about that very part of my proof too.

Btw, another question, if I have a space, I assume I can't just "take" a basis for it, i.e. simply assume it exists?

If one-point sets are closed in X, it suffices to take two distinct elements a and b from X, and apply the Urysohn lemma to the closed and disjoint sets {a} and {b}.

 

[radou]

Btw, the first attempt of my proof was the same, but I just "took" two closed disjoint sets in X, without showing that they exist. That's the point - I can't simply take two such sets. In a general toipological space, they need not exist, right?

 

[micromass]

Yes, that is correct!

About the basis thing. If you're given a topological space $$(X,\mathcal{T})$$ then you can always find a basis for this space, namely the topology $$\mathcal{T}$$ itself. But then it isn't really interesting to talk about a basis... You should only work with bases if they have an interesting description of interesting properties. For example, when working with second countable spaces, then there exists a basis with interesting properties, so it makes sense to work with such a basis.

 

[micromass]

Btw, the first attempt of my proof was the same, but I just "took" two closed disjoint sets in X, without showing that they exist. That's the point - I can't simply take two such sets. In a general toipological space, they need not exist, right?

Yes, in a general topology, such a thing will not exist. That's precisly why the T1 axiom is introduced.

 

[radou]

Yes, that is correct!

About the basis thing. If you're given a topological space $$(X,\mathcal{T})$$ then you can always find a basis for this space, namely the topology $$\mathcal{T}$$ itself. But then it isn't really interesting to talk about a basis... You should only work with bases if they have an interesting description of interesting properties. For example, when working with second countable spaces, then there exists a basis with interesting properties, so it makes sense to work with such a basis.

Yes, I see.

Yes, in a general topology, such a thing will not exist. That's precisly why the T1 axiom is introduced.

OK, thanks!