# Homework Help: Why not metrizable

1. Sep 16, 2007

### pivoxa15

1. The problem statement, all variables and given/known data
X={a,b}

Let U be a topology on X then

let U1={empty set, {a}, X}
U2={empty set, {b}, X}

why isn't U1 and U2 metrizable?

However {empty set, {a}, {b}, X} and {empty set, X} is metrizable.

3. The attempt at a solution
Can't see why.

2. Sep 16, 2007

### Hurkyl

Staff Emeritus
X is a rather small. You should be able to completely classify all metrics on X.

3. Sep 16, 2007

### HallsofIvy

Well, what do you know about metric topologies? In particular have you seen the proof that every metric space is "Hausdorf" (T2)- given any two points, p and q, there exist disjoint open sets, one containing p, the other containing q (Take $N_{\delta}(p)$ and $N_{\delta}(q)$ with $\delta$ equal to 1/3 the distance between p and q)? In U1, there is no open set containing b but not a and in U2 there is no open set containing a but not b.

In {empty set, {a},{b}, X} (the "discrete topology) you can take d(p,q}= 1 if $p\ne q$, 0 if p= q. Then $N_{1/2}(a)= {a}$ and $N_{1/2}(b)= {b}$.

I believe, however, that {empty set, X} (the "indiscrete topology") is NOT metrizable. There exist no open set containing a but not b and likewise no open set containing b but not a so it is not Hausdorf (it is not even T0- {a} and {b} are not closed).

Last edited by a moderator: Sep 16, 2007
4. Sep 16, 2007

### pivoxa15

Are T2 and T0 axioms of somthing? Here is the proof http://planetmath.org/?op=getobj&from=objects&id=5838

Are Hausdorff spaces all metrizable?

Are there extra conditions needed like this one?
Let (X,d) be a compact metric space, let (Y,U) be a Hausdorff space and let f:X->Y be a cts function that maps X ontp Y. Then (Y,U) is metrizable.
So an onto function must be found for a Hausdorff space to be metrizable.

Last edited: Sep 16, 2007
5. Sep 16, 2007

### morphism

http://mathworld.wolfram.com/SeparationAxioms.html

A topology satisfying the T2 axiom is usually called a Hausdorff topology.

edit: Try proving that every metric space is Hausdorff. It shouldn't be hard at all. If you can't do it, you really ought to look back at your definitions.

Last edited: Sep 16, 2007
6. Sep 17, 2007

### morphism

I just noticed this latest edit.

The answer is no, not all Hausdroff spaces are metrizable. For example the Sorgenfrey line (i.e. the real line with the topology generated by the basis of half-open intervals [a,b), a.k.a. the lower limit topology) is Hausdorff but not metrizable. It's easy to see that it's Hausdorff; it's not metrizable because it's separable (i.e. has a countable dense subset) but not second countable (i.e. does not have a countable basis) (and separable metric spaces are second countable).

Now what do you mean by "an onto function must be found for a Hausdorff space to be metrizable"? This makes no sense whatsoever. What does finding such a function have to do with metrizability?

Anyway, there are 3 well-known (and non-trivial) general metrization theorems: Urysohn's, Nagata-Smirnov, and Bing's. For Hausdorff spaces in particular, I think (but am not 100% sure) that a necessary and sufficient condition for their metrizability is the following: A Hausdorff space X is metrizable iff there exists a continuous function $f:X \times X\to\mathbb{R}$ such that f-1({0}) = { (x,x) : x in X }. Or something equally bizarre. I'm pretty sure this is in Willard, General Topology, but I do not have a copy on hand to check.

Last edited: Sep 17, 2007
7. Sep 17, 2007