Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

This is my understanding of the proof. Please let me know if I am correct.

If I take any two points in a trivial topology. I can find the minimum radius such that they both are in two different open sets. Now if I union them, then there is no guarantee that a third point will not be a part of the new open ball. So the union of the two points might not be another open set. So we cannot have a metric space for this topology and this topology is not in Hausdorff space.

If a topological space is Hausdorff why is it not a sufficient condition to be metrizable?

share|cite|improve this question
What do you mean by "the minimum radius such that they both are in two different open sets"? – Brad Feb 13 '14 at 17:25
And what do you mean by "the new open ball"? – MPW Feb 13 '14 at 17:30
up vote 3 down vote accepted

A one point space with the trivial topology is metrizable. We show that if there are $\ge 2$ points, it is not metrizable. Suppose to the contrary that it is. Let $x$ and $y$ be distinct points, and let $a=d(x,y)$. Then the open set of all $u$ such that $d(x,u)\lt a/2$ contains $x$ but not $y$, so the topology is not trivial.

share|cite|improve this answer

This look like at least two questions:

  1. Let $X$ be a topological space with trivial topology, i.e. only $\emptyset$ and $X$ are open (and closed). Every metric space is Hausdorff, so if $X$ is metrizable then for every $x\in X$ the set $\{x\}$ must be closed. This is only possible if $X$ is a one-point space (and in that case of course it is metrizable). - Another way to see this: Assume $x,y\in X$ are two distinct points in a metric space $(X,d)$. Then the open ball of radius $d(x,y)>0$ around $x$ is neither empty (contains $x$) nor the whole space (does not contain $y$). Hence the topology given by the metric is not the trivial one.

  2. A suitable examples for a non-metrizable Hausdorff space is the "long line"

share|cite|improve this answer

Metric spaces must satisfy some pretty strong properties which are not implied by Hausdorffness alone. Among these are

  • they must be first-countable (each point has a countable neighbourhood basis).

    For example, the Arens-Fort space is a Hausdorff space which is not first-countable.

  • they must be perfectly normal (the satisfy the normal separation axiom, and in addition all closed sets are the countable intersection of open sets).

    For example, the Niemytzki (or Moore) plane is a Hausdorff space which is not perfectly normal.

  • they are second-countable (have a countable basis) iff they are separable (have a countable dense set). (In arbitrary topological spaces, secound-countability implies separability.)

    For example, the Sorgenfrey line is a separable Hausdorff space which is not second-countable.

    (In full generality, in a metric space the least cardinality a dense set is equal to the least cardinality of a basis.)

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.