Topological spaces, uncountable subsets and separability - MathOverflow

Topological spaces, uncountable subsets and separability

2011-05-27T22:32:11Z

Hi, the following is a well known theorem

Let $M$ be a metric space. If every uncountable subset of $M$ has a limit point, then $M$ is separable.

Question: Is there a similar result for topological spaces?

I have almost no knowledge of topology so I can only hope that this is not trivial.

Answer by Shaun Ault for Topological spaces, uncountable subsets and separability

2011-05-27T23:07:00Z

The statement, "Let $M$ be a topological space. If every uncountable subset of $M$ has a limit point, then $M$ is separable," is false. Consider the first uncountable ordinal $\omega_1$, under the order topology (see http://en.wikipedia.org/wiki/First_uncountable_ordinal). $\omega_1$ is countably compact, hence also weakly countably compact (that is, every infinite subset has a limit point), but $\omega_1$ is not separable.

Answer by Robert Israel for Topological spaces, uncountable subsets and separability

2011-05-27T23:19:46Z

Consider the topological space $[0, \omega_1]$ where $\omega_1$ is the first uncountable ordinal. Every uncountable subset has a limit point, namely $\omega_1$, because the complement of any neighbourhood of $\omega_1$ is countable. However, it is not separable, since the supremum of a countable subset of countable ordinals is countable.

Answer by Stefan Geschke for Topological spaces, uncountable subsets and separability

2011-05-28T06:53:48Z

Every compact space satisfies "every infinite set has a limit point". (I am assuming here that by limit point you mean what I would call an accumulation point, i.e., a point $x$ such that evey neighborhood contains infinitely many elements of the set.) So in particular, in a compact space every uncountable set has a limit point. It follows that every compact space that is not separable (for instance a sufficiently high power of the closed unit interval or Robert Israel's example) shows that the theorem you mention does not hold for topological spaces in general.

The theorem you mention of course implies that compact metric spaces are separable.

Answer by Henno Brandsma for Topological spaces, uncountable subsets and separability

2011-05-28T11:37:27Z

The property "Every uncountable set has a limit point" is related to the Lindelöf property (every open cover has a countable subcover). For metrisable spaces these notions are equivalent, and in general Lindelöf implies the limit point property. And there are many Lindelöf spaces that are not separable, as the other examples show. For metrisable spaces, being Lindelöf, separable, second countable, etc. are all equivalent.