Topological spaces, uncountable subsets and separability - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-23T15:37:57Z http://mathoverflow.net/feeds/question/66240 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/66240/topological-spaces-uncountable-subsets-and-separability Topological spaces, uncountable subsets and separability Peter 2011-05-27T22:32:11Z 2011-05-28T11:37:27Z <p>Hi, the following is a well known theorem</p> <p><em>Let $M$ be a metric space. If every uncountable subset of $M$ has a limit point, then $M$ is separable.</em></p> <p>Question: Is there a similar result for topological spaces?</p> <p>I have almost no knowledge of topology so I can only hope that this is not trivial.</p> http://mathoverflow.net/questions/66240/topological-spaces-uncountable-subsets-and-separability/66243#66243 Answer by Shaun Ault for Topological spaces, uncountable subsets and separability Shaun Ault 2011-05-27T23:07:00Z 2011-05-27T23:07:00Z <p>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 <a href="http://en.wikipedia.org/wiki/First_uncountable_ordinal" rel="nofollow">http://en.wikipedia.org/wiki/First_uncountable_ordinal</a>). $\omega_1$ is countably compact, hence also weakly countably compact (that is, every infinite subset has a limit point), but $\omega_1$ is not separable.</p> http://mathoverflow.net/questions/66240/topological-spaces-uncountable-subsets-and-separability/66245#66245 Answer by Robert Israel for Topological spaces, uncountable subsets and separability Robert Israel 2011-05-27T23:19:46Z 2011-05-27T23:19:46Z <p>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.</p> http://mathoverflow.net/questions/66240/topological-spaces-uncountable-subsets-and-separability/66261#66261 Answer by Stefan Geschke for Topological spaces, uncountable subsets and separability Stefan Geschke 2011-05-28T06:53:48Z 2011-05-28T06:53:48Z <p>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.</p> <p>The theorem you mention of course implies that compact metric spaces are separable.</p> http://mathoverflow.net/questions/66240/topological-spaces-uncountable-subsets-and-separability/66271#66271 Answer by Henno Brandsma for Topological spaces, uncountable subsets and separability Henno Brandsma 2011-05-28T11:37:27Z 2011-05-28T11:37:27Z <p>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. </p>