Let T be an uncountable Hausdorff space. The following property of T will be referred to as "property P". If S is any uncountable subset of T, then the set of all points of S that are not limit points of S is at most countable. My question is: Does the posession of property P imply that T is second countable. The reverse implication is a well known classical theorem. .....My motive for asking this question comes from thinking about mathoverflow.net question No.66240, which is a rather similar question involving a somewhat weakened version of property P, and which received a "NO" answer. But the limit points of the sets S-the counterexamples for the "NO" answer-were never proved to be points of S. In the case where T was a set of ordinal numbers, if S was an uncountable set of countable ordinal numbers, its limit points need not belong to S at all. So I wondered whether, if one imposed property P-a requirement that sufficiently many of the limit points of S should belong to S-the answer might change from a "NO" to a "YES".
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asked Mar 16, 2014 at 19:02
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$\begingroup$ @Bjørn: The set of points in $[0,1]$ which are not limit points of $[0,1]$ is empty. Last I recall, empty sets are at most countable. So it's not a counterexample. $\endgroup$– Asaf Karagila ♦Commented Mar 17, 2014 at 0:19
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Given an example of a space $X$ with property $P$, let $Y$ be the topological sum of $X$ with any countable non-second countable space. Then $Y$ has property $P$ but is not second-countable.
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$\begingroup$ @ Santi Sparado: Thanks for your answer. Perhaps you could clarify one point. Let Z be the space having property P. Is Y the set union of X and Z and is the set union of any base of X with any base of Z, a base of Y? I am just trying to understand clearly what "topological sum" means $\endgroup$ Commented Mar 17, 2014 at 19:42
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$\begingroup$ I must apologize! My previous comment is topsy-turvy. X should be the space that has property P and Z should be the countable space that is not second countable. But I would still like to know what is meant by the "topological sum" of these two spaces. $\endgroup$ Commented Mar 17, 2014 at 20:05
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$\begingroup$ The disjoint sum of spaces $(X, \sigma)$ and $(Y, \tau)$ (where $X \cap Y=\emptyset$) is defined as the topology on $X \cup Y$ where a set $U$ is open if and only if $U \cap X \in \sigma$ and $U \cap Y \in \tau$. $\endgroup$ Commented Mar 17, 2014 at 20:35
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$\begingroup$ Thanks. That certainly proves that property P does not imply second countability and one can take any second countable topological space for X. $\endgroup$ Commented Mar 18, 2014 at 22:39