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In metric space there is a well-known result that $X$ is a separable metric space $\implies$Every uncountable set in $X$ has a limit point.My question is does the result hold if $(X,\tau)$ is arbitrary topological space?I think that it may not hold for Non-Hausdorff spaces where we do not have disjoint open sets around two points.

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  • $\begingroup$ @bof so what are the conditions on $X$ for the result that every uncountable set has a limit point to be true? $\endgroup$ Commented Mar 11, 2020 at 5:53
  • $\begingroup$ @bof So this result is not true for arbitrary topological space right? $\endgroup$ Commented Mar 11, 2020 at 6:06

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The Niemitzky plane (a.k.a. the Moore plane or the Moore-Niemitzky plane) is Hausdorff (in fact it is a Tychonoff space) which is separable because $\Bbb Q\times \Bbb Q^+$ is dense. And $E=\Bbb R\times \{0\}$ is an uncountable closed discrete subspace so $E$ has no limit point.

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See the example in the following link and take the uncountable set to the the real line. This set has no limit point since basic open sets containig a real number have only one point from the line: https://www.mathcounterexamples.net/a-separable-space-that-is-not-second-countable/

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