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Let be $X$ a topological space such that X is a Hausdorff, regular and separable space. If $U\subseteq X$ is open such that $U=int(cl(U))$, and $E\subseteq X$ is a countable dense set, I need to prove that $int(cl(U \cap E))\subseteq E$

dajc
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  • It's not true: take $X=\Bbb R$, $E=\Bbb Q$, and $U=(0,1)$. Check the statement of the result. – Brian M. Scott Apr 18 '15 at 23:57
  • I'm not sure this is true, take for instance $X=\mathbb{R}$ with Euclidean topology, $U=X$ and $E=\mathbb{Q}$. Then $$\mathrm{int}(\mathrm{cl}(U\cap E))=\mathrm{int}(X)=X=\mathbb{R}\not\subseteq \mathbb{Q}=E$$ – Uncountable Apr 18 '15 at 23:59
  • ¡Thanks for the help! Brian, I'm looking this post, but I don't understand how the injection of the first part maps the basis to $\mathbb{P}(D)$ – dajc Apr 19 '15 at 00:14
  • Ah, okay. So you really want $\operatorname{int}\operatorname{cl}(U\cap E)=U$, right? – Brian M. Scott Apr 19 '15 at 00:15
  • I understand that $int(cl(U \cap E)=U$, but I don't understand the part of the injection to say that the cardinality of the base if least or equal to $| \mathbb{P}(E) |$. It's obvious that the fuction that maps $U$ to $int(cl(U \cap E)$ is an injection, but I don't undestand why the image of this function is a subset of $\mathbb{P}(E)$, if the function is defined of this form. – dajc Apr 19 '15 at 00:19
  • The injection is the function that maps a regular open set $U$ to $U\cap E$. If $U$ and $V$ are distinct regular open sets, then $U\cap E\ne V\cap E$. – Brian M. Scott Apr 19 '15 at 00:25
  • The cardinality of the base is at most, not at least, the power set of $E$. Or did you make a typo, meaning is less than or equal to, when you wrote "least" instead of "less"? – Mirko Apr 19 '15 at 00:35

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