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Definition: A subset $A$ of a space $X$. Then $A$ is meager in $X$ if $A=\displaystyle\bigcup_{n\in N}A_n,$ where ${\rm int}(\overline{A_n})=\emptyset$, for all $n\in N.$ And $A$ is nowhere meager in $X$, if every non-empty relatively open subset of $A$ is not meager in $X$.

Problem: Suppose that $X$ is a topological space and $A\subseteq X$ nowhere meager. Thus $\overline{A}$ is a regular closed, that is, ${\rm int}(\overline{A})$ is dense in $\overline{A}$.

For to prove this, I try to see ${\rm int}\left(\overline{A}\right)$ intersects every non empty relatively open subset of $\overline{A}.$ In efect, Let be $V$ an open of $X$ such that $V\cap\overline{A}\neq\emptyset.$ Since $A$ is nowhere meager then $V\cap A$ is not meager in particular ${\rm int}\left(\overline{V\cap A}\right)\neq\emptyset.$ But I can not finish the proof. How do I conclude that ${\rm int}\left(\overline{A}\right)\cap V\cap\overline{A}\neq\emptyset$?

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    $\begingroup$ Are relative closed sets just sets who are closed under taking "relatives"? :) $\endgroup$ – Asaf Karagila Jul 24 '17 at 9:05
  • $\begingroup$ @Asaf Karagila: A subset $B\subset A$ is relative open/closed in $A$ if there exists an open/closed set $V\subset X$ such that $B=A\cap V$. $\endgroup$ – Mundron Schmidt Jul 24 '17 at 11:35
  • $\begingroup$ @MundronSchmidt. Karaglia is joking. The usual term is "relatively open." $\endgroup$ – DanielWainfleet Jul 25 '17 at 15:13
  • $\begingroup$ @AsafKaragila.Sorry for mis-spelling your name. $\endgroup$ – DanielWainfleet Jul 25 '17 at 16:20
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$\newcommand{\o}[1]{~\overline{#1}~}\newcommand{\i}{\operatorname{int}}$ First, consider $\i(\o A)\subset \o A$. Therefore $$ \i(\o A)\cap V\cap \o A=\i(\o A)\cap V. $$ Since $A\supseteq A\cap V$ we conclude $\o A\supseteq\o{A\cap V}$ and $\i(\o A)\supseteq\i (\o{A\cap V})$. Further we have $V\supseteq A\cap V$ and therefore $$ \i(\o A)\cap V\supseteq \i(\o{A\cap V})\cap(A\cap V). $$ Choose $x\in\i(\o{A\cap V})\neq\emptyset$ and we get an open neighbourhood $U$ of $x$ in $\o{A\cap V}$. Consider that $U\subset \i(\o{A\cap V})$. Further $x\in\o{A\cap V}$ implies that $U\cap (A\cap V)\neq\emptyset$ and hence $$ \i(\o{A\cap V})\cap(A\cap V)\supseteq U\cap (A\cap V)\neq \emptyset. $$

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  • $\begingroup$ Thanks for your help, the problem is solved. $\endgroup$ – Rigo Jul 24 '17 at 22:21
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By contradiction: Suppose $p\in \overline A$ \ $(\overline {Int(\overline A)}).$ Let $U=X$ \ $(\overline {Int(\overline A)}).$ Then $U$ is open and $U\cap A$ is relatively open in $A.$ And $U\cap A$ is not empty (Because $U$ is a nbhd of $p$ and $p\in \overline A$). So $U\cap A$ is not meager, so $$ Int(\overline {U\cap A})\ne \phi.$$ $$\text {We have }\quad Int (\overline {U\cap A}\subset Int(\overline A)\subset \overline {Int (\overline A)} = X \backslash U.$$ $$\text {We also have }\quad Int (\overline {U\cap A})\subset \overline {U\cap A}\subset \overline U.$$ Now $Int (\overline {U\cap A})$ is an open set which is disjoint from $U$, hence $Int(\overline {U\cap A})$ is disjoint from $\overline U$. But $Int (\overline {U\cap A})$ is also a subset of $\overline U.$ So $$Int (\overline {U\cap A})=\phi$$ which is a contradiction.

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