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 ${\rm int}\left(\overline{A}\right)\cap V\cap\overline{A}\neq\emptyset$?

  • 1
    $\begingroup$ 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

$\newcommand{\o}[1]{~\overline{#1}~}\newcommand{\i}{\operatorname{int}}$ 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 $$ \i(\o{A\cap V})\cap(A\cap V)\supseteq U\cap (A\cap V)\neq \emptyset. $$

  • $\begingroup$ Thanks for your help, the problem is solved. $\endgroup$ – Rigo Jul 24 '17 at 22:21

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.