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X is a topological space, $ A\subseteq B\subseteq X $, if $A$ is a nowhere dense subset of $B$ , then $A$ is a nowhere dense subset of the whole space $X$?

Is this right? I thind it's right, but I can't prove it. Maybe we need to add some contions: e.g. $B$ is a closed subset of $X$, and/or $A$ is (also) a closed subset of $X$.

David Chan
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  • I'm not sure and I'm too tired to try and figure it out right now. My gut says you're right, but I'm not going to commit without proof. I think you may find this post helpful for answering your question: http://math.stackexchange.com/questions/58190/understanding-of-nowhere-dense-sets – Mathemagician1234 Apr 28 '15 at 08:01

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If $A$ is nowhere dense in $B$, then $\operatorname{cl}_BA$ has empty interior in $B$, and therefore $B\setminus\operatorname{cl}_BA$ is a dense open set in $B$. There is then a $U$ open in $X$ such that $U\cap B=B\setminus\operatorname{cl}_BA$. Let

$$V=U\cup(X\setminus\operatorname{cl}_XB)\;;$$

clearly $V$ is open in $X$, and you should have no trouble showing that $V$ is dense in $X$. Finally, $V\cap A=\varnothing$, so $A$ must be nowhere dense in $X$.

Brian M. Scott
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  • you should have no trouble showing that V is dense in X,why? I just know another method of solving,but for your good answer, I can't prove this sentence. – David Chan Apr 29 '15 at 07:41
  • @David: Let $W$ be a non-empty open set in $X$. If $W\cap B=\varnothing$, then $W\subseteq X\setminus\operatorname{cl}_XB$, so $W\cap V\ne\varnothing$. And if $W\cap B\ne\varnothing$, then $W\cap U\supseteq W\cap(B\setminus\operatorname{cl}_BA\ne\varnothing$, wo $W\cap V\ne\varnothing$. Thus, every non-empty open set in $X$ has non-empty intersection with $V$, which is exactly what it means for $V$ to be dense in $X$. – Brian M. Scott Apr 29 '15 at 19:21
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If $\overline{B}$ has empty interior, then since $\overline{A} \subseteq \overline{B}$, $\overline{A}$ must have empty interior also (otherwise contradicting the nowhere-denseness of $B$).

Michael Cromer
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