Vtyjkyuk

Prove:
If $A$ is nowhere dense subset of a topological space $X$, then $X setminus A$ is
dense in $X$

This question is already appeared in this site but I'm mention again here is for check my proof.

Here's my try:

$A$ is nowhere dense means $textInt(overlineA)=phi$

In order to prove $X setminus A$ is dense, we prove every point of $X$ is either a point of $X setminus A$ or a limit point of $X setminus A$.

Let $x in X$ be arbitrary. If $x in X setminus A$, then we are done. So assume $x notin X setminus A$. That is $x in A$. In this case we prove $x$ is a limit point of $X setminus A$

Since $A subset overlineA$ implies $x in overlineA$ . Also we know $textIntA subset textInt(overlineA)(=phi,textby hypothesis)$, so $textIntA =phi$

So we write, $$x in overlineA=overlineA setminus phi =overlineA setminus textIntA=partial(A)=overlineA cap overlineXsetminus A $$

Hence $x in overlineXsetminus A$ and so $x$ is a limit point of $Xsetminus A$

Is this correct? Any suggestions must be appreciated!

Your proof is correct. In fact $X setminus A$ is dense in $X$ if and only if $textInt(A) = emptyset$ because $overlineX setminus A = X setminus textInt(A)$.