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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 $\text{Int}(\overline{A})=\phi$

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 $A \subset \overline{A}$ implies $x \in \overline{A}$ . Also we know $\text{Int}A \subset \text{Int}(\overline{A})(=\phi,\text{by hypothesis})$, so $\text{Int}A =\phi$

So we write, $$x \in \overline{A}=\overline{A} \setminus \phi =\overline{A} \setminus \text{Int}A=\partial(A)=\overline{A} \cap \overline{X\setminus A} $$

Hence $x \in \overline{X\setminus A}$ and so $x$ is a limit point of $X\setminus A$

Is this correct? Any suggestions must be appreciated!

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Your proof is correct. In fact $X \setminus A$ is dense in $X$ if and only if $\text{Int}(A) = \emptyset$ because $\overline{X \setminus A} = X \setminus \text{Int}(A)$.

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  • $\begingroup$ Thanks!.......... $\endgroup$ – user444830 Sep 6 '18 at 13:46

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