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This question already has an answer here:

Let $M$ be any subset of a Topological space $X$. We say $M$ is dense in $X$ if $\overline{M}=X$. Now I have to prove the question:

$M$ dense in $X$ $\Leftrightarrow$ for any subset $V \subseteq X : V \cap M \neq \emptyset $ where $V$ is open and not empty.

I tried to prove that by contradiction but didnt get far i hope you can help me out here.

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marked as duplicate by Eric Stucky, Misha Lavrov, Leucippus, Brevan Ellefsen, user91500 May 2 '17 at 8:56

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  • $\begingroup$ My only definition is; $M$ dense in $X \Leftrightarrow \bar{M} = X$ . And now i have to prove the above question $\endgroup$ – user439387 May 1 '17 at 17:09
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    $\begingroup$ Definition of $\bar{M}$ then? $\endgroup$ – Henno Brandsma May 1 '17 at 17:44
  • $\begingroup$ Since this is an introductory-level Q, you should state the def'n of the closure of M that you are starting from, as there are a few (equivalent ) ways to define it. $\endgroup$ – DanielWainfleet May 2 '17 at 4:31
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So we need to show:

$$\overline{M} = X \leftrightarrow \forall V \subseteq X \text{ open and non-empty } : V \cap M \neq \emptyset$$

Now it will depend on how you define $\overline{M}$. If you define it as the smallest closed subset that contains $M$ (one of the usual definitions) I'd go as follows:

Left to right: assume $\overline{M} =X$ and let $V$ be any non-empty open subset of $X$. Then $M \nsubseteq X \setminus V$, or otherwise the latter set would be a smaller closed subset than $X$ that contained $M$. So there is always a point of $M$ that is not in $X \setminus V$, or put equivalently: $M$ always intersects $V$, as required.

Right to left: suppose the right hand condition holds. Then let $C$ be a closed subset of $X$ with $M \subseteq C$. We want to show that $C =X$ (so $X$ is then the only (hence smallest) closed superset of $M$). If $C \neq X$, $V = X\setminus C$ is non-empty and open, but $V \cap M = \emptyset$, this contradicts the right hand condition. So $C = X$.

Another common definition is that $\overline{A}$ is the set of all adherence points of $A$, i.e. all $x \in X$ such that for any open set $O$ that contains $x$, $O \cap A \neq \emptyset$.

In that case the equivalence is more immediate:

Left to right. If $V \neq \emptyset$ is open then for $p \in V$ we have that $p$ is an adherence point of $M$, as $\overline{M} = X$, and so $V$ must intersect $M$.

Right to left: suppose $p \in X$. Then let $O$ be any open set that contains $p$. Then the right hand condition implies that $O \cap M \neq \emptyset$. So $p$ is an adherence point of $M$ and as $p \in X$ was arbitrary, $\overline{M} =X$

If your definition of closure is different, please let us know.

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  • $\begingroup$ Thank you very much that helped me a lot. $\endgroup$ – user439387 May 1 '17 at 17:58
  • $\begingroup$ @low77 Glad I could help. What definition of closure do you use? $\endgroup$ – Henno Brandsma May 1 '17 at 18:03
  • $\begingroup$ I am using the first definition $\endgroup$ – user439387 May 1 '17 at 18:04
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There is a lemma: $x\in\overline{A}$ if and only if for any open set $V$ s.t. $x\in V$, we have $V\cap A\ne\varnothing.$ This is rather easy to prove and it is useful here.

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