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Let $X$ be a metric space and $A$ a subset of $X$. $A$ is said to be dense if $\operatorname{Cl}(A)=X$. Prove that $A$ is dense iff the only superset of $A$ is $X$ iff the only open set disjoint from $A$ is the empty set iff $A$ intersects every non-empty open set iff $A$ intersects every open ball.

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What do you mean by "superset"? We know that $\mathbb{Q}$ is dense in $\mathbb{R}$, but $\mathbb{Q}\cup\{\pi\}$ is a superset of $\mathbb{Q}$ different from both $\mathbb{Q}$ and $\mathbb{R}$. I'm using the definition: $A$ is a superset of $B$ if $A \subseteq B$. Also, what did you attempt? –  dls Apr 18 '12 at 1:08
Presumably it should be "the only closed superset of $A$ is $X$". –  Robert Israel Apr 18 '12 at 1:10
@MiguelMoraLuna You have posted several questions of a similar nature: Namely here is question $X$, solve it for me. Please provide some context for your questions, what have you tried, etc. –  user38268 Apr 18 '12 at 9:57
Related meta thread: meta.math.stackexchange.com/questions/4001/… –  cardinal Apr 19 '12 at 23:57

1 Answer 1

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All of these follow almost immediately from the definition of $A$ being dense in $X$. For example, that $A$ is dense in $X$ is the same thing as saying that every point of $X$ is a limit point of $A$ or a point of $A$.

Recall that the closure of $A$ is the smallest closed set containing $A$. If $A$ is closed and dense in $X$, then $A$ has to be the whole space. So we may as well suppose that $A$ is not closed.

Let me give an example of how we can use this to prove one of your equivalencces: Suppose every non-empty open set $U$ intersects $A$. Now take a point $x \notin A$. Then by by assumption any open set $V$ about $x$ will intersect $A$, so it follows that $x$ is a limit point of $A$. Consequently $\overline{A} = X$. Conversely if $A$ is dense in $X$, this means that every point in $X$ is a point of $A$ or a limit point of it. So let $U$ be a non-empty open set in $X$. Suppose that $U \cap X = \emptyset$. Then this means that $X - U$ is a closed set containing $X$ that is strictly contained in $X$. But then this is a contradiction because the closure of $A$ is the whole space, so that there can be no closed sets properly contained in $X$ that contain $A$. This proves that every non-empty open set in $X$ must intersect $A$.

The rest of your equivalences can be proven using similar arguments.

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