"for x in X, every neighborhood of x contains $a \in A$" $\iff$ "the only closed subset of $X$ containing $A$ is $X$"?

Question

It's not exactly equivalent for dense set $A \subset X$ that:

"for any point $x \in X$, any neighborhood of $x$ contains at least one point from $A$"

$\iff$

"the only closed subset of $X$ containing $A$ is $X$"

Because doesn't the first one allow that $A$ can be larger than $X$ and that the complement of $X$ is some open non-empty set?

So in that case $X$ cannot equal $A$ and these two definitions for density are not equivalent?

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Wikipedia writes them as being equivalent. https://en.wikipedia.org/wiki/Dense_set

Answer 1

The statements are equivalent. Let $A$ satisfy the upper condition and suppose $F$ is a closed set of $X$ such that $A \subseteq F$. If $F$ were not equal to $X$, consider the open set $O=X\setminus F$. Let $x \in O$. Then $x$ has a neighbourhood $O$ that does not contain a point of $A$. This contradicts the property we assumed of $A$. So $F = X$.

Conversely, suppose the only closed set containing $A$ is $X$: now let $x \in X$. If $x$ would have a neighbourhood not containing any points of $A$, then there would be an open set $O$ such that $x \in O$ and $ O \cap A = \emptyset$, or $A \subseteq X\setminus O$. But then $X\setminus O$ would be a closed set containing $A$ not equal to $X$, contrary to assumption. So no such $x$ can exist and the first property is shown.

Answer 2

Let $F$ be a closed set s.t. $A\subset F$. So, $X=\overline{A}\subset F$ and then $F=X.$ Conversely, let $X$ be the only closed set containing $A$. Then obviously $\overline{A}=X.$

Here I used the definition: $A$ is dense $\iff \overline{A}=X$. This is trivially equivalent to the first condition by the characterization of a closure.