I am trying to prove the following proposition:
Let $X$ be a topological space, and $S$ a subset of $X$. then, an element $x$ of $X$ belongs to $\bar{S}$ (the closure of $S$) iff every open neighborhood of $x$ intersects $S$.
This is at page two of a primer on topological space and the definitions I should use to establish the result are those (standard) of the closure of a set and of an interior, plus – obviously – the definition of an open neighborhood of an element $x$ of the topological space.
In particular the last is given by the following: an open subset of $X$ that contains an element $x$ of $X$ is called an open neighborhood of $x$.
Approaching the problem, I frame it in the following way:
$ \forall x \in (X,\tau) ( x \in \bar{S} \Longleftrightarrow \forall A \in \tau (x \in A \Rightarrow A \cap S \neq \emptyset))$
My problems start with the "only if" condition [$\Longrightarrow$].
I assume that $x \in \bar{S}$ and that there is an open neighborhood $A^*$ of $x$ such that $A^* \cap S = \emptyset$ proceeding by contradiction. This implies that $x \not \in S$ and $x \in A^* \subseteq X\setminus S $. Now, here I stop, because I don't see how to relate this with the definition of $\bar S$, that is
$\bar S := \bigcap \{C: S \subseteq C \wedge X \setminus C \in \tau \}$
Beyond all of this, in terms of intuition, it seems to me that the all problem moves around the fact if $S$ is or not open, but I am not sure.
Any help or hint is more than welcome!
PS: I am aware that my post closely resembles this one, but still I did find some differences (or – at most – I could not see how it could help me).
Best Answer
We can see it by a few contrapositives. We can express that $x$ belongs to the intersection of all closed sets containing $S$ as
$$ x \in \overline{S} \iff \bigl(\forall U \in \tau\bigr)\bigl( S \subset (X\setminus U) \Rightarrow x \in X\setminus U\bigr) $$
by shifting the focus from the closed set $C$ to its complement $U = X\setminus C$. And now we take the contrapositive of the implication,
$$ x\in \overline{S} \iff \bigl(\forall U \in \tau\bigr)\bigl(\lnot(x \in X\setminus U) \Rightarrow \lnot(S\subset (X\setminus U)\bigr) $$
and rewrite the awkward negations,
$$x \in \overline{S} \iff \bigl(\forall U \in \tau\bigr)\bigl(x\in U \Rightarrow S\cap U \neq \varnothing\bigr).$$
And we're there.