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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).

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Kolmin
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  • "those (standard) of the closure of a set" One standard definition of the closure of a set is exactly (more or less, the neighbourhoods aren't necessarily required to be open) the property you shall prove. What is the definition of the closure of a set that you are working with? – Daniel Fischer Dec 17 '13 at 15:53
  • Fair enough, indeed the bood said that this was an alternative definition of the closure of set. The one from which I should start is at the end of the post, few lines before the PS. – Kolmin Dec 17 '13 at 15:55
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    Missing an intersection $\bigcap$ there in the post. So the closure is defined as the intersection of all closed sets containing $S$. – Daniel Fischer Dec 17 '13 at 15:57
  • More than right. I am gonna edit it right now. – Kolmin Dec 17 '13 at 15:58

1 Answers1

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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.

Daniel Fischer
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    Thanks a lot, that was very helpful. I am great at making complex things that should be easy. – Kolmin Dec 17 '13 at 16:20