# Let S be a subspace of topological space X. Show that the closure of S, the set of contact points, is indeed closed.

Let $S$ be a subspace of topological space $X$.

Show that the closure of $S$, the set of contact points, is indeed closed.

I need to prove that the closure is closed but I don't know how to go about it.

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What definition of closure do you use? – Michael Greinecker Feb 15 '12 at 10:50

If $x$ is not a contact point, then $x$ has a neighbourhood, we can assume open, $O$ that misses $A$ entirely. Now, for every $y \in O$, that same $O$ is a neighbourhood of $y$ that still misses $A$, and so all $y \in O$ are also not in the closure of $A$ (defined as the set of all "contact points", i.e. points whose every neighbourhood intersects $A$).

So starting from $x \in X \setminus \operatorname{Cl}(A)$, we have an open $O$ with $x \in O \subset X \setminus \operatorname{Cl}(A)$, showing that $x$ is in the interior of $X \setminus \operatorname{Cl}(A)$, and thus the latter set is open, and thus $\operatorname{Cl}(A)$ is closed.

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I think you mean given a subset $S$ of a topological space $X$, it's closure is always closed. The closure is closed because of the following definition:

Let $S \subset X$. The closure of $S$, which we denote by $\overline{S}$ is defined to be the intersection of all closed sets containing $S$.

Now because the intersection of an arbitrary number of closed sets is closed, $\overline{S}$ is closed. It is also the smallest one because if $A$ is another closed set containing $S$, then $A$ necessarily appears in

$$\bigcap \hspace{1mm} \bigg(\text{All closed sets that contain S}\bigg) = \overline{S}$$

so that $\overline{S} \subset A$.

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I guess you are in the following situation: A contact point of a set $S$ is defined to be a point such that every neighborhood of it meets $S$. The set of all contact points of $S$ is the closure of $S$ and you want to show that the closure is closed, i.e. the complement of an open set. So let $x$ be not a contact point. Then there is a neighorhood that doesn't meet $S$ and is therefore wholly in the complement of $S$. So every point in the complement of the closure has a neighborhood that lies in the complement. So the complement is an open set and the closure therefore closed.

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