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In my general topology textbook there is the following exercise:

Let $S$ be a dense subset of a topological space $(X,\tau)$. Prove that for every open subset $U$ of $X$: $$\overline{S \cap U} = \overline U$$


My approach:

My idea is to prove that $\overline{S \cap U} \subseteq \overline U$ and $\overline{S \cap U} \supseteq \overline U$, thus proving that $\overline{S \cap U} = \overline U$.

I had no trouble proving that $\overline{S \cap U} \subseteq \overline U$, but I'm having some trouble proving the opposite. I'm doing the following:

My proof:

Let $x \in \overline U$. We have that $\overline U = U \cup U'$, where $U'$ is the set of all limit points of $U$.

So we have that $x \in U \cup U'$. This means that $x \in U \vee x \in U'$

First case: If $x \in U$:

There are two possible scenarios:

  1. $x$ is also a member of $S$, this is $x \in S$
  2. $x$ is not a member of $S$, this is $x \notin S$

1:

We have that $x \in U \wedge x \in S$, so $x \in S \cap u$. Because $\overline{S \cap U}= (S \cap U) \cup (S \cap U)'$, then $x \in \overline{S \cap U}$.

2:

$S$ is dense in $X$. This means that $\overline S = S \cup S' = X$. So, if $x \notin S$, this implies that $x \in S'$.

Therefore, we know directly from the definition of limit point that:

$\forall A \in \tau$ such that $x \in A, \exists p \in S: p \in A \wedge p \neq x$

we have that $U \in \tau$, so $\exists p \in S: p \in U \wedge p \neq x$


This is the conclusion that I arrived, but I'm now sure how to proceed from now on. we know that $x \notin S \cap U$, because in situation 2 we have that $x \notin S$, so this leaves us with $x \in (S \cap U)'$. In other words I have to prove that:

$\forall A \in \tau$ such that $x \in A$, $\exists p \in S \cap U: p \in A \wedge p \neq x$.

How can I arrive at this conclusion?

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Let $x \in \overline{U}$. To show that $x \in \overline{U \cap S}$ take any open neighbourhood $O$ of $x$. Then $O$ intersects $U$ (as $x \in \overline{U}$) so $U \cap O$ is a non-empty open set and so intersects the dense set $S$. Hence

$$\emptyset \neq (U \cap O) \cap S = O \cap (U \cap S)$$

which shows (as $O$ was arbitrary) that $x \in \overline{U \cap S}$, as required.

The other inclusion is trivial as $U \cap S \subseteq U$ we immediately have $\overline{U \cap S} \subseteq \overline{U}$, by monotonicity of the closure.

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  • $\begingroup$ How do we know that $x \in \overline{S \cap S}$ from $O \cap (U \cap S)$? I do understand that $O$ is an arbitrary open neighborhood, but what guarantees that that intersection is not empty because there is another element of O other than x? $\endgroup$ – Eduardo Magalhães Aug 2 at 10:23
  • $\begingroup$ @EduardoMagalhães $O$ is an arbitrary open neighbourhood of $x$ and we start with $x \in \overline{U}$. So by definition of the closure, $O \cap U \neq \emptyset$. In the end we've shown $O \cap (U \cap S) \neq \emptyset$, so $x \in \overline{U \cap S}$ by that same definition of the closure.applied in the reverse direction. $\endgroup$ – Henno Brandsma Aug 2 at 10:25
  • $\begingroup$ But imagine that $\tau$ is the trivial topology, this is $\tau = \{X, \emptyset \}$, Then the only neighborhood of any set is $X$. So we have that $O = X$. $O \cap (U \cap S) = X \cap (U \cap S) = (U \cap S) $, so every element in $(U \cap S)$ will be in the intersection, and not necessarily only $x$, but correct me if I'm wrong, I'm still learning $\endgroup$ – Eduardo Magalhães Aug 2 at 10:32
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    $\begingroup$ @EduardoMagalhães In the indiscrete topology we already know $U=X$ or $U=\emptyset$, all non-empty subsets $S$ are dense and so in the former case we have $\overline{U}=X=\overline{U \cap S} = \overline{S} = X$, in the latter case both sets are empty. In your reply: $U \cap S = S$ which is dense and thus has closure $X$. I never claimed BTW that $x \in O \cap U$ in my proof, it's irrelevant and need not be true. We just need it to be a non-empty open set. $\endgroup$ – Henno Brandsma Aug 2 at 10:36
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    $\begingroup$ @EduardoMagalhães I use (for any space $(X, \tau)$, any subset $A$ of $X$, any $x \in X$: $$x \in \overline{A} \iff \forall O \in \tau: (x \in O \to O \cap A \neq \emptyset)$$ I don't use derived sets $A'$ etc. $\endgroup$ – Henno Brandsma Aug 2 at 10:54
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Let us show $\overline{S \cap U} \supseteq U$. Taking closures then gives $\overline{U} \subseteq \overline{\overline{S \cap U}}= \overline{S \cap U}$.

Let $u \in U$. Suppose to the contrary that $u \notin \overline{S \cap U}$. Then there is an open neighborhood $G$ of $u$ such that $G \cap S \cap U = \emptyset$. However, $S$ is dense and thus intersects every non-empty open subset. Hence, $G \cap U= \emptyset$, which is impossible since $u \in G \cap U$.

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  • $\begingroup$ if $A \subset B$ is it true that $\overline A \subset \overline B$? $\endgroup$ – Eduardo Magalhães Aug 1 at 13:13
  • $\begingroup$ @EduardoMagalhães Yes, it is. $\overline B$ is a closed set with $A\subseteq\overline B$. This justifies to conclude that $\overline A\subseteq\overline B$. $\endgroup$ – drhab Aug 1 at 13:54

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