# Proving that $\overline{S \cap U} = \overline U$ if $U$ open and $S$ dense [duplicate]

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?

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.

• 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? – Eduardo Magalhães Aug 2 at 10:23
• @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. – Henno Brandsma Aug 2 at 10:25
• 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 – Eduardo Magalhães Aug 2 at 10:32
• @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. – Henno Brandsma Aug 2 at 10:36
• @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. – Henno Brandsma Aug 2 at 10:54

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 $$u \in G \cap U$$.

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