1. ## Closure Problem

Hello everyone.

I am trying to prove that if $A \subseteq X$ is dense then for any open $T$ we have $T \subseteq Cl(A \cap T)$. I tried assuming not, and I arrive at the statement $t \in A^c$ and I can't see why that would be a contradiction.

Does anyone know of a good proof of this? Also, is it necessary that the T be open?

2. First if I am correct is assuming that $Cl$ is the closure operator, then I am puzzled by your confusion.
Here are some facts.
$\left[ {\forall A\,,\,A \subseteq Cl(A)} \right]$, and $Cl(A \cap B) \subseteq Cl(A) \cap Cl(B)$.

Because A is dense in X you also have $Cl(A) = X$.

Put all that together.

3. Plato, you proved the opposite inclusion. You proved that:

$Cl(A \cap T) \subseteq Cl(T)$

I want to show that $T \subseteq Cl(A \cap T)$ if T is open.

As an example if $A = \mathbb{Q}$ and $T = (0,1)$ t hen $(0, 1) \subseteq Cl(\mathbb{Q} \cap (0,1)) = [0, 1]$

4. Originally Posted by mpetnuch
Plato, you proved the opposite inclusion.
No I did not. If T is open then the statement follows as a matter of definitions. If there is a t which is not in $Cl(A \cap T)$ then realize that t is a limit point of A. But some open set O contains t and is a subset of T. Think about that contradiction.

You asked about T being open. Let $T = \left\{ {\sqrt 2 } \right\}\quad \& \quad A = Q$. It is clear that the statement is false using this example.

5. Originally Posted by Plato
No I did not. If T is open then the statement follows as a matter of definitions. If there is a t which is not in $Cl(A \cap T)$ then realize that t is a limit point of A. But some open set O contains t and is a subset of T. Think about that contradiction.
I am sorry, but I still don't see it. Since $A$ is dense means that $A$ intersects every non-empty open set. In particular $A \cap T \ne \emptyset$. I see this is the reason why it fails if $T$ is not open.

Also since $A$ is dense means that every point is a limit point of $A$, so in particular $t$ is a limit point of $A \forall t \in T$. And since $T$ is open, we can find an open neighborhood of $T$, $U$, such that $t \in U \subseteq T$.

Now, you asked me to assume that $t \notin Cl(A \cap T)$ and consider that open set $U$ defined above. Since $t$ is a limit point of $A$ implies that $A \cap U\backslash\{t\}\ne \emptyset$. But I still don't see the contradiction.

6. Because T is open there is an open set containing t which is a subset of T. Any open containing t must contain must contain a point of A distinct from t. Hence the second point must be in $(A \cap T)$.