Two proofs of cartesian product in topological space Let $(X \times Y, \tau)$ be cartesian product of topological spaces $(X, \tau_X)$, $(Y, \tau_Y)$. Let $ A \subset X$, $ B \subset Y$.
A) Prove that $\overline{A\times B}= \overline{A} \times \overline{B}$
B) Prove that $(A \times B)^d = (A^d \times \overline{B}) \cup (\overline{A} \times B^d)$
What is $d$?
Let $a$ in topological space $(X,\tau)$ be a limit point of A set. $A \subset X$, if every neighbourhood of $a$ contains element from $A$ set which differs from $a$. Set of limit points of $A$ is $A^d$. Let
$A^n = (...(A^d)^d...)^d$
be set obtained from A by repeating n times transition to set of limit points.
Well, I have done almost half of A part
$ A \subset \overline{A}$, $B \subset \overline{B}$, $A \times B \subset \overline{A} \times \overline{B}$
$\overline{A} \times \overline{B}$ is closed in this topology.
$\overline{A} \times \overline{B} = \overline{A} \times \overline{Y} \cap \overline{X} \times \overline{B}$
$ \overline{A\times B}$ is the smallest closed substed containing $A\times B$, so
$\overline{A\times B} \subset \overline{A} \times \overline{B}$
I dont have an idea about second part of A  and whole B.
 A: I assume you know that for any space $X$ and any subset $C$, $x \in \overline{C}$ iff every open set that contains $x$ intersects $C$. It even suffices to check this for open sets from a certain fixed base of $X$ as well.
I'll do b) first: 
Left to right inclusion:
Suppose that $(x,y) \in (A \times B)^d$. As $(A \times B)^d \subset \overline{A \times B} = \overline{A} \times \overline{B}$, $x \in \overline{A}$ and $y \in \overline{B}$. 
So suppose that $x \notin A^d$, otherwise $(x,y) \in A^d \times \overline{B}$ and we are done. Then the only way that $x \in \overline{A}$, is that $x$ is an isolated point of $A$, i.e. there exists some open set $U$ such that $\{x\} = U \cap A$. Now let $V$ be any neighbourhood of $y$. Then $U \times V$ is an open neighbourhood of $(x,y)$ that intersects $A \times B$ in some point $(a,b) \neq (x,y)$. But $x = a$ by the previous, so we know that $b \neq y$. In short, every open neighbourhood of $y$ intersects $B$ in a point different from $y$, so $y \in B^d$ and $(x,y) \in \overline{A} \times B^d$.
(Alternatively we could assume $b \notin B^d$, so $y$ is an isolated point of $B$ as well, but then $(x,y)$ is an isolated point of $A \times B$, contradiction.)
Right to left inclusion:
Suppose that $(x,y) \in A^d \times \overline{B}$. Let $U \times V$ be any basic open neighbourhood of $(x,y)$, then $V \cap B \neq \emptyset$, as $y \in \overline{B}$; say it intersects in $b$. Also, there is some $a \neq x$ in $A \cap U$ as $x \in A^d$. Then $(a,b) \neq (x,y)$ by virtue of the first coordinates being different, so $U \times V$ intersects $A \times B$ in a point distinct from $(x,y)$, so $(x,y) \in (A \times B)^d$.
The other part of that inclusion is entirely symmetrical.
As to a):
You're right that $\overline{A} \times \overline{B}$ is closed (and contains $A \times B$ so $\overline{A \times B} \subset \overline{A} \times \overline{B}$.
So suppose $(x,y) \in \overline{A} \times \overline{B}$. Let $(x,y) \in U \times V$, a basic open neighbourhood of it. Then there is some $a \in U \cap A$ and some $b \in V \cap B$, so $(a,b) \in (U \times V) \cap (A \times B)$. As this is true for all basic neighbourhoods of $(x,y)$, we conclude that $(x,y) \in \overline{A \times B}$, as required. 
