Prove $(A\times B)'=(A'\times cl(B))\cup(cl(A)\times B')$ Let $(X,T)$ and $(Y, T`)$ be two topological spaces and let $A$ be a subset of $X$, and $B$ a subset of $Y$. Then
$$(A\times B)'=(A'\times \operatorname{cl}(B))\cup(\operatorname{cl}(A)\times B')$$
where $\operatorname{cl}$ is the closure and the ($'$) is the set of limit points.
 A: Here is an outline of a proof.
Suppose $(x,y)\in A'\times \operatorname{cl}(B)$ and let $U\times V$ be a product topology basis element containing the point $(x,y)$. Use the fact that $x\in A'$ and $y\in \operatorname{cl}(B)$ to show $((U\times V)\cap (A\times B))\setminus\{(x,y)\}\neq\emptyset$ and conclude $(x,y)\in(A\times B)'$. Then you know $A'\times \operatorname{cl}(B)\subseteq(A\times B)'$ and by the same reasoning, $\operatorname{cl}(A)\times B'\subseteq(A\times B)'$ as well, so
$$(A'\times \operatorname{cl}(B))\cup(\operatorname{cl}(A)\times B')\subseteq (A\times B)'.$$
Conversely, let $(x,y)\in (A\times B)'$. If $U$ is an open set containing $x$ and $V$ is an open set containing $y$, then $U\times Y$ and $X\times V$ are product topology open sets containing $(x,y)$. Use this to show $U\cap A\neq\emptyset$ and $V\cap B\neq\emptyset$, so $x\in \operatorname{cl}(A)$ and $y\in \operatorname{cl}(B)$.
Assume now that $x\notin A'$ and $y\notin B'$. Use this assumption to show that $(x,y)\notin (A\times B)'$. Conclude that if $(x,y)\in (A\times B)'$, then either $(x,y)\in A'\times \operatorname{cl}(B)$ or $(x,y)\in \operatorname{cl}(A)\times B'$. Hence, we conclude
$$(A'\times \operatorname{cl}(B))\cup(\operatorname{cl}(A)\times B')=(A\times B)'.$$
A: Assume $a \in A'$ and $b \in cl(B)$, and let $U$ be a neighborhood of $(a,b) \in X \times Y$, then there are neighborhoods $V\subseteq X, W\subseteq Y$ of $a$ and $b$ respectively, such that $V\times W \subseteq U$. Since $a \in A'$, there is $\bar{a} \in (A \cap V) \setminus \{a\}$, and since $b \in cl(B)$ there is $\bar{b} \in (B \cap W)$. The point $(\bar{a}, \bar{b})$ lies then in $U \setminus \{(a,b)\}$. Since the neighborhood $U$ was arbitrary we proved that $(a, b)$ is limit point for $A\times B$.
This settles the inclsion $A' \times cl(B) \subseteq (A \times B)'$; by simmetry we also get $cl(A) \times B' \subseteq (A \times B)'$.
Now let $(a,b) \in (A\times B)'$, then 


*

*either for every nieghborhood $U$ of $(a,b)$ there is a point in $U \cap ((A \setminus \{a\}) \times B)$, and in such a case $a \in A'$, $b \in cl(B)$, as we can set $U$ to be any product of neighborhoods $V$ of $a$ and $W$ of $b$,

*or there is a neighborhood $U$ such that $U \cap ((A \setminus \{a\}) \times B)= \emptyset$.
In the latter case we have $U \cap (A \times B) = U \cap (\{a\} \times B)$, that is to say, all of the points of $U \cap (A \times B)$ have $a$ as first coordinate. It follows that $a \in cl(A)$ and $b \in B'$.
To see $b \in B'$, first note that we can assume wlog $U$ to be of the form $U= V \times W$ for $V$ and $W$ neighborhoods of $a$ and $b$ resp. 
Every $W'$ such that $W'\subseteq W$ has to be such that $(A\times B) \cap ((\{a\} \times W')\setminus \{(a,b)\}) \neq \emptyset$, hence $W' \cap B \setminus \{b\} \neq \emptyset$. 
