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Let $X$ and $Y$ be topological spaces and $A\subset X$ and $B\subset Y$. How can I show that $$Fr(A\times B)=Fr(A)\times \overline{B}\cup \overline{A} \times Fr(B)?$$

I struggle with taking the boundary of a product. My attempt to show this is this straight forward, but wrong, way:

$Fr(A\times B)=\overline{A\times B}\backslash (A\times B)^\circ=(\overline{A}\times \overline{B}) \backslash (A^\circ \times B^\circ)=\overline{A}\backslash A^\circ\times \overline{B}\backslash B^\circ =Fr(A)\times Fr(B)$

I suspect this

$$\overline{A\times B}\backslash (A\times B)^\circ=(\overline{A}\times \overline{B}) \backslash (A^\circ \times B^\circ)$$

may be wrong, and I am almost sure this

$$(\overline{A}\times \overline{B}) \backslash (A^\circ \times B^\circ)=\overline{A}\backslash A^\circ\times \overline{B}\backslash B^\circ $$

is wrong. How do I do it right? And where does $\overline{B}\cup \overline{A}$ come from? I think it is strange that the answer is a product of three, not two, sets. I believe my problem is my rather poor understanding of a product of two topological spaces all together.

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    It is not a product of three sets. Think of paratheses: It must be read as $Fr(A \times B) = (Fr(A) \times \bar{B}) \cup (\bar{A} \times Fr(B))$. However, with the "neighbourhood-definition" for *boundary* it is quite straightforward, right? – 2017-01-09
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    @M.U. Of course! That makes much more sense! :-) – 2017-01-09
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    @M.U. I still don't see it. I mean, it is intuitively correct, especially when drawing a $A\times B$ rectangular in an $X\times Y$-plane, but I am still not able to write out the proof, even when using the neighborhood-definition. Do you have an additional hint? – 2017-01-09

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You want to prove that $Fr(A \times B) = (Fr(A) \times \bar{B}) \cup (\bar{A} \times Fr(B))$.

By defintion $x \in Fr(Z)$ iff any neighborhood of $x$ contains at least one element of $Z$ and of $Z^c$.

So let $x \in Fr(A) \times \bar{B}$, say $x = (a,b)$, and let $N$ be a neighborhood of $x$. By definition of the product topology $N$ contains an open set of the form $U \times V$, where $U$ is open in $X$ and likewise $V$ is open in $Y$, s.t. $x \in U \times V$. That is to say that $a \in U$ and $b \in V$. But $a \in Fr(A)$, i.e. there exist elements $a_1, a_2 \in U$ such that $a_1 \in A$ and $a_2 \notin A$. Furthermore $b \in \bar{B} = B \cup L(B)$, where $L(B)$ is the set of limit points of $B$. I am sure you can now find elements in $N$ which are contained in $A \times B$ and some which are not, right? If you have found them you have proven that $x \in Fr(A \times B)$.

The other case is obviously the same thing just swapping $A$ and $B$. So you have:

$$Fr(A \times B) \supseteq (Fr(A) \times \bar{B}) \cup (\bar{A} \times Fr(B))$$

The other "direction" ($\subseteq$) is probably easier. Go by contradiction and you should finish easily.