Topology on cartesian product and product topology. Let X and Y be sets.  Does it have every topology on cartesian product X$\times$Y must be product topology ?
 A: It goes a bit further: one can show that for virtually any pair $X,Y$ of sets there are topologies on $X \times Y$ which cannot be expressed as the product topology on topologies on $X$ and $Y$, respectively.
For one example, consider the topology on $\mathbb{Z} \times \mathbb{Z}$ where the basic open sets are the "diagonals" $$U_k = \{ \langle n , n + k \rangle : n \in \mathbb{Z} \}$$ for each $k \in \mathbb{Z}$.  (This means that $U \subseteq \mathbb{Z} \times \mathbb{Z}$ is open iff for each $\langle m,\ell \rangle \in U$ all pairs of the form $\langle n , n+(\ell-m) \rangle$ is also in $U$.)
To show that there is no topologies $\mathcal{O}_1 , \mathcal{O}_2$ on $\mathbb{Z}$ such that the topology above is the product topology of $\langle \mathbb{Z} , \mathcal{O}_1 \rangle \times \langle \mathbb{Z} , \mathcal{O}_2 \rangle$, consider the following:

Given $\langle m,\ell \rangle \in \mathbb{Z} \times \mathbb{Z}$, consider the open set $U_{\ell-m} \ni \langle m,\ell\rangle$.    Note that as the only "rectangles" which are subsets of $U_{\ell-m}$ are the singletons $\{ \langle n , n + (\ell-m) \rangle \} = \{ n \} \times \{ n + (\ell-m) \}$, it follows that if $A , B \subseteq \mathbb{Z}$ are such that $m \in A$, $\ell \in B$ and $A \times B \subseteq U_{\ell-m}$, then $A = \{ m \}$ and $B = \{ \ell \}$.  It thus follows that if the topology above were the product topology of two topologies on $\mathbb{Z}$, then these two topologies must be discrete.  But note that the product of two discrete spaces is discrete, and the topology above is clearly not discrete.


Another (similar/simpler) example, let $X = \{ 0,1 \}$, and consider the topology $$\mathcal{O} = \{ \varnothing , \{ \langle 0,0 \rangle \} , \{ \langle 1,1 \rangle \} , \{ \langle 0,0 \rangle , \langle 1,1 \rangle \} , X \times X \}$$ on $X \times X$.
If $\mathcal{O}_1$ and $\mathcal{O}_2$ were topologies on $X$ such that $\mathcal{O}$ is the product of these two topologies, then as $\{ \langle 0 , 0 \rangle \}$ is open, it follows that the singleton $\{ 0 \}$ is open in both $\mathcal{O}_1$ and $\mathcal{O}_2$.  Similarly, $\{ 1 \}$ is open in both $\mathcal{O}_1$ and $\mathcal{O}_2$.  Note that by definition of the product topology it would follow that $\{ \langle 0,1 \rangle \} = \{ 0 \} \times \{ 1 \}$ is open in $\mathcal{O}_1 \times \mathcal{O}_2$, but it is clearly not open in $\mathcal{O}$.
This second example can be extended for any pair of sets $X , Y$ which contain at least two points.  (It should be noted that if either $|X| = 1$ or $|Y|=1$ then any topology on $X \times Y$ is the product of topologies on $X$, $Y$.)
