Show that $\mathcal{P}(X)\times\mathcal{P}(X)=\mathcal{P}(X\times X)$

Question

Let $X$ be a set, and let $X$ be equipped with the discrete topology (i.e. $\mathcal{T}=\mathcal{P}(X)$). Show that the product topology inherited by $X\times X$ is the same as the discrete topology on $X\times X$. I believe this amounts to showing that, $$\mathcal{P}(X)\times\mathcal{P}(X)=\mathcal{P}(X\times X).$$ To do this we must show that either set contains the other.

Let $U\in\mathcal{P}(X)\times\mathcal{P}(X)$. Then $U=V\times W$, for some $ V,W\in\mathcal{P}(X)$. I'm having some trouble showing that $V\times W$ is in $\mathcal{P}(X\times X)$. Any help would be appreciated.

Answer 1

Your supposition is mistaken; $\wp(X\times X)$ is not the same as $\wp(X)\times\wp(X)$.

If you want actual equality as sets, then, as Rahul as mentioned, the elements of $\wp(X\times X)$ are sets of pairs of elements of $X$, while the elements of $\wp(X)\times\wp(X)$ are pairs of sets of elements of $X$.

If you just want them to be naturally in bijection with one another... well, that doesn't work either. In general, $\wp(X)\times\wp(Y)$ is naturally in bijection with $\wp(X\amalg Y)$ (here $\amalg$ is disjoint union), which is rather different.

Other people have already pointed out how to show that $X\times X$ is discrete, so I'll skip that...

Answer 2

Let (a,b) be any point in X×X.
Then {a}, {b} are open within X.
Thus {a}×{b} = {(a,b)} is open within in X×X.
Consequently, since every singleton in X×X is open, X×X is discrete.

Answer 3

Careful: if $(X, \tau_X)$ and $(Y, \tau_Y)$ are topological spaces, then the product topology on $X \times Y$ is not $\tau_X \times \tau_Y$. Rather, it is the topology generated by $\tau_X \times \tau_Y$, by which I refer to the set $\{ U \times V \mid U \in \tau_X,\ V \in \tau_Y \}$, rather than the cartesian product of $\tau_X$ and $\tau_Y$.

That is, an open set in $X \times Y$ is one which is a finite intersection of, or arbitrary union union of, sets of the form $U \times V$ for $U \in \tau_X$ and $V \in \tau_Y$. In fact, since unions commute with cartesian products, the product topology is actually just the set of unions of sets of the form $U \times V$ as above.

Thus $$\tau_{X \times Y} = \left\{ \bigcup_{i,j} U_i \times V_j \ \middle|\ U_i \in \tau_X \text{ and } V_j \in \tau_Y \text{ for all } i,j \right\}$$

The discrete topology on $X \times Y$ is indeed given by $\mathcal{P}(X \times Y)$. As such, what you need to prove is not that $\mathcal{P}(X) \times \mathcal{P}(Y) = \mathcal{P}(X \times Y)$, which is false in general, but rather that every element of $\mathcal{P}(X \times Y)$ can be expressed as a union of sets of the form $U \times V$ for $U \subseteq X$ and $V \subseteq Y$.

Edit: For your question, read all '$Y$'s in the last paragraph as '$X$'s.