Proving that a closed set in the product topology need not be the product of closed sets Theorem
Let $X_1 $and $X_2$ be topological spaces and let $X_1 \times X_2$ have the product
topology . Then a closed subset of
$X_1 \times X_2$ need not be the product of closed set.
Do I give a counter example or a proof?
I am not to sure how to the proof here.
If F $\subset X_1 \times X_2$ then we
could say F$\subset X_1$ and
maybe F $\subset X_2$ . But this is not
very accurate…
Any help would be appreciated.
I would like to know how to do it .Thanks

Consider the circle $\mathbb{S}^1 = \{(x,y) \in \mathbb{R}^2 \ \vert \ x^2 + y^2 = 1 \} \subset \mathbb{R}^2$. Clearly $\mathbb{S}^1$ is closed in $\mathbb{R}^2 = \mathbb{R} \times \mathbb{R}$, but $\mathbb{S}^1$ is not the product of any two closed sets of $\mathbb{R}$. Indeed, suppose $\mathbb{S}^1 = A \times B$ where $A, B \subset \mathbb{R}$ are closed subsets. Since $(0, 1)$ and $(1, 0$) are in $\mathbb{S}^1$, then $1 \in A$ and $1 \in B$, therefore $(1, 1) \in A \times B = \mathbb{S}^1$, which is a contradiction (what we have actually proven is stronger: $\mathbb{S}^1$ can not be expressed as a cartesian product of any subsets of $\mathbb{R}$). Therefore no such $A$ and $B$ can exist and you have a counterexample.
