Question:
Suppose that $E ⊂ \mathbb{R}^n$ and $F ⊂ \mathbb{R}^m$ are both connected. How can I show that $E × F$ is connected in $\mathbb{R}^{n+m}$.
My Progress:
Let $U ⊂ \mathbb{R}^n$ be open with respect to the usual topology. Fix any real numbers $a_1, . . . , a_m$ for $m < n$ and consider the set
$U_{a_1,...,a_m} := ${ $(x_1, . . . , x_{n−m}) ∈ \mathbb{R}^{n−m} : (a_1, . . . , a_m, x_1, . . . , x_{n−m}) ∈ U$ }.
I have shown that $U_{a_1},...,a_m$ is open in $\mathbb{R}_{n−m}$ for any choice of ${a_1, . . . , a_m}$ because if I pick a point $(b_1,\cdots,b_{n-m})\in U_{a_1,\cdots,a_m}$, then $(a_1,\cdots,a_m,b_1,\cdots,b_{n-m})\in U$. Since $U$ is open in the box topology $\mathbb{R}^m\times \mathbb{R}^{n-m}$, there exists open $U_1$ in $\mathbb{R}^m$ and open $U_2$ in $\mathbb{R}^{n-m}$ such that $$(a_1,\cdots,a_m,b_1,\cdots,b_{n-m})\in U_1\times U_2 \subseteq U. $$ However, every point of $(a_1,\cdots,a_m)\times U_2$ is of the form $(a_1,\cdots,a_m,x_1,\cdots, x_{n-m})$. So in particular, $$(b_1,\cdots,b_{n-m})\in U_2\subseteq U_{a_1,\cdots, a_m}.$$ Hence $U_{a_1,\cdots, a_m}$ is open.
But how can I extend this result to answer my question above?
