I was working on Croom's Principles of Topology with the product topology; and came across this problem:
Knowing that a subset $A$ of $\mathbb R$ is compact if and only if it is closed and bounded, use product space ideas to prove the same characterization of the compact subsets of $\mathbb R^n$.
Thus I'm supposed to show that a subset $A:=\prod_{i=1}^n A_i$ of $\mathbb R^n$ is compact iff it is closed and bounded. And sadly I am completely lost. I think I could "decompose" $A$ into each coordinate spaces, and work on $A_i$ to find out the properties for $\mathbb R^n$. However I can't see any way to do so.
Croom uses the box topology for the term "product topology" for finite product spaces, but it doesn't really matter since I'm working on $\mathbb R^n$.
If $A \subseteq \Bbb R^n$ is compact, it is closed and bounded: this is true in any metric space (the first even for just Hausdorff spaces, for the second consider the cover by sets $B(a,n), n \in \Bbb N$ for some $a \in A$, which has a finite subcover etc.). If we start by assuming $A$ is closed and bounded we can find closed intervals $A_1,\ldots A_n \subseteq \Bbb R$ such that $A \subseteq \prod_{i=1}^n A_i$. The $A_i$ are compact by what we know about $\Bbb R$, and so Tychonoff tells us that $\prod_{i=1}^n A_i$ is compact and hence so is $A$, being a closed subspace of it. Hence the equivalence.