How can you prove that a Cartesian product of compact sets is compact?

[AxiomOfChoice]

I'm talking about $E \times F$, where $E,F \subseteq \mathbb{R}^d$. If you know $E$ and $F$ are compact, you know they're both closed and bounded. But how do you define "boundedness" - or "closed", for that matter - for a Cartesian product of subsets of Euclidean $d$-space?

The only idea I've had is viewing $E\times F$ as a subset of $\mathbb{R}^{2d}$. If this is a legitimate thing to do, boundedness is certainly preserved. Also, since $E$ and $F$ were both closed, any sequence of points in $E\times F$ that converges necessarily converges to a point $(x,y) = (x_1,x_2,\ldots,x_d,y_1,y_2,\ldots,y_d)$. Does this look right?

 

[wofsy]

I'm talking about $E \times F$, where $E,F \subseteq \mathbb{R}^d$. If you know $E$ and $F$ are compact, you know they're both closed and bounded. But how do you define "boundedness" - or "closed", for that matter - for a Cartesian product of subsets of Euclidean $d$-space?

The only idea I've had is viewing $E\times F$ as a subset of $\mathbb{R}^{2d}$. If this is a legitimate thing to do, boundedness is certainly preserved. Also, since $E$ and $F$ were both closed, any sequence of points in $E\times F$ that converges necessarily converges to a point $(x,y) = (x_1,x_2,\ldots,x_d,y_1,y_2,\ldots,y_d)$. Does this look right?

This is avoiding your question a little but if you think of compact to mean that every open cover has a finite sub-cover then one should be able to argue this from inverse images of the projection maps onto the two factors E and F.

But I think you can use the continuity of the projections to do it the close and bounded way as well.

 

[g_edgar]

You are quite right. If $E, F \subset \mathbb{R}^d$, then $E \times F \subset \mathbb{R}^{2d}$ .