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

  1. Sep 2, 2009 #1
    I'm talking about [itex]E \times F[/itex], where [itex]E,F \subseteq \mathbb{R}^d[/itex]. If you know [itex]E[/itex] and [itex]F[/itex] 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 [itex]d[/itex]-space?

    The only idea I've had is viewing [itex]E\times F[/itex] as a subset of [itex]\mathbb{R}^{2d}[/itex]. If this is a legitimate thing to do, boundedness is certainly preserved. Also, since [itex]E[/itex] and [itex]F[/itex] were both closed, any sequence of points in [itex]E\times F[/itex] that converges necessarily converges to a point [itex](x,y) = (x_1,x_2,\ldots,x_d,y_1,y_2,\ldots,y_d)[/itex]. Does this look right?
    Last edited: Sep 2, 2009
  2. jcsd
  3. Sep 3, 2009 #2
    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.
  4. Sep 3, 2009 #3
    You are quite right. If [itex]E, F \subset \mathbb{R}^d[/itex], then [itex]E \times F \subset \mathbb{R}^{2d}[/itex] .
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