Question

Topology question: Can you help me prove that Descartes' product of closed sets is a closed set? A1 subset of Rk and A2 subset of Rm are those closed sets and A1 x A2 subset of R(k+m).

Answer 1

descartes product is cartesian product. so A1 subset of R^k, and A2 subset of R^m , show A1xA2 is closed if A1 and A2 are closed .

Answer 2

Well, yes. I tried the proof by contradiction, but it looks like I'm missing something. I get that R^k x R^m is an open set but from that I can't deduce that some subset of that is a closed set. Do you know if I should try some other approach?

Answer 3

Let F = A × B. Let x = (x1,...,x_r+m) be in the closure of F. There exists a sequence xk = (xk1,...,xk_r+m) in F such that xk -> x. Thus |xk1 - x1| + ••• + |xkn - xn| -> 0. Since

0 ≤ |xki - xi| ≤ |xk1 - x1| + ••• + |xkn - xn|

for all i, the squeeze theorem implies xki -> xi for all i. Hence xi is in Fi for each i , i = 1,2 . Therefore x is in F. As x is arbitrary, F is closed.

Answer 4

did that make sense. maybe you can explain it to me

Answer 5

honestly i ripped it off from here
http://answers.yahoo.com/question/index?qid=20130202174546AABsUDF

Answer 6

Yes it makes sense, although I've been looking for a proof by definition of an open set or something like that, something that doesn't use arrays, but this is fine as well.

Answer 7

Show that \(R^{k+m}\backslash (A_1\times A_2)\)is open