Posted by Mark The problem is that we have now constructed a closed subset of an infinite dimensional product, the complement of which is open. But the complement is different from X-alpha, the entire dimension, for ALL alpha, which violates the "equals X-alpha in all but finitely many dimensions" part of the definition of product topology. What am I missing? (I'm not asking about the proof of 3b, which I have. Which is somewhat odd, to be able to prove something for which you have a counterexample...)
on October 7, 2007, 5:09 pm
I'm having a problem with the product topology on 3b. If we consider the closed unit square in R-omega, [-1,1]x[-1,1]x..., we get something that (I think) fits in the definition of closed subset along each A-alpha, and according to 3b, is closed. 193
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