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You are not logged in. #1 20080628 12:28:14
Vector spacesMore generally, an uncountable vector space over a countable field is not finite dimensional. #2 20080629 20:37:10
Re: Vector spacesGood proof! I think I'd have tried to show that there is an infinite set of real numbers which are linearly independent over the rationals. Square roots of primes or something. Why did the vector cross the road? It wanted to be normal. #3 20080629 21:06:12
Re: Vector spacesIndeed, there are many infinite sets of real numbers which are linearly independent over . The set of all √p where p is prime is an example; another example isThere is a rather deep theorem which states that every vector space (finite or infinitedimensional) has a basis. The proof, however, makes use of the axiom of choice (or rather Zorn’s lemma, which is equivalent to the axiom of choice) so the result may not be acceptable to mathematicians who don’t accept the axiom of choice in set theory. Fortunately they are only a minority. #4 20080630 13:25:01
Re: Vector spaces
"The Axiom of Choice is obviously true, the WellOrdering Principle is obviously false, and who knows with Zorn's Lemma." "In the real world, this would be a problem. But in mathematics, we can just define a place where this problem doesn't exist. So we'll go ahead and do that now..." 