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#1 2008-06-27 14:28:14

JaneFairfax
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Registered: 2007-02-23
Posts: 6,868

Vector spaces

More generally, an uncountable vector space over a countable field is not finite dimensional. smile


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#2 2008-06-28 22:37:10

mathsyperson
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Registered: 2005-06-22
Posts: 4,900

Re: Vector spaces

Good 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.

Your way is nicer though. smile


Why did the vector cross the road?
It wanted to be normal.

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#3 2008-06-28 23:06:12

JaneFairfax
Member
Registered: 2007-02-23
Posts: 6,868

Re: Vector spaces

Indeed, 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 is

There is a rather deep theorem which states that every vector space (finite- or infinite-dimensional) 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. tongue


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A: Click here for answer.

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#4 2008-06-29 15:25:01

Ricky
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Registered: 2005-12-04
Posts: 3,791

Re: Vector spaces

There is a rather deep theorem which states that every vector space (finite- or infinite-dimensional) 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.

"The Axiom of Choice is obviously true, the Well-Ordering 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..."

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