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

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

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

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.

Why did the vector cross the road?

It wanted to be normal.

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

Indeed, there are many infinite sets of real numbers which are linearly independent over

. The set of all √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 Zorns lemma, which is equivalent to the axiom of choice) so the result may not be acceptable to mathematicians who dont accept the axiom of choice in set theory. Fortunately they are only a minority.

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

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 Zorns lemma, which is equivalent to the axiom of choice) so the result may not be acceptable to mathematicians who dont 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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