Talk:Dimension theorem for vector spaces

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Could someone write about how to prove the uniqueness of the dimension for infinite cardinals? - Gauge 06:02, 18 December 2005 (UTC)[reply]

I know its been a year and a half since anyone replied, but done. jbolden1517^Talk 15:19, 12 May 2007 (UTC)[reply]

Cardinality

I think the whole thing with going back and forth between infinite and finite violates the spirit of cardinality. It should be a single case. Actually looking at it the infinite case is false. jbolden1517^Talk 17:48, 12 May 2007 (UTC)[reply]

I do not know if there is a nice proof that works for both the finite and infinite case. I do not agree that a case distinction violates the "spirit of cardinality". Finite cardinals behave very differently from infinite one, this is why there are two (often quite separate) fields of mathematics dealing with them, combinatorics and set theory.

You deleted my argument for the case of finite dimensional spaces. But the argument

"Since the set J has smaller cardinality than the set I, the union of finite sets indexed by J still has smaller cardinality than I"

works only if I is infinite.

--Aleph4 00:18, 27 May 2007 (UTC)[reply]

I changed your proof a lot. The first line just asserts that as the RAA hypothesis. So its true by given. jbolden1517^Talk 00:38, 27 May 2007 (UTC)[reply]

What do you mean by "RAA"? Which first line asserts what?

Please explain why the cardinality of I is bigger than the cardinality of the union of the E_j, j in J.

--Aleph4 14:45, 29 May 2007 (UTC)[reply]

I don't really know where to write this, but this proof is wrong, and quite obviously so, too :
just consider the case where b_1 is the sum of the a_i. Then E_1 is I, and there's no i_0 such that ...

This theorem is actually not so easy to prove.
I'm not sure there exist a simple elegant proof, that's what I was looking for here.
The proof I know is no fun.
I also know it can be seen as a consequence of the Jordan-Holder theorem (see Bourbaki), but I ghess that would be overkill —Preceding unsigned comment added by 88.183.30.174 (talk) 15:45, 17 February 2008 (UTC)[reply]

I again divided the proof into two cases, one for finite dimension, the other for infinite dimension. As I have pointed out before, the proof from the article that works for infinite dimensions does not work in the finite-dimensional case. --

Aleph4 (talk) 17:44, 17 February 2008 (UTC)[reply]

The proof for the infinite case does indeed look wrong ( $b_{1}=\sum _{i\in I}a_{i}$ , as someone mentioned). Could someone replace it with a correct proof? If not, it should be removed.

AC in Proof?

In the proof it is assumed that there is a relation between the cardinalities of the two bases (that one is greater than the other). It seems to me that this relies on the Well-Ordering Principle. However I'm not familiar with the Ultra-filter Lemma and its applications... is this where it is used? There is a proof here: http://www.dpmms.cam.ac.uk/~tf/cam_only/cleverblass.pdf that relies only on propositional compactness and Hall's Matching Theorem.

You (who?) are right to the extent that trichotomy (any two cardinals are comparable) is equivalent to the axiom of choice, as stated in that article. So indeed the very start of the proof (assume one basis bigger than the other) is in blatant contradiction with the claim in the lead that the theorem only needs the ultrafilter lemma. Marc van Leeuwen (talk) 05:43, 20 March 2011 (UTC)[reply]