Proof that every basis has the same cardinality

[member 587159]

Hello all.

I have a question concerning following proof, Lemma 1.

http://planetmath.org/allbasesforavectorspacehavethesamecardinalitySo, we suppose that A and B are finite and then we construct a new basis ##B_1## for V by removing an element. So they choose ##a_1 \in A## and add it to ##S_1##. How do we know for sure that ##a_1## is not yet in B? Can we say this because we suppose that m < n, thus there is certainly such an element?(to derive a contradiction)

Thanks!

 

[Erland]

It doesn't matter if ##a_1## is already in ##B##. Say ##a_1=b_1##. Then, certainly, ##B_1=\{a_1,b_2,\dots,b_m\}(=B)## spans ##V##. This is just a special case of the argument in the proof.

 

[fresh_42]

Hello all.

I have a question concerning following proof, Lemma 1.

http://planetmath.org/allbasesforavectorspacehavethesamecardinalitySo, we suppose that A and B are finite and then we construct a new basis ##B_1## for V by removing an element. So they choose ##a_1 \in A## and add it to ##S_1##. How do we know for sure that ##a_1## is not yet in B? Can we say this because we suppose that m < n, thus there is certainly such an element?(to derive a contradiction)

Thanks!

It doesn't matter whether ##a_1 \in B## or not. We only need a ##b_1## with non-zero coefficient to replace ##b_1## by ##a_1## and the fact that it spans ##V##. Let ##a_1 = \beta_1 b_1 + \dots \beta_m b_m##. If all ##\beta_i = 0 ## then ##a_1= 0## which we ruled out. So without loss of generality, let ##\beta_1 \neq 0##. Thus ##b_1 = \beta'_1 a_1 + \beta_2' b_2 + \dots \beta'_m b_m## and it can be replaced. We don't need to bother linear dependencies here.

Remark: My difficulty with the proof is in the first part, where ##A## and ##B## are infinite. This part is IMO only true, if ##|\infty| = |\infty| ## regardless which class of infinity is meant. However, what is if ##A## is uncountably infinite and ##B## countable? Why can't this happen?