# I Proof that every basis has the same cardinality

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1. Oct 6, 2016

### Math_QED

Hello all.

I have a question concerning following proof, Lemma 1.

http://planetmath.org/allbasesforavectorspacehavethesamecardinality

So, 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!

2. Oct 6, 2016

### 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.

3. Oct 6, 2016

### Staff: Mentor

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 infinte and $B$ countable? Why can't this happen?