Q:

Proving that every basis of a vector space has the same cardinality. 

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! 

A:

We don't know that.  But even if $a_1\in B$, we still have $|(B\setminus\{b_1\})\cup\{a_1\}|\le|B|<|A|$. But your idea to use the assumption $|B|<|A|$ to pick an element of $A$ that is not in $B$ to begin with is also a good way to make the proof work.