I need to prove that if $|A|<|B|$ and both $A$ and $B$ are linearly indepdent, then it is not the case that: $B\subseteq A$ (ie. there is $a\in B$, such that $a\notin sp(A)$. No advanced theorems...
So $A$ and $B$ are both linearly independent that is
$\sum_{a\in A} \lambda_a a=0$
$\sum_{b\in B} \lambda_b b=0$
Only if all the lambdas are zero. Since $|A|<|B|$, there are more elements in $B$ than in $A$. What is the best way to continue the proof from here? Inductively it is straight forward to prove if for finite case:
Suppose $|B|=1$:
There is nothing in $sp(A)$ so the case follows trivially.
Now assume that the case holds for $|B|=n$.
Now add an element $b^*$ to $B$, since it's linearly independent, this new element is
$\sum_{b\in B} \lambda_b b\neq b^*$
Also add an element $a^*$ to $A$, so:
$\sum_{a\in A} \lambda_a a\neq a^*$
Now if $a^*=b^*$ there is an element in $B$ that is not in $A$. And if $a^*$ is that element, then $b^*$ is in $B$, but not in $A$.
Is this completely backwards? Would it be more appropriate to use induction on $|A\triangle B|$? What about the infinite case?