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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?

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  • $\begingroup$ The linear independence is a red herring. If $|B|$ is greater than $|A|$, then $B$ cannot be a subset of $A$ by cardinality considerations alone. $\endgroup$ – Asaf Karagila Mar 21 '18 at 19:26
  • $\begingroup$ @AsafKaragila Apologies, it should have been $span(A)$ fixed now! $\endgroup$ – Dole Mar 21 '18 at 19:28
  • $\begingroup$ What you’ve written after “Now assume that the case holds for $|B|=n$.” is not clear. In particular, it is not clear what you are aiming to show with that assumption or what you have shown. At that point in your proof by simple induction your aim is to show that the result holds when $|B|=n+1$. Typically, after your inductive assumption you would say “Now assume that $A$ and $B$ are each linearly independent sets and that $|A|<|B|=n+1$. [You have to fill stuff in here!] ... Therefore the span of $B$ is not a subset of the span of $A$. Maybe you can try again? $\endgroup$ – Steve Kass Mar 21 '18 at 19:36

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