If $S=\{\alpha_1, ..., \alpha_s\}$ is an independent set of vectors and $T=\{\beta_1, ..., \beta_t\}$ be any finite set such that $S$ is dependent on $T$, then $s \leq t$.

In my textbook a proof by induction is done on the size of the independent set $S$. The base case is $s = 1$ and the induction is proof of the following conditional statement:

$[\text{Theorem is true for when the size of $S'$ is less than $s$}] \implies [\text{Theorem is true for when the size of $S'$ is $s$}]$.

That is, we assume the antecedent (the induction hypothesis) of the conditional above and also assume the antecedent of the consequent (since the theorem is a conditional) to deduce $s \leq t$. The idea of this I understand.

However, while the induction hypothesis applies to the size of any set $S'$, the author sets this $S'$ as $S' = \{\alpha_1, ..., \alpha_{s-1}\}$ which is necessarily a proper subset of $S$. Why is it the case that setting $S'$ as such comes without a loss of generality?

Thank you in advance.


1 Answer 1


This is a proof by induction. At this step of the proof, it is being assumed that the statement holds for any subset $S'$ of $S$ with less than $s$ elements. But using that assumption doesn't require that use it that $S'$ is any subset $S'$ whatsoever of $S$ with less than $s$ elements. If you can do it using one specific subset with less than $s$ elements, no harm is done.

If what you wanted was to prove that any subset $S'$ of $S$ with less than $s$ elements had a certain property, then, yes, it would be wrong to prove it only for a certain specific set.

  • 1
    $\begingroup$ Thank you! That makes perfect sense - if it is true for all independent sets with less than s elements then it is true for that particular set. The deduction continues from there until $s \leq t$ is arrived. $\endgroup$ Sep 9, 2021 at 12:21

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