# Understanding the proof of a theorem on linear independence and set size

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?

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.
• 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. Sep 9, 2021 at 12:21