# Suppose $V$ is a vector space generated by a subset $S$, and $U$ is a proper subspace of $V$. Might one also say that $S$ generates $U$?

To reiterate:

Suppose $$V$$ is a vector space generated by a subset $$S$$, and $$U$$ is a proper subspace of $$V$$. Might one also say that $$S$$ generates $$U$$, since $$U \subsetneq \text{Span}(S)$$? Or, is it the fact that $$U \neq \text{Span}(S)$$ that implies we cannot say this?

Most certainly not. For example, take $$S=\{(1,0,0), (0,1,0)\}$$, and $$U=\{(x, 0, 0)|x\in\mathbb R\}$$. Then, clearly, $$U$$ is the subspace of $$V=\{(x,y,0)|x,y\in\mathbb R\}$$, however, $$S$$ does not generate $$U$$.
In fact, it is impossible for $$U$$ to be generated by $$S$$. By definition, the vector space, generated by $$S$$, is the smallest vector space that contains all vectors contained in $$S$$. Therefore, by definition, there does not exist any smaller vector space that also contains all of $$S$$.
In other words, if there exists some proper vector subspace $$U$$ such that $$S\subset U$$, then $$V$$ is, by definition, not generated by $$S$$.
We have $$V= Span(S)$$. If we also have that $$U= Span(S)$$, then $$U=V$$.
Hence, if $$U$$ is a proper subspace of $$V$$, then $$U \subsetneq Span(S).$$