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Prove that if $S$ is a subspace of a vector space $V$, then $\mathrm{span}(S) = S$.

What I tried: I considered using the properties of vector spaces or maybe using an example where $S \subseteq \mathbb{R}^2$, but those strategies didn't amount to much. Any thoughts?

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  • $\begingroup$ What is your definition of span? $\endgroup$
    – Aloizio Macedo
    Jan 12 '16 at 17:29
  • $\begingroup$ just a normal spanning set i.e. a set of elements where you can take a linear combination of them and get S $\endgroup$
    – Raton
    Jan 12 '16 at 17:32
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$\mathrm{Span}(S)$ is the set of finite linear combinations of elements in $S$. In particular, every element in $S$ is in $\mathrm{Span}(S)$. Thus $S\subseteq\mathrm{Span}(S)$. Now, set $u\in\mathrm{Span}(S)$. Then $u=a_1s_1+...+a_ns_n$ for some scalars $a_i$ and $s_i\in S$. As $S$ is a subspace, it is closed under plus and product by scalar. Then $u\in S$. This proves $\mathrm{Span}(S)\subseteq S$.

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Note that the span of a set $S\subset V$ is the intersection of all subspaces of $V$ that contain $S$, i.e. $$\operatorname{Span}(S) = \bigcap_{S\ \subset\ T\ \leqslant V}T. $$ The intersection of subspaces is again a subspace, so $$\bigcap_{S\ \subset\ T\ \leqslant V}T$$ is a subspace containing $S$, which implies that $$\operatorname{Span}(S)\subset \bigcap_{S\ \subset\ T\ \leqslant V}T. $$ But $\operatorname{Span}(S)$ itself is a subspace containing $S$, therefore $$\operatorname{Span}(S)\supset \bigcap_{S\ \subset\ T\ \leqslant V}T. $$

It follows immediately that if $S\leqslant V$, then $\operatorname{Span}(S)=S$.

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