Let $(V, +, \times)$ be a linear space over field $F$. Prove: $S \subseteq V$ is subspace of $V$ iff $L(S) = S$ ($L(S)$ is linear span of set S) I've tried to solve this problem from Apostol's Calculus Volume 2, but there is no solution for this one, so i don't know how good my solution is.
Let $(V, +, \times)$ be a linear space over field $F$.
Here's what I've done so far.
1.($\Rightarrow$) Since $S$ is a subspace of $V$, closure axioms must hold(among others). So, $$(\forall a,b\in F)(\forall x,y\in S) (a\times x\in S) \wedge (x+y\in S) \wedge ((ax + by)\in S)$$
Since associativity of addition operator also must hold, we can trivially generalize this to $$(\forall a_i\in F)(\forall x_i\in S)\sum_{S}(a_i\times x_i)\in S$$
This is exactly Linear Span of $S$, so every element in $L(S)$ is also in $S$, hence $L(S)\subseteq S$. But trivial lemma(easy to prove) says that $S\subseteq L(S)$ so, in conjunction, we have $L(S)=S$

2.($\Leftarrow$) If $S\subseteq V \wedge S \neq \emptyset$ then $L(S)$ is subspace of V(it's easy to show that closure axioms hold). Since $L(S)$ is subspace of $V$, and $S=L(S)$, we conclude that $S$ is subspace of $V$.
Have i made any mistakes?
 A: Your proof is correct, but could use a few improvements. 
For instance
"Since associativity of addition operator also must hold, we can trivially generalize this to"
could be replaced by
"Since addition is associative, by induction we can generalize to say that for any coefficients $a_i \in F$ ($ = 1, \ldots, n$) and elements $x_i \in S$, we have
$$
\sum_i a_i x_i \in S.
$$
Since $L(S)$ is exactly the set of linear combinations of elements of $S$, we've shown that every element of $L(S)$ is in $S$. "

Similarly, when you write
"But trivial lemma(easy to prove) says that S⊆L(S)S⊆L(S) so, in conjunction, we have L(S)=S"
you might instead say "For any $x \in S$, we can form the linear combination $1 \times x \in L(S)$; this shows $S \subset L(S)$. Hence $S = L(S)$." 
Even when things are trivial, sometimes it's just as easy to write them out rather than trying to sound all mathematical and superior. :)
Finally, I'm a big fan of clear writing, and having lots of symbolic things like quantifiers and $\wedge$ symbols tends to make stuff harder to read, at least for me. I'd rewrite the first bit like this:


*

*Assume $S$ is a subspace. Any linear combination of one item in $S$, such as $a_1 s_1$ must be in $S$ because it's closed under scalar multiplication. We'll now show that all combinations of $n$ elements are in there, by induction. Consider a linear combination $\sum_{i=0}^n a_i x_i$ of $n > 1$ elements in $S$. Write it as  $(\sum_{i=0}^{n-1} a_i x_i) + (a_n x_n)$, The second term is in $S$ by closure under scalar multiplication; the first term is in $S$ by the inductive hypothesis, and their sum is in $S$ because $S$ is closed under addition. This completes the induction proof. 

