Prove that every vector space has a basis. I am going to use Zorn's lemma for this also here is a necessary definition regarding totally ordered subsets: one element will be contained in the other.
proof: A basis for $V$ is a maximally linearly independent set. Let $X$ be equal to all linearly independent subsets of $V$. On, $X$ define $A\leq B$ means $A\subset B$. Let $S$ be a totally ordered subset of $X$, let $T = \bigcup_{A_{s}\in S} A_s$ then $A_s\subset T$ for all $s\in S$. Does $T\in S$? i.e. is $T$ a linearly independent set?. Let $$O = \alpha_1 v_1 + \alpha_2 v_2 + \ldots + \alpha_n v_n \ \ \text{where} \ \ v_i\in T = \bigcup_{A_{s}\in S} A_s$$ Then there exists $S_i$ such that for all $i\in A_{s_{i}}$ all $i$ $A_{s_{1}},A_{s_{2}},\ldots, A_{s_{n}}$ (these will have one contained in the other then we would take the larger). So, there exists $A_{s_{k}}$ such that $A_{s_{k}}\supset A_{s_{i}}$ all $i$ just compare two at a time. Therefore, $V_i \in A_{s_{k}}$ all $i$, but $A_{s_{k}}$ is a linearly independent set. So, $\alpha_i = 0$ all $i$. Hence, $T\in X$ so $T$ is an upper bound for $S$. Hence every totally ordered subset of $X$ has an upper bound, hence by Zorn's lemma, there exists a maximal element and this would be the basis.
I am not sure if my proof is right, I apologize if it is a bit messy. If it is right, any suggestions on cleaning it up would be greatly appreciated.
Best Answer
The proof is very hard to understand. Here is why:
Either the elements of $S$ are $s$ or $A_s$, it's confusing to mix those apart.
You have a confusing typo, when you write "Does $T\in S$?" you mean $T\in X$.
$0$ and $O$ are two different things.
$i$ seems to be some index, but then you say that $i\in A_{s_i}$ which makes absolutely no sense. Is it some finite indexing? Is it a vector in $A_s$? What is it?
Where did $V_i$ come from? Perhaps you meant $v_i$? Remember that mathematics is case sensitive.
Other than that, if I understood the underlying idea you were trying to use, this is fine. Here's a simpler and cleaner version of what you wrote:
Let $X$ be all the linearly independent subsets of $V$, and let $\leq$ be the $\subseteq$ order on $X$. Let $S$ be a totally ordered subset of $X$, then we claim that $T=\bigcup S$($=\bigcup_{A\in S}A$) is an upper bound of $S$ and $T\in X$.
$T$ is a linearly independent set, since if $\alpha_1\cdot v_1+\ldots+\alpha_n\cdot v_n=0$ for some scalars $\alpha_i$ and $v_i\in T$, there are some $A_i\in S$ such that $v_i\in S_i$. But $S$ is linearly ordered so there is some $j$ such that $v_i\in A_j$ for all $1\leq i\leq n$. And because $A_j$ is linearly independent it has to be that $\alpha_i=0$ for all $i$. Therefore $T$ is linearly independent and $T\in X$.
It is an upper bound of $S$ since for every $A\in S$, it is true that $A\subseteq T$, so $A\leq T$.
We have shown that given an arbitrary totally ordered subset of $X$, it has an upper bound in $X$, so by Zorn's lemma there is a maximal linearly independent set, which is a basis. $\square$