Every vector space has a basis

Theorem: assuming Zorn’s lemma, every vector space has a basis.

Pf: let $V$ be a vector space and consider

$$
\mathcal L := \{S \subseteq V : S \text{ is linearly independent}\}.
$$
Every vector space has $\emptyset$ as a linearly independent subset so $\mathcal L$ is guaranteed to be non-empty.

Step 1: $(\mathcal L, \subseteq)$ is a poset

I know that $(\mathcal P(V), \subseteq)$ is a poset, and $\mathcal L \subseteq \mathcal P(V)$, so $(\mathcal L,\subseteq)$ is also a poset.

Step 2: $(\mathcal L, \subseteq)$ has a maximal element

Let $C$ be a chain in $\mathcal L$, so $C$ is a collection of linearly independent sets that are totally ordered w.r.t. $\subseteq$, i.e. they’re nested in each other. Note $\{\emptyset\}$ is a chain in $\mathcal L$ so I know at least one such chain exists (meaning I’m not assuming the existence of an impossible object).

This nesting suggests a good candidate for an upper bound will be the limit of the sets in $C$, i.e. take

$$
U = \bigcup_{S \in C} S.
$$

I’ll now verify that this is an upper bound for $C$ in $\mathcal L$. Clearly $S \subseteq U$ for all $S \in C$ so I just need to verify that $U \in \mathcal L$.

Let $x_1,\dots,x_n\in U$. Because $U$ is a union, for each $x_j$ there must be some $S_j \in C$ such that $x_j\in S_j$. This resulting collection $\{S_1,\dots, S_n\}$ is totally ordered w.r.t. $\subseteq$, and as a finite total ordering it has a maximal element (I prove this for partial orders in my post here). Let $S_k$ be this maximal element. Then $S_j\subseteq S_k$ for each $j$ which means $x_1,\dots,x_n \in S_k$. $S_k$ is linearly independent, so this means that $\{x_1,\dots,x_n\}$ is too. This was an arbitrary finite subset of $U$ so $U$ is linearly independent, which means $U\in\mathcal L$. Thus every chain in $\mathcal L$ has an upper bound in $L$.

I have assumed that I can use Zorn’s lemma, and the previous step showed that I meet the conditions for it, so there is a maximal element in $\mathcal L$, say $M$.

Step 3: this maximal element is a basis

$M \in \mathcal L$ by construction so I know it’s linearly independent. Thus I need to show that $\text{span } M = V$.
Suppose there is some $x \in V \,\backslash\, \text{span } M$. I want to show that $M^* := M \cup \{x\} \in \mathcal L$ so let $x_1,\dots,x_n \in M^*$. I will show that this arbitrary subset $\{x_1,\dots,x_n\}$ of $M^*$ is linearly independent.

If $x \notin \{x_1,\dots,x_n\}$ then $\{x_1,\dots,x_n\}\subseteq M$ which means it’s linearly independent by $M \in \mathcal L$. So WLOG I can assume $x \in \{x_1,\dots,x_n\}$, and again without losing any generality I can just take $x=x_1$.

Now suppose $\sum_{i=1}^n \alpha_i x_i = 0$ for some $\alpha \in \mathbb R^n$. Assume $\alpha_1 \neq 0$. This means I can divide through by $\alpha_1$ to get
$$
x_1 = -\sum_{i\geq 2} \frac{\alpha_i}{\alpha_1}x_i
$$
which means $x = x_1 \in \text{span } M$. But that’s a contradiction, so it must be that $\alpha_1 = 0$. Then I have
$$
\sum_{i\geq 2}\alpha_i x_i = 0.
$$
But now $\{x_2,\dots,x_n\}$ is in $M$, and is thereby linearly independent, so $\alpha_2=\dots=\alpha_n=0$. This means $\alpha=0$, and therefore no matter what $\{x_1,\dots,x_n\}$ is linearly independent. This means $M^* \in \mathcal L$. But $M \subseteq M^*$ and yet $M$ in maximal in $\mathcal L$, so this is a contradiction. Therefore there cannot be any $x \in V \,\backslash\,\text{span } M$ which means $\text{span } M = V$, and therefore $V$ is a basis.