Vector Space has Basis
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Theorem
Let $K$ be a division ring.
Let $V$ be a vector space over $K$.
Then $V$ has a basis.
Proof
The result follows from Vector Space has Basis between Linearly Independent Set and Spanning Set.
It suffices to find a linearly independent subset $L \subseteq V$ that is contained in a spanning set $S \subseteq V$.
By Empty Set is Linearly Independent, $L$ can be taken to be the empty set.
Or if $V$ is nonzero, by Singleton is Linearly Independent, $L$ can be taken to be any singleton of $V$.
$S$ can be taken to be $V$, since $V$ trivially spans itself.
Therefore, $L$ and $S$ exist and $L \subseteq S$ so $V$ has a basis $B$ with $L \subseteq B \subseteq S$.
$\blacksquare$