Vector Space has Basis

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 $L \subseteq B \subseteq S$.

$\blacksquare$

Also see

• Finitely Generated Vector Space has Finite Basis