Dimension theorem

Let $B_1=\{v_1,\dots,v_m\}$ and $B_2=\{w_1,\dots,w_k\}$ both be bases of $V$. Since $B_1$ spans $V$ (it is a basis) and $B_2$ is linearly independent (it is a basis), the exchange lemma proved above applies directly with $B_1$ playing the role of the spanning set and $B_2$ the independent set, giving $k\leq m$.

Symmetrically, since $B_2$ spans $V$ and $B_1$ is linearly independent, apply the exchange lemma again with the roles swapped — $B_2$ as the spanning set, $B_1$ as the independent set — giving $m\leq k$.

Combining the two inequalities $k\leq m$ and $m\leq k$ forces $k=m$: the two bases have exactly the same number of vectors.

Since this argument places no restriction on which particular finite bases $B_1,B_2$ were chosen, every finite basis of $V$ has this same common size. This justifies defining $\dim(V)$ to be that common size — it is an invariant of the space $V$ itself, never depending on which basis was used to compute it.