Vector space/Linear subspace/Dimensions/Compare/Fact/Proof

Proof

Set ${}n=\dim _{}{\left(V\right)}$. Every linearly independent family in ${}U$ is also linearly independent in ${}V$. Therefore, due to the basis exchange theorem, every linearly independent family in ${}U$ has a length ${}\leq n$. Suppose that ${}k\leq n$ has the property that there exists a linearly independent family with ${}k$ vectors in ${}U$, but no such family with ${}k+1$ vectors. Let ${}{\mathfrak {u}}=u_{1},\ldots ,u_{k}$ be such a family. This is then a maximal linearly independent family in ${}U$. Therefore, due to fact, it is a basis of ${}U$.

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