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

Proof

Set $n={\dim }_K\left(V\right)$. 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 length $\le n$. Suppose that $k\le n$ has the property that there exists a linearly independent family with $k$ vectors in $U$ but no such family with $k+1$ vectors. Let $𝔲=u_1,\dots ,u_k$ 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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