# Hoffman and Kunze, Linear algebra chapter 2 Theorem 5

Theorem : If $W$ is a subspace of a finite dimensional vector space $V,$ every linearly independent subset of $W$ is finite and is part of a (finite) basis for $W.$

Proof: Suppose $S_0$ is a linearly independent subset of $W.$ If $S$ is a linearly independent subset of $W$ containing $S_0,$ then $S$ is also a linearly independent subset of $V;$ since $V$ is finite-dimensional, $S$ contains no more than $\dim V$ elements.

....and the proof goes.

I don't understand the whole point of introducing $S$ into the proof. Wouldn't just $S_0$ suffice?

What would go wrong if the author wrote it this way: Suppose $S_0$ is a linearly independent subset of $W.$ Since $S_0\subset W \subset V,$ $S_0$ is also a linearly independent subset of $V;$ since $V$ is finite-dimensional, $S_0$ contains no more than $\dim V$ elements.

You're totally right. The set $S$ is not necessary in any way, and your version of the proof is correct and simpler. There is no point to introducing $S$; this was just an oversight by the authors.