Let us dispose first of an extreme case: if \(V\) consists only of the 0-vector, we may take the empty set as a basis for \(V\).

Now suppose \(V\) contains a nonzero vector. We use the following two facts in combination to build a basis of \(V\).

1. Given a linearly independent subset \(S\) of \(V\), there are two possibilities:
   1. \(S\) spans \(V\), in which case it is a basis of \(V\);
   2. \(S\) does not span \(V\) in which case we can add to \(S\) a vector from \(V\) outside \(\span(S)\) to obtain a larger linearly independent subset of \(V\).
2. This process of adding vectors to \(S\) to obtain larger linearly independent sets stops after at most \(n\) steps, because \(\RNrSpc{n}\) itself is spanned by \(n\) vectors.

Here are the details: Pick any nonzero vector in \(V\), call it \(\Vect{a}_1\), and form \(\EuScript{B}_1\DefEq \Set{ \Vect{a}_1 }\).

We know that \(\EuScript{B}_1\) is linearly independent. If \(\span( \EuScript{B}_1 )=V\), then \(\EuScript{B}\DefEq \EuScript{B}_1\) is a basis for \(V\). If \(\span( \EuScript{B}_1 ) \neq V\), there are vectors in \(V\) which do not belong to \(\span(\EuScript{B}_1)\). Pick any one such, call it \(\Vect{a}_2\), and form \(\EuScript{B}_2\DefEq \EuScript{B}_1\cup \Set{ \Vect{a}_2 }\).

We know that \(\EuScript{B}_2\) is linearly independent. If \(\span( \EuScript{B}_2 )=V\), then \(\EuScript{B}\DefEq \EuScript{B}_2\) is a basis for \(V\). If \(\span( \EuScript{B}_2 ) \neq V\), there are vectors in \(V\) which do not belong to \(\span(\EuScript{B}_2)\). Pick any one such, call it \(\Vect{a}_3\), and form \(\EuScript{B}_3\DefEq \EuScript{B}_2\cup \Set{ \Vect{a}_3 }\); etc.

The crucial fact now is that, for some \(k\leq n\), we must have \(\span( \EuScript{B}_k) = V\). To see this, we argue by contradiction: Suppose \( n< k \) and \(\EuScript{B}_k\) is a linearly independent subset of \(V\). Then \(\EuScript{B}_k\) is, in particular, a linearly independent subset of \(\RNrSpc{n}\); a contradiction to the fact any linearly independent subset of \(\RNrSpc{n}\) has at most \(n\) vectors. Thus the process described above stops after \(k\leq n\) steps, in which case \(\EuScript{B}\DefEq \EuScript{B}_k\) is a basis of \(V\), and \(\EuScript{B}\) contains at most \(n\) vectors.