> [!NOTE] Lemma > Let $V$ be a [[Real Vector Space|real vector space]] so that $V$ is not [[Finite Dimensional Real Vector Space|finite dimensional]]. Then $V$ contains an infinite set of [[Linearly Independent Subset of Real Vector Space|linearly independent]] elements. **Proof**: Certainly $V\neq \{ 0_{V} \}$ so choose $v_{1}\in V$ and set $W=\langle v_{1} \rangle$. Since $V$ is not finite dimensional, $W\neq V$, so we may chose some element $v_{2}\in W$. By [[Span of Linearly Independent Subset of Real Vector Space is Strongly Minimal]], the set $\{ v_{1},v_{2} \}$ is linearly independent. We proceed inductively: after $m$ steps, $\{ v_{1},\dots,v_{m} \}$ is a linearly independent set. Since $V$ is not finite dimensional $W=\langle v_{1},\dots,v_{m}\rangle \neq V$, so we may choose some element $v_{m+1}\not\in W.$