Dimension of a Vector Space lower bounds Spanning Set size
# Statement
Let $V$ be a Vector Space with dimension $\dim V = n$ for some $n \in \bar{\mathbb{N}}$. Then if $S \subset V$ s.t. $|S| < n$, $\text{span} S \subsetneq V$.
# Proof
First suppose $V$ is a Finite-Dimensional Vector Space with $\dim V = n$ for some $n \in \mathbb{N}$. Since $\dim V = n$, there exists a Vector Space Basis $\mathbf{a}{1}, \dots, \mathbf{a}{n} \in V$. Suppose $\text{span} S = V$. Then, because Spanning Set Size bound Linearly Independent Set Size, any Linearly Independent Set must have size less than $n$. But since $\mathbf{a}{1}, \dots, \mathbf{a}{n}$ is a Vector Space Basis, it is Linearly Independent $\unicode{x21af}$. Therefore $\text{span} S \subsetneq V$. $\checkmark$
Now suppose $V$ is an Infinite-Dimensional Vector Space. Then if $S \subset V$ is s.t. $\text{span} S = V$, then we must have that $|S| = \infty$. Otherwise, because A Vector Space is Finite-Dimensional iff it has a Finite Spanning Set, $V$ would be finite-dimensional, a contradiction. $\blacksquare$