# Size of linear independent set = dimension of vector space

Let $V$ be a vector space of dimension $dimV =n$, over a field F. $S$ is a linearly independent set. Suppose I am equipped with the fact that the size of any linearly independent set must be $\le$ the size of any spanning set, and with the fact that a basis is a minimal spanning set(with a proof akin to this one). Then, how can I prove that $\#S \le n$, and that if $\#S=n$ then $S$ is a basis? I wouldn't use any other theorems that aren't absolutely elementary beyond the two I am equipped with.

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1. Let $S$ be a linearly independent set, and let $\mathscr{S}$ be the set of all spanning sets in $V$. One the one hand, $$\forall T \in \mathscr{S}, \; |S| \leq |T|,$$ and on the other hand, $$n := \dim V = \min_{T \in \mathscr{S}} |T|.$$ So, what can you conclude?
2. Suppose that $|S| = n$. Assume by contradiction that $S$ is not a basis, i.e., not spanning, so that there exists some non-zero $x \in V$ such that $x \notin \operatorname{span}S$. Is $S^\prime := S \cup \{x\}$ linearly independent? Moreover, what is $\left|S^\prime\right|$?
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