Dimension of Proper Subspace Less Than its Superspace

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Theorem

Let $G$ be a vector space whose dimension is $n$.

Let $H$ be a subspace of $G$.

Then $H$ is finite dimensional and $\dim \left({H}\right) \le \dim \left({G}\right)$.

If $H$ is a proper subspace of $G$, then $\dim \left({H}\right) < \dim \left({G}\right)$.

Every linearly independent subset of the vector space $H$ is a linearly independent subset of the vector space $G$.

So the set of all natural numbers $k$ such that $H$ has a linearly independent subset of $k$ vectors has a largest member $m$, and $m \le n$.

Now, let $B$ be a linearly independent subset of $H$ having $m$ vectors.

If the subspace generated by $B$ were not $H$, then $H$ would contain a linearly independent subset of $m + 1$ vectors.

This would contradict the definition of $m$.

Hence $B$ is a generator for $H$ and is thus a basis for $H$.

Thus $H$ is finite dimensional and $\dim \left({H}\right) \le \dim \left({G}\right)$.

Now, if $\dim \left({H}\right) = \dim \left({G}\right)$, then a basis of $H$ is a basis of $G$ by Basis of Vector Space is Linearly Independent and a Generator, and therefore $H = G$.

$\blacksquare$