Dimension of Proper Subspace is Less Than its Superspace

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 $$\dim \left({H}\right) < \dim \left({G}\right)$$.

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

Therefore, it has no more than $$n$$ elements by Linearly Independent Subset of Finitely Generated Vector Space is Finite.

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 follows by Linearly Independent Subset also Independent in Generated Subspace.

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 $$H = G$$.