Linear subspace and dimension If $Y$ is a proper linear subspace of a finite dimensional linear space $X$, I want to show that $Y$ is also finite dimensional and $\dim(Y)< \dim(X)$.
 A: Hint:
If $\,Y=\{0\}\,$ there's nothing to prove, otherwise take $\,0\neq y_1\in Y\,$ . If $\,\operatorname{Span}\{y_1\}=Y\,$ we're done otherwise there exists $\,Y\ni y_2\notin \operatorname{Span}\{y_1\}\,$ . 
Then $\,\{y_1,y_2\}\,$ is linearly independent (why?), so if $\,\operatorname{Span}\{y_1,y_2\}=Y\,$ we done, otherwise there exists $\,Y\ni y_3\notin \operatorname{Span}\{y_1,y_2\}\,$ , so then $\,\{y_1,y_2,y_3\}\,$ is lin. independent (again, why?) ...
Continue as above, but note all this happens within $\,X\,$ and here a linearly independent set has only a finite number of elements...complete and end the argument.
A: Let $n=\dim X$. The lineraly independant families of $Y$ are also linearly independant for $X$  then they have at most $n$ elements. The set of cardinals of this linearly independant families of $Y$ is a subset of $\mathbb{N}$ bounded above by $n$ so it has a maximum say $p$ and $p\leq n$.
Let $(e_1,\ldots,e_p)$ a linearly independant family of $Y$ and $x\in Y$ so $(e_1,\ldots,e_p,x)$ is linearly dependant then there's a linear combination:
$$\alpha_1 e_1+\cdots+\alpha_p e_p +\beta x=0$$
and we can see easily that $\beta\neq 0$ (Why?) then 
$$x=\frac{-1}{\beta}(\alpha_1e_1+\ldots+\alpha_pe_p)$$
hence the family $(e_1,\ldots,e_p)$ spans $Y$ so it's a basis for $Y$ and then
$$p=\dim Y\leq \dim X=n$$
and if $p=n$ then $Y=X$ so by the hypothesis $Y$ is a proper subspace of $X$ we find the desired result.
