Show every subspace of $\mathbb R^n$ is closed with respect to the usual metric? How do I see that every subspace of $\mathbb R^n$ is closed with respect to the usual metric $p(x,y) = x^Ty$ ?
I've seen some sweet results regarding Hilbert Spaces $\mathcal H$, especially that for a subspace $S$, $(S^{\bot})^\bot = S$, and this is true (in the case $\mathcal H = \mathbb R^n$) if $S$ is closed with respect to the metric given above.
I already know that $\mathbb R^n$ is complete.
 A: Every subspace is the kernel of a linear application (even of a projection !) which is continuous as we're in finite dimension, so that the subspace is closed.
A: Every subspace is also complete (since it is isomorphic to $\Bbb R^m$ for some $m$), and complete implies closed since every Cauchy sequence in $\Bbb R^m\subseteq\Bbb R^n$ converges in that same $\Bbb R^m\subseteq\Bbb R^n$ we have that it is closed by definition (i.e. every convergent sequence of points in the set converges in the same set)
A: Every subspace is the orthogonal complement of its orthogonal complement, so it is enough to show that the orthogonal complement of a subset of $\mathbb R^n$ is closed. 
Now the orthogonal complement of a subset $S$ of your vector space is the intersection of the orthogonal complements of the elements of $S$. Since the intersection of closed subsets is closed, this means that it is enough to show that the set $v^\perp$ of vectors orthogonal to a vector $v\in\mathbb R^n$ is closed.
Now $v^\perp=\{w\in\mathbb R^n:\langle w,v\rangle=0\}$ or, equivalently, $v^\perp$ is the kernel of the linear map $f_v:w\in\mathbb R^n\mapsto \langle w,v\rangle\in\mathbb R$. This means that it is enough to show that every linear map $\mathbb R^n\to\mathbb R$ is continuous.
You probably know how to do that.
