Subspaces of Hilbert Spaces of finite dimension Given a Hilbert space $H$ of finite dimension, why is any subspace of this space closed? I tried bashing out an answer using an arbitrary Cauchy sequence $\{ f_1 , f_2, \ldots \} \subset S \subset H $ and trying to show its limit $f \in S$. I keep getting stuck and  suspect there's an easy answer that I'm missing. Could someone enlighten me on this? Thanks in advance!
 A: Let $S$ a subspace of $H$, and $\{e_1,\dots,e_d\}$ an orthonormal basis of $S$. We can complete it as a basis of $H$. By Gram-Schmidt process, we can assume that this gives an orthonormal basis $\{e_1,\dots,e_d,f_1,\dots,f_N\}$ of $H$. Then we notice that $S=\operatorname{Span}(f_j,1\leq j\leq N)^{\perp}$, and the orthogonal of a set is closed. 
A: Let $H$ be a finite dimensional Hilbert space and $V$ a subspace of $H$. If $V=0$, it's obvious that $V$ is closed. Suppose $V \ne 0$, and let $v_1,\ldots, v_m$ be an orthonormal basis of $V$, with $1 \le m \le \dim H$. Let $x \in \overline{V}$ and $(x^n)_n \subset V$ a convergent sequence whose limit is $x$. Thanks to the Cauchy-Schwarz inequality we have for every $1 \le k \le m$:
$$ 
\left|\langle v_k,x\rangle_H-\langle v_k,x_n\rangle_H\right|=\left|\langle v_k,x-x_n\rangle_H\right| \le \|v_k\|_H\|x-x_n\|_H=\|x-x_n\|_H.
$$
Hence 
$$
\lim_n\langle v_k,x_n\rangle_H=\langle v_k,x\rangle_H \quad \forall\ 1 \le k \le m.
$$
It follows that
$$
x=\lim_n x_n=\lim_n\sum_{k=1}^m\langle v_k,x_n\rangle_Hv_k=\sum_{k=1}^m\langle v_k,x\rangle_Hv_k,
$$
i.e. $x \in V$. Thus $\overline{V} \subset V$, and $V$ is closed.
A: A subspace of a finite dimensional vector space is always a finite intersection of hyperplanes.
Under the Hilbert space topology hyperplanes are closed (in fact they are the zero sets of linear forms).
A: While I personally prefer Davided's answer, let me show you another more crude way to do it. 
Fix an orthonormal basis $e_1,\ldots,e_n$. Each element in your Cauchy sequence is then
$$
f_j=\sum_k f_{kj}e_k,
$$
for numbers $f_{kj}$. As $\|f_j-f_i\|^2=\sum_k|f_{kj}-f_{ki}|^2$, it is easy to see that each sequence (of numbers) $\{f_{kj}\}_j$ is Cauchy, $k=1,\ldots,n$. 
Now you can take convergent subsequences one by one, as we only have $n$ sequences, and so there exist numbers $f_{k,0}$ with $f_{kj}\to f_{k,0}$. It is easy to see then that
$$
f_j\to\sum_kf_{k,0}e_k
$$
in $H$.
A: I think one can also argue as follows:
(i) Complete subsets of complete metric spaces are closed.
(ii) Every finite dimensional normed space is complete (see here for proof)
$S$ and $H$ are finite dimensional hence by (i) and (ii), $S$ is closed.
