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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!

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It is interesting to know every subspace of such space are closed. –  Paul Aug 2 '12 at 7:23
More generally, every finite-dimensional subspace of a linear normed space is closed, see PlanetMath or elsewhere. –  Martin Sleziak Aug 2 '12 at 8:49

5 Answers 5

up vote 2 down vote accepted

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.

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Very nice answer! –  Rudy the Reindeer Aug 2 '12 at 9:13

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.

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Notice your argument is better than the previous one as it doesn't use that $H$ is finite dimensional. –  chango Nov 3 '12 at 10:49
Actually yes it does. How else could we have chosen the basis $v_1, \dots, v_m$? –  Joshua Ciappara May 29 '13 at 1:40

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).

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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.

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Did you want to write complete metric space instead of complete topological space? (I might be wrong, but I think that you need at least uniformity to speak about completeness.) This is proved at PlanetMath, too. –  Martin Sleziak Aug 2 '12 at 9:17
@MartinSleziak Thank you for the comment. I think the easiest way to fix it is, yes, to replace "topological" with "metric" or "normed". But one can also define Cauchy sequences in spaces with countable neighbourhood bases. –  Rudy the Reindeer Aug 2 '12 at 9:20
No, a uniformity is not a topology (but it gives one). It's a uniform structure - a "system of neighborhoods" of the diagonal (Martin mentioned that word already, otherwise I wouldn't have). A topological space admits a uniformity iff it's $T_{3 \,1/2}$. In a metric space the uniform structure is generated by pairs of points such that $d(x,y) \lt \varepsilon$. In a topological (abelian) group you get a uniformity from a countable neighborhood base $\{U_n\}_{n=1}^\infty$ by considering the pairs of points $x,y$ such that $x-y \in U_n$. –  t.b. Aug 2 '12 at 9:47
@t.b. Ok. But I think my answer is correct now. (i) is true and a Hilbert space of course is also a metric space. No? –  Rudy the Reindeer Aug 2 '12 at 9:51
I was only referring to your first comment. Your answer is correct. –  t.b. Aug 2 '12 at 9:51

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 $H$.

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