What is the definition of closed subspace?

Question

I am trying to understand what is intended with closed subspace, I took the following guess:

A closed subspace $M$ of a Hilbert space $H$ is a subspace of $H$ s.t. any sequence $\{x_n\}$ of elements of $M$ converges in norm to $x \in M$ ,i.e. , $\|x_n - x\| \rightarrow 0$ as $n \rightarrow \infty$.

Is this correct?

Answer 1

That is correct. It is a subspace that is closed in the sense in which the word "closed" is usually used in talking about closed subsets of metric spaces.

In finite-dimensional Hilbert spaces, all subspaces are closed. In infinite-dimensional spaces, the space of all finite linear combinations of the members of an infinite linearly independent set is not closed because it fails to contain infinite linear combinations of is members. E.g. suppose the $n$th member of a basis of $\ell^2$ is $$ (0,0,0,\ldots,0,0,\ \underset{\uparrow}1,\ 0,0,\ldots) $$ where the arrow indicates the $n$th component, for $n=1,2,3,\ldots$. Then the set of all linear combinations of finitely many of these does not contain the point $$ \left(1,0,\frac12,0,\frac13,0,\frac14,0,\frac15,\ldots\right) $$ but one can find a sequence of vectors in that subspace that converges to this point.

(Note that I had to choose the coordinates in such a way that the sum of their squares is finite; otherwise it wouldn't be in the Hilbert space $\ell^2$ at all.)

Answer 2

The norm on $\mathcal H$ induces a metric by $d(f,g) = \|f-g\|_{\mathcal H}$. This metric induces a topology by taking the collection of open balls (sets of the form $\{g\in\mathcal H : \|f-g\|_{\mathcal H}<\varepsilon\}$ where $f\in\mathcal H$ and $\varepsilon>0$) as a basis. So a set $M$ is closed in $\mathcal H$ if its complement $\{f\in \mathcal H: f\notin M\}$ is open. Recall that if $V$ is a vector space and $W\subset V$, then $W$ is a subspace if and only if $W$ contains the zero vector and is closed under addition and scalar multiplication (i.e. if $u,v\in W$ then $u+v\in W$ and if $c$ is a scalar then $cv\in W$) - and Hilbert spaces are vector spaces. Therefore $M$ is a "closed subspace" if and only if $M$ is a subspace, and the complement of $M$ is open with respect to the metric topology.

An equivalent definition for a set $M$ to be closed in a metric space is that if $(x_n)$ is a sequence of elements in $M$ that is convergent, then the limit is in $M$ (i.e. $M$ contains all of its limit points). In the context of our Hilbert space, if for every sequence $(x_n)$ in $M$ with a limit $x\in\mathcal H$, we have $x\in M$, then $M$ is closed.

As Michael Hardy pointed out, a subspace of a finite-dimensional vector space is always closed, so this distinction only need be made for infinite-dimensional Hilbert spaces.