# Having trouble showing a subspace of $\ell^2$ is closed.

Let $$M =\{ (x_n)_{n \in \mathbb{N}} \in \ell^{2} \mid x_{2j} = 0, \text{ for all } j \in \mathbb{N}\}$$.

$$M$$ is a subspace of the Hilbert space $$\ell^{2}$$ and I'm supposed to show it's closed.

What I've thought so far is this:

let $$X_{k} = \big((x_n^k)_{n \in \mathbb{N}}\big)^{k \in \mathbb{N}}$$ be a sequence in $$M$$ which is Cauchy in $$\ell^{2}$$. So $$\lim_{k \to \infty}X_k = (x_n)_{n \in \mathbb{N}}.$$ We know that $$X_{2j}^k = 0$$, for all $$j \in \mathbb{N}$$. We also know that the distance (with norm $$\|\cdot\|_{2}$$) between any two elements in $$X_{k}$$ will tend to zero as we let $$k$$ approach infinity. I thought that I might argue that since that distance is zero for any two elements, then the elements with odd indices will tend to zero as $$k$$ approaches infinity. But since the elements with even indices are still equal to zero, this sequence $$(x_{n})_{n}$$ is also in $$M$$.

If I have concluded falsely anywhere, please tell me, and any hints/tips or full answers are very much appreciated.

• You know that $x^k_{2j}=0$ for every $k$ and $j$, not $X_{2j}=0$. Nov 5 '15 at 17:51
• You can use "\ell" to write $\ell$ Nov 5 '15 at 17:57

Your idea is not wrong, it just needs to be written down more carefully. I think there are a few points that you are slightly misunderstanding.

To prove that your subspace $M$ is closed we need to prove that every convergent sequence in $M$ converges to a point of $M$.

So, as you do, let's take a sequence $X_k = \left((x_n^k)_{n \in \mathbb N}\right)^{k \in \mathbb N} \subset M$ which is convergent (or, equivalently, being $\ell^2$ complete, Cauchy). Let's denote by $(x_n)_{n \in \mathbb N} \in \ell^2$ the limit of this sequence (as you did).

All we are left to prove is that $(x_n)_{n \in \mathbb N} \in M$, which means we need to prove that $x_{2j}=0$ for every natural number $j$ (note that we don't care about the behavior of the odd components). But, as suggested in the other answer, the components of the elements in the sequence in $\ell^2$ converge (in the sense of $\mathbb R$) to the components of their limit, that is $x_n^k \to x_n$, for every $n$.

In particular, $x_{2j}^k \to x_{2j}$ for every $j$, but since the sequences $\{x_{2j}^k\}_{k \in \mathbb N}$ are identically zero by hypothesis (recall $(x_n^k)_{n \in \mathbb N} \in M$, so $x_{2j}^k=0$ for every $j$ and $k$), so we must have that their limits are also zero, that is $x_{2j}=0$ for every natural number $j$. Thus $(x_n)_{n \in \mathbb N} \in M$.

We have hence proved that the limit of every convergent sequence in $M$ belongs to $M$, and so $M$ is closed.

Note that we did not need to use the notion of Cauchy sequence at all.

• Thanks, that made a lot of sense. I'm struggling a bit with handling sequences with two indices, like $k$ and $n$. It's reassuring at least that my thought process wasn't completely off-track :) Nov 5 '15 at 20:02
• @Auclair you're welcome. I tried to explain every single step the best I could, let me know if there's anything you don't understand. I see that even though you have been on the website for a while you tend not to accept answers. The reason the feature exists is that once the user has understood and agreed with one of the answers the question is not considered unanswered anymore by the system. So it's advisable to accept answers, generally speaking, most of all for future use of other people with the same questions that will be able to understand what is an acceptable answer. Nov 5 '15 at 20:20
• Yes I know, It's entirely my fault. Sometimes I remember to accept answers, sometimes I get so excited by understanding the problem that I completely forget. I'll try and do a better job of it :) Nov 5 '15 at 21:33
• @Auclair when you'll have a minute, spend some time upvoting/accepting the answers that addressed your questions in the past, if nothing, it's good karma :) (obviously if an answer doesn't satisfy you you don't need to!) Nov 5 '15 at 22:21

If $X_{k} = \bigl((x_{n}^{k})_{n \in \mathbb{N}}\bigr)^{k \in \mathbb{N}}$ converges to $X= (x_{n})_{n \in \mathbb{N}}$ in $\ell^2$, then $\lim_{k\to\infty}x_n^k=x_n$ for all $n$.

Another way is the following. Let $\phi_n\colon\ell^2\to\mathbb{R}$ be the linear functional defined by $\phi_n\bigl((x_{n})_{n \in \mathbb{N}}\bigr)=x_n$. $\phi_n$ is continuous and $M=\bigcap_{n\in\mathbb{N}}\{X\in\ell^2:\phi_{2n}(X)=0\}$.

We know that in a Hilbert space $S^\perp$ is always a closed subspace for any subset $S$ of $\mathcal H.$ It is very easy to see that $$M=\{e_{2i-1};i\geq 1\}^\perp$$