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

Question

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.

Answer 1

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.

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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.

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

Answer 2

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 $M=\bigcap_{n\in\mathbb{N}}\{X\in\ell^2:\phi_{2n}(X)=0\}$.

Answer 3

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 $$M=\{e_{2i-1};i\geq 1\}^\perp$$