Show that $Y$ is a closed subspace of $\ell^2$ Show that $Y = \{x\mid x=(\xi_j) \in \ell^2, \xi_{2n}=0, n \in \mathbb{N}\}$
is a closed subspace of $\ell^2$.
I have read the post here and I am still struggling to figure out how to show this. I know that for a subspace to be closed, we have to have that the space contains all its limit points. 
I also think that $Y=\{\xi_1, 0, \xi_3, 0, \ldots\}$.
Do I need the fact that $x=(\xi_j)\in \ell^2$ to say that $Y\in \ell^2$? I am really lost and my book isn't helping me...
 A: Let $x$ be a limit point of $Y$, with $(x_n)\subset \ell^2$ a sequence converging to $x$. Hence for all $\varepsilon >0$ there exists a $N\in \mathbb{N}$ such that $n>N$ implies that $\|x-x_n\|<\varepsilon$. Thus for all $n>N$ we have that
$$\sum_{i=1}^\infty|x^{(i)}-x_n^{(i)}|<\varepsilon^2.$$
Now assume that for some $i=2k$ for $k\in \mathbb{N}$ that $x^{(i)}\neq 0$ (i.e $x\notin Y$). You should be able to reach a contradiction from here. As $X$ is a metric space this proves that $Y$ contains all of its limit points, hence it is closed.
A: Perhaps the neatest way to do this is to note that the operator $P_{2k}:\ell^2 \to \mathbb{R}$ defined by $P_{2k}x = \xi_{2k}$ for $x = (\xi_k)$ is continuous and so $\ker P_{2k}$ is a closed set. As a result $Y = \bigcap_{k \in \mathbb{N}} \ker P_{2k}$ is closed as an intersection of closed sets.
A: The typical way to do these guys is to take a cauchy sequence (of sequences) in the space, here with even numbered terms being zero, and show it converges to an element of the set of sequences with even numbered terms being zero. This is analogous to proving the set contains its accumulation points.
Let $a_i(n)$ be such a cauchy sequence. Note that $\ell^2$ is complete, thus we know a cauchy sequence of $\ell^2$ sequences will converge to an $\ell^2$ sequence. Thus, given an $\epsilon>0$, we can find a big enough $n$ for which
$$
\sum_{i=0}^\infty|a_i(n)-a_i|^2<\epsilon^2
$$
where $a_i$ is the limit in $\ell^2$ that we must prove is also $0$ at even terms. 
But from the estimate above, it is certainly true that at any $i$ we will have
$$
|a_{2i}(n)-a_{2i}|<\epsilon\implies|a_{2i}|<\epsilon
$$
since the choice of epsilon was arbitrary, we know the limiting sequence vanishes at the even indices.
