# How can I show that this set is a closed subspace

Let $$E:= \{(x_n) \in l^{2}(\mathbb N ) \mid x_{2k}=x_{2k+1}\}$$

How can I show that $$E$$ is a closed subvector space of $$l^{2}(\mathbb N )$$ ?

I tried to write $$E$$ as the kernel of a continuous linear form but I ended with $$T:l^{2}(\mathbb N) \rightarrow \mathbb R : (x_n) \rightarrow \sum_{n=0}^\infty |x_{2k}-x_{2k+1}|$$

which is not linear.

• just use sequential defn of closed – mathworker21 Jun 16 at 14:59

You didn't quantify whether it is for all $$k\in\mathbb{N}$$ or for a single $$k\in\mathbb{N}$$.
For each $$k\in\mathbb{N}$$, note that $$(x_n)\in\ell^2(\mathbb{N})\mapsto x_{2k}-x_{2k+1}\in\mathbb{F}$$ is continuous and linear. So its kernel is a closed subspace of $$\ell^2(\mathbb{N})$$.
If your $$E$$ is $$\{(x_n)\in\ell^2(\mathbb{N})\mid x_{2k}=x_{2k+1}\quad\forall k\in\mathbb{N}\}$$, then we take intersection of closed subspace, so it is also a closed subspace.
I assume you can show it is a subspace. As to why it is closed define a sequence $$(x^k)_{k=1}^\infty\subseteq E$$ which converges to $$x$$. We have to show that $$x\in E$$. By definition of convergence in $$l^2$$ we have $$\sum_{n=1}^\infty |x^k_n-x_n|^2\to 0$$ when $$k\to\infty$$. This of course implies that for each $$n\in\mathbb{N}$$ we have $$|x^k_n-x_n|\to 0$$. That means we have convergence in every coordinate. Can you finish from here using the fact that the vectors in the sequence are in $$E$$?