Given two subspaces $M= \{ a \in \ell^\infty (\mathbb{N}) | a_{2n}=0 \}$ and $N= \{b \in \ell^\infty (\mathbb{N}) | b_{2n-1}=nb_{2n} \} $, is the direct sum $M\dot{+} N$ a closed subspace of $\ell^\infty (\mathbb{N})$?
The original question was for the $\ell^p (\mathbb{N})$ space, where I showed that the direct sum is not closed since it contains a dense subset of $\ell^p (\mathbb{N})$, but it is not the whole $\ell^p (\mathbb{N})$. I'm kinda stuck for the $p= \infty$ case, because $\ell^\infty (\mathbb{N})$ is not separable.
Any help/hints are appreciated.
So I came up with an answer:
The set of all sequences with the compact support is in $M \dot{+}N$ and sequences with the compact support are dense in $c_0(\mathbb{N})$. If $M \dot{+}N$ was closed then $c_0(\mathbb{N}) \subset M \dot{+}N$ would need to hold.
Take the sequence $c_n=\frac{1}{\sqrt{n}}$, $c\in c_0(\mathbb{N})$ and assume $c \in M \dot{+}N$. Then the sequences $a \in M, b \in N$ exist, s.t. $c=a+b$.
$$c_{2n}=b_{2n}=\frac{1}{\sqrt{2n}}$$ $$b_{2n-1}= \sqrt\frac{n}{2} \notin \ell^\infty (\mathbb{N})$$ which is a contradiction.