X is a normed linear space and A,B be subsets of X. Prove that,
(a) A+B is open if A is open or B is open.
(b) A+B is closed if A is closed and B is compact.
(c) Consider $X=c_0$, $A=\{(\alpha_n)_n \in c_0 : \alpha_m=m\alpha_{m+1} \forall m \text{ even }\}$ , $B=\{(\alpha_n)_n \in c_0 : \alpha_m=0 \forall m \text{ odd }\}$ and show that A,B are closed but A+B is not.
My attempt:
(a)Let B be open then since,$x \mapsto c+x$ is a homeomorphism, we have, $A+B = \cup_{a \in A} (a+B)$ thus being union of open sets is open.
(b)Let, $z_n=a_n+b_n$ be a sequence in $A+B$ such that it converges to some $z \in X$ . To show, $z \in A+B$ . Compactness of $B \implies \exists \{b_{n_k}\}$ converging to some $b \in B$ . Consider, the corresponding, $z_{n_k}=a_{n_k}+b_{n_k}$, then $z_{n_k} \to z $ i.e. $a_{n_k}= z_{n_k}-b_{n_k}$ is also convergent, $A$ closed implies, $a_{n_k} \to a \in A$ . Thus,$a=z-b$ i.e. $z=a+b \in A+B$.
(c) I couldn't proceed on this.
Thanks in advance for help!