Proof check of sum of a compact and closed set of real numbers is closed Let $A$ be a closed and $B$ be a closed and bounded set in $\mathbb R$ , then  we have to show that $A+B:=\{a+b:a\in A , b\in B \}$ is closed in $\mathbb R$ . 
My Proof : Let $\{a_n+b_n\}$ be a convergent sequence in $A+B$ , where $\{a_n\}\in A , \{b_n\}\in B$ , with limit $x$ , we have to show that $x \in A+B$ . Since $B$ is closed-bounded , there is a subsequence $\{b_{r_n}\}$ of $\{b_n\}$ such that $\{b_{r_n}\}$ converges with $\lim \{b_{r_n}\}=l$ (say) ; then since $B$ is closed , $l\in B$. Also $\lim \{a_{r_n}+b_{r_n}\}=x$ , thus $\lim \{a_{r_n}\}=x-l$ , and since $A$ is closed , so $x-l \in A$ , thus $x=(x-l)+l \in A+B$ . Am I correct ? 
 A: In the above proof, things become easy if we take $z_n=a_n+b_n \in A+B $ . Let $\lim z_n=z \in \mathbb{R}$   Since $B$ is compact, $(b_n)$ has a convergent subsequence, let's say $(b_{n_r})$. Let $\lim b_{n_r} = l \in B $ . So $ a_{n_r}=z_{n_r}-b_{n_r} $. So $(a_{n_r})$ is also convergent and let $\lim a_{n_r}=m .$  As $A$ is closed, $m\in A$ . So $z=m+l \in A+B $  which completes the proof.
A: Let $(a_n)_n$ be a sequence in $A$ and let $(b_n)_n$ be a sequence in $B$ such that $a_n+b_n=c+d_n$ where $\lim_{n\to \infty}d_n=0.$ Then $(d_n)_n$ is bounded. And $(b_n)_n$ is bounded, as $B$ is bounded. So $(a_n)_n=(c+d_n-b_n)_n$ is bounded. So, as $A$ is closed, there exists a closed bounded $A^*\subset A$ such that $A^*\supset Cl(\{a_n\}_n).$
The function $f(x,y)=x+y$ is continuous from $\mathbb R^2$ to $\mathbb R$, and $A^*\times B$ is compact, so $f(A^*\times B)$ is compact, so $f(A^*\times B)$ is closed. We have $$c\in Cl(\{a_n+b_n\}_n)\subset Cl(f(A^*\times B))=f(A^*\times B)\subset f(A\times B)=A+B.$$ 
Notation: $(a_n)_n$ is the sequence $(a_n)_{n\in  \mathbb N},$ and $\{a_n\}_n=\{a_n: n\in \mathbb N\}.$
