$A + B = \{\,x + y \mid x ∈ A,\, y ∈ B\,\}$ is closed in $ [0,\infty)$ for any closed $A, B \subseteq [0,∞)$ If $A, B$ are closed subsets of $[0,∞)$, then
$$A + B = \{\,x + y \mid x ∈ A,\, y ∈ B\,\}$$
is closed in $ [0,\infty)$.
If $A ,B$ are closed sets in $\mathbb R$ , then i know counterexamples but considering only non negative reals ,i can't able to get a counterexample which doesn't implies it must be true.
So i want to prove or disprove , I know it is not true in genral but under this particular situation i want to know it is TRUE /FALSE
 A: Take a sequence $(x_n)\subseteq A+B$ convergent to $x\in [0,\infty)$. We want to show that $x\in A+B$.
It follows that $x_n=a_n+b_n$ for some sequences $(a_n)\subseteq A$ and $(b_n)\subseteq B$. Since $A,B$ are subsets of $[0,\infty)$ then $$0\leq a_n\leq x_n$$
$$0\leq b_n\leq x_n$$
In particular, since $x_n$ is bounded then so are $a_n$ and $b_n$ (this is exactly the place where the proof would fail for whole $\mathbb{R}$ case). Therefore they have convergent subsequences, say $a_{n_k}$ and $b_{n_k}$ (we can choose indexes in such a way that they coincide). It follows that
$$x=\lim_{n=0}^{\infty} x_n=\lim_{k=0}^{\infty} x_{n_k}= \lim_{k=0}^{\infty}\big(a_{n_k}+b_{n_k}\big)=\lim_{k=0}^{\infty}a_{n_k}+\lim_{k=0}^{\infty}b_{n_k}\in A+B$$
the last "$\in$" because $A,B$ are closed.
A: If $A$ and $B$ are compact, $A + B$ is compact because it is the image of the compact set $A \times B$ under the continuous function $(a,b) \mapsto a+b$.
If $A$ and $B$ are closed subsets of $[0,\infty)$, then for all $N > 0$ we have 
$$(A + B) \cap [0, N] = ((A \cap [0,N]) + (B \cap [0,N])) \cap [0,N]$$ which is closed, therefore $A + B$ is closed.
A: Consider the compactification $X = [0,+\infty]$. We can extend the addition function to $X \times X$ by defining $+\infty + x = x + +\infty = +\infty$ for all $x \in X$, and one can check this is still continuous as a map $p$ from $X \times X \to X$. Then, if $A,B \subseteq [0,+\infty)$ are closed, then their closures in $X$ are either $A$ resp $B$ or $A \cup \{+\infty\}$ or $B \cup \{+\infty\}$ respectively, depending on whether $A$ and $B$ were bounded or not.
In either case $p[\overline{A} \times \overline{B}]$ is compact hence closed in $X$ and so $A + B = p[\overline{A} \times \overline{B}] \cap [0,+\infty)$ is also closed in $[0,+\infty)$.
