Prove $[a,b] + [c, d) = [ a+c, b+d)$ Prove that $[a, b] + [c, d) = [a+c, b+d)$ where $a,b,c,d \in \mathbb{R}$. 
Though a general proof is what I'm after, a specific example could be 
$ A = [3,5] , B = [-5, -3) $
Then $ A + B = [3, 5] + [-5, -3) $
which gives $[-2, 2)$
I'm not sure how to go about proving that this is the case though.
Perhaps if I have $-2 \leq x < 2$ and I want to show that I can choose $a \in A$ 
and $b \in B$ such that $a + b = x$.
Maybe a contradiction would work for this? I'm not sure how to go about it though.
Some context;
$ A = [3,5] , B = [-5, -3) $
$A + B = \{ a + b : a \in A, b \in B\}$, find $A + B$. 
Are infimum and supremum of $A + B$ elements of $ A + B$
 A: First, let's prove that $a+c$ is the greatest lower bound (even more, the minimum value) of $[a, b]+[c, d)$: if $x \in [a,b]$, then $x \geq a$, and if $y \in [c, d)$, then $y \geq c$. Therefore, for any $x \in [a, b]$ and $y \in [c, d)$, we have $x+y \geq a+c$. Also, $a+c \in [a,b]+[c, d) $, because if you take $a$ from $[a, b]$ and $c$ from $[c, d)$, you get $a+c \in [a, b]$.
Next, let's prove that $b+d$ is an upper bound of $[a, b]+[c, d)$: if $x \in [a,b]$, then $x \leq b$, and if $y \in [c, d)$, then $y < d$. Therefore, $x+y<b+d$ for any $x \in [a,b]$ and $y \in [c, d)$.
So now we have $[a,b]+[c, d) \subseteq [a+c, b+d) $. Now, if $z$ is an arbitrary element of $[a+c, b+d)$, we want to be able to choose $x \in [a, b]$ and $y \in [c, d)$ such that $x+y=z$. If we take $x=b$, $y=z-b$, we can cover some (not all) elements of $[a+c, b+d)$. More precisely, we can cover all $z\in [a+c, b+d)$ such that $z-b \in [c,d)$, i.e. all $z \in [a+c, b+d)$ such that $z \in [b+c, b+d)$. 
We're still "missing" the part $[a+c, b+c)$. So let's pick an arbitrary $z$ from that interval, take $y=c$ and $x=z-y$. We have $x \in [a,b]$ so the condition is satisfied. This completes the proof.
