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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$

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    $\begingroup$ What does $[a,b] + [c, d)$ mean? $\endgroup$ – zoli Feb 25 '17 at 11:49
  • $\begingroup$ @zoli it's the addition of two intervals, I'm not sure how it could be unclear given the context of the question but I can edit as needed $\endgroup$ – baxx Feb 25 '17 at 11:51
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    $\begingroup$ I've seen so far only closed intervals in interval arithmetic (except infinities as end points) . Could you pls. direct me to an article where open or half open intervals are treated? $\endgroup$ – zoli Feb 25 '17 at 12:21
  • $\begingroup$ @zoli no sorry - other than just referring to the question I've asked. I can add some more context to what this came from though $\endgroup$ – baxx Feb 25 '17 at 12:24
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    $\begingroup$ Thank you. (Take a look at this article: fab.cba.mit.edu/classes/S62.12/docs/Hickey_interval.pdf or the wikipedia article and you will see what my problem is. I seem to be undereducated te a little. $\endgroup$ – zoli Feb 25 '17 at 12:26
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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.

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