# Showing $a_3\le a_1+b_2\le b_3$ where $(a_1,b_1)$ is the image of $f(x)$ and $(a_2,b_2)$ is of $g(x)$ and $(a_3,b_3)$ is of $f(x)+g(x)$

For $x\in\mathbb{N}$, let the image $f(x)$ for be $(a_1,b_1)$ and the image $g(x)$ be $(a_2,b_2)$ and the image $h(x)=f(x)+g(x)$ be $(a_3,b_3)$.

We know that $a_1+a_2<f(x)+g(x)$ for any $x\in(a,b)$ therefore $a_1+a_2\le a_3$ since $a_3$ which is the infimum of $h(x)$ is greater or equal to any of it's lower bound.

Similarly, we know that $b_3\le b_1+b_2$.

But how would we show that $a_3\le a_1+b_2\le b_3$ ?

• What's known about these functions (i.e. we've used $a_1+a_2\leq h(x)$, but that hasn't been stated before, nor has anything else about them)? Also, what $x$ are we referring to in the first paragraph (later, it seems $x$ is any element of $(a,b)$)? Sep 9 '13 at 11:03
• I edited so that $x\in \mathbb{N}$ Sep 9 '13 at 11:04
• And nothing else is known about these 2 functions except that their images are bounded. Sep 9 '13 at 11:07

We should assume that, infimum of $f$ the domain is $a_1$ and supremum of $g_1$ over the same domain is $b_2$.
We know that $-g(x)\geq -b_2$ so we have $f(x)=h(x)-g(x)\geq a_3-b_2$ but then because $a_3-b_2$ is a lower bound for $f$ and $a_1$ the infimum of $f$, i.e. the greatest lower bound, we have $a_1\geq a_3-b_2$.
Similarly we have $-f(x)\leq -a_1$ therefore $g(x)=h(x)-f(x)\leq b_3-a_1$ and in a similar way, we should have $b_2\leq b_3-a_1$