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 $a_3\le a_1+b_2\le b_3$ ?

  • $\begingroup$ 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 $(a,b)$)? $\endgroup$ Sep 9 '13 at 11:03
  • $\begingroup$ I edited so that $x\in \mathbb{N}$ $\endgroup$
    – MathsMy
    Sep 9 '13 at 11:04
  • $\begingroup$ And nothing else is known about these 2 functions except that their images are bounded. $\endgroup$
    – MathsMy
    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 $b_2\leq b_3-a_1$

Note that without Supremum and infimum assumption we cannot compare these bounds and indeed there are counterexamples for it.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.