Prove that if $A,B\subseteq\mathbb R$ are non-empty and bounded from above, and $A+B=A$, it follows that $B=\{0\}$. Seems Wrong?

Question

A question from a basic textbook on real analysis:

Let $A,B\subseteq\mathbb R$ be non-empty and bounded from above. Prove that if $A+B = \{a+b : a\in A,\,b\in B\} =A$, it follows that $B = \{0\}$.

That seems wrong to me. For example, consider A = B = $\mathbb R_{\le0}$. $A+B$ is therefore $\mathbb R_{\le0}$ as well.

Am I missing something here?

You are right. The statement will be correct if you also require $B\subseteq [0,\infty)$ though. — Mark, Aug 25 '23 at 18:14
@Medo
It's called Infinitesimal Calculus by Michael Hochman et al. It was printed in Hebrew though, so I doubt you'll find it :) — Dima Bikov, Aug 25 '23 at 18:18
You are quite right. If $A=\mathbb R_{\lt c}$ and $\sup B=0$ then $A+B=A$. To get a correct statement, delete "from above". — bof, Aug 25 '23 at 18:33

score 1 · Answer 1 · answered Aug 25 '23 at 18:40

1

Your counterexample is correct. But if we suppose that both $A$ and $B$ are also lower-bounded (and so they lie in some bounded real intervals), then this claim holds. Or am I wrong?

answered Aug 25 '23 at 18:40

Juraj Hartman

119

Prove that if $A,B\subseteq\mathbb R$ are non-empty and bounded from above, and $A+B=A$, it follows that $B=\{0\}$. Seems Wrong?

1 Answers1