Confirmation about one simple inequality

Question

Suppose A & B are sets of reals.Then is it true that inf{A+B}$\leq$inf{A}+inf{B}.I consider all the cases like A contains all positive and B contains all positive,then alternate positive and negative combinations, and it seems true to me.Is it true?

Answer 1

Actually, we have $$\inf(A+B)=\inf A+\inf B \qquad\text{(and likewise for the supremum)}.$$

Indeed, let $a\in A,\; b\in B$. By definition ofthe infimum, we have $\inf A\le a$, $\;b\le\in B$, whence $\;\inf A+\inf B \le a+b$,and be definition of the infimum, we deduce instantly $$\inf A+\inf B\le\inf(A+B) .$$ Conversely, consider $a\in A, b\in B$; we have $\inf(A+B)\le a+b$. I'll denote $m=\inf(A+B)$.

Thus $m-a\le b$, and by definition of the infimum, $m-a\le \inf B$, so $\;m-\inf B \le a$, and, again by definition $$m-\inf B\le \inf A,\quad\text{whence }m=\inf(A+B)\le \inf A+\inf B.$$