$\inf(A+B) \geq \inf(A)+\inf(B)$, proving inequality holds?

Question

We have.... $x \in A + B ⇒ x = a + b, a \in A, b \in B$ We know that $a \geq \inf(A)$ and $b \geq \inf(B)$. Hence, $x \geq \inf(A)+\inf(B) ⇒ \inf(A+B) \geq \inf(A)+\inf(B)$

I made the proof this far and now I am trying to find an examples of where the strict inequality holds.

So If I set $A = \{0,1\}$ and $B = \{1,0\}$ ..Then $A+B = \{1,1\}$ Which doesn't work.

How would you prove the inequality holds?

Could you do set $A = \{0,2\}$ and set $B = \{2,0\}...A+B = \{2,2\}$

$\inf(A+B) = 2 > \inf(A) + \inf(B)$?

But im not sure this is the way to go about it?

Answer 1

For sets $A,B\subset\mathbb R$ we actually always have $\inf(A+B) = \inf A+\inf B$.

It is other circumstances where we have an inequality, for example when we are looking at functions $f,g:X\to\mathbb R$, then we have $$\inf_{x\in X}(f(x)+g(x))\geq \inf_{x\in X}f(x)+\inf_{x\in X}g(x).$$

You can find a strict inequality here with $f:\mathbb R\to\mathbb R, x\mapsto \operatorname{sgn}(x)$ and $g := -f$ for example.

Answer 2

The strict inequality never holds, in fact, we have $\inf(A+B)=\inf A+\inf B.$ Let's prove this as follows: Let $r>0$ be given. Then there are $a\in A$ and $b\in B$ such that $a<\inf A+r$ and $b<\inf B+r$ so $a+b<\inf A+\inf B+2r$ but $\inf(A+B)\leqslant a+b$ so $\inf(A+B)<\inf A+\inf B+2r$ and since $r$ is arbitrary then $\inf(A+B)\leqslant\inf A+\inf B$