Suppose that $A$ and $B$ are non-empty subsets of $\textbf{R}$ which are bounded below.
Let $A+B = \{x\in\textbf{R}: x = a + b\,\,\text{for some}\,\,a\in A,b\in B\}$. Show that $A + B$ is bounded below and
$$\inf(A+B) = \inf(A) + \inf(B)$$
What about products?
MY (NEW) ATTEMPT
If $a\in A$ and $b\in B$, then $a\geq\inf(A)$ and $b\geq\inf(B)$. Consequently, $a + b\geq\inf(A) + \inf(B)$, which means that the set $A + B$ is bounded below. Since it is not empty, it admits an infimum. We shall prove that $\inf(A + B) = \inf(A) + \inf(B)$.
Given that $\inf(A) + \inf(B)$ is an lower bound to $A+B$, it results that $\inf(A) + \inf(B) \leq \inf(A + B) $.
Let us assume by contradiction that $\inf(A + B) > \inf(A) + \inf(B)$.
Then I get stuck. Could someone provide a solution based on contradiction?
On the other hand, if we consider the set $AB = \{x\in\textbf{R}:x = ab\,\,\text{for some}\,\,a\in A,b\in B\}$, then $\inf(AB)\neq\inf(A)\inf(B)$ in general.
Indeed, consider $A = [0,1]$ and $B = [-1,0]$. Then $\inf(AB) = -1$, $\inf(A) = 0$ and $\inf(B) = -1$. Thus $\inf(AB)\neq\inf(A)\inf(B)$.
Could someone please verify if the wording of my proof is good enough?
This question has already been asked and here, but I'd like to know if there is another approach to the first part.