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$$\text{COMPARATIVE EXAMPLE}$$

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So I've been told that in order to show that
$$\sup{(A+B)}=\sup{A}+\sup{B}$$
for non-empty and bounded above sets $A,B\subseteq\mathbb{R}$ one must show that $\sup{A} +\sup{B}$ is an upper bound of $A+B$, and that anything else isn't, so if $\sup{A}+\sup{B}$ is an upper bound, then for $\alpha\in A$, and $\beta\in B$, then $\alpha \leq \sup{A}$ and $\beta\leq\sup{B}$. Thus, we have that
$$\alpha+\beta\leq \sup{A}+\sup{B},$$
so $\sup{A}+\sup{B}$ is an upper bound. Now, if we let an $\epsilon>0$ be given, then
$$\sup{A}+\sup{B} - \epsilon \le\sup{A}+\sup{B}.$$
From here we must show that this is not an upper bound of $A+B$. Now notice that
$$\sup{A}+\sup{B}-\epsilon=\sup{A}-\frac{\epsilon}{2}+\sup{B}-\frac{\epsilon}{2},$$
so
$$\sup{A}-\frac{\epsilon}{2}<\sup{A},$$
so there exists an $\alpha\in A$ such that
$$\alpha>\sup{A}-\frac{\epsilon}{2},$$
and similarly
$$\sup{B}-\frac{\epsilon}{2}<\sup{B},$$
so there exists a $\beta\in B$ such that
$$\beta>\sup{B}-\frac{\epsilon}{2},$$
as desired.

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$$\text{QUESTION}$$

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How can I use this template to show that $\inf{(A+B)}=\inf{A}+\inf{B}$?