Prove that $\inf_{n\in\mathbb{N}}\sup_{k\geq n}(a_k+b_k)\leq \inf_{n\in\mathbb{N}}\sup_{k\geq n}a_k+\inf_{n\in\mathbb{N}}\sup_{k\geq n}b_k$ Let $(a_n)_{n\in\mathbb{N}}$ and $(b_n)_{n\in\mathbb{N}}$ be two sequences of $\color{red}{\overline{\mathbb{R}}:=\mathbb{R}\cup\{-\infty,\infty\}}$. Suppose that the sum $a_n+b_n$ exists for all $n\in\mathbb{N}$. Prove that if the sum $\inf_{n\in\mathbb{N}}\sup_{k\geq n}a_k+\inf_{n\in\mathbb{N}}\sup_{k\geq n}b_k$ exists, then
$$\inf_{n\in\mathbb{N}}\sup_{k\geq n}(a_k+b_k)\leq \inf_{n\in\mathbb{N}}\sup_{k\geq n}a_k+\inf_{n\in\mathbb{N}}\sup_{k\geq n}b_k$$
I want to prove the above inequality without using the notion of limits! Remember that the following operations ARE NOT defined: $(\infty)-(\infty)$, $(-\infty)+(\infty )$, $(\infty )+(-\infty )$ and $(-\infty)-(-\infty )$.
The order relation $\leq $ is defined in the following way: given any $x,y\in\overline{\mathbb{R}}$ we say that $x\leq y$ if, and only if, at least one of the following propositions is true:

*

*If $x,y\in\mathbb{R}$, then $x\leq y$ (here $\leq$ is the usual order of the real numbers)

*$y=\infty$

*$x=-\infty$
I tried to use the inequalities below, but I couldn't solve that problem.

*

*If the sum $\sup_{k\geq n}a_k+\sup_{k\geq n}b_k$ exists, then $\sup_{k\geq n}(a_k+b_k)\leq \sup_{k\geq n}a_k+\sup_{k\geq n}b_k$ for all $n\in\mathbb{N}$

*If the sum $\inf_{n\in\mathbb{N}}a_n+\inf_{n\in\mathbb{N}}b_n$ exists, then $\inf_{n\in\mathbb{N}}a_n+\inf_{n\in\mathbb{N}}b_n\leq \inf_{n\in\mathbb{N}}(a_n+b_n)$
Thank you for your attention!
 A: Let us denote
$$
\newcommand{\newsup}{\mathop{\smash{\mathrm{sup}}}}
\begin{align}
 A &= \inf_{n\in\mathbb{N}}\newsup_{k\geq n}a_k \, ,\\
 B &= \inf_{n\in\mathbb{N}}\newsup_{k\geq n}b_k \, , \\
 C &= \inf_{n\in\mathbb{N}}\newsup_{k\geq n}(a_k+b_k) \, .
\end{align}
$$
If $A = +\infty$ or $B = +\infty$ then $A+B=+\infty$ (since the sum is assumed to be defined), so that $C \le A+B$ holds.
Now consider the case that $A, B \in \Bbb R \cup \{ - \infty \}$. The idea is to show that
$$ \tag{*}
 \forall x, y \in \Bbb R: \left(x > A \text{ and } y > B \implies C \le x + y \right)\, .
$$
It is not difficult to see that $(*)$ implies $C \le A+B$:

*

*If $A = -\infty$  then $C < x + y$ for fixed $y \in \Bbb R$ and all $x \in \Bbb R$, so that necessarily $C = -\infty$.

*Similarly if $B= -\infty$.

*Otherwise both $A$ and $B$ are real numbers and $(*)$ implies $C \le (A+\epsilon)+( B+\epsilon)$ with arbitrarily small $\epsilon > 0$.

It remains to prove $(*)$: If $x > A$ and $y > B$ then, using the definition of the infimum,
$$
\exists n_1 \in \Bbb N: \newsup_{k\geq n_1}a_k < x \, , \\
\exists n_2 \in \Bbb N: \newsup_{k\geq n_2}b_k < y \, .
$$
It follows that
$$ 
a_k + b_k \le x+y
$$
for all $k \ge \max(n_1, n_2)$ and therefore
$$
 C \le \sup_{k\geq \max(n_1, n_2)}(a_k+b_k) \le x +y \, .
$$
