How to translate the following to predicate logic language?

Question

(1)- Define the notion: "a real sequence is bounded" in predicate logic language.

(2)- If $f$, $g$ are two bounded real sequences, then so is $f+g$.

The proof is trivial. But the predicate logic languages is not familiar with me. Thanks in advance.

Answer 1

How about defining the domain to be the set $S$ of all sequences, the predicate Rx: f is a real sequence in $S$

$R_f = \{f \in S: \text{ f is real-valued sequence}\}$, and defining the predicate being $Bx: x$ is a bounded sequence.

Then "a real sequence is bounded" would be $Rx \land Bx$.

1. If f, g are two bounded real sequences, then so is f + g.

$$\forall \,f \forall g\, \left((R_f \land R_g \land B_f \land B_g) \rightarrow (R_{f+g} \land B_{f+g})\right)$$

Answer 2

If $(a_n)_{n\in \mathbb N}$ is a real sequence then we can say $$ (a_n)_{n\in \mathbb N}\text{ is bounded}\ :\Leftrightarrow\ \exists M\in\mathbb R\colon\forall n\in\mathbb N\colon |a_n|<M.$$ Thius "The sum of two bounded sequences is bounded" translates as $$ \exists M\in \mathbb R\colon \forall n\in \mathbb N\colon |f(n)|<M\ \land\ \exists M\in \mathbb R\colon \forall n\in \mathbb N\colon |g(n)|<M\\\to \exists M\in \mathbb R\colon \forall n\in \mathbb N\colon |f(n)+g(n)|<M.$$