If we have a function $f:\mathbb{R}\to\mathbb{R}$ where $\forall x\in \mathbb{R}$ $\exists y\in \mathbb{R}$ such that $y\geq x$ and $|f(y)|\geq 1$ We call it type A.
A function $g:\mathbb{R}\to \mathbb{R}$ is defined where $\forall y \in \mathbb{R}$ $\exists x\in \mathbb{R}$ s.t $y\geq x$ and $|g(y)|\geq 1$ is of type B.
(user2661923 : start of question)
This is a point of confusion - I ask the OP to edit his question to resolve the confusion and (at the same time), remove this section of his query.
For type-B functions, does the OP intend that $\forall y, |g(y)| \geq 1$?
I am unable to find an interpretation of the OP's description of type-B functions that does not have the above constraint as a consequence.
(user2661923 : end of question)
Now my question is how do I prove this using contradiction/direct proof that if a function is type A then its type B and the converse (type B then type A).
I've done it with a counterexample, but I want to prove it without using a counterexample if thats possible.
$$f(x)=\begin{cases}0 \quad\text{if} \quad x\in\mathbb{N}\\2\quad \text{otherwise}\end{cases}$$
edits:
Type $B$ is where $\exists x$ real, such that for all $y$ real $y\geq x$ such that $|g(y)|\geq 1$
type $A$ is where $\forall x$ real, $\exists y$ real, with $y\geq x$ such that $|f(y)|\geq 1$
If a function of type $A$ then it s type $B$ and if a function is type $B$ then it is type $A$