Demonstration Written in Formal Language

Let's show that $B\rightarrow A$. Infact take $f$ of type $B$ and take $x\in\mathbb{R}$, you know that it exists $x_0\in\mathbb{R}$ such that $$\forall y\in \mathbb{R}(y\geq x_0\rightarrow|f(y)|\geq 1).$$ So take $y=\max{(x,x_0)}$, you have that $$y\geq x\wedge |f(y)|\geq 1.$$ The converse is not true, infact you know that intervals $[n,n+1)$ are a partition of $\mathbb{R}$ with $n\in\mathbb{Z}$, so for all $x$ it exists an unique $n\in\mathbb{Z}$ such that $x\in[n,n+1)$, so put $$f(x) = \begin{cases} 1 & \text{if } 3 \mid n,\\[2mm] 0 & \text{otherwise}, \end{cases} .$$ Clearly $f$ is of type $A$ since if you take $x\in\mathbb{R}$ you can always chose an interval of the form $[3n,3n+1)$ such that $3n>x$ and pick $y\in[3n,3n+1)$, so $$|f(y)|=1\geq 1.$$ Obviously $f$ is not of type $B$, since for every $x\in\mathbb{R}$ it will always exists an $y\in\mathbb{R}$ such that $y\geq x$ and $f(y)=0$.

Edit: I'm assuming that $\textbf{R}$ is $\mathbb{R}$