Does $\implies wf(y)<\epsilon$ , mean $|f(y)-f(x)|<\epsilon?$

Question

Right here, the poster claimed that for $\epsilon>0\;\text{and}\;x\in I,$ $$|y-x|<\delta\implies wf(y)<\epsilon$$ which implies that the $f$ is continuous at $x.$

Question: Does $\implies wf(y)<\epsilon$ , mean $|f(y)-f(x)|<\epsilon?$

Answer 1

The poster's proof is not even correct, because their set $I$ depends on $\epsilon$. It appears they are using $\omega f(X)$ to mean $\sup_{x\in X}f(x)-\inf_{x \in X}f(x)$. In this case, it would not be correct to fill in $|f(x)-f(y)|<\epsilon$, because then the backward implication would be false: this would not guarantee that for any $y,y' \in (x-\delta,x+\delta) \cap [a,b]$, we have $|f(y)-f(y')|<\epsilon$ (the best you could put there is $2\epsilon$). It would also not be correct to put $|f(x)-f(y)|<\frac{\epsilon}{2}$ (which would fix the backward implication), because then the forward implication would be false.