In the textbook during the steps to expand the convergance of probability the following is provided.
the condition for converges in probability is given as $\lim_{n\rightarrow \infty}\mathbb{P}(|X_n-X|>\epsilon)=0$
$\mathbb{P}(|X_n-c|>\epsilon)=\mathbb{P}(X_n>c+\epsilon)+\mathbb{P}(X_n<c-\epsilon)$
I dont understand why the inequality is not expanded as $X_n>\epsilon - c$ and $X_n<c-\epsilon$. It seems to be something so trivial that I can't find anywhere (or perhaps the correct query word).
How does it work?
The condition $|X_n-c|\le \epsilon $ means $X_n$ is within distance $\epsilon$ of $c$. In other words $X_n \in [c-\epsilon, c+ \epsilon]$.
Hence the converse $|X_n-c| > \epsilon $ means $X_n \notin [c-\epsilon, c+ \epsilon]$. In other words either $X_n > c+ \epsilon$ or $X_n < c- \epsilon$.