Why converge in probability introduces epsilon in its definition? According to wiki:
a.s. converge: $$\Pr(\lim_{n\rightarrow \infty}X_n=X)=1$$
converge in probability: $$\lim_{n\rightarrow \infty}\Pr(|X_n-X|>\epsilon)=0$$
What about $\lim_{n\rightarrow \infty}\Pr(X_n=X)=0$? (Compared to a.s. converge, I think it is more intuitive to have such a definition.)
Why converge in probability introduce the $\epsilon$ in its definition?
 A: After some research and rethinking, I believe I figure it out.
Translating the limit language to plain words helps undertanding.
a.s. converge: for any $\epsilon > 0$, there exists large enough N so that for all n>N, $Pr(|X_n-X|>\epsilon)=0$.
Converge in distribution: for any $\epsilon > 0, \delta > 0$, there exists large enough N so that for all n>N, $Pr(|X_n-X|>\epsilon)<\delta$.
The difference is for any $\epsilon > 0$, a.s. converge gaurantees there exists big enough N so that it is almost impossible that $|X_n-X|>\epsilon$, however, converge in distribution leave open possbility of $|X_n-X|>\epsilon$.
So it is farely natural to have a $\epsilon$ in the brackets, just as in a.s. converge. Converge in probability is  a weaker version of a.s. converge.
As @Kavi Rama Murthy comments, to drop the $\epsilon$ excludes such cases where the sequence of r.v. converges to X apparently.
What confused me is just the notation of limit: there actually a $\epsilon$ in the definition of a.s. converge.
