Prove $${X_n}\buildrel P \over \longrightarrow {\delta _C} \Leftrightarrow {X_n}\buildrel D \over \longrightarrow {\delta _C}$$ where the distribution ${\delta _C}$ is characterized by $P\left( {{\delta _C} = C} \right) = 1$.
Hint: $\left| {a - b} \right| > d \Leftrightarrow a - b > d$ or $b - a < d$.
I know that $({X_1},{X_2}...,{X_n})$ converges in probability to $X$ if $$\forall \varepsilon \mathop {\lim }\limits_{n \to \infty } P\left( {\left| {{X_n} - X} \right| > \varepsilon } \right) = 0$$ Also, $({X_1},{X_2}...,{X_n})$ converges in distribution to $X$ if there is a ${F_n}\left( x \right)$ such that $$\mathop {\lim }\limits_{n \to \infty } {F_n}\left( x \right) = F\left( x \right)$$ for every point at which $F$ is continuous.
But I confess that making progress past the definitions is hard. I tried following the hint and wrote $$\mathop {\lim }\limits_{n \to \infty } P\left( {\left| {{X_n} - {\delta _C}} \right| > \varepsilon } \right) = 0$$ wich opens up to both $\mathop {\lim }\limits_{n \to \infty } P\left( {{X_n} - {\delta _C} > \varepsilon } \right) = 0$ and $\mathop {\lim }\limits_{n \to \infty } P\left( {{\delta _C} - {X_n} < \varepsilon } \right) = 0$. I thought about leaving ${{X_n}}$ alone on both of these equations. I think that if I can manage to organize them in such a way that this limit $\mathop {\lim }\limits_{n \to \infty } {F_n}\left( x \right) = {\delta _C}$ is respected, the question would be done, but I'm stuck here.
Also, on a side note, I'm not really sure how knowing $P\left( {{\delta _C} = C} \right) = 1$ helps me. What does this mean? Is this a one-valued density?