TITLE: A sequence of nonconstant i.i.d. random variables converges with probability zero
QUESTION [5 upvotes]: Proove:  $X_{n} iid, X_{n}$ not constant a.s. $\iff P(X_{n}$ $converges)=0$
My idea for "$\Rightarrow$": 
$X_{n}$ not constant a.s. $\iff \forall$ c $\in \mathbb{R}$, $\varepsilon$ > 0:  $P(|X_{n}-c|>\varepsilon)>0$
--> $X_{n}$ doesn't converge in Probability, thus $X_{n}$ doesn't converge. --> $P(X_{n}$ $ converges)=0$
$"\Leftarrow"$:
$P(X_{n}$ $converges)=0 \iff \forall X \in \mathbb{R}$ exists a $\varepsilon>0: |X_{n}-X|>\varepsilon$
--> $P(X_{n}\not=c)=1, c\in \mathbb{R}$, so $X_{n}$ is not constant a.s.
Are there any mistakes or wrong assumptions in this proof? Or can I take this? I got a hint that I can use Borel-Cantelli but I don't where to use it.
Thanks for any kind of help.

REPLY [4 votes]: First, you wrote that if $X_n$ is not constant a.s., then $\forall c\in\mathbb{R},\epsilon>0; P(|X_n-c| > \epsilon) > 0$. This is not true. Change the $\forall$ to an $\exists$ and it will be true.
Here's one approach:
$\implies)$ So $X_n$ is not constant a.s. That means there exist $c\in\mathbb{R}$ and $\epsilon>0$ such that $P(X_n>c+\epsilon)>0$ and $P(X_n<c-\epsilon)>0$. From the second Borel-Cantelli lemma, this means that with probability 1, $X_i>c+\epsilon$ for infinitely many $i$ and with probability 1, $X_i<c-\epsilon$ for infinitely many $i$. Hence, with probability 1, $X_n$ doesn't converge.
$\impliedby$) Prove the contrapositive. Suppose $X_n$ is constant a.s., so with probability 1, $X_n=c$ for some real $c$. Then $P(X_n\ne c)=0$ and by countable additivity,
$$
P(X_j \ne c\text{ for some $j$}) \le \sum_j P(X_j\ne c) = 0.
$$
Thus, $P(X_n\ne c\text{ for some $j$}) = 0$, which means of course $P(X_n=c\text{ for all $n$})=1$. In particular, $P(X_n\to c)=1$.