$P(\{ \lim_{n \to \infty} X_n \text{ exists}\})=0$ for i.i.d. sequence such that $P(\{ X_1=t\})<1$ $\forall t \in \mathbb{R}$ Suppose that $\{ X_n\}_{n\geq 1}$ is an i.i.d sequence such that $P(\{ X_1=t\})<1$ for all $t \in \mathbb{R}$. I want to show that 
$$P(\{ \lim_{n \to \infty} X_n \text{ exists}\})=0.$$
I have tried to prove this by contradiction. Suppose that $P(\{ \lim_{n \to \infty} X_n \text{ exists}\})>0$. Then there exists an $\omega$ for which $\lim_{n \to \infty} X_n(\omega)$ does not exist. Thus, the sequence $X_n(\omega)$ oscillates, so there exist $t_1, t_2 \in \mathbb{R}$ such that $X_n(\omega)\in \{t_1, t_2\}$ infinitely often. Then
$$P(\{\omega: X_n(\omega) \in \{t_1, t_2\} \, \text{ i.o.} \}) $$
$$  =P(\{ \omega: \exists \text{ subsequences } X_{n_j}(\omega)=t_1 \text{ and } X_{n_i}(\omega)=t_2   \, \text{ i.o.}\})  $$
$$ =P(\{\omega: X_{n_j}(\omega)=t_1  \, \text{ i.o.}  \}) P(\{\omega: X_{n_i}(\omega)=t_2  \, \text{ i.o.}  \})<1, $$
where the last equality follows from the fact that $X_1, X_2, \ldots$ are independent.
I am stuck here and I am not sure if what I have done is even correct. Any help is really appreciated. Thank you.
 A: hint
You can't say that there are $t_1,t_2$ such that $X_i \in \{t_1,t_2\}$ infinitely often. You don't even know if there are atoms in the distribution... it may be the case that the probability $X_i$ ever takes any particular value is zero. 
But if you can show that there are $t_1<t_2$ such that almost surely, $X_i\ge t_2$ infinitely often and $X_i \le t_1$ infinitely often, you'll have it.
In fact,  it suffices to simply show that $P(X_i\ge t_2)$ and $P(X_i \le t_1)$ are greater than zero, but of course you must prove this (naturally, this involves Borel-Cantelli).
A: From the condition of $P(X = t) < 1$ for all $t \in \mathbb{R}$, we have
$$ P(X \ne t ) > 0 \qquad \forall t \in \mathbb{R} $$
Thus $X$ is not a deterministic, (constant). 
Hence, there exists $t_0 \in \mathbb{R}$ and $\epsilon > 0$ such that 
$$ P(X > t_0 + \epsilon) > 0, \quad \text{and} \quad P(X < t_0 - \epsilon) > 0$$
This is because if $X$ were a constant, for any $t \in \mathbb{R}$, and for any $\epsilon > 0$
$$P(X > t + \epsilon) = 0 \quad \text{or} \quad P(X < t -\epsilon) = 0.$$
Then let $A_{n} = \{ X_n > t_0 + \epsilon \}$
and $B_{n} = \{X_n < t_0- \epsilon \}$.
Since $X_i$'s are iid, we have
$$ \sum_{n=1}^\infty P(A_{n}) = \infty = \sum_{n=1}^\infty P(B_{n})$$
which implies (from Borel-Cantelli, zero-one rule) that
$$ P(A_{n}, i.o.) = 1 = P(B_{n}, i.o.)$$
That is, $P(|X_n - t_0| > \epsilon, i.o) = 1$
which tells us that 
$X_n$ does not converge a.s.
