Convergence in probability realated question Consider $X_n$ and $Y_n$ be two real-valued random sequences, if
$$P(X_n \neq Y_n) \rightarrow 0 \text{ as $n \rightarrow \infty$}$$
is it equivalent to say that
$X_n$ converges to $Y_n$ in probability?
Since the definition of convergence in probability says

for every $\varepsilon>0$, we have
$$\lim_{n \rightarrow \infty} P(|X_n - Y_n| \ge \varepsilon) = 0$$

hence, if we observe the event:
$$\{|X_n - Y_n| \ge \varepsilon\} = \{X_n - Y_n \ge \varepsilon \text{ or } X_n - Y_n \le -\varepsilon\}$$
This seems to me I can write
$\{X_n \neq Y_n\} = \{|X_n - Y_n| \ge \varepsilon\}$ for all $\varepsilon>0$
It seems to me I can say yes, but I'm not quite sure.

Consider the probability space $([0,1],\mathcal{B}([0,1]),ℙ=\text{Lebesgue})$ with random variables 
$$X_n(ω) := 1, \quad Y_n(ω) := 1-\frac{1}{n}$$
Then for all $n$, $ℙ(X_n ≠ Y_n) = ℙ([0,1]) = 1$, but for any $ε>0$, when $n>\frac{1}{\epsilon}$ we have $|X_n - Y_n| = \frac{1}{n} < ε$ (almost) surely so $ℙ(|X_n - Y_n| > ε) → 0$ as $n→ ∞$. Hence the conditions are not equivalent(although one implication does hold).
The reason this fails is that the sets are related as follows:
$$ [X_n ≠ Y_n] = \bigcup_{ε > 0} [ |X_n - Y_n| > ε ] $$
as opposed to an intersection(='for all') as you said.
