How to prove $P(|X_n-X_m|>\epsilon)\leq P(|X_n-X|>\epsilon/2)+P(|X_m-X|>\epsilon/2)$? Consider random variables $X, X_1, X_2, ...$ in a probability space $(\Omega, \mathcal F, P)$ such that
$X_n\stackrel{p}{\rightarrow} X$.
Let $n, m \in \mathbb N$. How can I prove that for some $\epsilon > 0$,

$P(|X_n-X_m|>\epsilon)\leq P(|X_n-X|>\epsilon/2)+P(|X_m-X|>\epsilon/2)$?

I was thinking something along the lines of using the triangle inequality:
$|X_n-X-(X_m-X)|>\epsilon \Rightarrow |X_n-X|+|X_m-X|>\epsilon$
Any help would be appreciated
 A: You're on the right track. Now argue that if $|X_n-X|+|X_m-X|>\epsilon$, then $|X_n-X|>\epsilon/2$, or $|X_m-X|>\epsilon/2$. (Hint: Suppose not.) The desired result then follows from $P(A\cup B)\le P(A)+ P(B)$.
A: Let us denote events:
$A_{n,m} = \{|X_n - X_m| > \epsilon\} = \{|X_n - X + X - X_m| > \epsilon\}$
$A_{n} = \{|X_n - X| > \epsilon/2\}$
$A_{m} = \{|X_m - X| > \epsilon/2\}$
Then we need to show that $P(A_{n,m}) \le P(A_n) + P(A_m)$.
Proof:
Now that's that out of the way.
$P(A_{n,m}) = P(|X_n - X_m| > \epsilon) = P(|X_n - X + X - X_m| > \epsilon)$
$\le P(|X_n - X| + |X - X_m| > \epsilon)$
since $A_{n,m} \subseteq \{|X_n - X| + |X - X_m| > \epsilon\}$
$\le P(|X_n - X| > \epsilon/2 \cup |X - X_m| > \epsilon/2)$
since $\{|X_n - X| + |X - X_m| > \epsilon\} \subseteq \{|X_n - X| > \epsilon/2 \cup |X - X_m| > \epsilon/2\}$
$\le P(|X_n - X| > \epsilon/2) + P(|X - X_m| > \epsilon/2)$
$= P(A_n) + P(|X - X_m| > \epsilon/2)$
$= P(A_n) + P(|X_m - X| > \epsilon/2)$
$= P(A_n) + P(A_m)$
And we are done!
