Convergence in probability- sign inequality

Question

We know, that $\lim_{ n\to\infty } P\left\{ \left| X_n-X\right| \ge \epsilon \right\}=0$ for each $\epsilon >0$. When I can write $\lim_{ n\to\infty } P\left\{ \left| X_n-X\right| > \epsilon \right\}=0$ and why? (another inequality sign)

Answer 1

Since $\epsilon>0$ is arbitrary, the equivalence of these assertions follows from the line of inclusions

$$\{|X_n-X|> \epsilon \} \subset \{|X_n-X|\ge \epsilon \}\subset \{|X_n-X|> \epsilon/2 \}$$

and the monotonicity of probability measures.