# Bounded in Probability and smaller order in probability

I wanted to prove that if $$X_n$$ is bounded in probability and $$Y_n = o_p(X_n)$$, then $$Y_n \rightarrow 0$$ in probability

I know the following definitions that is $$X_n$$ is bounded in probability meaning that $$P(|X_n|1-\epsilon$$

and I know that $$Y_n = o_p(X_n)$$ implies that $$Y_n/X_n \rightarrow0$$ in probability

$$|Y_n| >\epsilon$$ implies either $$|\frac {Y_n} {X_n}| >\frac {\epsilon} M$$ or $$|X_n| \geq M$$. [You can prove this by contradiction]. Hence $$P(|Y_n| >\epsilon) \leq P(|\frac {Y_n} {X_n}| >\frac {\epsilon} M)+P(|X_n| \geq M)$$. Can you finish the proof?
Some details: Let $$\eta_1$$ and $$\eta_2 >0$$. Choose $$\epsilon >0$$ such that $$\epsilon <\eta_1$$ and $$\epsilon <\eta_2 /2$$ . Note that $$|Y_n| >\eta_1$$ implies that $$|Y_n|>\epsilon$$. Now choose $$n_0$$ such that $$P(|\frac {Y_n} {X_n}| >\frac {\epsilon} M) <\eta_2 /2$$ for $$n \geq n_0$$. Now put these together to conclude that $$P(|Y_n| >\eta_1) <\eta_2$$ whenever $$n \geq n_0$$. This proves that $$Y_n \to 0$$ in probability.