If $X_n=O_P(1)$ and $Y_n=o_P(1)$, prove $X_nY_n=o_P(1)$

Question

We say $X_n=O_P(1)$ if $X_n$ is bounded in probability. We say $Y_n=o_P(1)$ if $Y_n$ converges in probability to $0$.
My attempt:
Since $X_n=O_P(1)$ and $Y_n=o_P(1)$, we have $$ \forall\ \epsilon>0,\ \exists\ M \text{ and }\ n_0 \text{ such that } n\geq n_0\ \text{implies}\ P(|X_n|\leq M)\geq 1-\epsilon \text{ and} \lim_{n\to \infty}P\left(|Y_n|>\frac{\epsilon}{M}\right)=0. $$ By the definition of limits, $$ \forall\ \delta>0,\ \exists\ N_0\text{ such that } n\geq N_0\text{ implies } P\left(|Y_n|\leq \frac{\epsilon}{M} \right)\geq 1-\delta. $$ If we take $N=\max{(n_0,N_0)}$, then $n\geq N$ implies $$ P(|X_n|\leq M)\geq 1-\epsilon \text{ and } P\left(|Y_n|\leq \frac{\epsilon}{M } \right)\geq 1-\delta $$ Note that $$ \begin{aligned} P(|X_nY_n|\leq \epsilon) &=P(|X_n|\leq M,\ |X_nY_n|\leq \epsilon) +P(|X_n|>M,\ |X_nY_n|\leq \epsilon)\\ &\geq P(|X_n|\leq M,\ |X_nY_n|\leq \epsilon)\\ &\geq P\left(|X_n|\leq M,\ |Y_n|\leq \frac{\epsilon}{M} \right)\\ &\geq P(|X_n|\leq M)+P\left(|Y_n|\leq \frac{\epsilon}{M} \right)-1\\ &\geq 1-\delta-\epsilon,\quad \text{for all }n\geq N \text{ and any } \epsilon,\ \delta>0. \end{aligned} $$

I'm confused how we can conclude that $\lim_{n\to\infty}P(|X_nY_n|\leq \epsilon)=1$ and then get $X_nY_n \stackrel{P}{\rightarrow}0$.

Answer 1

Perhaps change the step

" $\forall\ \epsilon>0,\ \exists\ M \text{ and }\ n_0 \text{ such that } n\geq n_0\ \text{implies}\ P(|X_n|\leq M)\geq 1-\epsilon \text{ and} \lim_{n\to \infty}P\left(|Y_n|>\frac{\epsilon}{M}\right)=0.$ "

to

$\forall\delta >0, \exists\ M \text{ and }\ n_0 \text{ such that } n\geq n_0\ \text{implies}\ P(|X_n|\leq M)\geq 1-\delta$ and $\forall \epsilon>0$, we have $\text{ and} \lim_{n\to \infty}P\left(|Y_n|>\frac{\epsilon}{M}\right)=0.$

Following your steps, you will get that for all $\epsilon>0$, and $\delta>0$, there holds that $P(|X_nY_n|\leq \epsilon) \geq 1-2\delta,\quad \text{for all }n\geq N$. This looks like to be the definition of $X_nY_n = o_p(1)$.