This question is related to this.
Introduce the asymptotic notation Big Oh in probability. Suppose that the real sequence $a_n\to 0$ as $n\to\infty$. Then $$\frac{O_p(a_n)}{O_p(1)}=O_p(a_n).$$
Why is this true? It looks quite intuitive, but hard to prove. The intuition is that both denominator and numerator are stochastic terms which are bounded in probability, and thus the ratio should also be bounded with high probability.
My attempt
Let $X_n=O_p(a_n), Y_n=O_p(1)$, that is, given $\epsilon>0$, for all sufficiently large $n$, there are $M_1,M_2>0$ such that $$P(\lvert X_n\rvert\geq M_1 \lvert a_n\rvert)\leq\epsilon\text{ and } P(\lvert Y_n\rvert\geq M_2)\leq\epsilon.$$ I need to show that $$P\bigg(\bigg\lvert \frac{X_n}{Y_n}\bigg\rvert\geq M \lvert a_n\rvert\bigg)\leq\epsilon$$ for some $M>0$ and all sufficiently large $n$. I tried to use $\{\lvert X_n\rvert\geq M \lvert a_n\rvert\lvert Y_n\rvert\}=\{\lvert X_n\rvert\geq M \lvert a_n\rvert\lvert Y_n\rvert,\lvert Y_n\rvert\geq 1\}\cup \{\lvert X_n\rvert\geq M \lvert a_n\rvert\lvert Y_n\rvert,\lvert Y_n\rvert< 1\}$ but it clearly doesn't work.
Any suggestions?
If you think it is easier, does $1/O_p(1)=O_p(1)$ holds? If yes, $O_p(a_n)/O_p(1)=O_p(a_n)O_p(1)=O_p(a_n).$