If $X_n$ is $O_P(1)$ and $Y_n$ is $O_P(1)$, is $X_n Y_n$ also $O_P(1)$?
In this note we have that if, say, $Y_n$ were $o_P(1)$ then $X_n Y_n$ is $o_P(1)$. I have seen this many times but I have not seen an equivalent statement using big $O$ (suggesting only bounded in probability rather than convergent in probability, as little $o$ says). I feel like the statement should hold for big $O$ too but I can't get a proof, I've found none, and the fact that this is presented using little $o$ (and my struggle to get a proof) suggests this might not be true, yet I can't find a counterexample.
Let $\epsilon>0$.
Then $M$ and $N$ exist with $P(|X_n|>M)\leq\frac12\epsilon$ and $P(|Y_n|>N)\leq\frac12\epsilon$ for every $n$, since $X_n$ and $Y_n$ are both $O_P(1)$
Now observe that for every $n$: $$|X_nY_n|>MN\implies |X_n|>M\vee|Y_n|>N$$ imlying that: $$\{|X_nY_n|>MN\}\subseteq\{|X_n|>M\}\cup\{|Y_n|>N\}$$ and consequently: $$P(|X_nY_n|>MN)\leq P(|X_n|>M)+P(|Y_n|>N)\leq\epsilon$$
This can be done for every $\epsilon>0$ so proved is now that $X_nY_n$ is $O_P(1)$.