One rule for small op and big Op is
$$ (1+o_p(1))^{-1} = O_p(1) $$
(on page 13 of Vaart, A. W. van der. (1998). Asymptotic statistics. Cambridge University Press.)
I am curious whether it is true to hold
$$ (1+o_p(1))^{-1} \overset{?}{=} 1+o_p(1) $$
My thought is that suppose $X_n\rightarrow_p 0$, then $X_n=o_p(1)$. Consider the function $g(x)=1/(1+x)$, then by the continuous mapping theorem (https://en.wikipedia.org/wiki/Continuous_mapping_theorem), we have
$$ g(X_n)\rightarrow_p g(0) = 1\,. $$
Is that correct, or am I missing something?
If so, why not use the tight result $(1+o_p(1))^{-1} = 1+o_p(1)$?
If not, is there any counterexample that $(1+o_p(1))^{-1} = O_p(1)$ holds but not $(1+o_p(1))^{-1} = 1+o_p(1)$?
PS: I have checked the proof of the rule in another question, Op and op Convergence Property Related Question, and it seems that the answer also applies the continuous mapping theorem. No explanation for the difference between $O_p(1)$ and $1+o_p(1)$.
