asymptotic order in probability

Question

Consider the definition of small o in probability.

Let $X_n$ be a sequence of random variables and let $X_n=1+o_p(1)$. That is, for any $\delta,\epsilon>0: P(\lvert X_n-1\rvert\geq \delta)\leq \epsilon$ for sufficiently large $n$. The sequence converges in probability to one.

This led me to conjecture that there should be a constant $1>M>0$ so that $P(\lvert X_n\rvert\geq M)\to1$ as $n\to+\infty$. Or equivalently, $P(\lvert X_n\rvert < M)\to 0$ .

Question. If it is true, then how to prove it?

Answer 1

For any $\epsilon>0$, $$ \mathsf{P}(|X_n-1|\le \epsilon)=\mathsf{P}(1-\epsilon\le X_n\le 1+\epsilon)\to 1 $$ as $n\to\infty$, which implies that for any $M<1$, $$ \mathsf{P}(|X_n| \ge M)\ge \mathsf{P}(X_n \ge M)\to 1. $$