If a sequence of random variable is bounded in probability(Op(1)), then will its supremum be finite with probability 1(P (supn|Xn|I am practicing for my large sample theory course, and stumbled upon this problem and got stuck. New to the subject so kind of stumped how to proceed.
 A: No, boundedness in probability is not a sufficient assumption. Here is a counterexample :
Let $(X_n)_{n\in\mathbb N}$ be a sequence of i.i.d. standard normal random variables. It is clear that $(X_n)$ is $O_p(1)$ (it follows from, e.g., Chebyshev's inequality). We also have that, for any $M>0$, $\mathbb P(|X_n|\ge M)>0$, hence, for all $M>0$
$$\sum_{n\in\mathbb N}\mathbb P(|X_n|\ge M) = \infty $$
Therefore, by the second Borel-Cantelli lemma, it follows that, for any $M>0$
$$\mathbb P\left(\bigcap _{n\geq 1}\bigcup _{j\geq n}\{|X_j|\ge M\}\right) = \mathbb P\left(|X_n|\ge M \text{ infinitely often} \right)  = 1$$
Since this is true for all $M>0$, consider a sequence $(M_k)$ that diverges to infinity. By countable intersection of sets of probability $1$, we have that
$$\mathbb P\left(\bigcap_{k\ge1}\{|X_n|\ge M_k \text{ infinitely often}\} \right) = 1$$
And it is clear that the event $ \bigcap_{k\ge1}\{|X_n|\ge M_k \text{ infinitely often}\}$ is included in $\{\sup|X_n|=\infty\}$. From all that it follows that
$$\mathbb P(\sup|X_n|<\infty) = 1 -\mathbb P(\sup|X_n|=\infty) = 0\ne1 $$
