Please verify if my solution to this exercise has any mistake (Specially in the second part):
Let $(X_n)$ be a sequence of iid random variables with distribuction uniform $U[0,1]$. Show that $X_{(n)} \to 1$ in probability.
Let $0<\epsilon\leq1$. Then $$P(|X_{(n)} - 1|> \epsilon) = P(|1 - X_{(n)}| > \epsilon) = P(1 - X_{(n)} > \epsilon) = P(X_{(n)} < 1 - \epsilon) = F_{X_{(n)}}(1-\epsilon) = {(F_{X_n}(1-\epsilon))}^n = (1-\epsilon)^n \to 0$$
If $\epsilon>1$, then $P(|X_{(n)}-1|>\epsilon)=0$, since $X_{n}$ is never greater than $1$ nor less than $0$, then neither is $X_{(n)}$, so the distance from $X_{(n)}$ to $1$ can't be greater than $1$.
Thanks.
Your answer is definitely right.
Actually we only care the case when $\epsilon$ is very small rather than large $\epsilon$, because you want to prove when $n \rightarrow \infty$, there is no difference between $X_{(n)}$ and $1$ in almost every point of domain (except the subset of domain whose measure is 0), so you will try to use smaller $\epsilon$ as possible.