Let there be a sequence of iid r.v.'s $\{X_{n}\}_{n\geq1} \sim U(0,1)$, defining another sequence as $Z_{n} = max\{X_{1}, ...,X_{n}\}$, show that $Z_{N} \xrightarrow[]{P}1$.
So my thoughts are, \begin{align} \mathbb{P}(Z_{n} \leq a) = \mathbb{P}(\max\{X_{1},...,X_{n}\} \leq a) = \mathbb{P}(X_{1} \leq q, ..., X_{n} \leq a) = \mathbb{P}(X_{n} \leq a)^{n} \end{align} Now defining the convergence criteria and developing results $\forall \epsilon >0$,
\begin{align} lim_{n \xrightarrow[]{} \infty} \mathbb{P}(\lvert Z_{n} -1\rvert > \epsilon) &= lim_{n \xrightarrow[]{} \infty} 1 - \mathbb{P}(\lvert Z_{n} -1\rvert \geq \epsilon) \\ &= lim_{n \xrightarrow[]{} \infty}1-\mathbb{P}(-\epsilon \leq Z_{n}-1\leq\epsilon)\\ &= lim_{n \xrightarrow[]{} \infty}1-\mathbb{P}(1-\epsilon \leq Z_{n}\leq 1+\epsilon)\\ &= lim_{n \xrightarrow[]{} \infty}1-\epsilon^{n}I_{\epsilon \in[0,1]} -I_{\epsilon \in(1,\infty)} \\ &= 0 \quad \forall \epsilon >0 \end{align} Therefore $Z_{N} \xrightarrow[]{P}1$.
I'd like to know if my line of thought is correct, I'm self-learning the theme. Thanks in advance!