Convergence in Probability Clarification The following is from the textbook and I would like to clarify certain things which I am confused about. 
Consider a sequence of independent random variables $X_n$ that are uniformly distributed in the interval $[0,1]$, and let
$$Y_n=\min{(X_1,...,X_n)}.$$
The sequence of values of $Y_n$ cannot increase as $n$ increases, and it will occasionally decrease (whenever a value of $X_n$ that is smaller than the preceding values is obtained). Thus, we intuitively expect that $Y_n$ converges to zero. Indeed, for $\epsilon>0$ we have using the independence of the $X_n$, 
$$P(|Y_n-0|\geq \epsilon)=P(X_1\geq \epsilon, ... , X_n\geq \epsilon)=P(X_1\geq\epsilon)...P(X_n\geq \epsilon)=(1-\epsilon)^n.$$
Prior to continuing, I am confused about the $0$ in $P(|Y_n-0|\geq \epsilon)$ and how $(1-\epsilon)^n$ was found. 
After this I understand the remaining, but would appreciate the clarification of the above. Thank you!
In particular, 
$$\lim_{n\to\infty} P(|Y_n-0|\geq \epsilon)=\lim_{n\to\infty} (1-\epsilon)^n=0. $$
Since this is true for every $\epsilon > 0$, we conclude that $Y_n$ converges to zero, in probability. 
 A: You are looking at $|Y_{n}-0|$ because you are attempting to prove that $Y_{n}$ converges to $0$ in probabilty. This means, you must show
$$
\lim_{n\to\infty}P(|Y_{n}-0|\geq \epsilon)=0. 
$$
Now, $|Y_{n}-0|\geq\epsilon$ iff $|Y_{n}|\geq \epsilon$ iff $Y_{n}\geq \epsilon  $ since the $X_{i}$'s are uniformly distributed on $[0,1]$ and, hence, positive. Now, $Y_{n}\geq \epsilon$ iff $\min\{X_{1},\ldots, X_{n}\}\geq \epsilon$, which occurs iff each of the $X_{i}$'s for $1\leq i\leq n$ is $\geq\epsilon$. This is how you get
$$
P(|Y_{n}-0|\geq\epsilon)=P(X_{1}\geq \epsilon,\ldots,X_{n}\geq \epsilon).
$$
Now, by independence, $P(X_{1}\geq \epsilon,\ldots,X_{n}\geq \epsilon)=P(X_{1}\geq\epsilon)\cdots P(X_{n}\geq\epsilon)$. Since each $X_{I}$ is uniformly distributed on $[0,1]$, the probability that $X_{i}\geq \epsilon$ is just $\frac{1-\epsilon}{1}=1-\epsilon$. Thus, since this term appears $n$ times, we get
$$
P(X_{1}\geq\epsilon)\cdots P(X_{n}\geq\epsilon)=(1-\epsilon)^{n}.
$$
Therefore,
$$
P(|Y_{n}-0|\geq\epsilon)=(1-\epsilon)^{n}.
$$
Thus, taking the limit, we get
$$
\lim_{n\to\infty}P(|Y_{n}-0|\geq\epsilon)=\lim_{n\to\infty}(1-\epsilon)^{n}=0.
$$
