Why is this family of open sets a cover for $(0,1)$? I'm dealing with some beginner examples of compact spaces. The definition I am given is 

A subset $A$ of a topological space $X$ is compact if every open cover for $A$ has a finite subcover for $A$

Then the book shows that $(0,1)$ is not compact. The main idea is that te family of open sets $\{ (1/n, 1): n\in \mathbb{N}, n>1\}$, which covers $(0,1)$ has no single finite subfamily which covers $(0,1)$, because, lets say that the finite subfamily is $\{(1/n_1,1), ... (1/n_r, 1))\}$ covers only $(1/N,1)$, with $N=\max(n_1, n_2, ... , n_r)$.
So I see that this is just applying the definition. But how can I understand that the family of open sets$\{ (1/n, 1): n\in \mathbb{N}, n>1\}$, covers $(0,1)$?
 A: For any $x\in(0,1)$ we need to show there is some $n$ such that $x\in(\frac{1}{n},1)$. 
Since $x\neq 0$, we can find such $n$ by finding $n>\frac{1}{x}$ since $x\neq 0$, then $\frac{1}{n}<x$.
You might be confusing "covers" with "covers distinctly." The $x$ can be in more than one of the sets. Indeed, in this case, it is in infinitely many.
A: Hint: The sequence $(1/n)$ tends to $0$ for $n \to + \infty$. So for $x>0$ and for $n$ large enough, $x>1/n$.
A: Observe that $0<x<1$ if and only if there exists $n\geq 1$ such that $1/n<x<1$. The "if" is trivial. For the "only if", you can take explicitly $n=\lfloor 1/x\rfloor+1$, as this yields $n>1/x$. Therefore
$$
(0,1)=\bigcup_{n\geq 1}\left(\frac{1}{n}, 1\right)
$$
Note that cover only requires $\subseteq$. But while we're at it...
A: An easier way of proving  $(0, 1)$ is not compact is to show $\mathbb{R}$ is not compact since $(0, 1) \cong \mathbb{R}$ (assuming you've already seen this proof and proved that continuous surjections preserve compactness). The set $\displaystyle \bigcup_{n = 1}^\infty (-n, n)$ is an open cover of $\mathbb{R}$ and does not reduce finitely. Same idea as the original example but a little easier to visualize.
