Open cover of $(0,1)$ Suppose the open interval $(0,1)$ is given absolute value topology. Form $\{U_n \}$, $n=1,2,3…$, where $U_n=(\frac{1}{n+1},1)$. Prove that $\{U_n \}$ is an open cover of $(0,1)$ for all $n∈N$. Show that no finite number of $U_n$ cover $(0,1)$, even though any finite number of the $U_n$ may be omitted and what remains still give an open cover of $(0,1)$
here is what I got 
Suppose $X=(0,1)$ and $U_n=(\frac{1}{n+1},1)$ for $n∈N$. Notice that
$\cup _{n=1}^\infty U_n = (\frac {1}{2},1) \cup (\frac {1}{3},1) \cup (\frac {1}{4},1)....=(0,1)$
This shows that  $\{U_n \}$ is an open cover of $(0,1)$ for all $n∈N$. 
I don't know how to show the second part of this problem.
 A: Your proof that the $U_n$ cover $(0,1)$ isn't really that informative. You need to say why the union is the whole of $(0,1)$, and for that you need to show that $\bigcup_{n=1}^{\infty} U_n\subset (0,1)$ and also $\bigcup_{n=1}^{\infty} U_n\supset (0,1)$. One is easier than the other.
To show that no finite set of the $U_n$s can cover $(0,1)$, suppose $\mathcal{U}=\{U_i\mid i\in I\}$ for some finite subset $I$ of $\mathbb{N}$. Now suppose that $N$ is the maximum of the set of natural number $i$ such that $U_i$ is in our assumed cover $\mathcal{U}$ - so $N=\max I$ - (why does this maximum exist?). Can you see how $U_N$ contains $U_i$ for all $i\in I$? What does this say about $\cup\mathcal{U}$?
A: Your proof of why it is actually a cover is not satisfactory, IMHO. Provide an argument. Key fact: for every $x>0$ we can find $n \in \mathbb{N}$ such that $\frac{1}{n} < x$ (why?).
Note that the $U_n$ are nested: if $n < m$, then $\frac{1}{m} < \frac{1}{n}$ and so $U_n \subset U_m$ (if $x$ is bigger than $\frac{1}{n}$, it's certainly bigger than $\frac{1}{m}$..).
So a finite subcover has the same union as the one with the largest index. And no $U_n$ covers $(0,1)$, as $\frac{1}{2n} \notin U_n$, e.g.
