Find an open cover of the interval $(-1,1)$ that has no finite subcover Find an open cover of the interval $(-1,1)$ that has no finite subcover.
I have no idea how to do this problem. Can someone help me go step by step until I get this concept? 
Thanks you!
 A: The idea is that you can reach $1$ or $-1$ infinitely slowly: consider  open sets of the form $U_n = (-1, 1 - \frac 1n)$ and notice that $$\bigcup_{n = 1}^{\infty}U_n = (-1,1).$$ This equality is not difficult to show and I'll leave it for you to check in case you have not seen it before. Double inclusion is the way I would prove it, feel free to ask me if you can't find a convincing argument. 
This implies that $\mathcal{U} = \{U_n : n \in \mathbb{N}\}$ is an open cover of the interval $(-1,1)$, or, in other words, each element of $\mathcal{U}$ is an open set and the union of all the elements in $\mathcal{U}$ contains $(-1,1)$.
Now I claim that no finite sub-cover can cover the whole interval. A sub-cover is a subset of $\mathcal{U}$, and a sub-cover is said to be finite if it contains only a finite number of element of $\mathcal{U}$. 
To prove the claim, notice that the sets are increasing, meaning that $$U_n \subset U_{n+1}. \tag 1$$ Then we can argue by contradiction as follows: suppose that there exists a finite sub-cover $\mathcal{A} \subset \mathcal{U}$ with that property, say $\mathcal{A} = \{U_{n_i} : i = 1,\dots, N\}.$ Since $n_N$ is the largest index that appers in $\mathcal{A}$, by property $(1)$ $$\bigcup_{i = 1}^NU_{n_i} = U_{n_N} \subsetneq (-1,1).$$
This contradicts the fact that $\mathcal{A}$ covers $(-1,1)$, hence proving the claim. 
A: Here is a hint:
Take a sheet of paper and draw a long, straight, horizontal line on it.  On the left end of the horizontal line, mark $-1$, and on the right end, mark $1$.
Now, mark $0$ as the midpoint of the line.
Put parentheses on either side of $0$, but not anywhere near $-1$ or $1$.  Let the area enclosed in the parentheses signify an open interval (but you haven't yet labeled any end points).  Now draw another pair of parentheses that encloses both $0$ and the first pair, but is nowhere near $-1$ and $1$.  Can you keep doing this infinitely many times, where you the left and right parentheses are not on $-1$ or $1$?  Hopefully from the picture, you can see that yes, you can.
That is an open cover with no finite subcover.  Now the challenge for you is to come up with a formula for these open intervals you drew, and show that they are an open cover and have no finite subcover.
