5

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!

Giovanni
  • 6,321
ematth7
  • 719
  • Hint: Think about the sequence $1-1/n,n=1,2,\dots$ – zhw. Oct 25 '15 at 02:28
  • 2
    I probably should ask, what exactly is my problem asking. – ematth7 Oct 25 '15 at 02:34
  • The problem is asking for a family of sets $U_1,U_2,\dots$ such that $\bigcup U_k\supseteq(-1,1)$, and such that no finite union of $U$s equals $(-1,1)$. – Akiva Weinberger Oct 25 '15 at 03:15
  • The answers below will help you solve the problem, but won't help in learning how to solve such problems. The context of the problem is the key. You will have learned that all closed, bounded sets do have this property. So if this set does not have the property it is because it is not closed or it is not bounded. Well it is bounded but not closed and the points $-1$ and $+1$ are therefore the problem. This is what you must exploit and so the very first step is to think about what is happening at these endpoints to cause this property to fail. – B. S. Thomson Oct 25 '15 at 22:17
  • @B.S.Thomson: Both user46944 and I made clear that the endpoints are the problem. Seeing what goes "wrong" when finding a finite sub-cover of an open set is exactly what helps solving such problems. Moreover, once the OP has seen this trick, he or she might try to modify the argument for the closed interval $[-1,1]$ and obviously fail. I can't think of anything more instructive than a counterexample in which the only thing that is used is that one of the endpoints is not contained in the set. – Giovanni Oct 26 '15 at 02:32

2 Answers2

5

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.

Giovanni
  • 6,321
2

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.

layman
  • 20,191