# An open cover of $\{x:x>0\}$ with no finite subcover

Two parts. Second one doesn't really warrant its own thread but I want to confirm it.

First: Find an open cover of $\{ x: x>0\}$ with no finite subcover. I don't have any answer to this one.

Second: Find an open subcover of $(1,2)$ with no finite subcover. I believe an answer to this is $\bigcup_{n=1}^\infty \left( 1 +\frac{1}{n}, 2-\frac{1}{n} \right)$, but I'm not sure.

-

How about $\bigcup_{n=1}^{\infty}(0,n)$ for the first one. Your cover for the second one works.

-
So I guess that means I'm just confused about the definition of a cover. Is the thing being covered (x>0) just a subset of the cover? I for some reason thought that the cover had to include every element of the (x>0) in some way. –  Nick Sep 24 '13 at 5:07
@user71352 answer means the cover $\bigcup_{n\in\mathbb{N}} (0,n)$, which contains every $x>0$. –  Zephos Sep 24 '13 at 5:10
@maujet I don't get at all how that contains every x>0. That is strictly integers, unless I'm missing something pretty major here. –  Nick Sep 24 '13 at 5:11
Maujet is correct. I am referring to the union over all such intervals. I'll edit it to be more clear. –  user71352 Sep 24 '13 at 5:12
@Nick, I think you just need to take a step back. Fix an arbitrary positive real number $x$. Can you see that for each $n>x$, $x\in(0,n)$? –  Zephos Sep 24 '13 at 5:16

For the second one. Your answer is correct. A cover $\mathcal{U}$ of the set $(1,2)$ is a collection of open sets whose union contains $(1,2)$. The following equality is clear (we start the union at two to avoid the empty case). $$\bigcup_{n=2}^\infty \left(1+\frac{1}{n},2-\frac{1}{n}\right)=(1,2)$$ So the family $\mathcal{U}$ of open sets of the form $\left(1+\frac{1}{n},2-\frac{1}{n}\right)$ for $n\geq 2$ gives a cover of $(1,2)$. Now suppose you have a finite subcover. That is, a subfamily $\mathcal{V}$ made of members of $\mathcal{U}$ whose union also contains $(1,2)$. Since the set is finite and all of the sets are of the form $(1+1/n,2-2/n)$ then you can find one of this sets that contains all of the others (the one with the biggest value of $n$), say $(1+1/n^*,2-2/n^*)$. But then the union of all the elements of $\mathcal{V}$ is $(1+1/n^*,2-2/n^*)$ which does not contain the interval $(1,2)$ and gives a contradiction.

For the first one: Use the same idea, how about sets of the form $(1/n,\infty)$. They are open and the union of all of this set is the set $\{x \mid x>0\}$.

The proofs you are for asking are related to compactness. You should have a look at the concept. Note that $(0,1)$ and $(0,\infty)$ are very similar. But $[0,1]$ is not. You can infinitely approach $(0,1)$ from the inside with sets of the form $(0,1-1/n)$, the same happens when you approach $(0,\infty)$ with sets of the form $(0,n)$. But if you ever reach $[0,1]$ with sets of the form $[0,1/n]$ the moment you do it you can stop and have a finite subcover. This is in fact Borel's original insight that is reproduced in Heine-Borel's theorem that characterizes the compact sets is $\mathbb{R}$ as those that are closed and bounded.

-