# Open cover for $(0, 1)$

Would the set $(-2, 3)$ be an open cover for $(0, 1)$? In that case wouldn't $(-1, 2)$ be a subcover for $(0, 1)$? There's only 1 set in the subcover collection, thus the collection is finite. However, this wouldn't be right because I know that $(0, 1)$ has no finite subcover because the set is non-compact. I think I am still confused about the definition of an open cover. So does the open cover have to be a collection of sets (i.e., more than one?)? In that case $(-2, 3)$ isn't an open cover for $(0, 1)$ because it's only 1 set?

$(-2,3)$ is NOT a cover of $(0,1)$, $\{(-2, 3)\}$ is, and $\{(-1,2)\}$ is not a subset of $\{(-2,3)\}$, so does not give a subcover.

More to the point, a set is compact if EVERY possible open cover has a finite subcover. Though in this case, the cover $\{(-2,3)\}$ itself gives a finite subcover of $(0,1)$, there are many open covers that does not have a finite subcover. Eg. $$\left\{\left(0, 1-\frac{1}{n}\right)\ \bigg\vert\ n\in\mathbb{N}\backslash\{1\}\right\}$$

• I'm curious, why is $\{1\}$ excluded here? $(0, 0)$ is just the empty set so no harm in including it, is there? – Björn Lindqvist Jul 5 '18 at 20:20
• No, there is no harm. Back then, I was studying measure theory, and the null sets felt like an annoyance. Don't get me wrong, they are an important building block for the subject - but dealing with them meant adding several lines to each proof... – Saibal Jul 5 '18 at 22:26

Being able to choose finitely many open sets from a fixed open cover does not ensure compactness.

The definition of compact requires that for any open cover (i.e. collection of open sets $\{U_i\}$ such that $\cup_i U_i$ contains your set) it is possible to extract finitely many $U_i$.

In your case, for $(0,1)$, the open cover $$U_n:=(0,1-1/n)$$ does not allow to extract finitely many open sets such that the union still covers $(0,1)$.

$\{(-2,3)\}$ is AN open cover of $(0,1)$. Your confusion comes from the definition of compact. A set is compact if EVERY open cover has a finite subcover, not just if one particular open cover has a finite subcover. So, for example, the collection of sets $(\frac 1 n,1 - \frac 1 n)$ for each $n\in \mathbb N$ is another open cover of $(0,1)$, but there's no finite subcover, hence $(0,1)$ is not compact.

• Ehh, true, but I didn't think the distinction between a one element collection and the collection itself to be significant enough. I'll go ahead and edit for technicality though. – Alan Dec 15 '14 at 6:28

Every open set is an open cover for itself. However, for a compact set A every open cover should have a finite cover: there should be no collection of countably infinite open sets such that a finite number of sets from this collection forms an open cover for the set A.

For the given example, $(0,1)$ has an open cover $\{(0+1/n,1-1/n):n=3..\infty\}$ No finite subset of this collection is an open cover for $(0,1)$

In order to show that the set is compact you need to show that arbitrary open cover (which may be finite, countable, or of any carnality) has a finite subcover. In your case, you only checked one special case. A set is not compact does not imply any open cover of the set does not have a sub cover (be careful with quantifiers).