0
$\begingroup$

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?

$\endgroup$
3
$\begingroup$

$(-2,3)$ 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. $$\left\{\left(0, 1-\frac{1}{n}\right)\ \bigg\vert\ n\in\mathbb{N}\backslash\{1\}\right\}$$

$\endgroup$
  • $\begingroup$ I'm curious, why is $\{1\}$ excluded here? $(0, 0)$ is just the empty set so no harm in including it, is there? $\endgroup$ – Björn Lindqvist Jul 5 '18 at 20:20
  • $\begingroup$ 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... $\endgroup$ – Saibal Jul 5 '18 at 22:26
1
$\begingroup$

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 $(0,1)$.

$\endgroup$
1
$\begingroup$

$\{(-2,3)\}$ 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.

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Alan Dec 15 '14 at 6:28
0
$\begingroup$

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 $(0,1)$

$\endgroup$
0
$\begingroup$

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).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.