I have a doubt about covers, finite subcovers and its relation with compactness.

Let $X$ be a topological space. We define an open cover of $A\subset X$ as an union of a collection of open subsets of $X$ that contains $A$.

We define a finite subcover taking a finite number of those sets that still cover $A$.

If a set has a finite open subcover, then it's compact.

Question: $(-1,0)\cup(-1/2,2)$ is a finite open cover of $(0,1)$. Regarding the definition of compactness, $(0,1)$ is compact. What am I not understanding correctly?

  • $\begingroup$ For a set $A$ to be compact you must be able to take any collection of open sets $\{U_i: i\in I\}$ ($I$ is an arbitrary index set) such that $A\subset \cup_i U_i$ and then out of these arbitrary $U_i$ you must find finitely many that still do the job of covering $A$—it isn’t enough to find any finite subcover, you must extract one from the original arbitrary open cover $\endgroup$ – Nap D. Lover Jan 21 '18 at 18:41
  • $\begingroup$ Just to give a concrete example, consider the cover $\{(0,a): 0<a<1\}$ of $(0,1)$. If there was a finite subcover, then it would have the form $\{(0,a_1),\dots,(0,a_n)\}$ where we can impose the order $a_1<\dots<a_n$, and thus the union of these sets is $(0,a_n)$. But $a_n<1$, so $(0,a_n)$ is a proper subset of $(0,1)$ and does not cover it. Thus no finite cover can exist, and so $(0,1)$ is not compact. $\endgroup$ – Santana Afton Jan 21 '18 at 18:48
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    $\begingroup$ "If a set has a finite open subcover, then it's compact." No. It's not the set that has the subcover. It is the open cover that has the finite subcover. A set is compact if every open cover has a finite subcover. $\endgroup$ – fleablood Jan 21 '18 at 18:49

Your definition of compactness is wrong. A subset $A\subset X$ is compact if, given any open cover of that set, we can extract a finite open subcover. That is, to check compactness of $A$, you have to take an arbitrary collection $(U_i)_{i\in I}$ of open sets that cover $A$ and prove that you find finitely many $i_1,\dots,i_n\in I$ such that $A\subset \bigcup_{k=1}^n U_{i_k}$. By your definition, every subset would be compact since it can be covered by $X$ itself.

  • $\begingroup$ A finite open subcover. $\endgroup$ – Nap D. Lover Jan 21 '18 at 18:38
  • $\begingroup$ okay, so let's take $(-3,0)\cup (-1,0)\cup(-1/2,2)$ as a cover of $(0,1)$. Then we take the subcover in my question. It means that $(0,1)$ is compact? $\endgroup$ – Alure Jan 21 '18 at 18:39
  • $\begingroup$ @Alure No, you cannot just choose an open cover of your liking, you have to show that it works for any open cover. $\endgroup$ – asdq Jan 21 '18 at 18:42
  • $\begingroup$ @Alure you are still missing the fact that to prove this directly you need to start with an arbitrary open cover. One you know nothing about. $\endgroup$ – Nap D. Lover Jan 21 '18 at 18:42
  • $\begingroup$ @LoveTooNap29 Thanks, I edited it. $\endgroup$ – asdq Jan 21 '18 at 18:42

"If a set has a finite open subcover, then it's compact."

This is incorrect.

The correct definition is "If every open cover of a set, has a finite subcover, then the set is compact."

"(−1,0)∪(−1/2,2) is a finite open cover of (0,1)"

Right, but so what?

it is not the only open cover of $(0,1)$. For $(0,1)$ to be compact. Every open cover has a finite subcover. The is only one open cover. There are others and if $(0,1)$ is not compact, at least one open cover will not have an open cover. But if every possible open cover, any open cover anyone in the universe can think of has a finite subcover, then $(0,1)$ is compact.

Suppose I said: A country is "rich" if every single person has over \$500,000. And the I said: America is not rich because not every single person has over \$500,000. And you said: What about Malcolm Forbes? He has over \$500,000. So isn't America a "rich" country. The answer to that is Malcolm Forbes is not every single person. To be "rich" everyone must have more than \$500,000.

So the question comes down to: Can we think of an open cover of $(0,1)$ that has no finite subcover?

In the comment Santana Afton had a good example of an open cover of $(0,1)$ that has no finite subcover.

Let $0 < a < 1$ and let $U_a = (0,a)$. I claim that $\{U_a|a \in (0,1)\} = \{ (0,a)|0 < a < 1\}$ is an open cover.

Pf: If $x \in (0,1)$ then there is a $y$ so that $0 < x < y < 1$ so $x \in (0,y)= U_y$. So $x$ is "covered" by $\{U_a\}$. That is true for all $x$ so all of $(0,1)$ is covered. And as $(0,a)$ is open it is an open cover.

I claim it has no finite subset that covers $(0,1)$. If there is a finite subset $\{(0, a_n)\}$ for some finite number of $a_i$ then there is a maximum value of $A = \max (a_i)$ so $a_i \le A < 1$ for all $\{(0,a_n)\} $. So if $A < x < 1$ then $x \not \in (0, a_i)$ for any of the finite number of $a_i$. So $x$ is not covered by $\{(0, a_n)\}$.

So no finite subset of $\{(0,a)\}$ will cover $(0,1)$. So $\{(0,a)\}$ has no finite subcover.

So $(0,1)$ is not compact.


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