About covers, subcovers and compactness

Question

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?

Answer 1

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.

Answer 2

"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.