# Open Cover and Compactness

I'm having trouble when thinking about compactness.
We knew that compactness of $$A\subseteq\mathbb{R}$$, can be defined as: Every open cover of $$A$$ has a finite subcover.
My question is:
An open cover is a collection of open sets.
So, when we are talking about the finite subcover, can I have a number of countably infinite intervals, here?
Because a finite subcover means a finite open sets, and every open set in $$\mathbb{R}$$ is characterized as disjoint union of a countably open intervals.
Any help will be appreciate.

• I disagree with the downvote. It seems like a reasonable thing OP wants to clear up, and not really subject to the usual "show some work" objection. Commented Jul 6, 2021 at 9:27
• @coffeemath . I agree . This Q is a request for clarification of a definition.... & English can be very unclear when a string of modifiers precedes a noun as one of the modifiers may be modifying the next modifier instead of modifying the noun. Commented Jul 7, 2021 at 3:35
• +1 to compensate for the downvote. Commented Jul 7, 2021 at 7:07

Here we need to distinguish two quantities that we are counting:

• The number of open sets in the open cover
• The number of intervals that each open set can be decomposed into

Now certainly as you have suggested, every open set in $$\mathbb{R}$$ can be decomposed into countably many open intervals. This is related to the second quantity in the above.

However, the word finite in the definition of compact refers to the first quantity in the above. In other words, by finite subcover, we mean that the open sets are chosen from the original open cover and the number of such sets are finite.

So whether the open sets can be further decomposed into open intervals is irrelevant in the definition of compactness.

Edit: In response to the OP's follow up question, I would like to use a rather trivial example to emphasize that it is possible for some open set in a particular finite open cover to be a disjoint union of infinitely many open intervals. Consider the set $$[0,1]$$ in $$\mathbb{R}$$. This set is compact. Now consider this specific open cover $$\mathcal{O} = \lbrace (-1,2), [0,1]-\mathcal{C} \rbrace,$$ where $$\mathcal{C}$$ is the Cantor set. There are two open sets in $$\mathcal{O}$$, so $$\mathcal{O}$$ is a finite open cover of $$[0.1]$$. But the set $$[0,1] -\mathcal{C}$$ is a disjoint union of countably many open intervals. In particular, the number of open intervals is not finite.

In addition, if we write out $$[0,1] - \mathcal{C} = \bigcup_i I_i$$ explicitly, where each of the $$I_i$$ is an open interval. Then actually each of the $$I_i$$ is not an element of the open cover $$\mathcal{O}$$.

• The last sentence here might better be phrased "It may be that one or each of the finite number of open sets of the open cover happens to be expressible as a countably infinite union of open sets." [It seems a bit misleading to say there can be countably infinite intervals in the finite subcover. I know what you mean, any of the open sets in the finite cover might in turn be an uncountable union of smaller sets, but the important thing is the open sets in the subcover are finite in number. Commented Jul 6, 2021 at 9:06
• If you consider the usual topology on $\mathbb{R}$, every open sets can be decomposed into countably many open intervals. But I do agree on the point that the formatting of my answer might lead to confusion of the two ‘counting quantities’. I will edit my post accordingly. Commented Jul 6, 2021 at 9:16
• Now it is very clear [+1] Commented Jul 6, 2021 at 9:25
• Thank you very much for the response Ken Hung and coffeemath. But, so my question is still, can one or more of the elements of the finite subcover consists of a countably infinite open intervals? And so, the subcover consists of infinite number of open intervals. Because, I used to think that there are only finite open intervals (when we say finite subcover in $\mathbb{R}$), here. Moreover, the number of elements in the subcover is the same as the number of the open interval that covers, usually. Commented Jul 6, 2021 at 11:50
• You seem to be using the transitive law of vague language: if $A$ "consists of" $B$, and if $B$ "consists of" $C$, then $A$ "consists of" $C$. That is not a valid proof technique. Try rewording your last comment without using "consists of" in a vague manner. Instead, stick with set theoretic language: this object "is an element of" that set; this set "is a subset of" that set; and so on. Commented Jul 6, 2021 at 19:13

It might help to write out the definition of compactness in a form commonly used in applications: Let $$\Lambda$$ be an arbitrary set. If $$\{O_\lambda \mid \lambda \in \Lambda\}$$ is a collection of open sets indexed by $$\Lambda$$ such that $$A \subset \bigcup_{\lambda \in \Lambda}O_\lambda$$, then there exists a finite subset $$\{\lambda_1, \dots, \lambda_n\} \subset \Lambda$$ ($$n \geq 0$$) such that $$A \subset \bigcup_{i = 1}^{n}O_{\lambda_i}$$.

• Thank you @Mason, but: 1. $n$ should be strictly greater than 0, here, instead of $n \ge 0$. 2. The question is, can open set $\lambda_1$, let's say, be a disjoint union of countably infinite open intervals? Commented Jul 7, 2021 at 3:39
• @ZeroToZero $n = 0$ is allowed, though it doesn't make a difference. In the definition, each $O_{\lambda}$ is allowed be any open set in $\mathbb{R}$. There is no restriction. This same definition of compactness is used for topological spaces where there may be no such thing as an interval. Commented Jul 7, 2021 at 4:09
• But you wrote $\lambda_1$ as your finite subset, @Mason. So, it doesn't make sense. But, I agree you can just use $\{\lambda_0,\lambda_1,...,\lambda_n\}$ as your finite subset, instead. Commented Jul 7, 2021 at 4:24
• @ZeroToZero If $n = 0$, then $\{\lambda_1, \dots, \lambda_n\} = \{\} = \emptyset$. It's a common notation. Similar to how $\sum_{n = 1}^{0}a_n = 0$. Commented Jul 7, 2021 at 4:47
• Ok, @Mason ,thanks. Commented Jul 7, 2021 at 4:58