# Why is an open interval not a compact set?

I learned that every compact set is closed and bounded; and also that an open set is usually not compact.

How to show that a concrete open set, for example the interval $(0,1)$, is not compact? I tried to show that $(0,1)$ has no finite sub cover.

• If you define a compact set to be closed and bounded, then I don't see what the question is. – Igor Rivin Dec 16 '13 at 16:22
• Are you using the "every open cover has a finite sub-cover" definition of compactness? – Omnomnomnom Dec 16 '13 at 16:25
• I just want to know why a compact set is closed whereas an open set is not? – user114873 Dec 16 '13 at 16:26
• why don't open intervals satisfy that? This is what I want to know. – user114873 Dec 16 '13 at 16:31
• I just want to know why (0,1) has no finite sub cover? – user114873 Dec 16 '13 at 17:11

## 6 Answers

I think the following may be a source of confusion: the statement "$(0,1)$ has no finite sub cover" doesn't make any sense. You first have to choose a cover of $(0,1)$ by open sets. Then this may or may not have a finite sub-cover.

If $(0,1)$ were compact, any such cover would (by the definition of compact in terms of open covers) have to have a finite subcover. In fact, $(0,1)$ is not compact, and so what this means is that we can find some cover of $(0,1)$ by open sets which does not have a finite subcover.

Omnomnomnom gives an example of such a cover in their answer, and there are lots of others; here's one: the cover $\{U_2, \ldots, U_n , \ldots\}$ where $U_n = (1/n, 1-1/n).$

Here is a cover which is finite, and hence does have a finite subcover (namely, itself): $\{(0,1)\}$.

Here is another: $\{(0,1/2), (1/3,1)\}$.

Here is an infinite cover which admits a finite subcover $\{(0,1), (0,1/2), \ldots, (0,1/n) , \ldots \}$.

Hopefully these examples help to clarify what the definition of compactness in terms of finite subcovers is about.

To be sure, there exist sets that are open and closed and bounded. For example, if we take the space $X = [0,1] \cup [2,3] \cup [4,5]$ under the typical topology of $\mathbb{R}$, then $[2,3]$ is a closed, open, compact proper subset of $X$.

As for open intervals, consider the example of $X = (0,1)$. Defining $U_n = (1/n,1)$, we note that $\{U_1,U_2,\dots\}$ is an open cover of $X$ (since $X \subset \bigcup_{n=1}^\infty U_n$) that has no finite subcover.

(Assuming you're talking about $\Bbb{R}^1$)

Consider the open interval $(a,b)$. Let $d=|b-a|/2$, and let $c=(a+b)/2$, the midpoint of $(a,b)$. Take the collection of open intervals: $\{(c-\frac{n}{n+1}d,c+\frac{n}{n+1}d)\}_{n=1}^\infty$. This collection covers $(a,b)$ (its union equals $(a,b)$), but no finite sub-collection covers $(a,b)$.

For what it's worth:

A subset of the euclidean space $\mathbb R^n$ is compact if and only if it is closed and bounded. This is a possible definition of compactness of sets like these.

Regarding the concepts of open, closed, bounded: You will have to look up their definitions. Some examples of subsets of $\mathbb R$:

• The empty set is open, closed and bounded.
• The set $\mathbb R$ is open, closed and not bounded.
• The interval $(0,1)$ is open, not closed, bounded.
• The interval $(0,\infty)$ is open, not closed, not bounded.
• The interval $[0,1]$ is not open, closed, bounded.
• The interval $[0,\infty)$ is not open, closed, not bounded.
• The interval $[0,1)$ is not open, not closed, bounded.
• The set $\{0\}\cup (1,\infty)$ is not open, not closed, not bounded.

In the general realm of topology, these concepts are not really too related to each other. For example, in a finite set with the discrete topology every set is compact which are both open and closed. A compact set is not guaranteed to be closed unless you are in a Hausdorff space. In a topological set with the trivial topology, everything is compact, and here the only closed sets are the empty set and the set itself.

You seem to be confused about the concept of (sub)cover. Let me see if I can clear it up for you.

Suppose we have a set $X$ with some topology on $X,$ a set $A\subseteq X,$ and a set $\mathcal C$ of subsets of $X.$

• We say that $\mathcal C$ "covers $A$" (or "is a cover of $A$") if every point of $A$ lies in some $\mathcal C$-set (i.e.: $A\subseteq\bigcup\mathcal C$).
• We say that $\mathcal C$ "is an open cover of $A$" if $\mathcal C$ is a cover of $A$ consisting entirely of open sets.
• If $\mathcal C$ is a cover of $A,$ then a subcover is some $\mathcal S\subseteq\mathcal C$ such that $\mathcal S$ covers $A$.

It is inappropriate, then, to talk about "subcovers of $(0,1)$" because $(0,1)$ is not a set of subsets of the real line.

Let's consider $X=\Bbb R$ (with the usual topology), $A=(0,1).$ Now, there are open covers of $(0,1)$ with finite subcovers--for example, given any collection $\mathcal C'$ of open subsets of $\Bbb R,$ we have that $\mathcal C:=\mathcal C'\cup\bigl\{(0,1)\bigr\}$ is an open cover of $(0,1),$ and $\bigl\{(0,1)\bigr\}$ is a finite subcover. This is not enough to make $(0,1)$ compact, though. For that, we would need to know that every open cover of $(0,1)$ admits a finite subcover, which is not the case. Consider for example $$\mathcal C:=\left\{\left(\frac1{n+1},1\right)\mid n\text{ is a positive integer}\right\}.\tag{\star}$$ Suppose $\mathcal S\subseteq\mathcal C$ is finite. If $\mathcal S$ is empty, then it is clearly not a cover of $(0,1).$ Otherwise, there is some positive integer $m$ such that every element of $\mathcal S$ is a subset of $\left(\frac1m,1\right)$ (why?), so that $$\bigcup\mathcal S\subseteq\left(\frac1m,1\right)\subsetneq(0,1),$$ and so $\mathcal S$ does not cover $(0,1).$ Thus the open cover $\mathcal C$ defined in $(\star)$ has no finite subcover, and so $(0,1)$ is not compact.

In many topologies, open sets can be compact. In fact, the empty set is always compact.

Also, any topology with $\subseteq$-minimal non-empty open sets will have open compact subsets. For example, consider the topology on the real line having the empty set and the supersets of $(0,1)$ as open subsets of the real line. This is not the usual topology, of course, but you can check that

• arbitrary unions of open subsets of the real line are again open subsets of the real line,
• finite (in fact, arbitrary) intersections of open subsets are again open, and
• the empty set and real line are open.

So, this is indeed a topology on the real line. In this topology, $(0,1)\cup F$ is open, compact, and non-closed for any finite subset $F$ of the real line.

• Thank you for the great explanation! Could you give an example of a topology where opens sets are compact? – Konstantin Jan 8 '17 at 20:21
• @Konstantin: You mean other than the example I gave in my answer? – Cameron Buie Jan 8 '17 at 22:35