Why is an open interval not a compact set?

Question

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.

Answer 1

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.

Answer 2

(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 $(a,b)$.

Answer 3

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.

Answer 4

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.

Answer 5

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.

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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.

Answer 6

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.