How to prove every closed interval in R is compact?

Question

Let $[a,b]\subseteq \mathbb R$. As we know, it is compact. This is a very important result. However, the proof for the result may be not familiar to us. Here I want to collect the ways to prove $[a,b]$ is compact.

Thanks for your help and any link.

Answer 1

If you're looking for a proof that doesn't use Heine-Borel, here are some hints:

• Notice that $[a,b]$ can be rewritten as $[a,\frac{a+b}{2}]\cup[\frac{a+b}{2},b]$.
• Use contradiction, i.e, assume that some open cover of $[a,b]$ has no finite subcover and wlog no finite subcollection covers $[a,\frac{a+b}{2}]$.
• Keep going
• You will need the nested intervals theorem.

Answer 2

If you're wanting a proof, take the following.

Let $\mathscr{U}$ be an open cover of $[a,b]$ and let $$b'=\sup\{x\in[a,b]:[a,x]\text{ is covered by finitely many elements of the open cover}\}.$$ Clearly $b'\leq b$. Now, if $b'<b$, then it's contained in some open set from the cover implying $b'$ is contained in some open interval. This shows that there is an element greater than $b'$, say $b''\in[a,b]$ that is covered by finitely many open sets from the cover, a contradiction to $b'$ being the supremum. Hence $b'=b$ and $[a,b]$ is compact.

Answer 3

Of course, Heine-Borel tells us that $[a,b]$ is compact. But how do you prove Heine-Borel? By showing $[a,b]$ is compact first, in general. Here is a proof which only requires the lub property of $\mathbb{R}$.

Edit: this is the same argument as Clayton's, which appeared way before while I was typing. As I give more details, I think I'll leave this answer here. Note in particular that you need to verify that the sup is in the set $S$. Showing it is equal to $b$ is not enough.

Note: there is nothing special about $\mathbb{R}$ here, actually. It just turns out that the order topology coincides with the usual topology. And for a totally ordered set equipped with the order topology, the lub property is equivalent to the fact that every closed interval $[a,b]$ is compact. The proof below establishes the direction you are interested in. For the converse, see this thread.

Proof: the case $a=b$ is trivial, so assume $a<b$ and take an open cover $$ [a,b]\subseteq \bigcup_{i\in I}U_i. $$ In particular, the latter covers $[a,x]$ for every $x\in [a,b]$. So consider the set $S$ of all $x\in[a,b]$ such that $[a,x]$ is covered by finitely many $U_i$'s. There exists $i_0$ such that $a\in U_{i_0}$, so $a$ belongs to $S$. Hence $S$ is a non-empty subset of $\mathbb{R}$ bounded above by $b$. By the least upper bound property, we can define $$ x_0:=\sup S\in [a,b]. $$ We will first prove by contradiction that $x_0=b$. So assume $x_0<b$. Note that $x_0>a$, as there exist $i_0$ and $\epsilon>0$ such that $[a,a+\epsilon]\subseteq U_{i_0}$, whence $x_0\geq a+\epsilon$.

Take $i_0$ such that $x_0\in U_{i_0}$. Then take $\epsilon>0$ such that $a\leq x_0-\epsilon<x_0<x_0+\epsilon\leq b$ and $$ [x_0-\epsilon,x_0+\epsilon]\subseteq U_{i_0}. $$ As $x_0-\epsilon$ is not an upper bound of $S$, there exists $x_0-\epsilon\leq x_1\leq x_0$ such that $x_1$ belongs to $S$. So $[a,x_1]$ can be covered by finitely many $U_i$'s, say $$ [a,x_1]\subseteq \bigcup_{j=1}^nU_{i_j}\quad\Rightarrow\quad [a,x_0+\epsilon]\subseteq \bigcup_{j=1}^nU_{i_j} \cup U_{i_0}=\bigcup_{j=0}^nU_{i_j}. $$ It follows that $x_0+\epsilon$ belongs to $S$, contradicting the fact that $x_0$ is an upper bound of $S$.

Therefore $\sup S=b$ and the same argument shows that $b$ belongs to $S$. Indeed, we have $[b-\epsilon,b]\subseteq U_{i_0}$ for some $i_0$ and some $\epsilon>0$. And then some $x_1$ in $[b-\epsilon,b]$ belongs to $S$, yielding a finite subcover for $[a,b]=[a,x_1]\cup[b-\epsilon,b]$.

So $b$ belongs to $S$, i.e. there exists a finite subcover $$ [a,b]\subseteq \bigcup_{j=1}^nU_{i_j}. $$ QED.

Answer 4

As I remember, a long time ago, when I was a student, at an exam my teacher asked me to prove this. I answered to him, that this result is so simple, that I forgot how to prove it. :-)

But now, being a mature mathematician, when I am asked about the proof, I answer the following. Clearly, a two-point set $\{0,1\}$ is compact. Tychonov theorem implies that Cantor set $\{0,1\}^\omega$ is compact too. At last, a segment is compact as a continuous image of Cantor set. :-)

Answer 5

Here is a different proof. Alexander's lemma. (It states: given a subbase for the topology, $X$ is compact iff every cover by elements of the subbase has a finite subcover.) Now use the sets $$ [0,a), 0<a<1,\qquad (b,1], 0<b<1 $$ as your subbase $\mathcal B$.

Suppose $\mathcal A \subseteq \mathcal B$ is a cover of $[0,1]$. Since $1 \in [0,1]$, there is at least one set of the form $(b,1]$ in $\mathcal A$. Let $b_0 = \inf\{b : (b,1] \in \mathcal A\}$. Then, since $b_0 \in [0,1]$ there is $[0,a_1) \in \mathcal A$ such that $b_0 \in [0,a_1)$. And since $a_1>b_0$, by definition of inf there is $(b_1,1] \in \mathcal A$ with $b_1<a_1$. So in fact $[0,1] = [0,a_1) \cup (b_1,1]$. Not only is there a finite subcover, there is a subcover of just two sets.

Answer 6

Perhaps the shortest, slickest proof I know is by Real Induction. See Theorem 17 in the aforelinked note, and observe that the proof is six lines. More importantly, after you spend an hour or two familiarizing yourself with the formalism of Real Induction, the proof essentially writes itself.

Answer 7

For the sake of completeness, here I present a classical and wordy proof. If you find it too verbose, you may just read the boxed statements and the numbered formula $(1)$.

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We first prove the following lemma.

Lemma 1. Suppose that $(F_n)$ is a sequence of non-empty closed bounded subsets of $\Bbb{R}$ satisfying $F_1 \supset F_2 \supset F_3 \supset \cdots$. Then $\bigcap_{n} F_n$ is non-empty.

Proof. Note that $x_n = \max F_n \in F_n$ is a monotone decreasing sequence bounded below. So it converges to some $\alpha \in \Bbb{R}$. Since each $F_n$ is closed, it follows that $\alpha \in F_n$ for all $n$, which completes the proof. (Of course, Bolzano-Weierstrass theorem also gives a direct proof.)

Now fix any open cover $\mathcal{U}$ of $I = [a, b]$ and define the function $d : \Bbb{R} \to [0, \infty)$ as follows:

$$ d(x) := \sup \{ \delta > 0 : \text{there exists a finite subfamily of } \mathcal{U} \text{ which covers } B(x, \delta) \}. \tag{1}$$

Intuitively, $d(x)$ measures how difficult to cover a neighborhood of $x$ with elements of $\mathcal{U}$. In one extreme, $d(x) = 0$ implies that no neighborhood of $x$ can be covered by finitely many members of $\mathcal{U}$. It is also clear that $d(x) > 0$ on $I$, since a sufficiently small neighborhood of each $x \in I$ can be covered by a single element of $\mathcal{U}$.

The importance of the function $d$ can be glimpsed by the following remark:

Remark. If $\alpha := \inf_{I} d(x) > 0$, then it is clear that $I$ is covered by finitely many open balls $B(x_1, \alpha/2), \cdots, B(x_n, \alpha/2)$. Since each $B(x_i, \alpha/2)$ is covered by only finitely many elements of $\mathcal{U}$, we obtain a finite subcover of $\mathcal{U}$.

To make use of this observation, we prove that $d(x)$ is lower semi-continuous:

Lemma 2. for any $a \geq 0$, the inverse image $d^{-1}((a, \infty)) = \{ x \in \Bbb{R} : d(x) > a \}$ is open.

Proof. Assume $d(x) > a$. Then there exists $\delta > 0$ such that $d(x) \geq \delta > a$ and $B(x, \delta)$ is covered by finitely many members of $\mathcal{U}$. Pick $\epsilon \in (0, \delta - a)$. Then for any $x' \in B(x, \epsilon)$ we have $ B(x', \delta-\epsilon) \subset B(x, \delta)$ and thus $d(x') \geq \delta-\epsilon > a$. This proves that $B(x, \epsilon)$ is contained in the inverse image $d^{-1}((a, \infty))$, and hence the proof is completed.

A crucial property of a lower semi-continuous function $f$ on a compact set is that $f$ attains a minimum. Thus assuming that $I$ is compact, the function attains a minimum on $I$, which must be positive. Though the compactness of $I$ is the goal of the proof rather than a part of assertion, we can directly show that $d$ attains a positive infimum:

Theorem 3. $d$ has the positive infimum on $I$. Consequently, $I$ is compact.

Proof. Assume not. That is, $\inf_{I} d(x) = 0$. Then for each $n$, the set $$ F_n = \{ x \in I : d(x) \leq \tfrac{1}{n} \} $$ is non-empty, closed and bounded. Since $F_1 \supset F_2 \supset \cdots$, Lemma 1 shows that $\bigcap_{n} F_n$ is also non-empty. Let $x_0$ be an element of this intersection. Since $d(x_0) \leq 1/n$ for all $n$, we have a contradiction that $d(x_0) = 0$. Therefore we must have $\inf_{I} d(x) > 0$.

Since every open cover $\mathcal{U}$ of $I$ has a finite subcover by Remark, it follows that $I$ is compact.

Answer 8

In short: Every closed interval is the continuous image of the Cantor space, and therefore compact.

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1. The Cantor space $2^\omega$ is compact as a result of Tychonoff theorem (also by Koenig's theorem).

2. The Cantor space can be mapped continuously on the interval $[0,1]$ using the function: $$(x_n)\mapsto\sum_{n\in\omega}\frac{x_n}{2^{n+1}}$$ It is not difficult to show that this function is continuous and surjective (although not injective!).

3. Therefore $[0,1]$ is the continuous image of a compact space and so it is compact.

4. Let $[a,b]$ be an interval, if $a=b$ then this is a finite set and so it is compact; otherwise map $[0,1]$ to $[a,b]$ bijectively using a translation and rescaling $x\mapsto a+(b-a)x$.

5. Every closed interval is the continuous image of $[0,1]$ which is compact; therefore $[a,b]$ is compact.

Answer 9

I provide not the fastest proof, but certainly beneficial if you work it out.

1. Prove Bolzano-Weierstrass theorem.
2. Show $[a,b]$ is sequential compact.
3. Prove that in $\mathbb{R}^n$ with standard topology sequential compact spaces are totally bounded.
4. From 3 arrive that in $\mathbb{R}^n$ sequential compact implies compact.

Your question is solved now. In fact you proved Heine-Borel using a detour. Here are related results:

5. (related) Prove that in $\mathbb{R}^n$ compact implies limit point compact.
6. Prove that in $\mathbb{R}^n$ limit point compact implies sequential compact.

In conclusion, the following in $\mathbb{R}^n$ with standard topology are equivalent:

7. A subset is closed and bounded
8. A subset is open cover compact
9. A subset is sequential compact
10. A subset is limit point compact
11. A subset is complete and totally bounded

Answer 10

Here's a proof using a non-standard approach.

With the help of transfer principle, we could reformulate compactness as:

$A \subset \mathbb R$ is compact, iff for each $y \in ^\ast A$, there is $x \in A$ such that $x$ is infinitely close to $y$.

We could prove a bounded closed interval is compact by showing that every closed and bounded subset of $\Bbb R$ is compact.

Proof: Let $B$ be a closed and bounded subset of $\mathbb{R}$. It suffice to show that $\operatorname{st}(^\ast B) \subseteq B$. Since $B$ is closed, we only need to show that $\operatorname{st}(^\ast B) \subseteq \overline B$. For each $x \in \operatorname{st}(^\ast B)$, there is $y \in ^\ast B$ such that $\operatorname{st}(y) = x$. Appealing to the ultrapower construction, $y$ is actually a sequence $\{r_i\}_{i \in \Bbb{N}}$ such that $r_i \in B$ for all $i \in \Bbb{N}$. We can construct a new sequence of elements in $B$ that converge to $x$, which implies $x \in \overline B$. This construction is borrowed from an answer by Brian M. Scott. Given the free ultrafilter $\mathscr{U}$. Define $A_n = \{r_i:|r_i -x |<\frac{1}{n+1}\} \in \mathscr{U}$ for all $n \in \mathbb{N} $. With $\bf AC$, we have a choice function $f$ such that $A_n \mapsto a_n$ satisfying $a_n \in A_n$. Then the sequence $\{a_n\}_{n \in \Bbb N}$ is just what we need.