How to prove every closed interval in R is compact?

Solution 1:

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. :-)

Solution 2:

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.

Solution 3:

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.

Solution 4:

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

The Cantor space $2^\omega$ is compact as a result of Tychonoff theorem (also by Koenig's theorem).
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!).
Therefore $[0,1]$ is the continuous image of a compact space and so it is compact.
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 $x\mapsto a+(b-a)x$.
Every closed interval is the continuous image of $[0,1]$ which is compact; therefore $[a,b]$ is compact.

Solution 5:

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.