# proof of Heine-Borel theorem

To shorten the proofs, we will be using certain concepts and facts from general point-set topology, such as the very first assertion below. Even if you are not familiar with them, these assertions are all easily proven from the definitions (left as an exercise).

###### Compact implies closed and bounded.

Since ${\mathbb{R}}^{n}$ is a Hausdorff space, any compact subset $A$ has to be closed as well.
That $A$ is also bounded^{} is very easily seen by taking
the open cover ${\{{(-k,k)}^{n}\}}_{k\in \mathbb{N}}$.
Passing to a finite cover, we see that $A$ is contained
in a bounded set.
∎

###### Reduction for closed and bounded implies compact.

To prove that if $A$ is closed and bounded implies it is compact, it is only necessary to prove that the rectangle ${[a,b]}^{n}$ is compact. For general $A$, since $A$ is bounded, it is contained in some such compact rectangle ${[a,b]}^{n}$. Since $A$ is closed and contained in a compact set, $A$ is also compact. ∎

###### The case $n\mathrm{=}\mathrm{1}$: the closed interval is compact..

Let $\mathcal{C}$ be an arbitrary cover of $[a,b]$ by open sets in ${\mathbb{R}}^{1}$. Define

$$S=\{x\in [a,b]:\text{some finite subcollection of}\mathcal{C}\text{covers}[a,x]\}.$$ |

Set $t=supS$; then $$.

We first note that the supremum $t$ is attained: that is, there is a finite subcollection of $\mathcal{C}$ that covers $[a,t]$. Obviously we can choose one open set $U$ from $\mathcal{C}$ that covers the point $t$. This $U$ must also cover the open interval $(t-\u03f5,t+\u03f5)$ for some $\u03f5>0$. But by the definition of $t$, the interval $[a,t-\u03f5]$ has a finite subcover. When this finite subcover is put together with $U$, we obtain a finite subcover for $[a,t]$.

It is easy to see that $t$ cannot be less than $b$. For the same open set $U$ above covers the interval $(t,t+\u03f5)$, so we have a finite subcover for $[a,t+\u03f5/2]$. If $t$ is the supremum then it has go to be at the right endpoint of the interval $[a,b]$, i.e. $t=b$. ∎

For the case $n>1$, apply
Tychonoff^{}’s Theorem, in the finite case (http://planetmath.org/ProofOfTychonoffsTheoremInFiniteCase), which states:

If ${X}_{1},\mathrm{\dots},{X}_{n}$ are each compact, then ${X}_{1}\times \mathrm{\cdots}\times {X}_{n}$ in the product topology is compact.

Here we set ${X}_{i}$ to be the closed intervals; then the general closed rectangle in ${\mathbb{R}}^{n}$
is compact. It is not hard to see that the product topology for the closed rectangle is the same
as its subspace topology in the product topology ${\mathbb{R}}^{n}=\mathbb{R}\times \mathrm{\cdots}\times \mathbb{R}$.
And it is again a standard exercise to show that the product topology on ${\mathbb{R}}^{n}$ is the
same as the norm topology on ${\mathbb{R}}^{n}$ as a vector space^{}.

This completes^{} the proof of the Heine-Borel theorem.

## Proof by a bisection argument

There is another proof of the Heine-Borel theorem for ${\mathbb{R}}^{n}$ without resorting to Tychonoff’s Theorem. It goes by bisecting the rectangle along each of its sides. At the first stage, we divide up the rectangle $A$ into ${2}^{n}$ subrectangles. Suppose the open cover $\mathcal{C}$ of $A$ has no finite subcover. Then one of the subrectangles — call it ${A}_{1}$ — must have no finite subcover by $\mathcal{C}$. We can subdivide ${A}_{1}$ into ${2}^{n}$ pieces; since ${A}_{1}$ has no finite subcover, one of the new subrectangles of ${A}_{1}$ also has no finite subcover. And continue dividing to get a nested sequence of rectangles $A\supset {A}_{1}\supset {A}_{2}\supset \mathrm{\cdots}$ whose side lengths approach zero, and possessing no finite subcover.

By the nested interval theorem, the “limit rectangle” ${\bigcap}_{i=1}^{\mathrm{\infty}}{A}_{i}$ must consist of a sole point $x$, and this obviously has a finite subcover by an open set $U\in \mathcal{C}$. But $U$ must contain a small rectangle with centre $x$, which for $i$ large enough, contradicts ${A}_{i}$ having no finite subcover.

Of course, in both proofs of the Heine-Borel theorem, the completeness of the reals (the least upper bound property) enters in an essential way.

Title | proof of Heine-Borel theorem |
---|---|

Canonical name | ProofOfHeineBorelTheorem |

Date of creation | 2013-03-22 12:57:40 |

Last modified on | 2013-03-22 12:57:40 |

Owner | stevecheng (10074) |

Last modified by | stevecheng (10074) |

Numerical id | 23 |

Author | stevecheng (10074) |

Entry type | Proof |

Classification | msc 54D30 |