Question 6.T.12: (Heine-Borel) A set E ⊆ R is compact if, and only if, it is ...

(i) Suppose E is compact. We shall first prove that it is bounded. For every n ∈ \mathbb{N}, let I_{n} = (−n, n). Clearly,

E ⊆ \overset{∞}{\underset{n=1}{\cup}} I_{n} = \mathbb{R}.

Since E is compact, there is a positive integer N_{1} such that

E ⊆ \overset{N_{1}}{\underset{n=1}{\cup}} (−n, n) = (−N_{1}, N_{1}),

which means E is bounded.

Now we shall prove that E is closed by proving that its complement E^{c} is open. Suppose x ∈ E^{c}. For every n ∈ \mathbb{N}, define

G_{n} = \left\{y ∈ \mathbb{R} : |y − x| > \frac{1}{n}\right\}

= \left[x − \frac{1}{n} , x + \frac{1}{n} \right]^{c}.

Being the complement of a closed interval, G_{n} is open for every n ∈ \mathbb{N}.

Furthermore,

\overset{∞}{\underset{n=1}{\cup}} G_{n} = \overset{∞}{\underset{n=1}{\cup}} \left[x − \frac{1}{n} , x + \frac{1}{n} \right]^{c}

= \left\{\overset{∞}{\underset{n=1}{\cap}} \left[x − \frac{1}{n} , x + \frac{1}{n} \right] \right\}^{c}

= \left\{x\right\}^{c} = \mathbb{R}\setminus \left\{x\right\}.

Since x ∉ E it follows that E ⊆ \cup^{\infty}_{n=1}  G_{n}and since E is compact, there is an N_{2} ∈ \mathbb{N} such that

E ⊆ \overset{N_{2}}{\underset{n=1}{\cup}} G_{n} = GN_{2} = \left[x − \frac{1}{N_{2}} , x + \frac{1}{N_{2}} \right]^{c}

⇒ \left[x − \frac{1}{N_{2}} , x + \frac{1}{N_{2}} \right] ⊆ E^{c}

⇒ \left(x − \frac{1}{N_{2}} , x + \frac{1}{N_{2}}\right) ⊆ E^{c}.

This implies E^{c} is open, and hence E is closed.

(ii) Conversely, suppose E is closed and bounded, and let \left\{G_{λ} : λ ∈ Λ\right\}  be an open cover of E. We shall prove that a finite number of open sets in \left\{G_{λ} : λ ∈ Λ\right\}  suffices to cover E.

Assume, on the contrary, that no finite sub-collection of \left\{G_{λ} : λ ∈ Λ\right\} covers E. Since E is bounded, there is a finite closed interval I_{0} = [a_{0}, b_{0}] which contains E. Divide this interval at its center into the two closed subintervals

\acute{I_{0}} = \left[a_{0}, \frac{a_{0} + b_{0}}{2} \right] , {I}^{\prime \prime }_{0} = \left[ \frac{a_{0} + b_{0}}{2} , b_{0} \right].

At least one of the sets E ∩ \acute{I_{0}} and E ∩ {I}^{\prime \prime }_{0}  cannot be covered by a finite number of open sets in \left\{G_{λ} : λ ∈ Λ\right\}, for if both could be so covered then the same would apply to

E = (E ∩ \acute{I_{0}} ) ∪ (E ∩ {I}^{\prime \prime }_{0} ),

which would violate our assumption. Suppose I_{1} = [a_{1}, b_{1}] is one of the subintervals \acute{I_{0}} or {I}^{\prime \prime }_{0} whose intersection with E cannot be covered by a finite sub-collection of \left\{G_{λ} : λ ∈ Λ\right\}.

Again subdivide I_{1} into two subintervals of equal length, \acute{I_{1}} and I^{\prime \prime }_{1} , and conclude that at least one of them, say I_{2} = [a_{2}, b_{2}] intersects E in a set which cannot be covered by a finite sub-collection of \left\{G_{λ} : λ ∈ Λ\right\}.

Continuing in this manner, we end up with a sequence of closed intervals I_{n} = [a_{n}, b_{n}] which satisfies

(a) I_{n+1} ⊆ I_{n} for all n ∈ \mathbb{N}

(b) For every n ∈ \mathbb{N}, E ∩I_{n} cannot be covered by a finite sub-collection of \left\{G_{λ} : λ ∈ Λ\right\}

(c) l(I_{n}) = \frac{1}{2} l(I_{n−1}) = \frac{b_{0} − a_{0}} {2^{n}} .

By Cantor’s Theorem 3.11 for nested intervals, the intersection \cap  I_{n} is not empty, and since

\inf \left\{l(I_{n}) : n ∈ \mathbb{N}\right\} = 0,

it contains exactly one point, say x. Since E ∩I_{n} is not empty for every n, we can choose a sequence (x_{n}) in E such that x_{n} ∈ I_{n} for every n ∈ \mathbb{N}. This implies

|x_{n} − x| ≤ l(I_{n}) = \frac {b_{0} − a_{0}} {2^{n}} → 0,

hence x = \lim x_{n}. But E is closed, so, by Theorem 3.20, x ∈ E. Since \left\{G_{λ} : λ ∈ Λ\right\} covers E, there is an open set G_{λ_{0}} in the cover such that x ∈ G_{λ_{0}} . There is therefore a positive number ε such that (x−ε, x+ε) ⊆ G_{λ_{0}} . Choose n so that

\frac {b_{0} − a_{0}} {2^{n}} = l(I_{n}) < ε,

from which we conclude that I_{n} ⊆ (x − ε, x + ε) ⊆ G_{λ_{0}} , and hence E ∩ I_{n} ⊆ G_{λ_{0}} . Thus a single element of the collection \left\{G_{λ} : λ ∈ Λ\right\} covers E ∩ I_{n}, thereby contradicting our choice of I_{n}, and proving that the assumption that no finite sub-collection of \left\{G_{λ} : λ ∈ Λ\right\} covers E is false.