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math104-f21:heine-borel

Heine-Borel theorem

\gdef\In{\subset}

The theorem states that, for the metric space Rn\R^n, a subset is compact if and only if it is closed and bounded.

We will prove this theorem for R\R first, then we show that if AXA \In X is compact and BYB \In Y is compact, then A×BX×YA \times B \In X \times Y is compact.

For any metric space XX, a compact subset KK is bounded, i.e there exists a real number M>0M>0, such that for any x,yKx,y \in K, d(x,y)<Md(x,y) < M. Suppose not, then for any x0Kx_0 \in K, and n>0n > 0, K\Bn(x0)K \RM B_n(x_0) is not empty. Consider the open cover n>0Bn(x0)\cup_{n>0} B_n(x_0), then it has no finite sub-cover.

Hence, we have proven that, compactness implies closed and bounded.

The converse in general is not true. Consider any infinite set XX with discrete metric, d(x,y)=δx,yd(x,y) = \delta_{x,y}. Then, XX is closed and bounded, but not compact, since X=xX{x}X = \cup_{x \in X} \{x\}, and {x}=B1/2(x)\{x\} = B_{1/2}(x) are open sets.

It is true in R,Rn\R, \R^n.

Heine-Borel for R\R

Proposition Any closed interval [a,b][a,b] is compact. Let {Uα}\{U_\alpha\} be an open cover of [a,b][a,b]. Suppose there is no finite subcover, then we can divide the interval in half and find out which half is causing the trouble. Let I0=[a,b]I_0 = [a,b], let I1[a,b]I_1 \In [a,b] be either [a,(a+b)/2][a, (a+b)/2] or [(a+b)/2,b][(a+b)/2, b], that does not admit finite subcover. Repeating this process, we get a nested sequence of closed intervals I0I1I_0 \supset I_1 …, the diameter In=1/2In1|I_n| = 1/2 |I_{n-1}|. Let In=[an,bn]I_n =[a_n,b_n], then ana_n is a bounded increasing sequence, and bnb_n is a bounded decreasing sequence. We have a limit nIn={x}\cap_n I_n = \{x\}. But {x}\{x\} is counted in certain UαU_\alpha, and there is a r>0r>0, such that Br(x)UαB_r(x) \In U_\alpha. Hence, if In<r|I_n| < r, we have InBr(x)UαI_n \In B_r(x) \In U_\alpha, contradicting with InI_n does not admit finite subcover.

Corollary Any closed bounded subset in R\R is compact. Since it is a closed subset, and contained in a compact subset [M,M][-M, M] for MM large enough.

Others

Proposition If {Kα}\{K_\alpha\} is a collection of compact sets, such that any finite subcollection of them has non-empty intersection, then αKα\cap_{\alpha} K_\alpha \neq \emptyset.

Proof: Suppose not, then αKαc=X\cup_{\alpha} K_\alpha^c = X. Choose any α0\alpha_0, then αα0KαcKα0\cup_{\alpha \neq \alpha_0} K_\alpha^c \supset K_{\alpha_0}. By compactness, we have a finite sub-cover of Kα0K_{\alpha_0} Kα0Kα1cKαnc K_{\alpha_0} \In K_{\alpha_1}^c \cup \cdots \cup K_{\alpha_n}^c This means Kα0Kα1Kαn= K_{\alpha_0} \cap K_{\alpha_1} \cap \cdot \cap K_{\alpha_n} = \emptyset get contradiction.

Proposition If EKE \In K is an infinite subset, then EE has a limit point in KK.

If not, then every point in xKx \in K has an open ball UxU_x containing at most one point in EE. By compactness of KK, KK is covered by finitely many such UxU_x, hence EE is also covered by such finitely many UxU_x, which contradicts with EE being infinite.

Corollary If (xn)(x_n) in KK is a sequence, then there exists a convergent subsequence. (Compactness implies sequential compactness.)

If the value set V={xn}V = \{x_n\} is finite, then there is a value which appears infinite many times in the sequence, and we can take the constant subsequence. If the value set is infinite, then we by proposition, there exists a xKx \in K, such that for any r>0r>0, Br(x)VB_r(x) \cap V is an infinite set, hence {nNxnBr(x)}\{n \in \N| x_n \in B_r(x) \} is infinite, thus xx is a subsequential limit of (xn)(x_n).

math104-f21/heine-borel.txt · Last modified: 2022/01/11 08:36 by pzhou