Compactness in the Complex Plane

Compactness in the complex plane In our text we had the following Theorem as Proposition 1.4.18, here we have a proof of it. Recall that a set K is compact if every open cover of K contains a finite subcover of K. Also we use the notation D(z, ε) = {w ∈ C : |w − z| < ε}, i.e. the open disk centered at z with radius ε. Theorem 1 The following conditions are equivalent for a subset K of C (or of R): (i) K is closed and bounded. (ii) Every sequence of points in K has subsequence which converges to some point in K. (iii) K is compact. Proof. We first prove (i)⇒(ii). Suppose K is closed and bounded, consider a sequence {zj }∞ j=1 in K, it is enough to show that this sequence contains a convergent subsequence, because since K is closed we already know (see proposition 1.4.8) that the limit of such sequence would be some point in K. Since K is bounded then there is a disk centered at the origin containing K, by possibly shrinking and translating K, we can assume that K is contained in the square S1 with vertices 0, 1, 1 + i, and i. Notice that the shrinking and translating do not change the convergence behavior of the sequence. Now, consider the following recursive procedure: Using a vertical segment and a horizontal segment divide S1 into four squares of the same size S11 (with vertices 0, 12 , 12 + 2i , 2i ), S12 (with vertices 12 , 1, 1 + 2i , 12 + 2i ), S13 (with vertices 1 + 2i , 1 + 2i , 1 + i, 12 + i), and S14 (with vertices 2i , 12 + 2i , 12 + i, i), see figure 1. Since the 2 sequence {zi }∞ i=1 is infinite, there must be one of these squares with an infinite number of points. If S1j is such square, pick an arbitrary point in it, call it w1 , and now only consider those points in the sequence {zj }∞ j=1 with indices j larger than the index of w1 . Now, repeat the same procedure using S1j instead of S1 and proceed recursively. In this way we construct ∞ a sequence {wi }∞ i=1 where the indices of the wj ’s in the sequence {zj }j=1 are in order, and moreover each wj is contained in a square of side 1/2j and every subsequent wk (with k ≥ j) is also contained in the same square. Therefore |wj − wk | < 1 for every j ≥ k 2j 1 Figure 1: proof of (i) ⇒(ii) which means that the sequence {wj }∞ j=1 is a Cauchy sequence and therefore it is convergent. Now we prove (ii)⇒(iii). Suppose that every sequence in K contains a convergent subsequence with limit in K. Consider an open cover C = {Uα }α∈A of K. We first prove the following. Claim 2 There is ε > 0 such that every z ∈ K satisfies that there is α = α(z) such that D(z, ε) ⊆ Uα . Suppose not, then for every ε, in particular for ε = n1 with n ∈ N, there is a point zn ∈ K such that for every α ∈ A we have that D(zn , n1 ) * Uα . Consider the sequence {zn }, by assumption there is a convergent subsequence {znk } → z ∗ . Since C is a cover we know there Figure 2: proof of the claim is α∗ ∈ A such that z ∗ ∈ Uα∗ , since Uα∗ is open there is δ > 0 such that D(z ∗ , δ) ⊆ Uα∗ . But since {znk } → z ∗ there is N such that |z ∗ − znk | < δ/2 whenever nk ≥ N. Pick nk ≥ N and also nk > 2δ . Notice that if w ∈ D(znk , 2δ ) then |z ∗ − w| = |z ∗ − znk + znk − w| ≤ |z − znk | + |znk − w| < 2 δ δ + = δ, 2 2 thus D(znk , n1k ) ⊆ D(znk , 2δ ) ⊆ D(z ∗ , δ) ⊆ Uα∗ which contradicts the fact that D(zn , n1 ) * Uα for every α ∈ A. This completes the proof of the claim. Now, for every z ∈ K let Dz = D(z, ε) the disk given by the claim. Observe that {Dz }z∈K is an open cover of K. Suppose we know how to extract a finite subcover {Dz1 , Dz2 , . . . , DzM } of K, then since each Dzj is contained in some Uαj we can deduce that {Uα1 , Uα2 , . . . , UαM } is also a finite subcover. So it is enough to show that {Dz }z∈K has a finite subcover. Let z1 ∈ K, if Dz1 covers K we are done, else there is z2 ∈ K\Dz1 . If {Dz1 , Dz2 } covers K we are done, else there is z3 ∈ K\ (Dz1 ∪ Dz2 ). If we continue this process we either obtain a finite cover or a sequence {zj }. Observe that |zj − zk | > ε for every j and k since zk is not in Dzj and biceversa. Thus the sequence {zj } cannot have a convergent subsequence which is a contradiction to our main assumption. Therefore the process must end at some point and we should have a finite subcover of K. Finally we prove (iii)⇒(i). Suppose K is compact. Consider the collection C = {D(0; n) : n ∈ N}, clearly C is an open cover of K (in fact of all C), thus, since K is compact, there is a finite subcover CF = {D(0, n1 ), D(0, n2 ), . . . , D(0, nM )} of K. If we assume that n1 < n2 < · · · < nM then clearly K ⊆ D(0, nM ), i.e., K is bounded. Figure 3: proof of (iii)⇒(i) Now, to prove that K is closed we prove instead that C\K is open. Let z ∈ C\K, for every w ∈ K consider the disk Dw = D(w, 12 |w − z|) and let C = {Dw : w ∈ K}. Clearly C is an open cover of K since every point w in K has a disk centered at it in C. Since© K is compact, there is ª a finite subcover CF = {Dw1 , Dw2 , . . . , DwM }. Now, let ε = min 12 |wj − z| : 1 ≤ j ≤ M . Observe that since 12 |z − wj | > ε then z is farther apart than ε from every point in Dwj . Indeed, if w ∈ Dwj (i.e., |w − wj | < 12 |wj − z|) then 1 1 |z − w| = |(z − wj ) + (wj − w)| ≥ |z − wj |−|wj − w| > |z − wj |− |z − wj | = |z − wj | > ε. 2 2 Thus D(z, ε) ∩ Dwj = ∅ and then D(z, ε) ⊆ C\ M [ j=1 Dwj ⊆ C\K which proves that C\K is open, i.e. K is closed. 3