368 CHAPTER 14. FUNDAMENTALS
ρ (θw,θu) ≥ ρ (θz,θu) ≡ d (z,u). Thus it is a metric space. To say that zn→ ∞ is to saythat ρ (θzn,θ∞)→ 0. From the diagram, this means limn→∞ |zn|= ∞.
Suppose you have {zn} a sequence in Ĉ. If the sequence has no subsequence whichconverges to ∞, then there exists ε > 0 such that for n large enough, ρ (θzn,θ∞)≥ ε whichcorresponds to {zn} remaining in a compact subset of C for all n large enough. Hence thereis a convergent subsequence which converges to some point of C. ■
The mapping θ is clearly continuous because it takes converging sequences, to con-verging sequences. Furthermore, it is clear that θ
−1 is also continuous. The notion ofconnected sets is the same as earlier. The usual definition of being connected if it is notseparated will be retained in this extended complex plane.
Example 14.12.3 Consider the open set S≡ {z ∈ C such that Im(z)> 0} . Let S∪{∞} ≡Ŝ. This is connected in Ĉ .
It is clear that θ (S∪{∞}) is half of S2 including the top point, obviously an arcwiseconnected set. Thus Ŝ is connected.
Definition 14.12.4 A connected set S ⊆ C is said to be simply connected if S isconnected and also Ĉ \S is connected in Ĉ. Equivalently θ (S) and θ
(Ĉ\S
)are connected
subsets of the Riemann sphere.
Example 14.12.5 Consider the set S ≡ {z ∈ C such that |z|> 1} . This is a connected set,but it is not simply connected.
Consider θ (S) . This is connected, but θ
(Ĉ\S
)is not.
Is there an easy way to see that an open connected set is NOT simply connected? Theanswer is yes, using the Jordan curve theorem. Suppose you have an open connected setΩ and a simple closed curve contained in Ω called Γ. If there is a point z ∈UΓ the insideregion of Γ which is not in Ω, then Ω cannot be simply connected because, letting VΓ be theunbounded component of ΓC, Ĉ\Ω⊆ V̂Γ∪UΓ disjoint open sets in Ĉ and there are pointsof Ĉ \Ω in each of these disjoint open sets. Therefore, Ĉ \Ω cannot be connected. HereV̂Γ ≡VΓ∪{∞}.
Example 14.12.6 Consider the set S ≡ {z ∈ C such that |z|> 1} . This is a connected set,but it is not simply connected. Indeed, {z : |z|= 1.5} is in S but its inside isn’t. From theoriginal definition, SC = {∞}∪{z : |z| ≤ 1} which is clearly not connected in Ĉ.
Example 14.12.7 Consider S ≡ {z ∈ C such that |z| ≤ 1} . This connected set is simplyconnected because Ĉ \S corresponds to a connected set on S2.
14.13 Exercises1. Suppose U is an open subset of C and f : U → R is analytic. Describe f on the
connected components of U .
2. Suppose f : C→ C is analytic. Let g(z) ≡ f (z̄) . Show that g′ (z) does not existunless f ′ (z̄) = 0.