Math 411: Honours Complex Variables - University of Alberta

Lemma 13.1. Let D ⊂ C be open and connected, and let S ⊂ D be discrete. Then
D \S is open and connected.
Proof. Let z ∈ D \ S. Since S is discrete in D, there exists ǫ1 > 0 such that
Bǫ1(z)∩S = ∅. Also, since D is open, there exists ǫ2 > 0 with Bǫ2(z) ⊂ D. Setting
ǫ := min{ǫ1,ǫ2}, we get Bǫ(z). This proves the openness of D \S.
Assume that D\S is not connected. Then there exist open sets U �= ∅ �= V with
U ∩V = ∅ and U ∪V = D \S. Let s ∈ S, and choose ǫ > 0 such that Bǫ(s) ⊂ D
and Bǫ(s)∩S = {s}. Set W := Bǫ(s)\{s}, and note that W is open and connected.
Since (U ∩W)∩(V ∩W) = ∅ and (U ∩W)∪(V ∪W) = W, the connectedness of
W yields that either U ∩W = ∅ or V ∩W = ∅ and thus W ⊂ U or W ⊂ V.
Set
SU := {s ∈ S : there exists ǫ > 0 such that Bǫ(s)\{s} ⊂ U}
and
SV := {s ∈ S : there exists ǫ > 0 such that Bǫ(s)\{s} ⊂ V}.
By the foregoing, we have S = SU ∪SV, and trivially, SU ∩SV = ∅ holds. Set
Ũ := U ∪SU and ˜ V := V ∪SV.
Then Ũ �= ∅ �= ˜ V are easily seen to be open and clearly satisfy Ũ ∩ ˜ V = ∅ and
Ũ ∪ ˜ V = D, which contradicts the connectedness of D.
Theorem 13.1 (Meromorphic Functions Form a Field). Let D ⊂ C be open and
connected. Then the meromorphic functions on D, where we define (f + g)(z) =
lim[f(w)+g(w)]
and (fg)(z) = lim[f(w)g(w)],
form a field.
w→z w→z
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