Suppose we are given in C: (i) a sequence (an) of distinct points such that the sequence (∣an∣) is increasing and tends to +∞; (ii) for each n, a rational function z→Rn(z) defined in C−{an} and having a pole at an; (iii) an increasing sequence (Bn) of open discs centred at 0, whose union is C, and such that none of the ak is on the frontier of any of the discs Bn. For each n, let Bn′ denote the intersection of Bn and the complement in C of the set of points an; and let Xn denote the set of all mappings
z→S(z)=P(z)+ak∈Bn′∑Rk(z)
of Bn′ into C, where P is the restriction to Bn′ of a function which is continuous in Bn and holomorphic in Bn. We define a metric in Xn by putting
dn(S1,S2)=z∈Bn′sup∣S1(z)−S2(z)∣.
It is easily verified that Xn is complete with respect to this metric. Finally, for n≤m, we define a mapping fnm:Xm→Xn such that if S∈Xm then fnm(S) is the restriction of S to Bn′. It is clear that the fnm are uniformly continuous and that (Xn,fnm) is an inverse system of uniform spaces. This being so, an element of the inverse limit X=lim←Xn can be canonically identified with a meromorphic function F in C, whose only poles are the points an, and which is such that for each n, F(z)−Rn(z) is holomorphic at an. The classical theorem of Mittag-Leffler asserts that X is not empty; by virtue of Theorem 1, we have only to verify the condition (MLnn) for all n. Let
Sn=Pn+ak∈Bn∑Rk be an element of
Xn, where
Pn is continuous in
Bn and holomorphic in
Bn; for any
m≥n, let
Qmn be the restriction of
ah∈Bm−Bn∑Rh to Bn′; this latter sum is a holomorphic function in some neighbourhood of
Bn, hence (by Taylor's theorem) for each
ε>0 there is a polynomial
Pmn such that
∣Qmn(z)−Pmn(z)∣≤ε in
Bn; if
Sm is the restriction of
Sn+Qmn−Pmn to
Bm′, we have
Sm∈Xm and
∣Sm(z)−Sn(z)∣≤ε in
Bn′. This completes the proof. \*
Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.
Nothing.