Lemma 1
Tag 02HB, permanent. Cite it as [Bourbaki, Tag 02HB].
Théories spectrales, chapter I, § 8. Algèbres de fonctions continues sur un espace compact
Let X be a compact subset of the plane and let O be a bounded connected component of X. The boundary of O is contained in X.
The closure of the set O in X is equal to X) where is its closure in . Since O is a connected component of X, we therefore have X) = O, which proves that .
Let X be a compact subset of . Let be the unbounded connected component of X, and let be the family of bounded connected components of X, the subsets being pairwise distinct.
Let E be a subset of X. We denote by the closure in of the set of functions , where is a rational function on all of whose poles belong to E. The algebra is a unital Banach subalgebra of which separates the points of X. Let be the identity function on X. The full closed subalgebra of generated by is equal to (Lemma 2 of I, p. 6). The elements of are holomorphic in the interior of X.
In particular, one has . We write . Let I(E) be the set of elements such that , and
The set is compact, being bounded and closed, its complement in being the union of the open set and of the open sets which meet E.
Bourbaki sets the proof directly after the statement, and the corpus does not separate them, so what is above is the statement and its proof together.
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