Abstract
If X is a compact subset of the complex plane, C(X) will denote the sup norm algebra of continuous complex-valued functions on X; R(X) will denote the uniformly closed subalgebra of C(X) generated by the rational functions with poles off X. Whenever X has an interior, R(X) is a proper subalgebra of C(X). Mergelyan [I ] has shown that it is still possible for R(X) to be strictly contained in C(X) even if X has no interior: R(X) C(X) whenever X is a set formed by deleting from the interior of the closed unit disc a sequence of open discs with mutually nonintersecting boundaries of finite total length in such a way as to leave no interior. In order to understand how this R(X) could be a proper subalgebra of C(X), it had been conjectured that any such R(X) must be antisymmetric, i.e., must contain no nonconstant real-valued function. This conjecture will be proved false by the construction of a counterexample. The strategy of construction is to mark off on the horizontal diameter of the unit disc a general Cantor set. The vertical lines determined by the endpoints of the intervals complementary to the Cantor set then serve as guidelines which will be covered almost everywhere by open discs to be deleted from the unit disc. In order to insure that the total length of the circumferences of the deleted discs is finite, it is necessary to alternate between selecting an interval complementary to the Cantor set, and selecting the discs which cover the associated pair of guidelines: at each stage we must select the interval in such a way (i.e., large enough) as to insure that a large portion of the length of the associated guidelines is already deleted by the discs which were selected at the preceding stages. The final step-that of showing that the algebra R(X) is not antisymmetric-involves proving that the Cantor function (extended as a constant in the vertical direction) is in R(X). This proof relies on a lemma due to Mergelyan [1], which may be stated in this form:
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