On Hausdorff operators in ZF$\mathsf {ZF}$

Abstract

AbstractA Hausdorff space is called effectively Hausdorff if there exists a function F—called a Hausdorff operator—such that, for every with , , where U and V are disjoint open neighborhoods of x and y, respectively. Among other results, we establish the following in , i.e., in Zermelo–Fraenkel set theory without the Axiom of Choice (): is equivalent to “For every set X, the Cantor cube is effectively Hausdorff”. This enhances the result of Howard, Keremedis, Rubin and Rubin [13] that is equivalent to “Hausdorff spaces are effectively Hausdorff” in . The Boolean Prime Ideal Theorem and the statement “For every infinite set X, the Stone space of the Boolean algebra is effectively Hausdorff” are mutually independent. In particular, the latter statement is not provable in . The Axiom of Choice for non‐empty subsets of () is equivalent to each of “Separable Hausdorff spaces are effectively Hausdorff” and “The Cantor cube is effectively Hausdorff”. The Principle of Dependent Choices in conjunction with the Axiom of Choice for continuum sized families of non‐empty subsets of does not imply the axiom of choice for partitions of . The latter independence result fills the gap in information in Howard and Rubin's book “Consequences of the Axiom of Choice”. The axiom of countable choice for non‐empty subsets of is equivalent to each of “Denumerable Hausdorff spaces are effectively Hausdorff”, “Denumerable T3 spaces are completely normal” and “Denumerable Tychonoff spaces are Urysohn”.