More on separation of a diagonal
Following the second author a completely regular \(T_1\)-space \(X\) is called functionally \(\Delta\)-paracompact if for every \(A\subset X^2\setminus\Delta_X\) closed in \(X^2\) there exists a locally finite cover \({\mathcal U}\) of \(X\) consisting of co-zero sets such that \(\bigcup\{U\times U\mid U\in{\mathcal U}\}\) does not meet \(A\). It is known from the second author's paper [ibid. 157, No.~2, 352--358 (2010; Zbl 1187.54020)] that every paracompact Hausdorff-space and every GO-space is \(\Delta\)-paracompact. It is also known that a completely regular \(T_1\)-space is \(\Delta\)-paracompact if and only if it is divisible in the sense of \textit{Á. Császár} [General topology. Bristol: Adam Hilger Ltd. Budapest: Akademiai Kiado. 488 p. (1978; Zbl 0366.54001)]. In this paper the following is shown. Theorem 1. Every first countable, countably compact, monotonically normal space is functionally \(\Delta\)-paracompact. A completely regular \(T_1\)-space \(X\) is called regular \(\Delta\)-paracompact if for every \(A\subset X^2\setminus\Delta_X\) closed in \(X^2\) there exists a locally finite open cover \({\mathcal U}\) of \(X\) such that \(\bigcup\{\text{cl\,} U\times\text{cl\,} U\mid U\in{\mathcal U}\}\) does not meet \(A\). Theorem 2. Every regular \(\Delta\)-paracompact space is collectionwise normal. It is known that functionally \(\Delta\)-paracompactness implies regular \(\Delta\)-paracompactness. Question 1. Is every regular \(\Delta\)-paracompact space functionally \(\Delta\)-paracompact? Obviously, every regular \(\Delta\)-paracompact space is \(\Delta\)-paracompact, i.e., has the property that for every \(A\subset X^2\setminus\Delta_X\) closed in \(X^2\) there exists a locally finite open cover \({\mathcal U}\) of \(X\) such that \(\bigcup\{U\times U\mid U\in{\mathcal U}\}\) does not meet \(A\). Theorem 3. Every \(\Delta\)-paracompact space is strongly collectionwise Hausdorff. Theorem 4. Every \(\Delta\)-paracompact pseudocompact space is countably compact. Every functionally \(\Delta\)-paracompact space is functionally \(\Delta\)-normal, i.e., has the property that for every \(A\subset X^2\setminus\Delta_X\) closed in \(X^2\) there exists a continuous function \(f: X^2\to[0,1]\) such that \(f[A]= \{1\}\) and \(f[\Delta_X]= \{0\}\). Example 1. There exists a functionally \(\Delta\)-normal space that is not \(\Delta\)-paracompact. A negative answer to the following question would also answer Question 1. Question 2. Is every regular \(\Delta\)-paracompact space functionally \(\Delta\)-normal? Obviously, every functionally \(\Delta\)-normal space is \(\Delta\)-normal in the sense of \textit{K. P. Hart} [Colloq. Math. 53, 49--56 (1987; Zbl 0633.54010)]. However, it is still not known whether the two concepts coincide. Question 3. Is every \(\Delta\)-normal space functionally \(\Delta\)-normal? By two interesting examples it is shown that functional \(\Delta\)-normality is quite different from normality. Example 2. There exists a functionally \(\Delta\)-normal space that is not normal. Example 3. There exists a normal space with a \(G_\delta\)-diagonal that is not \(\Delta\)-normal.
functionally \(\Delta\)-paracompact space
scientific article; zbMATH DE number 5775742