Compactness and completeness in partial metric spaces (Q2291584)

A symmetric mapping \(p \colon X^{2} \to [0, \infty)\) is a partial metric on \(X\) if the following conditions hold for all \(x\), \(y\), \(z \in X\): \(\bullet\) \(x = y \Leftrightarrow p(x, x) = p(x, y) = p(y, y)\); \(\bullet\) \(p(x, x) \leqslant p(x, y)\); \(\bullet\) \(p(x, z) \leqslant p(x, y) + p(y, z) - p(y, y)\). Several interesting examples of partial metrics are constructed. In particular it is shown that: \(\bullet\) There exists a Hausdorff compact partial metric space which is not complete; \(\bullet\) There exists a compact partial metric space \((X, p)\) such that the space \((X, r)\) is not complete for any partial metric \(r\) which is equivalent to \(p\); \(\bullet\) There exists a not compact, partial metric space \((X, p)\) which is completely regular, separable, perfect and pseudocompact. The authors also prove the following surprising proposition which gives a negative answer to a question formulated in [\textit{H. Suzhen} et al., Topology Appl. 230, 77--98 (2017; Zbl 1377.54029)]. Let \(X = \{x_n \colon n \in \mathbb{N}\}\) and \(p(x_n, x_m) = \max\{n, m\}\) for every \(n\), \(m \in \mathbb{N}\). Then the partial metric \(p \colon X^{2} \to [0, \infty)\) is not equivalent to any bounded complete partial metric on \(X\).