2021/01/25 by Will Brian, Brian, Will
Mathematics · #Advanced Topology and Set Theory #Advanced Banach Space Theory #Mathematical and Theoretical Analysis
paper · pdf · doi:10.48550/arxiv.2101.10088
Given a completely metrizable space X, let \mathfrakpar(X) denote the smallest possible size of a partition of X into Polish spaces, and \mathfrakcov(X) the smallest possible size of a covering of X with Polish spaces. Observe that \mathfrakcov(X) ≤ \mathfrakpar(X) for every X, because every partition of X is also a covering. We prove it is consistent relative to a huge cardinal that the strict inequality \mathfrakcov(X) < \mathfrakpar(X) can hold for some completely metrizable space X. We also prove that using large cardinals is necessary for obtaining this strict inequality, because if \mathfrakcov(X) < \mathfrakpar(X) for any completely metrizable X, then 0^† exists.