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Feb 28, 2022 at 15:56 vote accept Peluso
Feb 26, 2022 at 21:32 comment added Peluso How curious, not even one of the natural weakenings of the statement is true. Furthermore, if you weaken "perfect" to "countable pseudocharacter" the statement does not hold either: the square of Alexandroff-Urysohn double arrow, $\mathbb{A}^{2}$, is a compact, $T_4$ and first countable space with $|\mathsf{RO}(\mathbb{A}^{2})|<|\tau_{\mathbb{A}^{2}}|$. Perhaps there are other stronger and different combinations of properties that do imply the desired equality, we just don't know them yet.
Feb 26, 2022 at 14:47 history edited Anonymous CC BY-SA 4.0
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Feb 26, 2022 at 14:37 history answered Anonymous CC BY-SA 4.0