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May 12, 2020 at 10:22 comment added Dominic van der Zypen Right @gowers, in that case $E$ is the (only) maximal partite subset of itself. But the question is, even if $H=(V,E)$ is not partite, can we find $E_1 \subseteq E$ such that $(\bigcup E_1, E_1)$ is partite, but no proper superset of $E_1$ has this property?
May 12, 2020 at 7:25 comment added gowers I don't understand the question. What's to stop $E$ itself being partite, in which case all its subsets will be partite?
May 10, 2020 at 14:44 history asked Dominic van der Zypen CC BY-SA 4.0