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Jan 2, 2017 at 1:31 answer added user44191 timeline score: 2
Jan 1, 2017 at 23:24 comment added Harry Altman So combining this fact and the answers and comments below: The minimum number is $2^{|X|}$ if $X$ is uncountable, $|X|$ if $X$ is countable and nonempty, and $1$ if $X$ is empty.
Jan 1, 2017 at 3:29 comment added user44191 @SamHopkins It's true for finite $X$; the proof starts by checking it for $X$ "connected" as defined by Will Sawin's graph below, and then inducting on components of $X$ by using the fact that $xy \geq x + y$ if $x, y > 1$ if no component has one element, and using the "extra" map that maps everything to that one element if one does.
Dec 31, 2016 at 8:56 vote accept Dominic van der Zypen
Dec 30, 2016 at 16:07 comment added Sam Hopkins Is it false for finite $X$?
Dec 30, 2016 at 16:00 answer added Will Sawin timeline score: 7
Dec 30, 2016 at 15:12 answer added Goldstern timeline score: 5
Dec 30, 2016 at 14:27 answer added Stefan Mesken timeline score: 8
Dec 30, 2016 at 11:01 history asked Dominic van der Zypen CC BY-SA 3.0