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Jun 9 at 14:48 comment added Goldstern @bof Yes, in my whole answer the set X is infinite. For finite sets there is the 1935 paper by Webb, which also gives a single binary function.
Jan 1 at 14:13 comment added bof @Goldstern In that theorem of Sierpiński the given set $X$ is assumed to be infinite, right? If $X$ is a finite set with at least $3$ elements, you need $3$ selfmaps to generate them all by composition: two bijections and one that is not bijective.
Oct 15, 2016 at 20:28 comment added Goldstern Zusmanovich (2015): arxiv.org/abs/1408.2982 and sciencedirect.com/science/article/pii/S0723086915000894 . Łoś (1950): matwbn.icm.edu.pl/ksiazki/fm/fm37/fm3718.pdf
Aug 8, 2014 at 14:05 comment added Goldstern The preprint On the last question of Stefan Banach by Pasha Zusmanovich points out that the result in my answer was already proved by Jerzy Łoś in 1950 (Fund.Math 37, p. 84-86). Łoś uses this theorem of Sierpiński: Every countable set of unary functions on a given set X is included in the set generated by two unary functions.
Feb 11, 2014 at 13:00 comment added Goldstern As I just discovered, Trevor Evans showed this (at least for countable base sets) in 1989 in his paper "Embedding and representation theorems for clones and varieties", Bull. Austral. Math. Soc. 40. MR1012828 (90k:08004)
Dec 3, 2012 at 15:45 comment added Goldstern Reference: Martin Goldstern, A single binary function is enough. Contributions to general algebra 20, 35–37, Heyn, Klagenfurt, 2012. Abstract: (1) Every countable clone (on any base set) is contained in a 1-generated clone. (2) For a countable base set, the local clone generated by a single function f will be the full clone of all operations (this holds for many operations f). (3) But on an uncountable base set a finitely or countably generated local clone will never contain all operations. See Math Reviews MR2908433.
May 13, 2011 at 17:37 vote accept Joel David Hamkins
May 13, 2011 at 14:43 history edited Goldstern CC BY-SA 3.0
z, not x
May 13, 2011 at 13:27 comment added Joel David Hamkins Great! This seems very clear. (But I think you mean $(z,n+1)$ in the first paragraph.) I am inclined to accept this answer as the most sweeping general solution provided.
May 12, 2011 at 17:25 comment added Goldstern I forgot to mention that for finite sets there is a 1935 paper of Donald L. Webb (Zentralblatt 0012.00103), pointed out to me by Agnes Szendrei, which shows that there is a single binary operation generating all functions, a generalization of the Sheffer stroke: $f(x,x) = x+1$ modulo $n$, and $f(x,y) = 0$ otherwise.
May 12, 2011 at 17:08 history edited Goldstern CC BY-SA 3.0
fixed stupid error
May 12, 2011 at 17:02 history answered Goldstern CC BY-SA 3.0