It is "a well-known theorem of Cantor", said Sierpinski (circa 1920), that every countable total order can be imbedded in the rationals, and he proceeds to demonstrate that, assuming the continuum hypothesis, it is possible to construct a similar "universal order" of cardinal $\aleph_1$. I have two questions : 1) Where did Cantor prove his theorem 2) Can CH be weakened in the previous result ? For reference, Sierpinski construct an ordering on binary sequences indexed by countable ordinals (up to some ordinal $<\omega_1$), having the desired property than any countable Dedekind cut is separated by some element ; this is actually the same order than the one on surreal numbers born before day $\omega_1$ (as easily shown by the Gondor construction)
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asked Jul 11, 2012 at 21:53
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2 Answers
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Concerning question 2:
a. It is consistent that $2^{\aleph_0}= \aleph_2$ and there is a universal order of size $\aleph_1$.
b. It is consistent that $2^{\aleph_0}= \aleph_2$ and there is no universal order of size $\aleph_1$.
See Kojman+Shelah, JSL. preprint
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$\begingroup$ In his 1895/97 papers "Beiträge zur Begründung der transfiniten Mengenlehre", Cantor proved that any two countable dense orders without endpoints are isomorphic, according to this article by Charles Silver. The method used by Cantor also shows that every countable linear order embeds into the rationals. I do not have Cantor's paper here, so I am not sure if Cantor stated this result. $\endgroup$ Jul 11, 2012 at 22:50
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Sorry, almost all pertinent answers (even with the exact reference of the Sierpinski article) were given in the article Universal order type ; so the only question I have left is "where did Cantor state it ?"
answered Jul 11, 2012 at 22:28