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Nov 20, 2015 at 14:38 comment added Goldstern @PeterLeFanuLumsdaine: Second: You really think that if we let $\alpha_n:=n$, and $\alpha_{\omega+\beta}:= \aleph_{\beta}$, then the scale of alphas would be "more natural" than the scale of alephs? For "more natural" it is not enough for the definition to look nicer, it also has to be useful. Given that most theorems of set theory deal with infinite cardinals only, I would expect that most definitions and theorems would become more complicated. A characterisation (or definition) of GCH, for example, would now read "$2^{\aleph_\alpha}=\aleph_{\alpha+1}$ for all $\alpha\ge \omega$".
Nov 20, 2015 at 14:30 comment added Goldstern @PeterLeFanuLumsdaine: First: Even if you would use a different indexing for the alephs, this would not invalidate my point. There still are many other examples where you have to distinguish between 0 and nonzero limit ordinals in inductive definitions/proofs, such as the inductive definition of ordinal arithmetic.
Nov 20, 2015 at 0:16 comment added Peter LeFanu Lumsdaine @Goldstern: This simply shows that we have chosen the wrong indexing for the alephs. If we use that definition at all limit ordinals (and cardinal successor at successor ordinals as usual), then the resulting hierarchy starts by enumerating the finite cardinals, and than at stage $\omega$ it reaches the usual $\aleph_0$ and continues with the normal alephs from there. Very natural, no? It also removes the need to give special treatment to $n=0$ in various basic propositions about cofinalities, etc.
Jan 25, 2013 at 11:39 comment added Goldstern It makes sense to call 0 a limit ordinal, and some authors do so. But often you have to distinguish between 0 and nonzero limit ordinals in inductive definitions/proofs. For example: $\aleph_\delta$ is defined as $\bigcup_{\alpha < \delta} \aleph_\alpha$ for nonzero limit ordinals,whereas $\aleph_0$ is defined as $\omega$.
Jan 1, 2012 at 6:40 history edited Andrés E. Caicedo CC BY-SA 3.0
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Oct 22, 2010 at 17:53 comment added John Wiltshire-Gordon I've always felt that zero should be a limit ordinal. The statement of transfinite induction benefits from this point of view: we may omit the first hypothesis! By the way, is there a good reason to exclude zero from the definition of limit ordinal? It seems artificial to me.
Oct 22, 2010 at 15:15 comment added gowers I agree that that's three hypotheses, but they aren't given by adjectives or adjectival phrases. In theory they could be: one could say something like this (my definitions are made up but their meanings should be obvious). Let A be a set of ordinals. If A is properly rooted, successor-closed and limit-closed, then A contains all ordinals.
Oct 22, 2010 at 13:57 history answered Joel David Hamkins CC BY-SA 2.5