Timeline for Compact definition of ordinals

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Sep 18, 2019 at 21:34	comment	added	Alexander Kuleshov		Ahaha that's where my $\varepsilon-\delta$ habit failed, I always forget that there is no $m$ such that $n<m<n+1$... Thanks) In fact doesn't matter, still $\beta \in x$ !
Sep 18, 2019 at 21:27	comment	added	Nik Weaver		Almost --- for each $\beta < \alpha$ there exists $y \in x$ such that ${\rm rank}(y) \geq \beta$. (This covers the case where $\alpha$ is a successor too, though it wouldn't be any trouble to just consider that case separately.)
Sep 18, 2019 at 21:14	comment	added	Alexander Kuleshov		Thanks, Nik! Now I briefly looked through some definitions and rewrote "the ranks of elements of $x$ are cofinal in $\alpha$" in more familiar to me $\varepsilon-\delta$ manner: "for each ordinal $\beta < \alpha$ there exists $y \in x$ such that $rank(y)>\beta$". But $y$ is an ordinal by induction hypothesis, so $rank(y)=y>\beta\Rightarrow \beta\in y \in x\Rightarrow \beta \in x$ by transitivity. Therefore $x$ is is a set of all the ordinals $\beta < \alpha$ thus $x=\alpha$ by the property of (von Neumann) ordinals. Did I get it correctly?
Sep 18, 2019 at 20:03	history	answered	Nik Weaver	CC BY-SA 4.0