How can I show that,
if $\lambda$ and $\kappa$ are two ordinals (limit or successors) and if $\lambda < \kappa$,
then $\lambda\cap\kappa<\kappa$
Ordinal properties: if $\lambda<\kappa$ then $\lambda\cap\kappa<\kappa$
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elementary-set-theory
ordinals
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0What exactly you mean by $A\cap B$ for ordinals? For example if you define ordinals as equivalence classes then this doesn't make much sense. Anyway there's always an embedding $i:A\cap B\to B$, $i(x)=x$ that preserves an order (I assume that both $A$, $B$ share ordering, i.e. they are subsets of some well ordered set). Hence always $A\cap B < B$ (no mattter how $A$ and $B$ are related). – 2017-02-21
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0There is nothing precise written about intersection, which is why I suppose just the standard notion which corresponds to those and only those elements which are contained in both of the ordinals. – 2017-02-21
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0Freakish : ordinals aren't usually defined as equivalence classes; I think here the definition used is that of a transitive set well ordered by $\in$. – 2017-02-21
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1@Eurydice : $\lambda\in\kappa$, since $\kappa$ is an ordinal this implies $\lambda\subset \kappa$. Therefore... – 2017-02-21
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4isn't $\lambda\cap\kappa=\lambda$???, since $\forall \xi,\xi<\lambda\iff\xi<\lambda<\kappa$ – 2017-02-21
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1Yep, if we define ordinals as transitive sets, then this is trivial as @JulioMaldonadoHenríquez notes. – 2017-02-21