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so i understood that the hartogs set of a well ordered set $A$ is defined as $H(A)$ the minimal ordinal such that $H(A)\nleq A$ (there is no injection from $H(A)$ to $A$) and i also uderstood the proof to the existance of an ordinal like that.

the only thing that im having trouble to understand is, giver well ordered set $A$ can i find its Hartogs set? because it feels pretty much like hartogs set will equal $S(\alpha) = \alpha \cup \{ \alpha \}$ (when $\alpha$ is the ordinal that is isomerphic to $A$.

now of course i cant just say that because i feel like it, so ill be glad if some could please give me an example when it doesnt work, or maybe just explain why we need this. thank you!

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No, that's not true.

The Hartogs number is not the least ordinal which does not "order embed" into a set, because "a set" does not have any well-ordering on it (although sometimes we have a natural candidate, e.g. $\in$ in the case of an ordinal). Instead the Hartogs number is only about injections.

So $\omega+1=\omega\cup\{\omega\}$ certainly injects into $\omega$. Indeed, every countable ordinal does. So the Hartogs number of $\omega$ is by definition an uncountable ordinal, in fact the least uncountable ordinals, for obvious reasons. In other words, it is $\omega_1$.

How do we construct it? Well, we can't really construct it with "elementary operations". In some sense the Hartogs function is an elementary operation, although in the proof of existence we go through:

  1. Look at all well-ordered chains of subsets of $X$ up to isomorphism;
  2. define the natural well-order on the equivalence classes;
  3. argue this well-order does not embed into $X$, and that it is the minimal one;
  4. in the case we want to use von Neumann ordinals (which is the usual case in $\sf ZF$), find the matching ordinal.

We don't start adding elements one by one to an ordinal, or go through some transfinite recursion. We utilise, in some sense, much larger sets than $X$, whereas ordinals smaller than $X$ are, well, by definition smaller than $X$.

But if $X$ can be well-ordered, then we can say that $H(X)$ is the smallest ordinal strictly larger in cardinality than $X$.

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