This post follows the approach of Kelley General topology.

I ask three questions about $\in$ relation and definition of the ordinal. I think they are related and can be asked in the same post. If you don't think so please feel free to add a comment and I'll create new posts.

1.- We say that a set (class) $X$ is transitive (complete, saturated) if every element of $X$ is also a (sometimes proper) subset of it. Namely,

$$X\subseteq \{x:x\subset X\}.$$

2.- We refer as the epsilon class to the class

$$\mathcal E = \{(x,y):x\in y\}.$$

Now, an ordinal is defined as a transitive class $\alpha$ such that $\mathcal E$ is trichotomic on $\alpha$.

My first question is: How can I prove that $\mathcal E$ is transitive on an ordinal $\alpha$? Because $x\in y\in z$ doesn't imply, a priori, that $x\in z$.

My second question is about the epsilon class: To define ordinals, is $\mathcal E$ valid or we should consider $$ \mathcal E_\alpha =\{(x,y)\in \alpha\times \alpha : x\in y\} $$ (as Enderton does)?

And finally, if our theory is allowed to work with proper classes, should we add the condition be a set in the definition of ordinal? Because later we consider the class $$ \mathbf {Ord}=\{\alpha:\alpha\mbox{ is an ordinal}\}, $$ but if ordinals aren't sets this class doesn't exist. Moreover, Kelley proves that $\mathbf{Ord}$ is an ordinal, but then $\mathbf{Ord}$ should be an element of itself, which violates the Axiom of Regularity.


  • $\begingroup$ The definition of_transitive set_ is that every element is a proper subset. I thought that in symbols the definition reads $$ X= \{x:x\subset X\} .,$$ but probably I wrong. $\endgroup$ – Dog_69 Feb 9 '18 at 14:30
  • $\begingroup$ @bof: Thank you very much. I'm correcting the post now. Sorry for the mistake. I wonder if you can also answer the other questions. $\endgroup$ – Dog_69 Feb 10 '18 at 21:48
  • $\begingroup$ @bof: $alpha$ is an ordinal. I have just added the date to the question. $\endgroup$ – Dog_69 Feb 11 '18 at 12:40

1. Suppose $\alpha$ is an ordinal; we want to show that $\in$ is transitive on $\alpha.$ Suppose $x,y,z\in\alpha$ and $x\in y\in z;$ we have to show that $x\in z.$

Let $w=\{x,y,z\}.$ By the Axiom of Regularity, some element of $w$ is disjoint from $w;$ since neither $y$ nor $z$ is disjoint from $w,$ we must have $x\cap w=\emptyset.$ It follows that $x\ne z$ and $z\notin x;$ since $\in$ is trichotomic on $\alpha,$ the only remaining possibility is that $x\in z.$

Note that this argument depends heavily on the rarely used Axiom of Regularity. Without this axiom, we would have to define ordinals in a more complicated way, namely, as transitive sets which are well-ordered by $\in.$

2. As far as I can see, "$\mathcal E$ is trichotomic on $\alpha$" means exactly the same thing as "$\mathcal E\cap(\alpha\times\alpha)$ is trichotomic on $\alpha.$" What distinction do you see?

3. Of course $\mathbf{Ord}\notin\mathbf{Ord};$ only a set can be an element of anything, and $\mathbf{Ord}$ is not a set. For most people, an ordinal is a kind of set; although $\mathbf{Ord}$ is sort of like an ordinal, it is not an ordinal because it's not a set. In Kelley's peculiar system, $\mathbf{Ord}$ is an ordinal, is not a set, and is not an element of itself. The apparent contradiction with the definition $$\mathbf{Ord}=\{x:x\text{ is an ordinal}\}\tag1$$ is resolved by noting that $(1)$ really means $$\mathbf{Ord}=\{x: x\text{ is a set and }x\text{ is an ordinal}\}.$$ In my copy of Kelley (July 1957 printing), this is stated on p. 253:

For each $\beta,\ \beta\in\{\alpha:|A\}$ if and only if $\beta$ is a set and $B.$

(Here $B$ is a formula obtain from $A$ on substituting $\beta$ for $\alpha.$)

  • $\begingroup$ Very nice answer. very clear. Could you answer my other two questions. I would wish mark your answer to be correct. They are about if $\mathcal E$ needs to be a subset of $\alpha\times \alpha$ as Enderton says ($\alpha$ an ordinal) or not. And about the definition of the class $\mathbf{Ord}$, which turns out to be an ordinal (and then I think we would found $\mathbf{Ord}\in\mathbf{Ord}$). Thanks a lot. $\endgroup$ – Dog_69 Feb 11 '18 at 22:19
  • $\begingroup$ I expanded my answer to say something about your third question. Not sure what you mean by the second question. $\endgroup$ – bof Feb 11 '18 at 23:58
  • $\begingroup$ I have already mark it. My second question is a detail. If yo read Enderton's definition of a well-order on a set $A$, it is a class (actually a set because he works with ZFC) $<\subseteq A\times A$. My $\mathcal E$ is much bigger than any $\alpha\times \alpha$, more if ordinals are sets. So, it is worried more about be consistent with definitions more than other things. $\endgroup$ – Dog_69 Feb 12 '18 at 9:19
  • $\begingroup$ By the way, the statement you cite is the axiom of Clasification axiom-scheme. I know it, but I had not thought reason question 3 from it. :( $\endgroup$ – Dog_69 Feb 12 '18 at 9:26

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