Why is the class of all sets a stage?

I want to prove that the class of all sets $\mathbb{S}=\{x \mid x=x \}$ is a stage (p. 15) (and then that it is a limit thus that it is the successor of another stage).

One way to do it is to proof that $$\mathbb{S} = acc(H(\mathbb{S}))$$

where $H(S)$ is the history (p. 15) of a class $S$ and

$$acc(A) := \{x \mid \exists y \in A; \ x \in y \lor x \subseteq y \}.$$

I'm trying to figure out what the history of $\mathbb{S}$. Any hints on that? Is that even a good approach to proof that $\mathbb{S}$ is a stage?

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I have never heard the term "stage" in the context of set theory. Nor the term "history". – Asaf Karagila Oct 28 '12 at 18:40
I was trying to avoid all the definitions in my question. Do you think it is okay to give references when aksing a question which contains "special" definitions? – joachim Oct 28 '12 at 18:47
@joachim When the definitions are special, it is necessary. Sounds like Scott-Potter set theory. – Michael Greinecker Oct 28 '12 at 18:49
I guess that "all the world's a stage" – Marc van Leeuwen Oct 28 '12 at 18:49
joachim: I sense your last comment was directed at me. I think once you have to so-called "Axiom of Creation" (also p.19) there isn't a difference. The author mentions that without this axiom one cannot prove that every set is in a stage (or, rather, one cannot prove that stages exist), and it would seem that the desired result may not hold. – arjafi Oct 28 '12 at 19:36

Let $\mathbb{H}$ be the class of all stages that are sets. We show that $\mathbb{H}$ is a history and $\text{acc}(\mathbb{H})=\mathbb{S}$, which shows that $\mathbb{S}$ is a stage.

By Lemma 2.9 (c), every stage that is a set is hereditary and transitive. So to show that $\mathbb{H}$ is a history it suffices that for every stage $S$ sthat is a set, $S=\text{acc}(\mathbb{H}\cap S)$. But this follows directly from Lemma 2.9 (d). So $\mathbb{H}$ is a history.

We are now ready to show $\mathbb{S}=\text{acc}(\mathbb{H})$. Since $\text{acc}(\mathbb{H})$ is a class, we have trivially that $\text{acc}(\mathbb{H})\subseteq\mathbb{S}$. So let $a$ be any set. By the axiom of creation, there is a stage $S$ with $a\in S$. Hence $a\in\text{acc}(\mathbb{H})$ and since $a$ was arbitrary, we have $\mathbb{S}=\text{acc}(\mathbb{H})$.

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As Achim Blumensath already says on his p.11, $\mathbb{S}$, the class of all sets is not a set. So a fortiori it is not a stage (as a stage is a special kind of set). What you say you want to prove is indeed not provable in his (or any) version of Scott-Potter set theory!

[That's wrong: "or any" overshoots, as I was forgetting about the earlier version of Potter as @Michael Greinecker points out. But it is perhaps worth leaving the answer in place, as it at least will serve to point up a difference between Potter's smooth later version of Scott-Potter -- which I'd thought of as the "best buy" -- and Blumensath.]

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p. 19 bottom: $\mathbb{S}$ is the only stage that is a proper class – joachim Oct 28 '12 at 20:51
@PeterSmith In Potter's previous book Sets: An Introduction, he extends the cumulative hierarchy beyond the class of all sets. – Michael Greinecker Oct 28 '12 at 21:22
I stand corrected! Compare Potter's canonical Set Theory and Its Philosophy. He talks of levels rather than stages, but with the same definition (I think). And in his axiomatization now, for every level there is a later one -- which rules out the universe as counting as a level. [The surplus in the early version was a sop to category theorists, who didn't appreciate it, so it is not there in the later version. If I recall!] – Peter Smith Oct 28 '12 at 21:33