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Here Professor Blass describes the following cumulative hierarchy of sets:

Begin with some non-set entities called atoms ("some" could be "none" if you want a world consisting exclusively of sets), then form all sets of these, then all sets whose elements are atoms or sets of atoms, etc. This "etc." means to build more and more levels of sets, where a set at any level has elements only from earlier levels (and the atoms constitute the lowest level). This iterative construction can be continued transfinitely, through arbitrarily long well-ordered sequences of levels. This so-called cumulative hierarchy is what I (and most set theorists) mean when we talk about sets.

We want to agree on the following principles:

  1. For every level there is a succeeding level.
  2. For every sequence of levels: $l_1,l_2,l_3,\dots$ there is a level succeeding all levels $l_1,l_2,l_3,\dots$. One might call this level "limit level".

Question:

Why is the axiom of replacement true under this interpretation of the term "set" (set = anything that is formed at some level of this hierarchy)?

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    $\begingroup$ In The Iteractive Conception of Set (1971) George Boolos states: "We do not believe that the axioms of replacement or choice can be inferred from the iterative conception." But see George Boolos, Iteration Again (1989) for a different point of view. For another approach, see J.R.Shoenfield, Axioms of Set Theory into : Jon Barwise (editor), Handbook of Mathematical Logic (1982), page 321-on. $\endgroup$ – Mauro ALLEGRANZA Jan 11 '16 at 19:20
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    $\begingroup$ Please come up with a better title. $\endgroup$ – Asaf Karagila Jan 11 '16 at 19:27
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    $\begingroup$ @Paul: As for "why should we 'believe' Replacement?", this was more or less covered in the many discussions in the answers and comments of mathoverflow.net/questions/208711/who-needs-replacement-anyway $\endgroup$ – Asaf Karagila Jan 11 '16 at 19:41
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    $\begingroup$ @Paul: My point is that I think that you misread the question, and the edit exacerbates that misreading. $\endgroup$ – Asaf Karagila Jan 11 '16 at 20:00
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    $\begingroup$ @PaulTaylor I do agree with Asaf that that edit was a bit on the presumptuous side. (I don't know that I agree with him that just because the answers were deemed 'helpful', it means your edit reflects the author's intention. On the other hand, if it does, then contra Asaf I'm not sure it duplicates David Roberts's question; I'd need to think about that more.) $\endgroup$ – Todd Trimble Jan 12 '16 at 17:40
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For an argument that the iterative conception implies something weaker than unrestricted Separation (implied by unrestricted Replacement), i.e. $\Sigma_2$ Replacement, see Randall Holmes 2001 http://math.boisestate.edu/~holmes/holmes/sigma1slides.ps. (According to Professor Holmes, “this contain[s] an error, which Kanamori pointed out to me and which I know how to fix.”)

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  • $\begingroup$ Sadly, the link does not work. $\endgroup$ – djafe Jan 12 '16 at 18:18
  • $\begingroup$ Now it is working. $\endgroup$ – djafe Jan 12 '16 at 18:54
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On the Foundations of Mathematics mailing list some years ago, Arnon Avron argued that replacement is the way mathematicians naturally construct many sets. I quote one example from his article:

When asked to write a term denoting the set of singletons of elements of $\mathbb N$, I bet that at least 999 mathematicians (either in the broad sense, including first-year students, or in a narrower sense) out of 1000 would write: $$\{\{n\}: n\in {\mathbb N}\}$$ and not $$\{x\in P(P({\mathbb N})):\exists n\in {\mathbb N}. x=\{n\}\}.$$ This is not only because the former is shorter, but because it directly translates the definition in words of this set, and precisely reflects our intuition how this set is formed/constructed. In contrast, one has to think for a while in order to get the second definition correctly (and for many students it is even difficult at first to understand why this term is a correct description of this set. Anyone who have taught a basic course in set theory or discrete mathematics has experienced this). It is clear therefore that practically everyone relies on replacement for getting this set, and not on the powerset axiom.

According to this line of thinking, replacement is an intrinsic feature of any faithful description of the universe of sets, including the cumulative hierarchy.

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    $\begingroup$ Should that iterated power set be just a power set? $\endgroup$ – Steven Landsburg Jan 12 '16 at 1:16
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    $\begingroup$ @StevenLandsburg: Yes, I think so, but I'll leave it as it is because it's a quotation. $\endgroup$ – Timothy Chow Jan 12 '16 at 4:40
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    $\begingroup$ Hmm. In the standard von Neumann representation of natural numbers, $P(\mathbb N)\subseteq P(P(\mathbb N))$, so it is technically correct, but weird. $\endgroup$ – Emil Jeřábek Jan 12 '16 at 10:39
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    $\begingroup$ Technically correct - The best kind of correct ;-) $\endgroup$ – Johannes Hahn Jan 12 '16 at 18:01
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    $\begingroup$ cs.nyu.edu/pipermail/fom/2007-September/011918.html $\endgroup$ – Emil Jeřábek Jan 12 '16 at 18:07
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There is a wonderful blog post by Joel David Hamkins at Transfinite recursion as a fundamental principle in set theory which goes into great depth on this topic.

My answer to your question would then be "We should believe the axiom of replacement because we believe in recursion, even at the transfinite level."

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  • $\begingroup$ See my two comments to Paul Taylor under the main question. $\endgroup$ – Asaf Karagila Jan 11 '16 at 19:41

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