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In ZF with the axiom of infinity removed, is the axiom scheme of collection provable?

Note that Collection does follow from the axiom of Transitive Containment, which states that everything belongs to a transitive set. Mancini gave a model of ZF minus Infinity where Transitive Containment fails: see the end of Section 3 in

https://projecteuclid.org/euclid.ndjfl/1054837937

Mancini's model validates Collection, so this does not answer my question.

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    $\begingroup$ What is "the axiom of Transitive Containment"? $\endgroup$ – Wojowu Dec 12 '20 at 23:07
  • $\begingroup$ What is the axiom scheme of Collection? Is it provable in ZF? $\endgroup$ – bof Dec 12 '20 at 23:13
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    $\begingroup$ Perhaps see Kaye, Richard; Wong, Tin Lok. On Interpretations of Arithmetic and Set Theory. Notre Dame J. Formal Logic 48 (2007). projecteuclid.org/euclid.ndjfl/1193667707 $\endgroup$ – jeq Dec 12 '20 at 23:20
  • $\begingroup$ @bof en.wikipedia.org/wiki/Axiom_schema_of_replacement#Collection $\endgroup$ – Wojowu Dec 12 '20 at 23:26
  • $\begingroup$ @Wojowu, the axiom of Transitive Containment says that everything belongs to a transitive set. It is provable in ZF. But Mancini provided a model of ZF minus Infinity where it does not hold: projecteuclid.org/euclid.ndjfl/1054837937 $\endgroup$ – Paul Blain Levy Dec 13 '20 at 1:00
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ZF - Inf does imply Collection. Fix a set $X$ and a property $P$ (which can be formalized in terms of a formula and a parameter). Since we have separation, we may assume for all $x \in X,$ there is $y$ such that $P(x,y).$ Suppose $X$ is finite, with cardinality $n.$ A standard inductive argument shows there is a set $Y$ such that for all $x \in X,$ there is $y \in Y$ satisfying $P(x,y).$

Now suppose $X$ is infinite. Then $\omega$ exists, by replacing every element of $\mathcal{P}_{\text{fin}}(X)$ with its cardinality. Thus ZF holds, and this instance of Collection is justified by the standard argument.

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    $\begingroup$ One needs to be careful about that inductive argument. It is internal to the model. $\endgroup$ – Asaf Karagila♦ Dec 13 '20 at 16:32
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    $\begingroup$ What model? Elliot didn't mention any model. $\endgroup$ – Paul Blain Levy Dec 13 '20 at 22:58

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