Does it exist a "recursively infinite" set?

Question

I was inspired by this question:

Does it exist an infinite set whose elements are all infinite sets?

This question received many excellent answers but made me wonder about the following in the context of set theory (say ZFC):

We say that a set has the property $P$ if it is infinite and each of its element also has the property $P$.

So, in order to satisfy $P$, a set $A$ has to be infinite, and its elements also have to be infinite. And the elements of the elements of $A$ too. And the elements of the elements of the elements...

Is there a set satisfying $P$?

I couldn't construct an example by hand, nor prove such a set couldn't exist. I am not even completely sure the property $P$ is well-defined...

Answer 1

I am not sure whether this answers your question.

If the axiom of foundation is accepted then no set exists such that it is infinite and that all elements of its transitive closure are infinite.

This because by induction on $\overline{\in}$ it can be proved that:$$\forall x[x=\varnothing\vee\varnothing\overline{\in}x]\tag1$$

Here $\overline{\in}$ stands for the transitive closure of relation $\in$ and $\varnothing\overline{\in}x$ can be read as $\varnothing\in\mathsf{Tc}(x)$ where $\mathsf{Tc}(x)$ stands for the transitive closure of $x$.

So in $(1)$ it is stated that a set that is not empty has a transitive closure that contains the empty set (which is not infinite).

Answer 2

There is a confusion here. Recursion goes up, based on a well-founded relation, it does not go down. When you define something by recursion it is a step-by-step definition. From assuming you could define it so far, you define it one step further.

Your definition seems to be co-well-founded, you postulate the property of a set and require it implies the property for its elements.

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In $\sf ZF$ the $\in$ relation is well-founded, so this is not quite possible. But we can still get some sufficiently nice properties which will hold for the elements. For example being hereditarily ordinal definable, or constructible. Or being an ordinal. Or just being a set.

But in none of these cases you start by postulating the property for a set, and then require it implies the property for the elements of the set. Instead you start by construction some class, and taking its hereditary part, or you define something from bottom to top, and then you end up with a class satisfying your property.

When you tack on the requirement that the property implies that the set is infinite, then you quickly run into a contradiction. The bottom-up constructions start with the empty set—for a reason, too: it's the basis of the set theoretic universe—and go upwards. So at the root of every set, you will find the empty set. And so in every non-empty transitive set, you will find the empty set as an element.

If you require that $P$ is hereditary, then it implies that if $x$ has it, the transitive closure of $x$ has it. But if $x$ is non-empty, $\varnothing$ is an element of that transitive closure. So infinitude cannot be part of the definition of $P$.