Axiom of foundation allows sets consisting of a descending sequences plus some 'atom'?

Question

I am reading a text which describes how the Axiom of Foundation prevents sets that are built from a descending sequence such

$$X=\{x_0, x_1,\ldots\},\text{ with } x_1\in x_0, x_2\in x_1,\ldots$$

In the above case I can see where there is an inconsistency with the axiom of foundation. Namely for any candidate you select from $X$ -- say $x_j$ -- if you take the intersection of $x_j$ and $X$, you find that they contain an element in common, and therefore their intersection is non-empty. E.g., in the case of $x_j$ you find $x_j$ contains $x_{j+1}$, and also $X$ contains $x_{j+1}$.

Now, let's say you construct $Y$ such that $Y = \{ \{\} \}\cup X$.

Then the set $Y$ is allowed by the axiom of foundation (yes?).

In the case of $Y$, the empty set becomes a 'bottom element', but there is still the infinite sequence $x_0, x_1\ldots$.

If the preceding is correct, I guess I have understood the axiom. My question is why is descending sequence by itself ruled out, while a descending sequence with a 'bottom element' thrown in allowed ?

Sorry if this is too informal a question, or if it is answered elsewhere.. I couldn't find any other question that exactly addressed the point that has me a bit confused.

Answer 1

It's not even clear what is a "bottom element". Remember that $\in$ is not a transitive relation in general. Namely, if $x\in y$ and $y\in z$, then there's no reason to expect that $x\in z$ as well.

So unlike $\subseteq$, where the notion of "a bottom element" is clear, it is simply the intersection of all the sets in the decreasing sequence, it is completely unclear for $\in$. Do you mean that the "bottom element" is an element of all the elements of the sequence? Or just "the last one" or maybe none of them?

There is also an issue with the notion of an atom. When working in $\sf ZF$ we only have sets. So the only atom is $\varnothing$. So in some sense you're trying to have that $\varnothing$ is "below" all the elements of the sequence. Which makes sense, since if we think about the von Neumann hierarchy, it means that every set is somehow constructed from the empty set using power sets, unions and subsets. But well-foundedness tells us that this is not quite the same as working "backwards" from some given set $x$.

And finally, as I wrote in the comment and as Arthur explained well in his answer, the "bottom element" is of course definable in this sequence, so by using comprehension we can still construct a decreasing $\in$-sequence which is a set without an $\in$-minimal element.

Answer 2

The axiom of foundation says "$\forall x(\phi(x))$", for a suitable formula $\phi(x)$. Your $Y$ is indeed a set that does not violate $\phi$ by itself. But there is a subset of $Y$, obtained by using the axiom of comprehension and the formula $\psi(z)$ given by "$z \neq \{\}$". This set is the one you called $X$, and $X$ does violate $\phi$. Therefore, a universe that contains $Y$ and at the same time satisfies the axiom of comprehension cannot satisfy the axiom of foundation.