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I would like to prove that if $Χ$ is a non empty set with the property $(\forall m\in X)[m\cap X\neq\emptyset]$

then there exists a function $f:\mathbb{N}\rightarrow X$ such that $(\forall n)[f(n+1)\in f(n)]$.

My attempt to prove this is the following:

I may use the Axiom of Choice (or any equivalent form of the Axiom of Choice) so there exists a function $\epsilon: \mathcal{P}(X)\setminus \{\emptyset\}\rightarrow X$ such that $\epsilon(m)\in m$.

I can the define a function $h:X\rightarrow X$ such that $h(m)=\epsilon(m\cap X)$.

From the recursion theorem there exists a function $f:\mathbb{N}\rightarrow X$ such that $f(0)=\epsilon(X)$ and $f(n+1)=h(f(n))=\epsilon(f(n)\cap X)\in f(n)\cap X$ so $f(n+1)\in f(n)$. Is my proof correct?

What troubles me a bit is a question following this exercise that says “Is this an intuitively acceptable result?”. I don’t know what to say about that…

So overall, is my proof correct? Any comments on that last question?

Thank you in advance for your time and effort.

epsilon
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    Your proof is correct. The point of the comment is that one of the standard axioms of $\mathsf{ZFC}$ is foundation, that rules out the existence of decreasing $\in$-sequences, and thus of sets such as $X$. (But it is consistent with the other axioms that foundation fails, so one may assume that such $X$ exist after all.) Why one may think that such sets are counter-intuitive is that the usual picture of the set theoretic universe assumes one has all elements of a set "before" having the set. But here, we never "get started" (so never get to have $X$), since we have a decreasing $\in$-sequence. – Andrés E. Caicedo Dec 30 '13 at 21:55

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You haven't quite described $\epsilon$ adequately to do the trick. Rather, you must know that for any non-empty subset $A\subseteq X,$ we have $\epsilon(A)\in A.$ Once that's fixed, the rest of the proof looks fine.

The next question is basically asking you: "Do you think this sort of thing (an $\in$-decreasing sequence) should exist?" The existence of such a sequence is a direct consequence of the failure of the Axiom of Foundation (Regularity), so it's more or less asking if you think Foundation should or shouldn't hold.

Cameron Buie
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The first part was written as an answer to the first revision of the question.

The proof is okay. I'd point out that $\epsilon$ is from $\mathcal P(X)$ and not $\mathcal P(x)$; and that $\epsilon(m)\in m$, rather than $X$ itself as a variable (clearly $\epsilon(X)\in X$, because $\epsilon\colon\mathcal P(X)\to X$).

I also don't fully understand the necessity of "and I can pick $k\in X$." If $k$ is the same one from the next paragraph, where you define $f$, perhaps it is worth to move $k$ to that paragraph. I'd also recommend to be even more explicit and take $f(0)=\epsilon(X)$, but that's a personal preference.

Now, note that this function defines a decreasing sequence $f(0)\ni f(1)\ni\ldots$; one of the axioms of $\sf ZF$ is the axiom of foundation, as references by Cameron and Andres, which states that this situation cannot occur. That is, $\forall X(X\neq\varnothing\rightarrow\exists y(y\in X\land y\cap X=\varnothing))$.

This axiom is not needed if you just want to develop classical mathematics within set theory. However it is used heavily in modern set theory, and so comes the question -- do you feel this sort of axiom is natural or not? The following thread might interest you as well, Where is axiom of regularity actually used?

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Asaf Karagila
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Such a set $X$ does not exists as its existence contradicts the Axiom of Regularity:

$$ \forall x \big(x\ne\varnothing\longrightarrow\exists y\in x(x\cap y=\varnothing)\big). $$

Yiorgos S. Smyrlis
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    That doesn't really answers the question. The question didn't really specify the axiomatic settings, nor claims that such set exists. – Asaf Karagila Dec 30 '13 at 22:14