4
$\begingroup$

Is there exists a set $A$ such that if $F$ is a finite subset of A, then $F\in A$? Under ZFC axiom system. I guess the answer may not, but I’m not sure yet. Any help or hint will be appreciated, thanks!

$\endgroup$
  • 1
    $\begingroup$ Your question's title and your actual question seem to be different. In your question you ask whether there exists a set $\;A\;$ which contains as elements all its finite subsets. That's different from what is written in the title, so: which one do you want and what have you done so far about that? $\endgroup$ – DonAntonio Mar 26 at 15:03
  • $\begingroup$ @DonAntonio Yes, Antonio, I mean it contains as elements all its finite subsets. I edit title right now. Thanks! $\endgroup$ – Landau Mar 26 at 15:08
10
$\begingroup$

There are such sets, an example is $V_\omega$, the set of all sets which are hereditarily finite, meaning that $x\in V_\omega$ if and only if $x$ is finite, all elements of $x$ are finite, all elements of all elements of $x$ are finite, and so on.

If $y\subseteq V_\omega$ is finite, this means that for every $x\in y$ we have $x\in V_\omega$, that is elements of $y$ are hereditarily finite sets. Since $y$ is finite this means that $y$ itself is an hereditarily finite set, hence $y\in V_\omega$.

More generally let $\mathcal P_{\mathrm{fin}}(X)$ denote the set of all finite subsets of $X$ and let $\mathcal P^n_{\mathrm{fin}}(X)=\mathcal P_{\mathrm{fin}}(\mathcal P^{n-1}_{\mathrm{fin}}(X))$, where $\mathcal P^0_{\mathrm{fin}}(X)=X$ by convention. Then $\bigcup_{n\in\Bbb N}\mathcal P^n_{\mathrm{fin}}(X)$ is a set with the desired property regardless of $X$.

$\endgroup$
  • $\begingroup$ Thanks, Codenotti. By the last paragraph, if we replace “finite subsets” with “subsets whose cardinal are $\kappa$, the same claim is also true. Am I right? $\endgroup$ – Landau Mar 26 at 15:29
  • $\begingroup$ This construction looks like it's the Kleene star operation on some monoid. I can't quite pin down what that monoid would be, though. $\endgroup$ – Larry B. Mar 27 at 3:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.