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Let $ F$ be a non-empty family of non-empty sets such that for every set $A \in F$ , $\exists $ a proper subset $B \subset A$ such that $B \in F$ . I can prove that every set in such a family $F$ is infinite but I'm having trouble to determine whether the family $F$ itself is infinite or not ; I have an intuition that the family must also be infinite but am not able to prove ( or disprove) it ; please help . Thanks

Asaf Karagila
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Souvik Dey
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  • doesn't a similar method show both at once? or is this a more subtle question about the axiom of choice? – David Holden Aug 15 '14 at 08:56
  • Maybe answers on http://math.stackexchange.com/q/829188/75923 are useful here. – drhab Aug 15 '14 at 09:03
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    @David: drhab's answer uses some choice, Ishfaaq's answer doesn't. What it shows is that for every $n$, there are at least $n$ elements in the collection. In the absence of choice it is consistent that such collection exists where every chain is finite, but the collection itself, of course, cannot be finite. – Asaf Karagila Aug 15 '14 at 12:56
  • @Asaf, thanks for thar helpful clarification – David Holden Aug 15 '14 at 13:15

2 Answers2

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$F$ is non-empty. Then there is a set $A_1 \in F$. Now as per the conditions there is a propoer subset $A_2 \subset A_1$ such that $A_2 \in F$. You can continue to induct that there is a sequence $A_1, A_2, ...A_n, ..$ such that each of them are elements of $F$.

Or proceed by contradiction. Suppose there are but a finite number of elements in $F$. Say, $F = \{B_1, B_2, .., B_n\}$. Now there must be a set $C$ in the finite collection $F$ such that there is no other set in $F$ which is a proper subset of $C$. The absence of such a set $C$ immediately entails that $F$ is infinite. But then since $C$ is in $F$ there must be a proper subset of it which is also in $F$ leading to a contradiction. Hence $F$ must be an infinite collection.

Ishfaaq
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Order $F$ by inclusion. Then it has no minimal element. That makes it possible to construct a infinite chain. So $F$ must be infinite. Start with some $A_1\in F$. Then find $A_2\in F$ as a proper subset of $A_1$. Repeat this.

drhab
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