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Prove there exists a countable set $C$ containing an uncountable family $F$ of subsets such that if $A,B$ are in $F$ and $A \cap B \neq \emptyset $, then either $A \subset B$ or $B \subset A$.

I tried to show this by giving an explicit example, consider the set of integers $\mathbb{Z}$ and take it's power set $\mathcal{P} (\mathbb{Z} )$, I know that $\mathbb{Z} $ is countable and $\mathcal{P} (\mathbb{Z} )$ is uncountable , can you help?

Hanul Jeon
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    I think you want $A \cap B \neq \emptyset$ in the statement of what you want to prove. If one is contained in the other and not the empty set, the intersection is not zero. – Ross Millikan May 25 '21 at 21:17
  • You might have $A = {\emptyset}$ and $B = {\emptyset , ... }$ so that this statements makes sense. – Jan Safronov May 25 '21 at 21:17
  • @SubGenius: no, then $A \cap B = {\emptyset}$, which is not the same as $\emptyset$ – Ross Millikan May 25 '21 at 21:18
  • @SubGenius: no, each of them has $\emptyset$ as an element, so their intersection is the same as $A$, which is not empty. It is the difference between the empty set and the set containing the empty set. $A$ is the latter. – Ross Millikan May 25 '21 at 21:20
  • @RossMillikan but $A \cap B = x \in A \land x \in B$, but $\emptyset$ is both in A and B. $\emptyset \neq {\emptyset}$ – Jan Safronov May 25 '21 at 21:21
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    @SubGenius: No, $A \cap B={x|x\in A \wedge x \in B}$. The intersection has $\emptyset$ as an element, because it is an element of both $A$ and $B$. That means the intersection is not the empty set. It is the set which contains the empty set as its only element. – Ross Millikan May 25 '21 at 21:24
  • I am really sorry about this, but I was just informed this was a typo. It's actually $A \cap B \neq \emptyset$ –  May 25 '21 at 21:24
  • @RossMillikan ah right!, I forgot it was a set by itself, sorry for the inconvenience. – Jan Safronov May 25 '21 at 21:25
  • @Kthamil: you should correct the question using the edit button at the bottom so people don't have to find it in the comments. – Ross Millikan May 25 '21 at 21:29

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