Given families of sets $S_i$ s.t. for all $A\in S_{i+1}$ there is $B\in S_i$ s.t. $B\subseteq A$. Is there a $\subseteq$-increasing $A_i\in S_i$?

Question

Given an infinite sequence of sets $S_0,S_1,...$ These sets consist of sets, and it's known that if $A\in S_{i+1}$, then some $B\in S_{i}$ is a subset of $A$. Is there an infinite sequnce $A_0,A_1,...$ with $A_i\in S_{i}$, s.t. $A_i\subseteq A_{i+1}$? I've tried to apply the Hausdorff maximal principle to prove the result but couldn't see why the maximal chain must be infinite. I'd be grateful if you give the proof, or the link to it.

Answer 1

There's no reason to expect such sequence to exist.

For $n,m\in\Bbb N$, let $X_{n,m}=\{(n,i)\mid i<m\}\subseteq\Bbb{N\times N}$. Now consider $S_i=\{X_{i,m}\mid i\geq m\}$.

In effect this is a tree of sets where the root is $\varnothing$, which splits to $\{\{(i,0)\}\mid i>0\}$, then $\{\{(i,0),(i,1)\}\mid i>1\}$, and so on.

It is very easy to see that every branch in this tree is finite. So no such sequence exists.

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Alternatively, if this definition is a bit confusing, recall that every partial order can be realised as sets under $\subseteq$, and then simply take a tree of height $\omega$ without infinite branches and let $S_i$ be the $i$th level.