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Let $A$ be a set. Let $A^\omega$ denote the set of infinite sequences of members of $A$ (i.e., functions from $\omega$ to $A$). Define $\omega_n = \omega \setminus \{n\}$. Let $A^\omega_n$ denote the set of functions from $\omega_n$ to $A$ (i.e., infinite sequences missing an $n$-th element).

If $X \subseteq A^\omega$ is a set of sequences, call $X$ "$n$-cocomplete" iff for every $f \in A^\omega_n$, there is a $g \in X$ such that $f \subseteq g$ (i.e. , $g(m) =f(m)$ for all $m \in A^\omega_n$). (I.e., given an arbitrary sequence $f$ which is missing an $n$-th element, we can find a $g \in X$ which is $f$ with something filled in for the $n$-th element.)

Let $X_1,X_2,... \subseteq A^\omega$ be an infinite sequence of sets of sequences, where each $X_i$ is $i$-cocomplete.

Question: Is $\bigcap X_i$ nonempty?

Thank you!

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up vote 1 down vote accepted

Counterexample: Let $A$ is set of positive integers, and let define

$$X_n:=[\{\left< x_k \right> : x_n=2x_{n+1}\}$$

Then each $X_n$ is $n$-cocomplete. If $\bigcap X_n$ is not empty, there exists $\left< x_k \right>$ s.t. $x_{n+1}=\frac{x_n}{2}$ for all $n$, and $$0<x_n<1$$ for large $n$. It contradicts to $\left< x_k \right>$ is sequence of positive integers.

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Thank you! I think that does it. –  Nick Thomas Jan 20 '13 at 4:47
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