Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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!

share|cite|improve this question
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.

share|cite|improve this answer
Thank you! I think that does it. – Nick Thomas Jan 20 '13 at 4:47

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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