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

In my previous question, I asked about infinite-length formula in ZFC. But I am still confused over following:

Suppose you want to build a function from a set of sequences to a set that chooses $n$th element in each sequence in the set of sequences to form a new set with each chosen element as an element in the new set.

But how can I do this nicely in ZFC? Or is this impossible?

share|cite|improve this question
up vote 1 down vote accepted

If $A$ is a set of sequences of elements from $X$ (that is, $A\subseteq X^{\mathbb N}$), then just take

$$ \{ f(n) \mid f\in A \} $$

You can justify the existence of this by the axiom of replacement applied to $A$ and the function $f\mapsto f(n)$.

share|cite|improve this answer
You do not even need replacement. This is a subset of $X$ definable with parameters, so comprehension suffices. – Andrés E. Caicedo Mar 23 '14 at 0:03
@Andres: Sure, but since we have replacement, might as well use it, when it's closer to the intuition we want to capture. And it also works when there's no common $X$. – Henning Makholm Mar 23 '14 at 1:17
Not my style, the "might as well use it", it feels wasteful. Anyway, call it a aesthetic preference. – Andrés E. Caicedo Mar 23 '14 at 2:46

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.