# For given set $A$, show that there exist a set of 'every finite sequence on $A$'.

To be specific, I'll state some definitions first.

1. $0=\emptyset$
$1=\{\emptyset\}=\{0\}$
$2=\{\emptyset,\{\emptyset\}\}=\{0,1\}$
...
$n=\{0,1,2,....,n-1\}$

2. For given set $\text A$ and $\text B$,
$\text B^\text A=\{ f\in P(\text A \times \text B)$ | $f:\text A \to \text B \}$

3. Any function $f:n\to \text{A}$ is finite sequence.
Any function $f:\mathbb N \to \text A$ is infinite sequence.

In terms of above definitions, $$\bigcup_{n\in \mathbb N} A^n$$ is set of every finite sequence on $A$.

I want to prove this set exists. If I prove $J=\{\text A^0,\text A^1, \text A^2, ....\}$ exists, above set exsits by Axiom of union.

But I have no idea how to prove $J$ exists. I know each $\text A^n$ exist. but how can I construct such a big set?
I'm sure that I can't apply Axiom of pair infinitely many times.

This is exactly what the Axiom of Replacement is for. Once you know $\mathbb N$ exists, the Axiom of Replacement guarantees that $$\{ A^i \mid i \in \mathbb N \}$$ is a set.
Either use the Axiom Schema of Replacement, $$J=\{\,A^n\mid n\in\Bbb N\,\}$$ Or note that all $A^n$ are subsets of $P(\Bbb N\times A)$ and use the Axiom Schema of Separation from there.