Synopsis
I have two short and simple questions. One is about proof-verification (kinda), and the other is about proof-understanding. Thank you for your time.
Note the following definition:
For any set $A$, Define a sequence in $A$ to be a function from some natural number into $A$. Let $\text{Sq}(A)$ be the set of all sequences in $A$.
My textbook claims that $\text{Sq}(\omega)$ has cardinality $\aleph_0$, however only sketches out that this is the case since $\omega \times \omega \approx \omega$ and leaves the details to the reader.
So, as the reader, I've been trying to find a function from $\omega \times \omega$ onto $\text{Sq}(\omega)$. I was thinking perhaps that something like $F(m,n) = {}^m n$ would work since $\text{Sq}(\omega) = \bigcup \{{}^n\omega \mid n \in \omega \}$ but I doubt myself a lot and I was wondering if perhaps one of you could help boost my low mathematical self esteem (or tear it down further) by letting me know if this is the right direction to go.
Next, my textbook claims that $\text{Sq}(A)$ is countable for any countable set $A$. The author states that since there was a one-to-one function $g$ from $A$ into $\omega$, that there naturally is a one-to-one function from $\text{Sq}(A)$ into $\text{Sq}(\omega)$. Perhaps I'm stupid but can someone please explain why this is?
Your suggested $F$ is a map from $\omega\times\omega$ to $\{{}^mn:m,n\in\omega\}$, not to $\operatorname{Sq}(\omega)$: it assigns to each $\langle m,n\rangle\in\omega\times\omega$ a whole set of sequences in $\omega$, namely, all of them with domain $m$ and codomain $n$. I suspect that what was intended is for you to use the fact that $|\omega\times\omega|=\omega$ to show by induction on $m$ that ${}^m\omega$ is countable for each $m\in\omega$; $\operatorname{Sq}(\omega)$ is then the union of countably many countable sets.
For the second question, let $g:A\to\omega$ be one-to-one. We can use $g$ to define a function $G:\operatorname{Sq}(A)\to\operatorname{Sq}(\omega)$ as follows: if $\sigma\in{}^mA$, let $G(\sigma)\in{}^m\omega$ be given by
$$\big(G(\sigma)\big)(k)=g\big(\sigma(k)\big)$$
for each $k\in m$. Now just check that $G$ is one-to-one. If you think of $\sigma$ as essentially an $m$-tuple $\langle\sigma(0),\ldots,\sigma(m-1)\rangle$ in $A^m$, $G(\sigma)$ is the $m$-tuple in $\omega$ that you get by using $g$ to ‘push $\langle\sigma(0),\ldots,\sigma(m-1)\rangle$ across to $\omega^m$’ component-wise.