Prob 8, Sec 7 in Munkres' TOPOLOGY 2nd ed: How do we show these sets have the same cardinality?

Question

Here's Prob. 8. Sec. 7 in Topology by James R. Munkres, 2nd edition:

Let $X$ denote the two element set $\{0,1\}$; let $\mathscr{B}$ be the set of countable subsets of $X^{\omega}$. Show that $X^{\omega}$ and $\mathscr{B}$ have the smae cardinality.

Here $X^{\omega}$ denotes the set of all infinite binary sequences (i.e., the set of all the functions each with domain the set $\mathbb{N}$ of natural numbers and range a (non-empty) subset of $\{0,1\}$).

My effort:

Using the Schroeder Bernstein theorem, our aim is to show the existence of injective maps $f \colon X^{\omega} \to \mathscr{B}$ and $g \colon \mathscr{B} \to X^{\omega}$.

We can define $f$ as follows: $$f(s) \colon= \{s\} \ \mbox{ for all } \ s \in X^{\omega}.$$

How do we define our desired map $g$?

Answer 1

Suppose I have a countable sequence $C=(S_n)_{n\in\omega}$ of infinite binary sequences - say, $S_n=(a_i^n)_{i\in\mathbb{N}}$. I can view $C$ as an $\omega$-by-$\omega$ array of $0$s and $1$s. Can you see how to turn an $\omega$-by-$\omega$ array into an infinite sequence? (HINT: why is $\mathbb{N}^2\cong\mathbb{N}$?)

---

OK, fine, the above was talking about countable sequences of elements of $X^\omega$. But how does the set of countable subsets compare to the set of countable sequences?

Answer 2

You can't quite define an injection from $\scr B$ into $X^\omega$. The reason is that it is consistent with the failure of the axiom of choice that there is no such injection.

This means that you have, at some point, resort to using the axiom of choice. Luckily, we can fine a surjection from $X^\omega$ onto $\scr B$, by noting that $(X^\omega)^\omega$ and $X^\omega$ have the same cardinality. Therefore we can think of every element of $X^\omega$ as encoding a sequence of elements of $X^\omega$.

Can you see how does that help you in obtaining the surjection?