Proving Cardinality of Sets: $\{a_n: n \in \omega\}$

[evinda]

Hello! (Wave)

Suppose that $X$ contains a countable set. Let $b \notin X$. Show that $X \sim X \cup \{b\}$.
Prove that in general if $B$ is at most countable with $B \cap X=\varnothing$ then $X \sim X \cup B$.

Proof:We will show that $X \sim X \cup \{b\}$.
There is a $\{ a_n: n \in \omega \} \subset X$.
We define the function:
$$f: \{ a_n: n \in \omega \} \cup \{ b \} \overset{\text{bijective}}{\to} \{ a_n: n \in \omega \}$$
as follows:

$$f(b)=a_0\\f(a_n)=a_{n+1} \text{ for each } n \in \omega$$
We easily see that $f$ is 1-1 and surjective.

We define $g: X \cup \{b\} \to X$ as follows:

$$g(x)=x \text{ if } x \in X-(\{ a_n: n \in \omega \} \cup \{ b \})\\g(x)=f(x) \text{ if } x \in (\{ a_n: n \in \omega \} \cup \{b\})$$

Then $g$ is 1-1 and surjective.
From which point do we deduce that there is a $\{ a_n: n \in \omega \} \subset X$ ? (Thinking)

 

[Evgeny.Makarov]

From which point do we deduce that there is a $\{ a_n: n \in \omega \} \subset X$ ? (Thinking)

From the fact that $X$ contains a countable set.