Let $ \ M=\{A_{\alpha}: \alpha \in \mathcal{N} \} \ $ such that each $ \ A_{\alpha} \ $ is countable and for $ \ \alpha <\beta \in \mathcal{N} \ $ , $ \ A_{\alpha}\ $ is a proper subset of $ \ A_{\beta} \ $. Let $ \ B=\bigcup A_{\alpha} \ $
(a) Determine the cardinality of $ \ B \ $.
(b) Show that if $ \ C=\{c_n : n \in \omega \} \ $ is a countable subset of $ \ B \ $ ,
then $ \ C \subset A_{\gamma} \ $ for some $ \ \gamma \in \mathcal{N} \ $.
Answer:
(a)
We know that countable union countable sets is again countable.
Thus $ \ \bigcup A_{\alpha}=B \ $ is countable.
(b)
Choose $ \ \gamma_n \ $ such that $ \ c_n \in A_{\gamma_n} \ $.
If possible let , the given statement is false.
Then we shall get a contradiction .
We assumed that,
$ C=\{c_n: n \in w \} \ $ is countable subset of $ \ B=\bigcup A_{\alpha} \ $ but $ \ C \nsubseteq A_{\gamma} \ $
A contradiction need to produce right now.
But I am unable to do this.
help me doing this