$A$ = $\{X\subseteq \mathbb N : |X| = \aleph_0\land |\bar X| \lt\aleph_0 \} $

I need to find the cardinality of group A.

As I understood, I need to find a function that helps me determine the cardinality.

Any ideas?


marked as duplicate by Asaf Karagila cardinals May 12 at 16:07

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ I am not sure about it but I think that $A$ is the set of finite subsets of $\mathbb N$ $\endgroup$ – Youem May 12 at 15:40
  • 2
    $\begingroup$ I think that this problem is equivalent to counting the number of finite subsets of $\mathbb{N}$, since any subset is clearly uniquely determined by it's compliment. Try counting subsets of order 0, 1, 2, 3... and recall that a countable union of countable sets is countable. $\endgroup$ – jmacmanus May 12 at 15:42
  • $\begingroup$ @Youem Cofinite. $\endgroup$ – Fabio Somenzi May 12 at 15:42
  • $\begingroup$ Complement set of X** @PeterForeman $\endgroup$ – Gil May 12 at 15:47

Let $B_n = \{X\subset \mathbb N: |X| = n \}$, it is clear that $$|A| = \left|\bigcup_{n\in \mathbb N}B_n \right|.$$

Since $\left|\displaystyle\bigcup_{n\in \mathbb N}B_n \right| \ge \left|B_1\right| = |\mathbb N| = \aleph_0$ and $\left|B_n\right|\le \left|\mathbb N^n\right| = \aleph_0$, then $|A| = \aleph_0$


Not the answer you're looking for? Browse other questions tagged or ask your own question.