# find cardinality of A [duplicate]

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

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 StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); May 12 at 16:07

• I am not sure about it but I think that $A$ is the set of finite subsets of $\mathbb N$ – Youem May 12 at 15:40
• 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. – jmacmanus May 12 at 15:42
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 $$|A| = \aleph_0$$