Am I right about $ |\bigcup_n \mathcal{P} (B-A_n)| = {\aleph_0} $?

Question

I'm trying to solve that equation: $$ |\bigcup_n \mathcal{P} (B-A_n)| = ? $$ When it is known that:

$$ n \in \mathbb{N} - \{0,1\} $$ $$ A_n=[0,1/n] $$ $$ B=\{1/k | k \in \mathbb{N} - \{0\}\} $$

$A_n$ is not a subset of $B$.

So now it means that $|(B-A_n)|$ is finite.

It means that $|\mathcal{P}(B-A_n)|$ is finite too. Thus: $$ |\bigcup_n \mathcal{P} (B-A_n)| = {\aleph_0} $$

It looks staright-forward to me until it comes to the union. So I am not sure if it is still ${\aleph_0}$ or not. Thanks!

Answer 1

Now that everything has been clarified (see the comments) :

it is indeed clear how everything works until the union bit.

Not that it is not true in general that a union of finite sets has cardinal $\leq \aleph_0$, so here you have to use some specificities of your union to get to the result.

There are (at least) two possible approaches here, which both generalize in different directions :

• Note that $A_{n+1}\subset A_n$ so $B\setminus A_n \subset B\setminus A_{n+1}$, and so $P(B\setminus A_n)\subset P(B\setminus A_{n+1})$. Therefore you have an increasing union of finite sets. Now it can be shown that such a union necessarily has cardinality $\leq \aleph_0$; more precisely :

let $(I,\leq)$ be a totally ordered set, and $(A_i)_{i\in I}$ and $I$ indexed-family of finite sets such that $i\leq j \implies A_i\subset A_j$. Then $\bigcup_{i\in I}A_i$ is at most countable.

Here it's even easier, because $I$ itself is $\mathbb N$, and so is countable, so you can easily write down an explicit enumeration of the elements of the union. I just wanted to point out the more general framework (in fact, it can be generalized even more)

• The second approach is to forget about details and subtlety in your counting : you have a countable union of finite sets, therefore it is at most countable: $|\bigcup_n A_n|\leq \sum_n |A_n| \leq \sum_n \aleph_0 = \aleph_0\times \aleph_0 = \aleph_0$. This also has its advantages, and can be interesting at times (when you don't need precise estimates)

Note that both these approaches actually show a bound on the cardinality. In your case, however, it's easy to prove that the set is at least $\geq \aleph_0$, so by Cantor-Bernstein it has cardinality $\aleph_0$ precisely.