Let $X$ $\subseteq$ $\mathcal P \left({\mathbb{N}}\right)$. Determine the cardinalities of the following sets:

  1. $X = \{ A \subseteq \mathbb{N} |$ for every $B\subseteq\mathbb{N}$: $A\cap B=\emptyset$ or $A\cap B=A$ $\}$

  2. $X = \{ A \subseteq \mathbb{N} |$ both $A$ and $\mathbb{N}-A$ are infinite$\}$

  3. $X = \{ A \subseteq \mathbb{N} |$ for every ascending sequence $(a_n)_{n\in\mathbb{N}} \in \mathbb{N}$ there is an $a_n \in A$ with $n \in \mathbb{N}$$\}$

For the first one, I think that $|X|={|\mathbb{N}|}$ (because every $A$ consists of just one natural number). My guess for the second one would be $|X|={|\mathbb{N}|}$, because every $A$ would be of the form $\mathbb{N}-B$ with $B$ some infinite set. For the third one I have actually no idea.


The cardinality of the first set is $\mathbb{N}$ and the second set is $2^{\mathbb{N}}$

  • 3
    $\begingroup$ Hint for the third set: Show that for each $A$ the set $\mathbb{N}\setminus A$ contains only finite number of elements. $\endgroup$
    – njguliyev
    Aug 22 '13 at 13:32
  • 1
    $\begingroup$ Hint for the second set: Take all subsets of $\mathbb{N}$, take away the finite ones, and take away the cofinite ones. $\endgroup$
    – vadim123
    Aug 22 '13 at 13:34
  • $\begingroup$ You used terrible notation to write that down. $\endgroup$ Aug 22 '13 at 13:35
  • $\begingroup$ For example, let A be the set of all the odd natural numbers. As you can see, there are $|\mathbb{N}|$ odd natural numbers and the cardinality of it's complement is, again, $|\mathbb{N}|$. $\endgroup$
    – ABC
    Aug 22 '13 at 13:36
  • $\begingroup$ For the first one, you are right. Just empty $A$ also satisfies the the condition. $\endgroup$
    – user87690
    Aug 22 '13 at 14:20

For 3, there must be a finite number of numbers not in $A$. Otherwise the complement of $A$ is a sequence that violates the condition.


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