Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $2^{\mathbb{N}}$

share|cite|improve this question
Hint for the third set: Show that for each $A$ the set $\mathbb{N}\setminus A$ contains only finite number of elements. – njguliyev Aug 22 '13 at 13:32
Hint for the second set: Take all subsets of $\mathbb{N}$, take away the finite ones, and take away the cofinite ones. – vadim123 Aug 22 '13 at 13:34
You used terrible notation to write that down. – Trismegistos Aug 22 '13 at 13:35
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, $|\mathbb{N}|$. – ABC Aug 22 '13 at 13:36
For the first one, you are right. Just empty $A$ also satisfies the the condition. – user87690 Aug 22 '13 at 14:20
up vote 1 down vote accepted

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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