A friend thought of this problem and I found it interesting to think about so I want to share it with you. I am intrigued how you will solve the problem.  Find countably infinite many sets $U_i$ such that $$\mathbb{N}=\cup_{i=1}^{\infty}U_i,$$ with $U_i$ disjunct and countably infinite.
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The title says it. I thought of the following: we want $$\Bbb N = \dot {\bigcup_{n \geq 1} }A_n$$ We pick multiples of primes. I'll add $1$ in the first subset. For each set, we take multiples of some prime, that hasn't appeared in any other set before. Then $$\begin{align} A_1 &amp;= \{1, 2, 4, 6, 8, \cdots \} \\ A_2 &amp;= \{3, 9, 15, 21, 27, \cdots \} \\ A_3 &amp;= \{5, 25, 35, 55, \cdots \} \\ A_4 &amp;= \{7, 49, 77, \cdots \} \\ &amp;\vdots \end{align} $$  I'm heavily using the fact that there are infinite primes. I think these sets will do the job. Can someone check if this is really ok? Also, it would be nice to know how I could express my idea better, instead of that hand-waving. Alternate solutions are also welcome. Thank you!  Edit: the subsets must be also infinite.
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One more question about set theory:  $A\subseteq R$ is an infinite set of positive numbers.  Assume there is a value $k \in Z$ such that for any $B \subseteq A$:  $\sum_{i=0}^\infty b(i) \le k$ where $b \in B$  Show that A is of countable cardinality  Hint: Look at the sets $A(n) = \{a\in A | a&gt;\frac{1}{n}\}$  What I tried doing: I understand that when n goes to infinity, A(n) gets closer and closer to A. And if I could show that for all values of n, A(n) is countable, then I could show that the unification:  $A(1) \cup A(2) \cup ... \cup A(n) \cup A(n+1) \cup...=A$ is countable.   But why can you say that if there exists such a value k, such that the summary of all values of A(n) is less than k, then A(n) is countable?
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How many subsets of $\mathbb{N}$ have the same cardinality as $\mathbb{N}$?   I realize that any of the class of functions $f:x\to (n\cdot x)$ gives a bijection between $\mathbb{N}$ and the subset of $\mathbb{N}$ whose members equal multiples of n.  So, we have at least a countable infinity of sets which have the same cardinality of $\mathbb{N}$.  But, we could remove a single element from any countably infinity subset of the natural numbers and we still will end up with a countably infinite subset of $\mathbb{N}$.  So (the reasoning here doesn't seem quite right to me), do there exist uncountably many infinite subsets of $\mathbb{N}$ with the same cardinality of $\mathbb{N}$?    Also, is the class of all bijections $f: \mathbb{N} \to \mathbb{N}$ and a given countably infinite subset uncountably infinite or countably infinite?
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I am stuck on the following problem when I was trying to solve an entrance exam paper:  How many maps $\phi \colon \Bbb N \cup \{0\} \to \Bbb N \cup \{0\}$ are there with the property that $\phi(ab)=\phi(a)+\phi(b)$ for all $a,b \in \Bbb N \cup \{0\}$ ?  The options are as follows :   none  finitely many  countably many  uncountably many    Option 1 is not possible as if we take $a=1,b=0$ then we get $\phi(1)=0$ which is possible. But, I am not sure about the other options and could not decide which one holds good. Can someone explain?
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Let $\{E_n\}_{n\in\mathbb{N}}$ be a sequence of countable sets and let $S=E_1\times\cdots\times E_n\times\cdots $. Show $S$ is uncountable. Prove that the same statement holds if each $E_n=\{0,1\}$.  By the definition of Cartesian product of sets,   $$\displaystyle S=\Pi_{n\in\mathbb{N}} \{f\colon\mathbb{N}\rightarrow\cup_{n\in\mathbb{N}}E_n\mid\forall n, f(n)\in E_n\}$$  If $E_n=\{ 0,1\}$, then  $$\displaystyle S_{01}=\Pi_{n\in\mathbb{N}}\{0,1\}=E^{\mathbb{N}}$$, where $E=\{0,1\}$.  By a theorem, $\cup_{n\in\mathbb{N}} E_n$ is countable since the sequence is countable.  I'm not sure how to go on from here to show $S$ is uncountable. Can we say anything about the function $f$ that maps a countable $\mathbb{N}$ to another countable union of sequence of countable sets?
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What would be an example of a set $S$ so that $S$ is countable. However $S'$ is uncountable. In this $S'$ is the set of all the limit points of $S$.
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This is part of a bigger problem I'm trying to prove, but my argument relies of the validity of the following idea. Note that when I say countable, I don't mean finite -- I mean countable infinity.   Consider the set of natural numbers, $\mathbb{N}$. Now take a countable subset of them, call it $\mathsf{S}_1$. Then take a countable subset of $\mathbb{N}\setminus{\mathsf{S}_1}$ and call it $\mathsf{S}_2$. Then take a countable subset of $\mathbb{N}\setminus\{{\mathsf{S}_1}\cup{\mathsf{S}_2}\}$ and call it $\mathsf{S}_3$. Continue in this manner to construct $\mathsf{S}_i$'s such that ${\mathsf{S}_i}\cap{\mathsf{S}_j}=\emptyset$ for $i≠j$.  My question is this: is it necessarily the case that eventually there will be finitely many $\mathsf{S}_i$'s with $\mathbb{N}\setminus\bigcup_{i}{\mathsf{S}_i}={P}$ where $P$ is either empty or finite? My guess is yes, but I'm not how to prove it or how to find a counterexample.
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Let $U$ be an open set in $\mathbb R$. Then $U$ is a countable union of disjoint intervals.    This question has probably been asked. However, I am not interested in just getting the answer to it. Rather, I am interested in collecting as many different proofs of it which are as diverse as possible. A professor told me that there are many. So, I invite everyone who has seen proofs of this fact to share them with the community. I think it is a result worth knowing how to prove in many different ways and having a post that combines as many of them as possible will, no doubt, be quite useful. After two days, I will place a bounty on this question to attract as many people as possible. Of course, any comments, corrections, suggestions, links to papers/notes etc. are more than welcome.
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Does there exist countable/uncountable number of mutually disjoint subsets of $\mathbb R$, each of which is countable and dense in $\mathbb R$?  For finite number, I am able to choose irrationals say, $r_n=\sqrt2-\frac1n$, and get the finite number of mutually disjoint sets as the rationals in $s_1=(-\infty,r_1),s_2=(r_1,r_2),\dots$ But how is such a thing possible for countable/uncountable irrationals because they can not be arranged in increasing order? Is there any other way of doing this? Is the answer to this question in the affirmative? Please help.
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I've been trying to challenge myself but with no luck. Maybe one of you will have a better idea. How can it be proven that there can't be a set which contains its own powerset, using only russell's paradox? I've managed to prove it with other principles or axioms, such as that a set can not belong to itself, or that a powerset cannot belong to itself.
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I know that with choice, the countable union of countable sets is countable, making $\omega^\omega$ and $\epsilon_0$ both countable. Can we show this without choice? E.g. in the case that $\omega_1$ is a countable union of countable sets.
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This is a question from my proofs course review list that I have had trouble understanding.   I understand the concept of disjoint sets.  I'm not sure what they mean by countable. How would one prove a set is countable and furthermore, that the union of two countable sets is countable.  Please help me get started with the question by understanding this concept.
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Let $A_1,...A_k$ be countable sets, I need to prove that the union $\bigcup_{i=1}^{k}A_i$ is countable by explicitly building a surjection.  I can't assume that the sets are disjoint.  I can easily build an injection $g:\bigcup_{i=1}^{k}A_i\rightarrow \mathbb{N}\times\mathbb{N}$ by defining $g(a)=(f_{i}(a),i)$ where $f_i$ is the first bijection from the bijections $f_i,...f_k$ that maps $A_i\rightarrow\mathbb{N}$ such that $a\in A_i$.  However, this map $g$ obviously isn't surjective.  Any ideas for building such map?
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I got the answer as the set $\{5x + 1 : x \in \mathbb{N} ∪ \{0\} \}$. Can anyone confirm if this is correct? Thanks.
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Prove or Disprove: For any family of sets $\{A_n\}_{n\in\mathbb N}$  $$\bigcup_{n=1}^\infty\mathcal P \left({A_n}\right)\subseteq \mathcal P \left({\bigcup_{n=1}^\infty A_n}\right)$$  How do I approach proving this? I know how to unpack the definition of powersets ($\mathcal P \left({A}\right) = \{x | x \subseteq A\}$) but I'm not sure what else I can do. I've done powerset proofs before but none involving indexed family of sets.
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1) Stuck with this problem, can you please help? Show that if $X$ is connected and for a connected subspace $A$ of $X$ we have $X \setminus A = U \cup V$ where $U,V$ are open in $X \setminus A$ and disjoint, then the sets $A \cup U$ and $A \cup V$ are connected.  2) The following result appears in Encyclopaedia of mathematics by Hazewinkel page $375$:  Any compact subset of the Sorgenfrey line is countable.  I tried to prove this without any luck. Why is this true?
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Is the finite Cartesian Product of Countable sets countable?  If possible, could you give a bijection that would work for any example?  Any help would be appreciated.