Let $A$ be the set of all finite subsets of $\mathbb{Z}_+$, then is $A$ countable?   Defiene $I_j$ as the set consisting of all subsets of $\mathbb{Z}_+$ having $j$ elements, then i think $A=\bigcup I_j$ and since countable union of countable sets are countable thus $A$ is countable!
3225
Show that the set of all finite subsets of $\mathbb{N}$ is countable.   I'm not sure how to do this problem. I keep trying to think of an explicit formula for 1-1 correspondence like adding all the elements in each subset and sending that sum to itself in the natural numbers, but that wouldn't be 1-1 because, for example, the set {1,2,3} would send to 6 and so would the set {2,4}.  Multiplying all the elements in each subset and sending that product to itself in the natural numbers wouldn't work either since, for example, {2,3} would send to 5 and so would the set {1,5}. Any ideas?
108507
I don't understand a proof right at the beginning of this document found .   This proof is on why any base for the open sets in a second countable space has a countable subfamily that is a base.  My trouble comes in the second paragraph of the first proof. It says $\{B_{i_x}\mid x\in C_k\}$ is a countable set whose union is $C_i$. First, why is the set countable? If you take a set $B_{i_x}$ for any $x\in C_k$, what if $C_k$ is not countable itself? And why is the union $C_i$? Should it instead be $C_k$? I don't see $C_i$ mentioned anywhere in the second paragraph.
100753
Let $X$ be any uncountable set with the cofinite topology. Is this space first countable?  I don't think so because it seems that there must be an uncountable number of neighborhoods for each $ x \in X$. But I am not sure if this is true.
21587
I want to prove that $\mathbb{Q}$ is countable. So basically, I could find a bijection from $\mathbb{Q}$ to $\mathbb{N}$. But I have also recently proved that $\mathbb{Z}$ is countable, so is it equivalent to find a bijection from $\mathbb{Q}$ to $\mathbb{Z}$?
56539
One of my homework questions was to prove, from the axioms of ZF only, that a countable union of countable sets does not have cardinality $\aleph_2$. My solution shows that it does not have cardinality $\aleph_n$, where $n$ is any non-zero ordinal (not necessarily finite). I have a sneaking suspicion that my solution is actually invalid, but I can't find any reference which invalidates my conclusion.  I have read that it is provable in ZF that there are no cardinals $\kappa$ such that $\aleph_0 &lt; \kappa &lt; \aleph_1$, but I believe the conclusion of my proof does not preclude the possibility that the cardinality is incomparable to $\aleph_1$ or some such.   I think the weakest point in my solution is where I claim that the supremum of a countable set of countable ordinals is again countable. This is true, of course, but it sounds uncomfortably close to the claim "a countable union of countable sets is countable", which is well-known to be unprovable in ZF. Can anybody confirm that the ordinal version is provable in ZF though? If not, I think I can weaken the claim to "the supremum of any set of countable ordinals is at most $\omega_1$", and this establishes the weaker result that the cardinality of a countable union of countable sets is not $\aleph_n$ for any ordinal $n \ge 2$.
18368
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.
91996
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.
22260
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?
39733
Here is the proof I tried to weave while trying to prove this theorem:     Theorem. Every infinite subset of a countable set is countable.      Proof. Let $A$ be a countable set and $E\subset A$ be infinite. Then $A\thicksim\mathbb{N}$. This implies that there is a sequence $\left\{x_{n}\right\}_{n\in\mathbb{N}}$ where $x_{n}\in A$. Construct a sequence $\left\{n_{k}\right\}_{k\in\mathbb{N}}$ where $x_{n_{k}}\in A$, $n_{1}$ is the smallest positive integer such that $x_{n_{1}}\in A$, and $n_{k+1}&gt;n_{k}$. Then      $$ E=\bigcup_{k=1}^{\infty}x_{n_{k}}, $$      and $E$ is countable, because $E\thicksim\mathbb{N}$. $\blacksquare$   Is it convincing?
68243
I was reading about countably infinite sets and the definition goes as, "A set is countably infinite if its elements can be put in one-to-one correspondence with the set of natural numbers"(Source: Google). Well yes, this make sense. As the set $\mathbb{N}$ is countably infinite and set which can it can create a bijection with, would end up having the same number of elements and thus they too would be countably infinite. But this would be true iff $\mathbb{N}$ is countably infinite. How would you prove that? Or is the idea of countably infinite base on some assumption that $\mathbb{N}$ is countably infinite?
22378
In my book, it proves that an infinite subset of a coutnable set is countable. But not all the details are filled in, and I've tried to fill in all the details below. Could someone tell me if what I wrote below is valid?  Let $S$ be an infinite subset of a countable set $T$. Since $T$ is countable, there exists a bijection $f: \mathbb{N} \rightarrow T$. And $S \subseteq T = \{ f(n) \ | \ n \in \mathbb{N} \}$. Let $n_{1}$ be the smallest positive integer such that $f(n_{1}) \in S$. And continue where $n_{k}$ is the smallest positive integer greater than $n_{k-1}$ such that $f(n_{k}) \in S$. And because $S$ is infinite, we continue forever. Now consider the function $\beta : \mathbb{N} \rightarrow S$ which sends $k \rightarrow f(n_{k})$.   So in order for $S$ to be countable, $\beta$ would have to be a bijection. $\beta$ is injective since if $f(n_{k}) = f(n_{j})$, then $n_{k} = n_{j}$ because $f$ is injective. We also can conclude $k = j$ because the various $n_{i}$ chosen were strictly increasing.   Edit: $\beta$ also needs to be surjective. So for every $r \in S$ there exists a $q \in \mathbb{N}$ such that $\beta(q) = f(n_{q}) = r$. We know that $f^{-1}(r) \in \{n_{1}, n_{2}, ..., n_{k} ... \}$, so we can let $f^{-1}(r) = n_{d}$. Thus $\beta(d) = f(n_{d}) = r$.
50394
Denote by $\mathbb{N}$ the set of natural numbers and by $2^{\mathbb N}$ the set of all subsets of $\mathbb N$. Let $E$ be some subset of $2^{\mathbb N}$ such that for  every pair of elements in $E$, one is a subset of the other. Clearly, $E$ could be countable by considering $E=\big\{\{1,\cdots, n\}, n\in\mathbb N\big\}$. My question is the following: could we find an example that $E$ is uncountable? Many thanks for the answer!
48495
If you have an uncountable subset of $\mathbb{R}$, which is dense in $\mathbb{R}$, is its complement countable?   I'm trying to see if you can take the set of irrationals, remove a countable amount, and still have a dense subset in $\mathbb{R}$.If this is the case, this sets complement would still be countable.  My goal is to determine if you can have an uncountable subset of $\mathbb{R}$, which is dense in $\mathbb{R}$, yet its complement is uncountable.  If somebody could just let me know if this is possible or not I would appreciate it. Thanks.
83876
There is a theorem that states that the finite union of closed sets is closed but I was wondering if we have a set that consists of countable many subsets that are all closed if that set is closed. I really want to believe that the set is closed but I've been wrong in past so if anyone can supply me with an answer I would be very grateful.  Thank you.
24139
Let $f:A \to \mathbb R$ and suppose that $$ \sup\Big\{\sum_{a\in F}\lvert\, f(a)\rvert :  F\text{ is finite subset of }A\Big\} &lt; \infty $$ then the set $\{ a \in A : f(a) &gt; 0\}$ is countable.  My try: Define these element of ${ a ∈ A : f(a) &gt; 0}$, then $∑_{n=1}^\infty\lvert\, f(a_n)\rvert$ is determined, regardless of the method of arrangement.
100909
Possible Duplicate:          Those terms keep confusing me. Could somebody give a clear and unambiguous definition? How do we prove a set is countable, finite, ...? I know that there are more real numbers than there are integers. How do we prove this? What about other sets like $\mathbb{Z, Q, C}$?
56817
I have alot of questions.   Do Infinite sets have the same cardinality when there is a bijection between them? Are $\mathbb{N}$ and $\mathbb{Z}$ infinite sets? I assume they are, but why? Why does there not exist a bijection between them? Why do $\mathbb{N}, \mathbb{Z}, \mathbb{Q}$ have the same cardinality? I assume there exists a bijection between them, but how can I find this function?