I was wondering how can you prove that $\mathbb{N}^\mathbb{N} \sim  2^\mathbb{N}$ (where $\mathbb{N}^\mathbb{N}$ is the set of all functon $f:\mathbb{N}\rightarrow \mathbb{N}$).  I think I can show that $2^\mathbb{N} \sim n^\mathbb{N}$ for some $n\in\mathbb{N}$ with a tedious bijection, but I'm not sure what to do with $\mathbb{N}^\mathbb{N}$.  I apologize if the question lacks details/notation, I have only studied elementary set theory and we haven't mentioned $\mathbb{N}^\mathbb{N}$'s cardinality at all (I usually ended up finding injections/bijections to/from $2^\mathbb{N}$, but now I'm curious as to why the above correct).  Thanks in advance.
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My motivation for asking this question is that a classmate of mine asked me some kind of question that made me think of this one. I can't recall his exact question because he is kind of messy (both when talking about math and when thinking about math).   I'm kind of stuck though. I feel like the set $A^{\mathbb{N}} = \{f: \mathbb{N} \rightarrow A, f \text{ is a function} \}$ should have the same cardinality as the power set of A, if A is infinite. On the other hand, in this , it is stated that the sequences with real coefficients have the same cardinality as the reals.   It's easy to see that $A^{\mathbb{N}} \subseteq P(A)$, but (obviously) I got stuck on the other inclusion. Is there any general result that says anything else? References would be appreciated.  EDIT To clarify the intetion of this question: I want to know if there are any general results on the cardinality of $A^{\mathbb{N}}$ other that it is strictly less than that of the power set of A.   Also, I was aware that the other inclusion isn't true in general (as the post on here I linked to gave a counterexample), but thanks for pointing out why too. :)
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What is the simplest way of proving (to a non-mathematician) that the power set of the set of natural numbers has the same cardinality as the set of the real numbers, i.e. how to construct a bijection from $\mathcal{P}(\mathbb{N})$ to $\mathbb{R}$?
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I am having trouble getting started with this particular problem.  Let $A$ be a nonempty set, and let $\mathcal{B}$ be the set of all functions $f:A\to\{0,1\}$. Show that $\mathcal{B}\sim\mathcal{P}(A)$; that is, construct a bijection $\phi: \mathcal{B} \rightarrow \mathcal{P}(A)$.   Faced with this problem, I know that the power set contains all subsets of $A$. Now, does that mean all possible combinations of the functions which map to $\{0,1\}$? Also, does the fact that the characteristic function is associated with the power set have anything to do with the problem? Thanks.
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Question: Let $m,n\in\mathbb{N}$, prove that if $n|m$, $F_n|F_m$.  I've tried to use induction, but I don't really know where to start since there's $2$ numbers:  $n$ and $m\ \dots$ I did induction before with just $1$ number, like proving $1+2+\cdots+n=\frac{n(n+1)}{2}$, that's only dealing with one number $n$, but with this I have no clue how to do it.  Can anyone give me a little hint on how to start, I don't want you to do whole problem for me but can I have some hint so I know where to start?  Thanks for the help!
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I asked a similar question yesterday about well ordered sets, now I am having troubles with equivalence relations.   Could someone suggest an injection from a well known set of cardinality $2^{\aleph_{0}}$ to the set of all equivalence relations of $\mathbb{N}$?  Many thanks in advance!
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I came across this long proof on this site:    But I would like to know whether my direction can work.  Say we want to find the cardinality of all equivalence relations in $\mathbb{N}$. Since it is a subset of all relations in $\mathbb{N}$, I conclude it has a cardinality smaller or equal to $\aleph$. Now, define an injective function from $P(\mathbb{N})$ to the set of equivalence relations by matching each subset of $\mathbb{N}$ with the identity relation (which is an equivalence relation in $\mathbb{N}$.  Therefore the cardinality of all equivalence relations in $\mathbb{N}$ is greater or equal to $\aleph$ and using CSB we get the desired result.  Seems legit?
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Theorem. Let $X$ denote an arbitrary set such that $|X|=n$. Then $|\mathcal P(X)|=2^n$.      Proof. The proof is by induction on the numbers of elements of $X$.      For the base case, suppose $|X|=0$. Clearly, $X=\emptyset$. But the empty set is the only subset of itself, so $|\mathcal P(X)|=1=2^0$.      Now the induction step. Suppose $|X|=n$; by the induction hypotesis, we know that $|\mathcal P(X)|=2^n$. Let $Y$ be a set with $n+1$ elements, namely $Y=X\cup\{a\}$. There are two kinds of subsets of $Y$: those that include $a$ and those that don't. The first are exactly the subsets of $X$, and there are $2^n$ of them. The latter are sets of the form $Z\cup\{a\}$, where $Z\in\mathcal P(X)$; since there are $2^n$ possible choices for $Z$, there must be exactly $2^n$ subsets of $Y$ of which $a$ is an element. Therefore $|\mathcal P(Y)|=2^n+2^n=2^{n+1}$. $\square$      From the above explanation, I don't understand why the set that contains $\{a\}$ will contain $2^{|n|}$ elements when it should clearly be $2^{|1|}$.  The construction of a new set $S$ is the union of the old set with cardinality $n$ and an new element $\{a\}$, therefore the set that do not contain $\{a\}$ still has cardinality $n$ and the set that contains $\{a\}$ is just $\{a\}$, one element...  Can someone please elucidate?
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${ \mathbb{N} }^{ \left\{ 0,1 \right\}  }$ and ${ \left\{ 0,1 \right\}  }^{ \mathbb{N} }$ to be more specific, and is there a countable subset in each one of them? How do I find them?
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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!
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How would I show that $P(\mathbb{N})$ is equivalent to $2^\mathbb{N}$? The questions asks to form a theorem from the statement I just gave. Any help would be appreciated.
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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 understand that both $\mathbb{N}$ and $\mathbb{N}\times\mathbb{N}$ are of the same cardinality by the Shroeder-Bernstein theorem, meaning there exists at least one bijection between them. But I can't figure out what such a bijection would be. The paper that I'm reading gives an example of an injective function $f:\mathbb{N}\to\mathbb{N}\times\mathbb{N}$, $f(n)=(0,n)$, and an injective function $g:\mathbb{N}\times\mathbb{N}\to\mathbb{N}$, $g(a,b)=2^a3^b$. I was thinking perhaps if there were some way to combine these two, I could find a bijection, but I have no idea how to go about that or if it's even possible.  What is an example of a bijection between these two sets, and please explain the process by which you found it?
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Suppose that $\;f_1,\;f_2:\;A\to B$ such that $f_1$ is injective and $f_2$ is surjective.   I was trying to find out whether there exists $f_3:\;A\to B$ such that $f_3$ is bijective. Is it possible to construct it from $f_1$ and $f_2$?  I tried proving the statement non-constructively via a cardinality argument, but ended up hand-waving a lot (i.e., since the cardinality is the same, there exists a bijection, a bit of circular reasoning). But, I'd be more interested in a construction based argument, as I can't think of one.
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I'm trying to find a surjective function $f:\mathbb{N}\to \mathbb{Q}$;  I know that at least one such function must exist since $\mathbb{Q}$ is countable, but I haven't been able to find one.  Can someone show me one such function?  Best regards,  lorenzo
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So the idea that I am having is, saying that we let f be surjective, and we would prove that $f^{-1}$ is injective. What is confusing me is how I would word my proof and how I would represent it. I am starting it off with Let X and Y be non-empty sets, and let $f: X\rightarrow Y$ be a function. Now I'm stuck.
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Find the cardinality of the set of all finite subsets of $\mathbb{R}$.    I have proved that the set of all finite subsets of $\mathbb{N}$ is countable . But I cannot find the cardinality of the set in case of $\mathbb{R}$ .  My Attempt:   First I have considered the set $$A_k=\{\{a_1, a_2, ..., a_k\}| a_i \in \mathbb{R}  \ and\ a_1&lt;...&lt;a_k  \}$$  Then $S=$ The set of all finite subsets of $\mathbb{R}=\bigcup_{n=1}^{\infty}A_{n}$, and now in the case of $\mathbb{N}$, I could show that this $A_k$ is countable and consequently the set of all finite subset of $\mathbb{N}$ is countable.  But in this case I cannot say anything like this...In this case the set will be of the cardinality as that of $\mathbb{R}$ (I think). But cannot prove it.  I appreciate your help.  Thank you.
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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 $\mathbb{Z}$?