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I know that the set of functions $ f:\mathbb N \to \mathbb N$ is uncountable, but what if we consider only $f$ such that $f$ is increasing? I want to know if this set is countable D: and also the case of bijective and increasing (clearly if the firstone holds then this too). I think that it could be possible that it's countable

Martin Sleziak
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Miguel
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    A slight modification of this question (non-decreasing vs. increasing) was also asked as one part of this question. – Martin Sleziak Apr 04 '13 at 06:26
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    The "real" question has been answered already, but since you also asked about "bijective and increasing", notice that the only bijective increasing function $\mathbb N\to\mathbb N$ is the identity function. – Andreas Blass Apr 04 '13 at 13:14

5 Answers5

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This is essentially reversing Brian M. Scott's initial argument...

Let $\mathbb{N}^{\mathbb{N}\uparrow}$ denote the family of strictly increasing functions $\mathbb{N} \to \mathbb{N}$. Given any function $f : \mathbb{N} \to \mathbb{N}$ define $\Phi ( f ) : \mathbb{N} \to \mathbb{N}$ by $$\Phi ( f ) ( n ) = \sum_{i=0}^n ( f(i) + 1 )$$ Then $\Phi ( f )$ is strictly increasing, and the mapping $f \mapsto \Phi ( f )$ is one-to-one.

Therefore, $| \mathbb{N}^\mathbb{N} | \leq | \mathbb{N}^{\mathbb{N}\uparrow} |$. A quick application of the Cantor–Bernstein–Schröder Theorem then yields $| \mathbb{N}^{\mathbb{N}\uparrow} | = | \mathbb{N}^\mathbb{N} | = 2^{\aleph_0}$.

user642796
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  • @einpoklum: It's customary to make such additions in the comments. Not into the answers. – Asaf Karagila Apr 04 '13 at 10:34
  • @AsafKaragila: That's not what the FAQ says (Other people can edit my posts?) "If you are not comfortable with the idea of your contributions being collaboratively edited by other trusted users, this may not be the site for you." – Dietrich Epp Apr 04 '13 at 11:17
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    @Dietrich: The FAQ cannot be changed per site; but the mentality and behavior of the community does differ from one site to another. I want to think that in my nearly thousand days of activity I learned a few things about the community norm here. – Asaf Karagila Apr 04 '13 at 11:20
  • @AsafKaragila: I would think very-high-rep users would in general be more accepting of these kinds of edits... or am I wrong? – einpoklum Apr 04 '13 at 13:28
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    @einpoklum: I don't follow this logic. – Asaf Karagila Apr 04 '13 at 13:31
  • "$\Phi ( f ) ( n ) = \sum_{i=0}^n ( f(i) + 1 )$" does this actually sum all the sequences after adding 1 to each one so it will equal to a new 'combined' increasing sequence ? Can you also explain a bit about how to apply CBS here ? – GinKin Mar 11 '14 at 12:32
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    @GinKin: (I'm not exactly sure of the first part of your comment.) As a preliminary note, be warned that I include $0$ in $\mathbb{N}$. I set up the function $\Phi$ so that (1) it always outputs a strictly increasing function, and (2) it is one-to-one. Note that I cannot simply take $\Phi(f)(n)=\sum_{i=0}^nf(i)$, since if $f(n+1)=0$ then $\Phi(f)(n)=\Phi(f)(n+1)$. The "$+1$" is there to ensure that $\Phi(f)$ is strictly increasing. For one-to-one, note that given $\Phi(f)$ we can simply reverse the process to get $f$ back: $f(0) = \Phi(f)(0)-1$; $f(1) = \Phi(f)(1)-f(0)-2$, etc. [cont...] – user642796 Mar 11 '14 at 12:44
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    [...inued] For CBS, note that the function $\Phi$ witnesses that $| \mathbb{N}^{\mathbb{N}} | \leq | \mathbb{N}^{\mathbb{N}\uparrow} |$. As $\mathbb{N}^{\mathbb{N}\uparrow} \subseteq \mathbb{N}^{\mathbb{N}}$ we clearly have that $| \mathbb{N}^{\mathbb{N}\uparrow} | \leq |\mathbb{N}^{\mathbb{N}}|$. – user642796 Mar 11 '14 at 12:44
  • Thanks for the explanation. In the first part I asked what does the sum actually do here, does it sum all the sequences so each $f(i)$ represents a sequence ? – GinKin Mar 11 '14 at 12:52
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    @GinKin: Sorry about this long delay; was busy/sleeping. Okay, so $f$ itself is a sequence in $\mathbb{N}$ (or, equivalently, a function $\mathbb{N} \to \mathbb{N}$). So $f(i)$ is just a natural number, meaning that we are just summing up naturar numbers. – user642796 Mar 12 '14 at 06:03
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Let $e_1,e_2,e_3,\dots$ be any sequence of $0$'s and/or $1$'s. Let $a_1=1$ and $a_{i+1}=a_i+e_i+1$.

This shows that there are at least $c$ increasing sequences. So there are exactly $c$.

André Nicolas
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  • Why do you take $e_i$ to be binary sequences when the question asks about all increasing sequences of natural numbers ? – GinKin Mar 11 '14 at 12:24
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    @GinKin: From any binary sequence, we produce an increasing sequence of natural numbers. It is clear that two distinct binary sequences produce distinct increasing sequences. So we have produced an injective mapping from binary sequences to increasing sequences. To put it another way, the set of binary sequences has the same cardinality as a subset of the increasing sequences. Thus the cardinality of the set of increasing sequences is at least as big as the cardinality of the set of binary sequences. The set of binary sequences has cardinality $c$. (Cont) – André Nicolas Mar 11 '14 at 13:06
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    (Cont) So the set of increasing sequences has cardinality $\ge c$. But the set of increasing sequences also has cardinality $\le c$. Therefore the vardinality is exactly $c$. – André Nicolas Mar 11 '14 at 13:08
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It's uncountable. Suppose that $f_n$ is a sequence of strictly increasing sequences, let $g(0)=f_0(0)+1$; and $g(n+1)=\max\{g(n)+1, f_{n+1}(n+1)+1\}$. Clearly $g$ is not one of the $f_n$'s.

Asaf Karagila
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  • How does uncountability follow from this? – Git Gud Apr 04 '13 at 06:08
  • @GitGud: The same way it would follow from any standard diagonalization argument. Any function from $\Bbb N$ onto this set cannot be surjective. – Asaf Karagila Apr 04 '13 at 06:09
  • This doesn't answer what the cardinality is, or am I missing something ? – GinKin Mar 11 '14 at 12:37
  • @GinKin: That is correct. But if you read the actual question, then OP wanted to know "that this set is countable". I answered with a way to prove that it is not. – Asaf Karagila Mar 11 '14 at 12:53
  • The title actually asks for the cardinality. Also my question from yesterday was marked as duplicate of this one even though mine was asking for the cardinality and not if it's countable. – GinKin Mar 11 '14 at 12:57
  • @GinKin: Yes. The title says cardinality, and the question says "uncountable". There are other answers which answer both the title and the body of the question, and therefore answer your question as well. – Asaf Karagila Mar 11 '14 at 13:08
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Let $\mathscr{F}$ be the set of strictly increasing functions from $\Bbb N$ to $\Bbb N$: then $|\mathscr{F}|=2^\omega=|\wp(\Bbb N)|$, so $\mathscr{F}$ is certainly uncountable. One way to see this is to construct a bijection between $\mathscr{F}$ and ${}^{\Bbb N}\Bbb N$, the set of all functions from $\Bbb N$ to $\Bbb N$. This isn’t too hard.

Define a function $\mathscr{F}\to{}^{\Bbb N}\Bbb N:f\mapsto\hat f$ as follows: let $f\in\mathscr{F}$, and define $\hat f$ by setting $\hat f(0)=0$ and $\hat f(k)=f(k)-f(k-1)-1$ for $k>0$. It’s easy to check that this map is a bijection, so $|\mathscr{F}|=\left|{}^{\Bbb N}\Bbb N\right|$, which you probably already know is $2^\omega=|\wp(\Bbb N)|$.

Added: The inverse map is easy enough to write down as well: if $f\in{}^{\Bbb N}\Bbb N$, let

$$g:\Bbb N\to\Bbb N:n\mapsto n+\sum_{k\le n}f(k)\;.$$

then $$\hat g(n)=\left(n+\sum_{k\le n}f(k)\right)-\left((n-1)+\sum_{k\le n-1}f(k)\right)-1=f(n)\;,$$

i.e., $\hat g=f$.

Brian M. Scott
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  • May I ask, why do you use the notation $^\Bbb N\Bbb N$? – SBF Apr 04 '13 at 08:18
  • It is written there – gukoff Apr 04 '13 at 08:20
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    @Ilya: Because ${}^AB$ is a standard notation for the set of functions from $A$ to $B$, one that can’t be confused with exponentiation. That isn’t a problem here, but it is when you’re talking about cardinal numbers $\kappa$ and $\lambda$ and want to distinguish the set of functions ${}^\kappa\lambda$ from the cardinal number $\lambda^\kappa$. – Brian M. Scott Apr 04 '13 at 08:20
  • @BrianM.Scott: I see, I thought $B^A$ is a standard one, like $\Bbb R^{[0,\infty)}$ in stochastic processes usually denotes the space of real-valued functions on the domain $[0,\infty)$ – SBF Apr 04 '13 at 08:22
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    @Ilya: They’re both standard; it’s probably only people working in somewhat set-theoretic areas who have to worry about the potential confusion and who end up using ${}^AB$. – Brian M. Scott Apr 04 '13 at 08:24
  • What does $k$ stands for here? – GinKin Mar 10 '14 at 18:13
  • @Ilya, looks like Brian hasn't logged in in a long time, do you know what does the $k$ stands for in his answer ? – GinKin Mar 11 '14 at 12:17
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    @GinKin: as far as I can follow, Brian defines a transformation $\hat$ from the collection of function $\mathscr F$ to another collection $^\Bbb N\Bbb N$. Given any $f\in \mathscr F$, we hence obtain that $\hat f\in ^\Bbb N\Bbb N$ or equivalently $\hat f:\Bbb N \to \Bbb N$. Now, to define $\hat f$ as a function you need to define it for any natural number $k\in \Bbb N$ - that was is done in the answer, $k$ is just a generic argument of $\hat f$. – SBF Mar 11 '14 at 16:48
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A nice idea how to prove this was given in this answer by Srivatsan. I am reposting it here since the other question addresses several points, so this might be easily missed among solutions of other parts of that question.

Consider only maps $\varphi \colon \mathbb N\to\mathbb N$ such that $k$ is mapped either to $2k$ or to $2k+1$.

Every such map is increasing.

We have two possibilities how to choose $\varphi(k)$ for each $k$, so there is $2^{\aleph_0}=\mathfrak c$ such functions. (To make this argument more rigorous, you can show that you can obtain for each element of $\{0,1\}^{\mathbb N}$ a different function of this form.)

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Martin Sleziak
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