Is $Aut(\mathbb{N})$ countable?

Question

I'm trying to solve a problem from set theory. The problem is simple: is the set of automorphisms of natural numbers $Aut(\mathbb{N})$ countable?

It seems to me the answer is "not" and my attempt for a proof is following. Let us find any injective map $F:Aut(\mathbb{N})\rightarrow (0,1)$. I think it is possible to construct such a map explicitly. Let $Aut^{+}(\mathbb{N})$ be a subset of $Aut(\mathbb{N})$ which consist of automorphisms of $\mathbb{N}$ which map any element $n\in\mathbb{N}$ to $n+1,n+2,...$.

Comment: for any $m\in\mathbb{N}$ consider a map $p_{m}$ which sends any $n\in\mathbb{N}$ to $n+m\in\mathbb{N}$. The constructed map $p_{m}:\mathbb{N}\rightarrow \mathbb{N}$ is bijective which implies that it is an element of $Aut(\mathbb{N})$ and I think we can say that $Aut^{+}(\mathbb{N}) = \cup _{m\in \mathbb{N}} \:p_{m}$.

For any $n,m\in\mathbb{N}$ let $p_{m}(n)$ be an image of $n$ under the action of $p_{m}$ - map. Let us try to define $F$ as follows:

$F_{n}:p_{m}(n)\rightarrow \frac{n}{p_{m}(n)}$

For fixed $n\in\mathbb{N}$ defined above $F_{n}$ is a map from $Aut^{+}(\mathbb{N})$ to $(0,1)$ which sends a tuple $\{p_{1}(n),p_{2}(n),...\}$ to $\{\frac{n}{n+1},\frac{n}{n+2},...\}$. It is not surjective. We can consider

$F:p_{m}\rightarrow \cup_{n\in\mathbb{N}} F_{n}(p_{m})$

and it seems to be a good map but it is not injective and I have a trouble. Is it possible to continue this attempt for a proof or should I turn to another ideas?

Answer 1

If $(z_n)_{n\geq1}$ is a sequence of zeroes and ones, consider the map $\mathbb N\to\mathbb N$ which maps each set of the form $P_n=\{2n+1,2n+2\}$ into itself: if $z_n=0$, the map fixes the two elements of $P_n$, and if $z_n=1$ it swaps them.

This gives an injection from the uncountable set $\{0,1\}^{\mathbb N}$ into the group of bijections of $\mathbb N$.

Answer 2

EDIT: This is a side concern, but it looks like you're confusing two things: the set of injections (these are endomorphisms) from $\mathbb{N}$ to $\mathbb{N}$, and $Aut(\mathbb{N})$, which is the set of bijections (automorphisms) from $\mathbb{N}$ to $\mathbb{N}$.

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To show that $Aut(\mathbb{N})$ is uncountable, it is enough to find a surjective map from $Aut(\mathbb{N})$ to some uncountable set (such as $(0, 1)$), or an injective map from some uncountable set to $Aut(\mathbb{N})$. Remember:

• If $f: A\rightarrow B$ is surjective, then $A$ is at least as big as $B$.

• If $g: B\rightarrow A$ is injective, then $A$ is at least as big as $B$.

So you are trying to construct the wrong thing. (To see this, note that there is an injection from $\{1, 2, 3\}$ to $(0, 1)$, but surely $\{1, 2, 3\}$ is not uncountable!)

You can indeed construct an explicit surjection from $Aut(\mathbb{N})$ to $(0, 1)$; however, it's kind of messy. I would suggest that it's easier to construct a surjection from $Aut(\mathbb{N})$ to $\mathcal{P}(\mathbb{N})$, the set of all sets of natural numbers (which is also uncountable).

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You can also argue directly via a diagonal argument: suppose I had a countable list $\pi_i$ of permutations of $\mathbb{N}$. Do you see how to construct a permutation $\psi$ of $\mathbb{N}$ which is different from each of the $\pi_i$s? (HINT: each $\pi_i$ either swaps $2i$ and $2i+1$, or it doesn't . . .)

Answer 3

A sketch proof goes like this: every infinite subset of $\Bbb{N}$ is in bijection with $\Bbb{N}$. So every bijection corresponds to an infinite subset of $\Bbb{N}$ and the subset of $2^{\Bbb{N}}$ comprising all the infinite subsets is uncountable. And so the set of bijection is uncountable.

Answer 4

Let $P'(N)$ be the subset of $P(N)$ the set of subsets of $N$, such that the cardinal of the complementary of every element of $P'(N)$ is >1. You can embed $P'(N)$ in $Aut(N)$ as follows: if $E\in P'(N)$, consider an automorphism $f_E$ of $N$ such that $Fix(f_E)=E$. Remark that the cardinal of $P'(N)$ is the cardinal of $P(N)$. This implies that $card(Aut(N))\geq card(P(N))>card(N)$.

Answer 5

Judging from what you've written, I assume that by "automorphisms of $\mathbb{N}$" we mean "automorphisms of $\mathbb{N}$ as a set" and not, for example, as an arithmetic structure.

What your proof would show, if you could complete it, is that there is a subset of the set of automorphisms of $\mathbb{N}$ that is small enough to fit into the interval $(0, 1)$. That's definitely not what you want to prove. A good approach in the same vein might be to show that there's an injection from $(0, 1)$ into $Aut(\mathbb{N})$; that is, construct a new automorphism for each real number in that interval. Another option would be to build a surjection from a subset of $Aut(\mathbb{N})$ onto $(0, 1)$; I don't think your $Aut^+(\mathbb{N})$ will work for that - if I understand it correctly, it's countable, since the $p_m$ are already indexed by natural numbers.

Answer 6

Notation: For any set $A$ we write $P(A)$ for the power-set of $A, $ which is the set of all subsets of $A.$ Clearly the map $f(a)=\{a\}$ for each $a\in A$ is an injection from $A$ into $P(A).$

Theorem (Georg Cantor). There is no map $g$ from $A$ onto $P(A).$ Proof: For any $g:A\to P(A)$ let $s_g=\{a\in A: a\not \in g(a)\}.$ We have $s_g\in P(A).$

Now if $s_g=g(a)$ for some (any) $a\in A$ then

(1). $a\in s_g\implies a\in g(a)\implies a\not \in s_g$ by def'n of $s_g.$

(2). $ a\not \in s_g\implies a\not \in g(a)\implies a\in s_g$ by def'n of $s_g.$

In both (1) and (2) we have contradictions, so the assumption $s_g=g(a)$ is untenable. So $g$ is not onto , because $s_g\not \in \{g(a):a\in A\}.$

Corollary: Let $B$ be the set of bijections from $N$ to $N.$ There is no map from $N$ onto $ B . $

Proof: For $f\in B$ let $\phi (f)=\{n\in N:f(n)=n\}.$ Now for any $s\in P(N)$ there exists $f_s\in B$ such that $\phi (f_s)=s.$ Because there exists a bijection $h_s:N$ \ $s\to N$ \ $s$ such that $h_s(n)\ne n$ for each $n\in N$ \ $s $ ; so let $f_s(n)=n$ for $n\in s , $ and $ f_s(n)=h_s(n) $ for $n\in N$ \ $s.$

Now consider any $h:N\to B.$ If $h$ were onto then $\{\phi (h(n)):n\in N\}\supset \{\phi (f_s):s\in P(N)\}=\{s:s\in P(n)\}=P(N).$ This would mean that the composite function $\phi \cdot h $ is a surjection from $N$ to $P(N), $ which is impossible by the theorem of Cantor.