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I want to use Cantor's diagonalisation argument to prove that the set S of surjective functions of the form $\Bbb{N} \to \Bbb{N}$ is uncountable. The normal procedure is creating a matrix and filling it with elements of S and introducing a new surjective function $p$.The problem that I'm having is that I don't know in which way to change the diagonal elements so that $p$ should, but cannot be in the matrix.

Can someone help me, or at least give me some tips?

Zev Chonoles
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n00b1990
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    You can use other methods to show that $S$ is uncountable. The diagonal process doesn't fit every uncountability proof. – Asaf Karagila May 22 '13 at 18:29
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    As Asaf says, a naive diagonalizability argument won't work here - for instance, it could be the case that $S_i(i)=10$ for all $i$ in the enumeration you choose, in which case you'll never have an $n$ such that $p(n)=10$. – Steven Stadnicki May 22 '13 at 18:32
  • @Steven: One can use a variation of the diagonal argument, as in my answer below. – Asaf Karagila May 22 '13 at 18:59

5 Answers5

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Here is an idea for how to use the diagonalization method after all.

Suppose $f_n$ is a list of surjections. We define $g$ in the following way:

$$g(k)=\begin{cases}\frac k2 & k\text{ is even}\\1+\sum_{m\leq k}f_m(k) &\text{otherwise}\end{cases}$$

So $g$ is clearly surjective (its restriction to the even numbers is surjective). To see that $g$ is not one of the $f_n$'s, we need to show that whenever $n\in\Bbb N$ there is some $k$ such that $f_n(k)\neq g(k)$.

Let $n\in\Bbb N$, and let $k$ be any odd number such that $n\leq k$, then we have that $$f_n(k)\leq\sum_{m\leq k}f_m(k)<1+\sum_{m\leq k}f_m(k)=g(k).$$

Therefore $f_n(k)\neq g(k)$ so $f_n\neq g$, and so we have that $g$ is not on the list.

If you are willing to use something other than diagonalization then we can give the following argument:

For every infinite subset of $\Bbb N$, we can write the following surjection:

$$f_A(n) =|\{a\in A\mid a<n\}|$$

Then $f_A$ is surjective because $A$ is infinite, so for every $n$ we can find $k$ such that $f_A(n)=k$.

Note that if $A\neq B$ then $f_A\neq f_B$: if $n=\min A\triangle B$, and suppose that $n\in A$, then $f_A(n+1)>f_B(n+1)$.

Therefore there are at least $2^{\aleph_0}$ surjections. Moreover all of them are non-decreasing!

Asaf Karagila
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  • Thank you for the help, I appreciate it. Why do you you choose to divide by 2? Or can I just say $k-1$ instead of $k/2$? Also, why do you choose to use the $max$ function? – n00b1990 May 22 '13 at 21:01
  • @n00b1990: If you just take $k-1$ you won't assure that the function is surjective, because you have guarantee that any even integer is in its range. As to the $\max$ function, I just wanted to be sure that $g(k)$ is larger than all the values possible for $f_n(k)$. You could also use $g(k)$ as larger than $f_n(k)$ for all $n\leq k$, not just the last two. The important thing is that you assure that $g(k)$ is distinct from "enough" functions on the list, so you can be certain it isn't there. – Asaf Karagila May 22 '13 at 21:20
  • I'm sorry to bother you. It's just that I don't see why the proof holds. How do you guarantee that $g$ cannot be in the matrix? Also, why are you defining $k$ in terms of $n$, can't we just use $n$? Would you be kind enough to show me this proof in matrix form (if it's not too much effort I hope)? – n00b1990 May 23 '13 at 18:29
  • It's going to be difficult in matrix form, because I'd require things which are too complicated for MathJax to compile. Let me give the gist, again. First we want to be sure that $g$ is surjective. This is the difficulty to begin with. So we ensure that the restriction of $g$ to the even numbers is surjective. Now in order to use the diagonal argument we need to compensate for the fact that we already decided the values of the even numbers. How do we do that? (to be continued in the next comment...) – Asaf Karagila May 23 '13 at 18:34
  • We ensure that the odd coordinates are not different from the function with the same index, but from the one before that as well. It might be more transparent if we require $g$ to be different than all the $f_n(k)$ for $n\leq k$, but the idea is the same. The heart of the diagonal argument is to ensure your result is different than all those in the list. In the simple situation of the real numbers one just have to flip the bit in the diagonal position, but here we can't afford that much, and we need to be clever. – Asaf Karagila May 23 '13 at 18:35
  • @n00b1990: I changed the function a little bit, it might be easier to see now why $g$ is not on the list. – Asaf Karagila May 23 '13 at 18:44
  • Now I understand, there is only one more thing :). When $k$ is not even, can't we just say that $g(k) = 1 + f_k(k)$ instead of $1 +$ the sum of all diagonal elements? If we did it that way then $g$ would still be different than $f$, right? – n00b1990 May 23 '13 at 18:59
  • And what about $k = 0$. Then $k$ would be even but $g(k)$ would still be the same as $f_k(k)$? – n00b1990 May 23 '13 at 19:02
  • @n00b1990: But then $g$ would be distinct from all the odd-indexed functions, but we have no way to assure it is distinct from any of the even-indexed functions. We allow $g(k)=f_k(k)$ when $k$ is even, but then we assure that $g\neq f_k$ by showing that for some odd $n$ we have $g(n)\neq f_k(n)$. – Asaf Karagila May 23 '13 at 19:02
  • Can't we assure that it is distinct for all indexed functions by saying that if $k$ is even then $g(k) = k/2$ and if $k$ is odd then $g(k) = 1 + f_k(k)$. Wouldn't $g(k) = k/2$ also make it distinct from all even-indexed functions? – n00b1990 May 23 '13 at 19:06
  • @n00b1990: Suppose $f_0$ has the property that $f_0(k)=k/2$ for even $k$ and $0$ otherwise. How are you going to show that $g\neq f_0$? – Asaf Karagila May 23 '13 at 19:07
  • Let me state it another way: What if some even-indexed diagonal element, $f_4(4)$, is $0$. Then by your definition of $g(k)$ that would make $g(4)$ zero as well and then we can't say that $f_4(4) \neq g(4)$ because $f_4(4) = g(4)$. Therefore we have two rows $f_4$ and $g$ for which we don't know if they are equal or not? – n00b1990 May 23 '13 at 20:07
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Among the surjections $\mathbb N\to\mathbb N$ you have the maps $$f(n)=\begin{cases}k&\text{if }n=2k\\g(k)&\text{if }n=2k-1\end{cases}$$ where $g\colon\mathbb N\to\mathbb N$ is one of the uncountable arbitrary maps $\mathbb N\to\mathbb N$.

Hagen von Eitzen
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The following is essentially Asaf's construction except that I think it's a little closer to the standard "diagonal" method. I assume that, as in the usual picture of the diagonal method, we're given an infinite matrix and that we want a new sequence that differs from every row of that matrix, but that we also want our new sequence to contain every natural number at least once. Let me start by producing every second term of my new sequence; I'll fill in the omitted terms later. The key idea is to use a slightly different "diagonal". Instead of the "main diagonal", consisting of the $n$-th entry of the $n$-th row for all $n$, I'll use a "diagonal" of half the "slope". That is, in the $n$-th row, I'll concentrate on the $2n$-th entry, and I'll make sure my new sequence differs from all these. To be specific, I'll define the $2n$-th entry of my new sequence to be $1$ plus the $2n$-th entry in the $n$-th row. (So I'm defining the even-numbered terms of my new sequence here; the odd-numbered ones will be defined later.) Notice that, by doing this for every $n$, I've already ensured that my new sequence will differ from every row in the given matrix; specifically, it differs from row $n$ in position $2n$, and this difference is guaranteed no matter what I do with the odd-numbered positions in my new sequence. But now it's easy to finish the job and take care of the surjectivity requirement; just put $k$ into the $k$-th odd-numbered position (i.e., into position $2k+1$) for every $k$.

Andreas Blass
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  • Ok, I understand now. So all natural numbers are then positioned at all the odd-numbered positions and because of skipping one column by the $2n$ each time, you have assured that $g$ cannot be the same as some other row $f$. If so, then I get it now. – n00b1990 May 23 '13 at 19:54
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I'll show the set of bijections of $\mathbb N$ to $\mathbb N$ is uncountable, which is a stronger result.

Let $B$ denote the set of binary sequences, i.e., the set of sequences $(b_k)$ where $b_k\in \{0,1\}$ for each $k.$ It is well known that $B$ is uncountable.

Divide $\mathbb N$ into "couples" $C_k=\{2k-1,2k\},k=1,2,\dots$ We will be looking at bijections of $\mathbb N$ to $\mathbb N$ that for eack $k$ either equals the identity on $C_k$ or switches $2k-1$ with $2k.$

Here's how we do it: Let $b=(b_k)\in B.$ Define $f_b:\mathbb N\to \mathbb N$ as follows: For $k=1,2,\dots,$ if $b_k=0$ we let $f_b$ be the indentity on $C_k;$ if $b_k=1$ we let $f_b$ be the switching map on $C_k.$

Then each $f_b$ is a bijection of $\mathbb N$ to $\mathbb N$. Furthermore, the map $b\to f_b$ is injective. Since $B$ is uncountable, so is $\{f_b:b\in B\}.$ It follows that the set of bijections from $\mathbb N$ to $\mathbb N$ is uncountable.

zhw.
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$\begin{array}{ccc} f_1(1) && f_1(2) && f_1(3)...\\ f_2(1) && f_2(2) && f_2(3)...\\ f_3(1) && f_3(2) && f_3(3)...\\ \vdots \end{array} $

Form a function $g$ with $g(1)$ the least natural number other than $f_1(1)$, $g(2)$ the least natural number other than $f_2(2)$ which is not already used for $g(1)$ and so on.

If $g$ is a surjection, done. If $g$ is not a surjection, then there exist constants $k,c$ such that for all $n \geq k$, $f_n(n)=c$. All functions with exactly one member that maps to $c$ are in the sequence, and for any $n>k$ there are infinitely many such functions with $f(n)=c$, but only one can go in the $n$th place. Contradiction.

Asaf Karagila
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Angela Pretorius
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  • @AngelinaRichardson Thanks for the help. What does $k$ stand for? And how are there infinitely many functions? – n00b1990 May 22 '13 at 20:55
  • And how do you know if $g$ is a surjection? Or maybe we don't.. – n00b1990 May 22 '13 at 21:08
  • And what if the diagonal elements are the same? Then $g$ won't hold because that particular number can never occur in $g$ (by your definition)? – n00b1990 May 23 '13 at 18:33