Cantor's diagonal method in simple terms?

Question

Could anyone please explain Cantor's diagonalization principle in simple terms?

Answer 1

Here's the standard application explained in the simplest terms I can:

Theorem: There are more real numbers than there are integers.

Lemma: A real number has a decimal representation (that might not terminate), and all decimal representations create real numbers.

Proof of Theorem: Suppose there are as many integers as reals. Then we can list the reals in some order,

$r_1 = 3.141592...$

$r_2 = 2.718281...$

$r_3 = 1.000000...$

and so on for each $r_i$ where $i$ is an integer

There is a contradiction because I can construct a real that is not in the above list. Let $r' = 0.d_1d_2d_3d_4\ldots$ where $d_1$ is not equal to the 1st digit after the decimal of $r_1$, $d_2$ is not equal to the 2nd digit after the decimal of $r_2$, and so forth. For example, I could add 5 and take it mod 10. In the example this gives $r' = 0.665\ldots$. The contradiction is that $r'$ is not in the list, because for any $i$ I know that $r'$ and $r_i$ differ by at least $4*10^{-i}$ (since they differ in $d_i$).

The fact that the reals cannot be enumerated as such shows that they are larger than the integers. But, the important result of Cantor is simply that they cannot be enumerated.

Answer 2

Perhaps you will find my blog post "On a proof of Cantor's theorem" illuminating. In short, the right way to prove Cantor's theorem is to first prove Lawvere's fixed point theorem, which is more computer-sciency in nature than Cantor's theorem.

Given two sets $A$ and $B$, let $B^A$ denote the set of all functions from $A$ to $B$.

Theorem (Lawvere): Suppose $e : A \to B^A$ is a surjective map. Then every map $f : B \to B$ has a fixed point, i.e., $x \in B$ such that $f(x) = x$.

Proof. Consider the map $s:A \to B$ defined by $s(x)=f(e(x)(x))$. Because $e$ is onto, there is $y \in A$ such that $e(y)=s$. Then we have $e(y)(y)=s(y)=f(e(y)(y))$, therefore $e(y)(y)$ is a fixed point of $f$. QED.

Corollary (Cantor): There is no onto map from a set $A$ onto its powerset $\mathcal{P}(A)$.

Proof. First note that $\mathcal{P}(A)$ is isomorphic to $2^A$ where $2 = \lbrace 0, 1 \rbrace$. That is, the subsets of $A$ correspond to their characteristic functions $A \to \lbrace 0, 1 \rbrace$. Suppose $e : A \to 2^A$ is onto. By Lawvere's theorem every map $2 \to 2$ has a fixed point. But this is not the case for the map $f(x) = 1 - x$. Therefore no such $e$ can exist. QED.

The usual proof of uncountability of reals (using decimal expansions) is of the same form too. There is a map $f : \lbrace 0, 1, \ldots, 9\rbrace \to \lbrace 0, 1, \ldots, 9\rbrace$ on the decimal digits which has no fixed point, therefore there can be no onto map $\mathbb{N} \to \lbrace 0, 1, \ldots, 9\rbrace^\mathbb{N}$.