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# proof of Cantorβs theorem

The proof of this theorem is fairly simple using the following construction, which is central to Cantorβs diagonal argument.

Consider a function $F\colon X\to{\mathcal{P}}(X)$ from a set $X$ to its power set. Then we define the set $Z\subseteq X$ as follows:

$Z=\{x\in X\mid x\not\in F(x)\}$ |

Suppose that $F$ is a bijection. Then there must exist an $x\in X$ such that $F(x)=Z$. Then we have the following contradiction:

$x\in Z\Leftrightarrow x\not\in F(x)\Leftrightarrow x\not\in Z$ |

Hence, $F$ cannot be a bijection between $X$ and ${\mathcal{P}}(X)$.

Keywords:

diagonal argument

Major Section:

Reference

Type of Math Object:

Proof

Parent:

## Mathematics Subject Classification

03E17*no label found*03E10

*no label found*

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## Recent Activity

Jul 5

new correction: Error in proof of Proposition 2 by alex2907

Jun 24

new question: A good question by Ron Castillo

Jun 23

new question: A trascendental number. by Ron Castillo

Jun 19

new question: Banach lattice valued Bochner integrals by math ias

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new question: young tableau and young projectors by zmth

Jun 11

new question: binomial coefficients: is this a known relation? by pfb

Jun 6

new question: difference of a function and a finite sum by pfb

new correction: Error in proof of Proposition 2 by alex2907

Jun 24

new question: A good question by Ron Castillo

Jun 23

new question: A trascendental number. by Ron Castillo

Jun 19

new question: Banach lattice valued Bochner integrals by math ias

Jun 13

new question: young tableau and young projectors by zmth

Jun 11

new question: binomial coefficients: is this a known relation? by pfb

Jun 6

new question: difference of a function and a finite sum by pfb

## Comments

## infinite sets

You proved that |P(X)| is bigger than |X|, because you can't find any x so that F(x)=Z.

But what does it means for infinite sets |P(X)|=|X|+1 ?

It's like saying that natural numbers are more numerous than even numbers. You're right, but they have the same cardinal.