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MathML Test: The infinite in set theory

Introduction

The study of sizes of sets has been an important topic of set theory since its beginning. For finite sets, their numbers of elements can be counted and as a consequence, their sizes easily compared. The idea of Cantor was to extend the comparaison to infinite sets: two set "have the same size" iff there exists a one-to-one map from one onto the other. Clearly it is a relation of equivalence and a class of equivalence contains sets that "have the same size". In fact, we can get a class of representants ordered by the following relation: $X\le Y$ iff there is a one-to-one map from X to Y. Only the reflexivity is not trivial and is given by Cantor-Bernstein theorem. This class of representants is called cardinals whose first elements are the well-know natural numbers: 0, 1, 2... Then comes the cardinality of $\mathbb{N}$ denoted by ${\aleph }_0$ and the sets in this class are said countable. The great success of Cantor was to show the existence of greater sets (uncountable) and more precisely of an infinity of infinite cardinals. Later, the set theorists discovered several properties of cardinals and developped a beautiful arithmetic of infinite...

Cantor-Bernstein Theorem

Let X and Y be two sets. If there are $f:X\to Y$ and $g:Y\to X$ two one-to-one maps, then there exists a one-to-one map from X onto Y.

Proof: Define $X_0=X\backslash g\left(Y\right)$, for all $n\in \mathbb{N},X_{n+1}=\left(g\circ f\right)\left(X_n\right)$, $X_{\omega }=\bigcup _{n<\omega }{X}_n$ and finally

$h:\mid \begin{array}{c}X\to X\backslash X_0\\ x\mapsto \{\begin{array}{c}(g\circ f)\left(x\right)\text{ if}x\in X_{\omega }\\ x\text{ otherwise}\end{array}\end{array}$

h is one-to-one as $x\mapsto x$ and $g\circ f$ are one-to-one, and their ranges are respectively included in the two disjoint sets $X\backslash X_{\omega }$ and $X_{\omega }$. More over, by construction all the $X_i$ are disjoint and we have:

$\left(g\circ f\right)\left(X_{\omega }\right)=\bigcup _{n<\omega }\left(g\circ f\right)\left(X_n\right)=\bigcup _{n<\omega }{X}_{n+1}=X_{\omega }\backslash X_0$

Consequently, $h\left(X\right)=h\left(X\backslash X_{\omega }\bigcup X_{\omega }\right)=h\left(X\backslash X_{\omega }\right)\bigcup h\left(X_{\omega }\right)=\left(X\backslash X_{\omega }\right)\bigcup \left(X_{\omega }\backslash X_0\right)=X\backslash X_0$

So h is one-to-one onto.

Finally, because g is one-to-one and $g\left(Y\right)=X\backslash X_0$ we have $h^{-1}\circ g$ is one-to-one onto. Finally ${\left(h^{-1}\circ g\right)}^{-1}$ is a one-to-one map from X onto Y. □

Cantor-Bernstein Theorem allows to order sets by size. Thanks to the axiom of choice (AC), we can prove than every set is equipotent to a set of a particular class. Moreover, this class of sets, called cardinals, is well-ordered.

The first cardinals are 0, 1, 2, 3... ${\aleph }_0$, ${\aleph }_1$, ${\aleph }_2$, ... We can extend the operations on $\mathbb{N}$ to cardinals, for instance $\lambda \times \mu$ is the cardinal of the cartesian product and ${\lambda }^{\mu }$ is the cardinal of the set of maps from $\mu$ to $\lambda$. To avoid ambiguity, the latter set is denoted by ${}^{\mu }\lambda$. The existence of uncountable cardinals is given by the following theorem.

Cantor's Theorem

$\forall X,\mathrm{Card}\left(\wp \left(X\right)\right)>\mathrm{Card}\left(X\right)$

Proof: Trivially, $x\mapsto \left\{x\right\}$ is a one-to-one map from X to its powerset, so $\mathrm{Card}\left(X\right)\le \mathrm{Card}\left(\wp \left(X\right)\right)$.

Suppose there exists $f:X\to \wp \left(X\right)$ onto. Define $Y=\left\{x\in X|x\notin f\left(x\right)\right\}$ and take $x_0\in f^{-1}\left(Y\right)$.

If $x_0\in Y$ then $x_0\notin f\left(x_0\right)$ but $f\left(x_0\right)=Y$ and consequently $x_0\notin Y$. In the same way, if $x_0\notin Y$ then you get $x_0\in Y$. A contradiction. □

Note that $f:\{\begin{array}{c}\wp \left(X\right)\to {}^X\{0;1\}\\ P\mapsto {\chi }_P\end{array}$ is one-to-one onto, so Cantor's Theorem can also be written $2^{\mid X\mid }>\mid X\mid$.

Finally we get infinitely many infinite cardinals:${\aleph }_0<2^{{\aleph }_0}<2^{2^{{\aleph }_0}}<2^{2^{2^{{\aleph }_0}}}<\text{...}$

Let ${\beth }_0,{\beth }_1,{\beth }_2...$ denote these cardinals, then generalized continuum hypothesis (GCH) is that this sequence gives all the infinite cardinals. In fact, we can show than even the weaker assumption ${\beth }_1={\aleph }_1$, called the continuum hypothesis (CH) can nor be proved neither disproved from the axioms of set theory. Lots of other statements about cardinals have be shown to be undecidable. Nevertheless, in the next paragraph, we will see an interesting result that can be obtained if we assume the axiom of choice.

König's Theorem (AC)

Let ${\left({\lambda }_i\right)}_{i\in I}$ and ${\left({\mu }_i\right)}_{i\in I}$ be two families of cardinals indexed by a set I. If $\forall i\in I{\lambda }_i<{\mu }_i$ then we have:

$\sum _{i\in I}{\lambda }_i<\prod _{i\in I}{\mu }_i$

(By the way, these sum and product are defined as the cardinality of the sets $S$ and $P$ below. One can easily check the compatibility of this definition with the case where $I$ is finite)

Proof:

Define $S=\left\{\left(\alpha ,i\right)|\left(i\in I\right)\wedge \left(\alpha <{\lambda }_i\right)\right\}$ and $P=\left\{f\in {}^I\left(\bigcup _{i\in I}{\mu }_i\right)|\forall i\in I,f\left(i\right)\in {\mu }_i\right\}$.

$f:\mid \begin{array}{c}S\to P\\ \left(\alpha ,i\right)\mapsto \left(j\mapsto \{\begin{array}{c}\alpha \text{if}j=i\\ {\lambda }_i \mathrm{otherwise}\end{array}\right)\end{array}$ is clearly one-to-one, so we have $\mid S\mid \le \mid P\mid$

Suppose the equality. then there is $g:P\to S$ one-to-one. Define for $i\in I$, $X_i=\left\{g^{-1}\left(\left(\alpha ,i\right)\right)\left(i\right)|\left(\alpha ,i\right)\in S\right\}$.

For all $i\in I$ we have $\mid X_i\mid \le {\lambda }_i<{\mu }_i$ so by the axiom of choice, we can take $x_i\in {\mu }_i\backslash X_i$.

Then for $\left(\alpha ,j\right)=g(i\mapsto x_i)$ we have $x_j\in X_j$. A contradiction. □

Note that Cantor's Theorem becomes a corollary of König's Theorem as we have:

$2^{\mid X\mid }=\prod _{i\in X}2>\sum _{i\in X}1=\mid X\mid$

Miscellaneous formulae

The famous Shelah's inequality: ${{\aleph }_{\omega }}^{{\aleph }_0}\le 2^{{\aleph }_0}+{\aleph }_{{\omega }_4}$

Two theorems that use the arrow notation:

1. $\forall n,k\in \omega ,{\aleph }_0\to {\left({\aleph }_0\right)}_k^n$ (Ramsey's theorem)
2. Define for each $\kappa$, ${\exp }_0\left(\kappa \right)=\kappa$ and for all $n$,${\exp }_{n+1}\left(\kappa \right)=2^{{\exp }_n\left(\kappa \right)}$.

   Then, for all κ, ${\left({\exp }_n\left(\kappa \right)\right)}^{+}\to {\left({\kappa }^{+}\right)}_{\kappa }^{n+1}$ (Erdös-Rado Theorem).

   In particular, for $\kappa ={\aleph }_0$, ${{\beth }_n}^{+}\to {\left({\aleph }_1\right)}_{{\aleph }_0}^{n+1}$.

$\mathrm{cf}\left({{\aleph }_{\alpha }}^{{\aleph }_{\beta }}\right)>{\aleph }_{\beta }$

Hausdorff formula : ${{\aleph }_{\alpha +1}}^{{\aleph }_{\beta }}={{\aleph }_{\alpha }}^{{\aleph }_{\beta }}{.\aleph }_{\alpha +1}$