Past exam of the logic and set theory course of the University of Oxford / 2007 / B / 7

The Cantor-Schröder-Bernstein theorem says that injections $X\to Y$ and $Y\to X$ imply a bijection $X\to Y$. Thus mutual cardinal inequalities imply equal cardinalities, without a choice assumption.

Write $\mathfrak{c}=2^{{\aleph }_0}=|\mathbb{R}|$. Standard cardinal arithmetic gives $\mathfrak{c}\cdot \mathfrak{c}=\mathfrak{c}$ and ${\mathfrak{c}}^{{\aleph }_0}=\mathfrak{c}$. The correspondence $x+iy\leftrightarrow (x,y)$ is a bijection $\mathbb{C}\leftrightarrow {\mathbb{R}}^2$. HenceEquivalently, the real axis supplies the lower inequality and a coding of two real coordinates supplies the upper one; Cantor-Schröder-Bernstein then gives the equality.

The set of all functions from a set of size $\mathfrak{c}$ to another set of size $\mathfrak{c}$ has size ${\mathfrak{c}}^{\mathfrak{c}}$. Cardinal exponentiation and the law $({\kappa }^{\lambda }{)}^{\mu }={\kappa }^{\lambda \mu }$ giveThereforeThe lower bound can also be seen by taking characteristic functions of all subsets of $\mathbb{C}$. Cantor's theorem makes this strictly larger than $\mathfrak{c}$, unlike the continuous-function space in the next part.

The set $D=\mathbb{Q}+i\mathbb{Q}$ is countable and dense in $\mathbb{C}$. Restriction to $D$ is injective on continuous functions $\mathbb{C}\to \mathbb{R}$: if two such functions agree there, their equality set is closed and contains a dense set, so is all of $\mathbb{C}$. Equivalently a sequence of points of $D$ converging to any chosen point and continuity determine its value uniquely.

Thus the family has cardinality at mostFor the lower bound, every real number determines a different constant continuous function. Cantor-Schröder-Bernstein yieldsThe reduction to countably many values is the feature that separates this answer from unrestricted functions.

Every strictly increasing sequence is in ${\mathbb{N}}^{\mathbb{N}}$, whose cardinality is ${\aleph }_0^{{\aleph }_0}=\mathfrak{c}$. This gives the upper bound.

For each binary sequence $ϵ=({ϵ}_n{)}_{n\in \mathbb{N}}$, defineThen $a_{n+1}-a_n=2+{ϵ}_{n+1}-{ϵ}_n\ge 1$, so the sequence is strictly increasing. Distinct binary sequences give distinct increasing sequences. This injects a set of size $2^{{\aleph }_0}$ into the desired family, giving the matching lower bound. HenceOne may also identify an increasing sequence with its infinite range subset of $\mathbb{N}$; the explicit binary coding establishes the lower bound without discarding finite subsets by cardinal subtraction.

The family is a subset of all complex sequences, so its cardinality is at mostFor each $z\in \mathbb{C}$, the sequence $(z,0,0,\dots )$ belongs to the family because its displayed series has just one nonzero first-order term. These sequences are distinct as $z$ varies, giving a lower bound $\mathfrak{c}$. ThusThe source condition is convergence of ${\sum }_j{a}_j^2$, without absolute-value bars. The argument uses exactly that printed condition and makes no assertion that it is equivalent to square summability of complex sequences. The same upper and lower bounds would also work for the stronger absolute-square condition, but no alteration of the source is needed.

Inclusion gives $|\mathbb{R}\setminus \mathbb{Q}|\le \mathfrak{c}$. If $\kappa =|\mathbb{R}\setminus \mathbb{Q}|$, the disjoint decomposition of $\mathbb{R}$ givesThe irrational numbers are infinite, for instance $n+\sqrt{2}$ are pairwise distinct irrationals. Standard infinite cardinal addition gives ${\aleph }_0+\kappa =\max ({\aleph }_0,\kappa )$. Since Cantor's theorem gives $\mathfrak{c}>{\aleph }_0$, the equality forces $\kappa =\mathfrak{c}$. ThereforeThis is the permitted cardinal-arithmetic formal proof; it does not subtract infinite cardinals as though they were finite numbers.