Cantor-Bernstein-Schröder Theorem/Lemma/Proof 1

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Let $S$ be a set.

Let $T \subseteq S$.

Suppose that $f: S \to T$ is an injection.

Then there is a bijection $g: S \to T$.


Recursively define a sequence $\langle C_n \rangle$ in the power set of $S$ as follows:

$C_0 = S \setminus T$, the difference between $S$ and $T$.
$C_{n+1} = f \left[{C_n}\right]$, the image of $C_n$ under $f$.

Let $\displaystyle C := \bigcup_{n \mathop \in \N} C_n$.

Define a mapping $h: S \to T$ as follows:

$\displaystyle h \left({x}\right) = \begin{cases} f \left({x}\right) & : x \in C \\ x & : x \notin C \end{cases}$

$h$ is a mapping from $S$ to $T$

By Law of Excluded Middle, for each $x \in S$:

$x \in C$


$x \notin C$

Thus the construction produces a mapping on S.

It remains to be shown that:

$\forall x \in S: h \left({x}\right) \in T$

Let $x \in C$.


$h \left({x}\right) = f \left({x}\right) \in T$

Let $x \notin C$.

Then by Set is Subset of Union:

$x \notin C_0$

As $x \in S$:

$x \in S \setminus C_0$

by the definition of set difference.

By Relative Complement of Relative Complement:

$S \setminus C_0 = T$

Thus $x \in T$ by the definition of subset.

As $h \left({x}\right) = x$ in this case:

$h \left({x}\right) \in T$


$h$ is injective

Let $x, y \in S$.

Suppose that $h \left({x}\right) = h \left({y}\right)$.

By Law of Excluded Middle for Two Variables:

$\left({x \in C}\right) \land \left({y \in C}\right)$


$\left({x \notin C}\right) \land \left({y \notin C}\right)$


$\left({x \in C}\right) \land \left({y \notin C}\right)$


$\left({x \notin C}\right) \land \left({y \in C}\right)$

Let $x, y \in C$.

Then $x = y$ because $f$ is injective.

Let $x, y \notin C$.

Then $x = y$ by Identity Mapping is Injection.

Let $x \in C$ and $y \notin C$.

Then $x \in C_n$ for some $n$ by the definition of union.

By Set is Subset of Union:

$h \left({x}\right) = f \left({x}\right) \in C_{n+1} \subseteq C$

Thus $h \left({x}\right) \in C$.

$h \left({y}\right) = y$ by the definition of $h$.

Since $y \notin C$, this contradicts the assumption that $h \left({x}\right) = h \left({y}\right)$.

The argument for the case of $x \notin C$ and $y \in C$ is identical to the preceding.

Thus $h \left({x}\right) = h \left({y}\right) \implies x = y$ for all $x, y \in S$, so $h$ is injective.


$h$ is surjective

Let $y \in T$.

By Law of Excluded Middle:

$y \in C$


$y \notin C$

Let $y \notin C$.


$h \left({y}\right) = y$

Let $y \in C$.

Then as $y \notin C_0 = S \setminus T$:

$\exists n \in \N_{>0}: y \in C_{n+1}$


$\exists x \in C_n: y = f \left({x}\right)$

By the definition of union:

$x \in C$


$h \left({x}\right) = f \left({x}\right) = y$


$\forall y \in T: \exists x \in S: h \left({x}\right) = y$

Thus by definition, $h$ is surjective.



This article incorporates material from proof of Schroeder-Bernstein theorem on PlanetMath, which is licensed under the Creative Commons Attribution/Share-Alike License.