User:Inconsistency/Sandbox

Proof 1
By assumption there exist injections $f \colon S \hookrightarrow T$ and $g \colon T \hookrightarrow S$. Using these injections one attempts to construct partitions:
 * $\mathfrak S \subseteq \mathcal P\left(S\right)$ and $\mathfrak T \subseteq \mathcal P\left(T\right)$

as well as a bijection:
 * $B \colon \mathfrak S \hookrightarrow \hspace{-8pt} \rightarrow \mathfrak T$

and a family of bijections:
 * $\forall \mathfrak s\in\mathfrak S \colon \left(b_{\mathfrak s} \colon \mathfrak \hookrightarrow \hspace{-8pt} \rightarrow B\left(\mathfrak s\right)\right)$

Note that, since $\mathfrak S$ is a partitioning:

There exists a surjection $\mathbf{\iota \colon S \rightarrow \hspace{-8pt} \rightarrow \mathfrak S}$ defined by $\mathbf{\forall s\in S: s\in\iota(s)}$
 * Define $\iota := \left\{\langle s,\mathfrak s\rangle\in S\times\mathfrak S \colon s\in\mathfrak s\right\}$.


 * Let $\langle s,\mathfrak r\rangle,\langle s,\mathfrak s\rangle \in S\times\mathfrak S$.
 * Then $s\in\mathfrak r \wedge s\in\mathfrak s \implies \mathfrak r \cap \mathfrak s\ne\varnothing \implies \mathfrak r = \mathfrak s$, since elements of a partition are either disjoint or equal.
 * So $\iota$ is many-to-one.
 * Let $s\in S$ and assume $\not\exists \mathfrak s\in\mathfrak S \colon s\in\mathfrak s$.
 * Then $\displaystyle \left\{s\right\}\cap S = \left\{s\right\}\cap \bigcup \mathfrak S = \bigcup_{\mathfrak s \mathop\in \mathfrak S} \left\{s\right\}\cap\mathfrak s = \bigcup_{\mathfrak s \mathop\in \mathfrak S} \varnothing = \varnothing$ would imply $s\not\in S$.
 * So $\iota$ has to be left-total.
 * Thus $\iota$ is a function.
 * Now let $\mathfrak s \in \mathfrak S$ and assume $\not\exists s\in S \colon s\in\mathfrak s$.
 * Then $S\cap\mathfrak s = \varnothing$ would imply $\mathfrak s = \varnothing$ since $\mathfrak s\in\mathcal P\left(S\right)$.
 * So $\iota$ has to be right-total and therefore surjective.

And we can conclude: $\mathbf{s \mapsto b_{\iota\left(s\right)}\left(s\right)}$ is a bijection

Then by Composite of Injections is Injection: $(h_S := g\circ f) : S \hookrightarrow S$ and $(h_T := f\circ g) : T \hookrightarrow T$. In the following denote by $X$ either $S$ or $T$. Using $h_X$, one constructs a partitioning $\mathbb X$ of $X$ such that there exists a bijection between the classes of $S$ and those of $T$.

divided into equivalence classes

Now the Fundamental Theorem on Equivalence Relations implies that these classes form a partition of $X$.

First define $\preceq_X \subseteq X^2$ by


 * $\langle x_1,x_2 \rangle \in \preceq_X \hspace{8pt}\Leftrightarrow\hspace{8pt} \exists n\in\N: x_2 = h_X^n x_1$

Claim

 * $\preceq_X$ is a preorder relation.

Proof

 * (Reflexivity) Let $x \in X$ then $x = h_X^0 x$ $\implies x\preceq x$
 * (Transitivity) Let $x \preceq y$ and $y \preceq z$ then $y = h_X^m x$ and $z = h_X^n y$ $\implies z = h_X^{m+n}x \implies x\preceq z$

Now define $\sim_X \subseteq X^2$ by


 * $\langle x_1,x_2 \rangle \in \sim_X \hspace{8pt}\Leftrightarrow\hspace{8pt} x_1\preceq x_2 \vee x_2\preceq x_1$

Claim

 * $\sim_X$ is an equivalence relation.

Proof

 * (Reflexivity) and (Symmetry) are trivial consequences of the definition of $\sim_X$ and the reflexivity of $\preceq_X$.
 * (Transitivity) Let $x_1,x_2,x_3\in X$ such that $x_1,x_2\preceq x_3$. Then $x_3 = h_X^m x_1 = h_X^n x_2$ for $n,m \in \N$.
 * Wlog assume $m\le n$. Then $x_1 = h_X^{n-m} x_2 \underset{h_X\text{ injective}}\iff h_X^m x_1 = h_X^m h_X^{n-m} x_2 \iff x_3 = h_X^n x_2$.
 * Therefore $x_1 \preceq x_2$.

For $x\in X$ denote by $\left[\!\left[{x}\right]\!\right] := \left[\!\left[{x}\right]\!\right]_{\sim_X} \in X/\!\sim_X$ the equivalence class of $x$.

Claim

 * $\left[\!\left[{x}\right]\!\right]$ is a chain.

Claim

 * $\left[\!\left[{x}\right]\!\right]$ is a linearly ordered abelian group.

FALSCH! $\left[\!\left[{x}\right]\!\right]$ ist ein Untertyp von $\Z$! Defin $\left[\!\left[{x}\right]\!\right]$ durch wahl eines repr r Z = BxN Z == r S == h_X^n N == n -> -n

r, h_X == N ohne repetitionen r, h_X == Z/nZ mit -*- => Z/nZ <= [x] Aber genauso [x] <= Z Theorem: Alle Subtypen von Z haben diese Form. Lemma: f ist ein morphismus von typen. (trivial) Theorem: Seien S,T typen und S<=T, T<=S. Dann S==T.

Betrachte die totalordnung auf [x]. 1) [x] besitzt ein minimales element => [x]

General construction: Let v : Z -> X for some set Z be injective st h|_{Im(v)} bijective.

wie wählt man Z intelligent, sodass entweder NS == S oder N = 0 ???

Proof

 * By Equivalence Class is not Empty one can fix an element $r\in[x]$.
 * And define addition ...
 * This definition does not depend on the choice of $r$ since if $q$ is another representative:
 * $q = h_X^k r$ => (x +_q y) = ... = (x +_r y)
 * Note that this addition is commutative by construction.
 * Finally observe that $\forall x,y,a:x ...

Now [f]:[s]->[t] is a monomorphism of groups and is completely determined by [f](r). [g] analogously. But [f] monomorph => ord([s])<=ord([t]) and ... => ord([s])=ord([t]) => [s] \simeq [t]

Since Equivalence Classes are Disjoint these $\left[\!\left[{x}\right]\!\right]_X$ form a partition of $X$ and therefor every element of $X$ belongs to exactly one element of $X/\!\sim_X$ (namely $\left[\!\left[{x}\right]\!\right]_X$).

Now define $\widetilde{f} : S/\!\sim_S \to T/\!\sim_T$ by $f(\left[\!\left[{s}\right]\!\right]_S) = \left[\!\left[{f(s)}\right]\!\right]_T$ and $\widetilde{g}$ analogous.

Claim

 * $\widetilde{f}$ is well defined and injective.
 * $\widetilde{f}$ is surjective.

Proof

 * $s_1,s_2\in \left[\!\left[{s}\right]\!\right]_S \iff s_1\sim_S s_2 \underset{\text{wlog}}\iff s_1 \preceq s_2 \overset{n\in\N}\iff s_2 = h_S^n s_1 = (g\circ f)^n s_1 \underset{f\text{ injective}}\iff f(s_2) = f((g\circ f)^n s_1) = f\circ(g\circ f)^n s_1 = (f\circ g)^n\circ f\ s_1 = h_T^n (f(s_1)) \iff f(s_1) \preceq_T f(s_2) \iff f(s_1) \sim_T f(s_2)$


 * Let $\left[\!\left[{t}\right]\!\right]_T \in T/\!\sim_T$ then $\widetilde{g}\left[\!\left[{t}\right]\!\right]_S \in S/\!\sim_S$ and obviously $\widetilde{f}\widetilde{g}(\left[\!\left[{t}\right]\!\right]_S) = \left[\!\left[{f(g(t))}\right]\!\right]_T = \left[\!\left[{h_T(t)}\right]\!\right]_T = \left[\!\left[{t}\right]\!\right]_T$.



Claim

 * $\langle\left[\!\left[{x}\right]\!\right]_X, \preceq\rangle \simeq \langle\Z/n\Z, \le\rangle$ for some $n\in\N$

Proof

 * By Equivalence Class is not Empty one can fix $r\in\left[\!\left[x\right]\!\right]_X$.
 * Let $y \in \left[\!\left[x\right]\!\right]_X$ be arbitrary.
 * Now $r,y\in\left[\!\left[x\right]\!\right]_X \implies r\sim y \implies r \preceq y \vee y \preceq r$.
 * By an abuse of notation write $y = h_X^k r$ for $k\in\Z$ and keep in mind that for $k<0$ $y = h_X^k r$ means $r = h_X^{-k} y$.
 * Consider two cases:
 * $\exists l\ne k: y=h_X^l r$
 * Wlog assume $k<l$ then $h_X^{l-k} r = r \underset{h_X\text{ injective}}\iff h_X^kh_X^{l-k} r = h_X^kr \iff h_X^kr = h_X^kr$.
 * Define $n := l-k$ the smallest of such distances (which exists becaus $\N^2$ is well ordered). Then for any $y\in\left[\!\left[x\right]\!\right]_X : y = h_X^n y$.
 * Now define $o_r : \left[\!\left[x\right]\!\right]_X \to \Z/n\Z$ by $o_r(y) = k + n\Z$.
 * $\not\exists l\ne k: y=h_X^l r$
 * Define $o_r : \left[\!\left[x\right]\!\right]_X \to \Z = \Z/0\Z$ by $o_r(y) = k + 0\Z$.
 * Observe that $o_r$ is injective and surjective by construction and trivially order preserving.

Remark: This statement implies that all elements of $\left[\!\left[x\right]\!\right]_X$ are uniquely determined by $r\in\left[\!\left[x\right]\!\right]_X$ and $k\in\Z/n\Z$. For fixed $r\in\left[\!\left[x\right]\!\right]_X$ one can define addition by $y + z = h_X^{o_r(y)} + h_X^{o_r(z)} := h_X^{o_r(y)+o_r(z)}$ and $f,g,o_r$ become a morphisms of linearly ordered groups. Therefor $o_r \circ f\circ o_{r}^{-1} : \Z_m \to \Z_n$ and $o.g.o^-1 : \Z_n \to \Zm$ are both monomorphisms such that $(og.g.og-1)(of.f.of-1) = og.gf.of-1 = og.of-1 = id$.

Finally define the bijection $b:S\simeq T$ by


 * $b(s) = o_{f(r)}^{-1}o_r(s)$

where $r$ is a chosen representant of $\left[\!\left[{s}\right]\!\right]_S$