hcju/proofwikiqa huggingface dataset:
From Surjection by Restriction of Codomain, any $f: S \to \Img f$ is a surjection. The mapping $g: S \to \Img f$ where $\map g x = \map f x$ is therefore also clearly a surjection. The mapping $g: S \to \Img f: \map g x = \map f x$ is clearly unique, by Equality of Mappings. From Inclusion Mapping is Injection, $i_T: \Img f \to T$ is an injection. Likewise, the mapping $i_T: \Img f \to T : \map {i_T} x = x$ is also unique, by its own definition. {{qed}}
From the definition of Induced Equivalence, the mapping $f: S \to T$ induces an equivalence $\RR_f$ on its domain. As $f: S \to T$ is a surjection, $T = \Img f$ by definition. From Renaming Mapping is Bijection, the renaming mapping $r: S / \RR_f \to T$ is a bijection, where $S / \RR_f$ is the quotient set of $S$ by $\RR_f$. Hence: :$r \circ q_{\RR_f} = f$ :$r$ is the only mapping $r: S / \RR_f \to T$ that satisfies this equality. {{qed}}
From Factoring Mapping into Surjection and Inclusion, $f$ can be factored uniquely into: :A surjection $g: S \to \Img f$, followed by: :The inclusion mapping $i: \Img f \to T$ (an injection). ::$\begin{xy}\xymatrix@L+2mu@+1em {  S \ar[drdr]_*{g}    \ar@{-->}[rr]^*{f = i_T \circ g} & & T \\ \\ & & \Img f \ar[uu]_*{i_T} }\end{xy}$ From the Quotient Theorem for Surjections, the surjection $g$ can be factored uniquely into: :The quotient mapping $q_{\RR_f}: S \to S / \RR_f$ (a surjection), followed by: :The renaming mapping $r: S / \RR_f \to \Img f$ (a bijection). Thus: :$f = i_T \circ \paren {r \circ q_{\RR_f} }$ As Composition of Mappings is Associative it can be seen that $f = i_T \circ r \circ q_{\RR_f}$. The commutative diagram is as follows: ::$\begin {xy} \xymatrix@L + 2mu@ + 1em {  S \ar@{-->}[rrr]^*{f = i_T \circ r \circ q_{\RR_f} }    \ar[ddd]_*{q_{\RR_f} }    \ar@{..>}[drdrdr]_*{g = r \circ q_{\RR_f} } & & & T \\ \\ \\ S / \RR_f \ar[rrr]_*{r} & & & \Img f \ar[uuu]_*{i_T} } \end {xy}$ {{qed}}
{{begin-eqn}} {{eqn | l = \forall S: \O       | o = \subseteq       | r = S       | c = Empty Set is Subset of All Sets }} {{eqn | ll= \leadsto       | l = \forall S: \O       | o = \in       | r = \powerset S       | c = {{Defof|Power Set}} }} {{end-eqn}} {{qed}}
Take the general relation $\mathcal R \subseteq S \times T$. Let $X \subseteq S$, that is $X \in \powerset S$. Suppose $X = \O$. Then from Image of Empty Set is Empty Set: :$\map {\mathcal R^\to} X = \O \subseteq T$ Suppose $X = S$. Then from Image is Subset of Codomain: Corollary 1: :$\map {\mathcal R^\to} X = \Img {\mathcal R} \subseteq T$ Finally, suppose $\O \subset X \subset S$. From Image is Subset of Codomain, again: :$\map {\mathcal R^\to} X \subseteq T$ Now, from the definition of the power set, we have that: :$Y \subseteq T \iff Y \in \powerset T$ We defined $R^\to \subseteq \powerset S \times \powerset T$ such that: :$R^\to: \powerset S \to \powerset T: \map {\mathcal R^\to} X = \set {t \in T: \exists s \in X: \tuple {s, t} \in \mathcal R}$ By definition, there is only one $\map {\mathcal R^\to} X$ for any given $X$, and so $R^\to$ is many-to-one. We have shown that: :$\forall X \subseteq S: \map {\mathcal R^\to} X \in \powerset T$ So: :$\forall X \in \powerset S: \exists_1 Y \in \powerset T: \map {\mathcal R^\to} X = Y$ So: :$R^\to$ is defined for all $X \in \powerset S$ and therefore $R^\to$ is a mapping. {{qed}}
$\mathcal R^{-1}$, being a relation, obeys the same laws as $\mathcal R$. So Direct Image Mapping of Relation is Mapping applies directly. {{qed}}
Suppose $f^\to: \powerset S \to \powerset T$ is not an injection. Then: :$\exists Y \in \powerset T: \exists X_1, X_2 \in \powerset S: X_1 \ne X_2 \land \map {f^\to} {X_1} = \map {f^\to} {X_2} = Y$ There are two cases to consider:  :$(1): \quad$ Either $X_1$ or $X_2$ is the empty set :$(2): \quad$ Neither $X_1$ nor $X_2$ is the empty set. === Either subset is the empty set === {{WLOG}} assume that $X_1 = \O$. Then, from Image of Empty Set is Empty Set: :$Y = \map {f^\to} {X_1} = \O$ But that means: :$\map {f^\to} {X_2} = \O$ Now $X_1 \ne X_2$, so $X_2 \ne \O$. Thus: :$\exists x_2 \in X_2: \neg \exists y \in T: \map f {x_2} = y$ which means that $f$ is not even a mapping, let alone an injection. The same argument applies if $X_2$ is empty. {{qed|lemma}} === Neither subset is the empty set === Now, assume that neither $X_1$ nor $X_2$ is the empty set. As $X_1 \ne X_2$, there must be at least one element in either one which is not in the other. {{WLOG}} assume that $\exists x_1 \in X_1: x_1 \notin X_2$. Now suppose $\map f {x_1} = y \in Y$. However, as $\map {f^\to} {X_2} = Y$, there must be some $x_2 \in X_2$ such that $\map f {x_2} = y \in Y$. So: :$\exists x_1 \ne x_2 \in S: \map f {x_1} = \map f {x_2}$ which means, by definition, that $f$ is not injective. Thus, if $f^\to: \powerset S \to \powerset T$ is not an injection then neither is $f: S \to T$. Thus, by the Rule of Transposition, the result follows. {{qed}} Category:Injections Category:Direct Image Mappings 1wwgkycxtfjzr2g8406zchd8gyumae6
Let $f^\gets$ be the inverse image mapping of $f$. Let $Y \in \powerset T$. Let $X = \map {f^\gets} Y$. By Subset equals Image of Preimage iff Mapping is Surjection: :$\map {f^\to} X = Y$ As such an $X$ exists for each $Y \in \powerset S$, $f^\to$ is surjective. {{qed}}
Let $f: S \to T$ be a surjection. Then: :$\forall y \in T: \exists x \in S: f \paren x = y$ From the Quotient Theorem for Surjections, there is one and only one bijection $r: S / \mathcal R_f \to T$ such that $r \circ q_{\mathcal R_f} = f$. Each element of $S / \mathcal R_f$ is a subset of $S$ and therefore an element of $\powerset S$. Thus as $r: S / \mathcal R_f \to T$ is a bijection, and hence a fortiori also an injection: :$\forall X_1, X_2 \in \powerset S: \map r {X_1} = \map r {X_2} \implies X_1 = X_2$ Because $f$ is a surjection, every $y \in T$ is mapped to by exactly one element of the partition of $S$ defined by $\mathcal R_f$. Let $T = \set {y_1, y_2, \ldots}$. Let the partition defined by $\mathcal R_f$ be $\bigcup \paren {X_1, X_2, \ldots}$ where $\map r {X_n} = y_n$. Let $Y_r \in \powerset T$, such that $Y_r = \set {y_{r_1}, y_{r_2}, \ldots}$. Then: :$\map {f^\to} {X_r} = Y_r$ where $\displaystyle X_r = \bigcup \paren {X_{r_1}, X_{r_2}, \ldots}$. As $\set {X_1, X_2, \ldots}$ is a partition of $S$, $\forall Y_r \in \powerset T: X_r$ is unique. Thus $f^\to: \powerset S \to \powerset T$ is a surjection. {{Qed}}
Let $f: S \to T$ be a surjection. By definition, $f^\to$ is defined by sending subsets of $S$ to their image under $f$. That is: :$\forall X \subseteq S: \map {f^\to} X = \set{\map f x: x \in X} \subseteq T$ To prove that $f^\to$ is a surjection, we need to show that every subset of $T$ is the image under $f^\to$ of some subset of $S$. Let $Y \subseteq T$. Since $f$ is a surjection, by definition we have that: :$\forall t \in T: \exists s \in S: \map f s = t$ Hence (possibly requiring the Axiom of Choice), we can select for each $y \in Y$ some $s_y \in S$ such that $\map f {s_y} = y$. Define $X_Y$ to be: :$X_Y := \set {s_y: y \in Y}$ Then: {{begin-eqn}} {{eqn | l = \map {f^\to} {X_Y}       | r = \set {\map f {s_y}: s_y \in X_Y}       | c =  }} {{eqn | r = \set {\map f {s_y}: y \in Y}       | c =  }} {{eqn | r = \set {y : y \in Y}       | c =  }} {{eqn | r = Y       | c =  }} {{end-eqn}} Thus $f^\to: \powerset S \to \powerset T$ is a surjection. {{qed}} {{AoC||4}}
Let $S$ be a set. Let $\mathcal P \left({S}\right)$ be the power set of $S$. Let $f: S \to \mathcal P \left({S}\right)$ be a mapping. Let $T = \left\{{x \in S: \neg \left({x \in f \left({x}\right)}\right)}\right\}$. Then $T \subseteq S$, so $T \in \mathcal P \left({S}\right)$ by the definition of power set. We will show that $T$ is not in the image of $f$ and therefore $f$ is not surjective. Aiming for a contradiction, suppose that: : $\exists a \in S: T = f \left({a}\right)$ Suppose that: : $a \in f \left({a}\right)$ Then by the definition of $T$: : $\neg \left({a \in T}\right)$ Thus since $T = f \left({a}\right)$: : $\neg \left({a \in f \left({a}\right) }\right)$ By Rule of Implication: : $(1) \quad a \in f \left({a}\right) \implies \neg \left({ a \in f \left({a}\right) }\right)$ Suppose instead that: : $\neg \left({a \in f \left({a}\right)}\right)$ Then by the definition of $T$: : $a \in T$ Thus since $T = f \left({a}\right)$: : $a \in f \left({a}\right)$ By Rule of Implication: : $(2) \quad \neg \left({ a \in f \left({a}\right) }\right) \implies a \in f \left({a}\right)$ By Non-Equivalence of Proposition and Negation, applied to $(1)$ and $(2)$, this is a contradiction. As the specific choice of $a$ did not matter, we derive a contradiction by Existential Instantiation. Thus by Proof by Contradiction, the supposition that $\exists a \in S: T = f \left({a}\right)$ must be false. It follows that $f$ is not a surjection. {{qed}}
{{AimForCont}} $S$ is a set with a surjection $f: S \to \powerset S$. Then: {{begin-eqn}} {{eqn | l = \forall x \in S: \map f x       | o = \in       | r = \powerset S       | c = By Hypothesis }} {{eqn | ll= \leadsto       | l = \forall x \in S: \map f x       | o = \subseteq       | r = S       | c = {{Defof|Power Set}} }} {{end-eqn}} Now by Law of Excluded Middle, there are two choices for every $x \in S$: : $x \in \map f x$ : $x \notin \map f x$ Let $T = \set {x \in S: x \notin \map f x}$. As $f$ is supposed to be a surjection, $\exists a \in S: T = \map f a$. Thus: : $a \in \map f a \implies a \notin \map f a$ : $a \notin \map f a \implies a \in \map f a$ This is a contradiction, so the initial supposition that there is such a surjection must be false. {{qed}} {{LEM||3}}
The proof proceeds by induction. For all $n \in \N_{> 0}$, let $\map P n$ be the proposition: :There is no surjection from $S$ onto $\map {\PP^n} S$. === Basis for the Induction === $\map P 1$ is Cantor's Theorem. This is our basis for the induction. === Induction Hypothesis === Now we need to show that, if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true. So this is our induction hypothesis: :There is no surjection from $S$ onto $\map {\PP^k} S$. Then we need to show: :There is no surjection from $S$ onto $\map {\PP^{k + 1} } S$. === Induction Step === This is our induction step: Suppose that $\map P k$ is true. {{AimForCont}} that $f: S \to \map {\PP^{k + 1} } S$ is a surjection. Define the mapping $g: S \to \map {\PP^k} S$ as: :$\displaystyle \map g x = \bigcup \map f x$ This is actually a mapping into $\map {\PP^k} S$, as follows: :$\map f x \in \map {\PP^{k + 1} } S = \powerset {\map {\mathcal P^k} S}$ By the definition of power set: :$\map f x \subseteq \map {\PP^k} S$ Thus each element of $\map f x$ is a subset of $\map {\PP^{k - 1} } S$. Thus by Union of Subsets is Subset: :$\displaystyle \bigcup \map f x \subseteq \map {\PP^{k - 1} } S$ Therefore: :$\displaystyle \bigcup \map f x \in \map {\PP^k} S$ That is, $\displaystyle \map g x$ is a mapping into $\map {\PP^k} S$. Next we have that: :$\forall y \in \map {\PP^k} S: \set y \in \map {\PP^{n + 1} } S$ But by hypothesis $f$ is surjective, and so: :$\exists x \in S: \map f x = \set y$ Then: :$\displaystyle \map g x = \bigcup \set y = y$ As this holds for all such $y$, $g$ is surjective. But this contradicts the induction hypothesis. Thus we conclude that the theorem holds for all $n$. {{qed}}
We temporarily introduce the notation: :$a^n = \begin{cases} a : & n = 0 \\ \set {a^{n - 1} } : & n > 0 \end{cases}$ where $a \in S$. Thus $a^n$ consists of a single element $a^{n - 1} \in \map {\PP^{n - 1} } S$. Let there be a bijection $f: S \to \QQ^n$ where $\QQ^n \subseteq \map {\PP^n} S$. Then define: :$\AA^{n - 1} = \set {s^{n - 1} \in \map {\PP^{n - 1} } S: s^{n - 1} \ne \map f s}$ Each $\QQ^{n - 1} \in \QQ^n$ is the image of some element of $S$ under $f$. Let $\QQ^{n - 1} = \map f x$ for some $x \in S$. Let $x^{n - 1} \in \QQ^{n - 1} = \map f x$. Then $x^{n - 1} \notin \AA^{n - 1}$, and so $\AA^{n - 1} \notin \QQ^{n - 1}$. On the other hand, let $x^{n - 1} \notin \QQ^{n - 1} = \map f x$. Then $x^{n - 1} \in \AA^{n - 1}$, and so again $ A^{n - 1} \notin \QQ^{n - 1}$. Thus $\AA^{n - 1} \notin Q^n$. Therefore $Q^n$ is a proper subset of $\map {\PP^n} S$. Hence the result. {{qed}}
Let $X$ be the preimage of $\mathcal R$. Suppose $\mathcal R {\restriction_X} \subseteq X \times T$ is a mapping, but not a surjection. Then: :$\exists y \in T: \neg \exists x \in S: \tuple {x, y} \in \mathcal R$ Because no element of $S$ relates to $y$, no subset of $S$ contains any element of $S$ that relates to the subset $\left\{{y}\right\} \subseteq T$. Thus: :$\exists \set y \in \powerset T: \neg \exists X \in \powerset S: \map {\mathcal R^\to} X = \set y$ So we see that $\mathcal R^\to: \powerset S \to \powerset T$ is not a surjection. Now, suppose $\mathcal R {\restriction_X} \subseteq X \times T$ is not even a mapping. This could happen, according to the definition of a mapping, in one of two ways: :$(1): \quad \exists x \in X: \neg \exists y \in T: \tuple {x, y} \in \mathcal R$ :$(2): \quad \exists x \in S: \tuple {x, y_1} \in \mathcal R \land \tuple {x, y_2} \in \mathcal R: y_1 \ne y_2$ Because $X$ is already the preimage of $\mathcal R$, the first of these cannot happen here. For the second, it can be seen that neither $\set {y_1}$ nor $\set {y_2}$ can be in $\Rng {\map {\mathcal R^\to} {\powerset S} }$. Therefore $\mathcal R^\to: \powerset S \to \powerset T$ can not be a surjection. {{questionable|See talk page}} Thus, by the Rule of Transposition, the result follows. {{qed}} Category:Surjections Category:Direct Image Mappings 3go48kp22mh6x6rmyb0svhk154d6im9
Suppose $f: S \to T$ is a mapping, but not injective. Then: : $\exists x_1 \ne x_2 \in S: \map f {x_1} = \map f {x_2} = y$ Let: :$X_1 = \set {x_1}$ :$X_2 = \set {x_2}$ :$Y = \set y$ Then it follows that: :$\map {f^\to} {X_1} = \map {f^\to} {X_2} = Y$ Thus $f^\to: \powerset S \to \powerset T$ is not injective. So by the Rule of Transposition, the result follows. {{qed}} Category:Injections Category:Direct Image Mappings fh8xao0jy03y7omejf9wlya8ginzydw
A bijection is by its definition also a surjection. By Cantor's Theorem there is no surjection from $S$ to $\powerset S$. Hence the result. {{Qed}}
To show equivalence between two sets, we need to demonstrate that a bijection exists between them. We will now define such a bijection. Let $T = \left\{{\left\{{x}\right\}: x \in S}\right\}$. {{begin-eqn}} {{eqn | ll= \forall x \in S:       | l = \left\{ {x}\right\}       | o = \subseteq       | r = S       | c = {{Defof|Subset}} }} {{eqn | ll= \implies       | l = \left\{ {x}\right\}       | o = \in       | r = \mathcal P \left({S}\right)       | c = {{Defof|Power Set}} }} {{eqn | ll= \implies       | l = T       | o = \subseteq       | r = \mathcal P \left({S}\right)       | c = {{Defof|Subset}} }} {{end-eqn}} Now we define the mapping $\phi: S \to T$: : $\phi: S \to T: \forall x \in S: \phi \left({x}\right) = \left\{{x}\right\}$ $\phi$ is an injection: : $\forall x, y \in S: \left\{{x}\right\} = \left\{{y}\right\} \implies x = y$ by definition of set equality $\phi$ is a surjection: :$\forall \left\{{x}\right\} \in T: \exists x \in S: \phi \left({x}\right) = \left\{{x}\right\}$ So $\phi$, being both an injection and a surjection, is a bijection by definition. To show that $T \subset \mathcal P \left({S}\right)$, that is, is a proper subset of $\mathcal P \left({S}\right)$, we merely note that $\varnothing \in \mathcal P \left({S}\right)$ by Empty Set is Element of Power Set, but $\varnothing \notin T$. Thus $T \subseteq \mathcal P \left({S}\right)$ but $\mathcal P \left({S}\right) \not \subseteq T$. Hence the result. {{qed}} Category:Power Set Category:Subsets phjkamlgupbj8jjllksmk1viqo6xv8z
Let $S, T$ be sets. Let $\powerset S, \powerset T$ denote their power sets. Let $f: S \to T$ and $g: T \to S$ be injections that we know to exist between $S$ and $T$. Consider the relative complements of elements of $\powerset S$ and $\powerset T$ as mappings: :$\complement_S: \powerset S \to \powerset S: \forall X \in \powerset S: \relcomp S X = S \setminus X$ :$\complement_T: \powerset T \to \powerset T: \forall Y \in \powerset T: \relcomp T Y = T \setminus Y$ which follow directly from the definition of relative complement. Let $\alpha$ and $\beta$ denote the direct image mappings of $f$ and $g$, respectively. Consider the mapping $z: \powerset S \to \powerset S$ defined by the composition: :$z = \complement_S \circ \beta \circ \complement_T \circ \alpha$ === $z$ is an increasing mapping === Let $A, B \in \powerset S$ with $A \subseteq B$. We have: {{begin-eqn}} {{eqn | l = \map \alpha A       | o = \subseteq       | r = \map \alpha B       | c = Image of Subset under Relation is Subset of Image: Corollary 2 }} {{eqn | ll= \leadsto       | l = \map {\paren {\complement_T \circ \alpha} } A       | o = \supseteq       | r = \map {\paren {\complement_T \circ \alpha} } B       | c = Set Complement inverts Subsets }} {{eqn | ll= \leadsto       | l = \map {\paren {\beta \circ \complement_T \circ \alpha} } A       | o = \supseteq       | r = \map {\paren {\beta \circ \complement_T \circ \alpha} } B       | c = Image of Subset under Relation is Subset of Image: Corollary 2 }} {{eqn | ll= \leadsto       | l = \map {\paren {\complement_S \circ \beta \circ \complement_T \circ \alpha} } A       | o = \subseteq       | r = \map {\paren {\complement_S \circ \beta \circ \complement_T \circ \alpha} } B       | c = Set Complement inverts Subsets }} {{eqn | ll= \leadsto       | l = \map z A       | o = \subseteq       | r = \map z B       | c = Definition of $z$ }} {{end-eqn}} {{qed|lemma}} By the Knaster-Tarski Lemma: Power Set, there is a $\mathbb G \in \powerset S$ such that: :$\map z {\mathbb G} = \mathbb G$ From Relative Complement of Relative Complement we have that $\complement_S \circ \complement_S$ is the identity mapping on $\powerset S$. Thus we obtain: {{begin-eqn}} {{eqn | l = \relcomp S {\mathbb G}       | r = \map {\paren {\complement_S \circ z} } {\mathbb G}       | c =  }} {{eqn | r = \map {\paren {\complement_S \circ \complement_S \circ \beta \circ \complement_T \circ \alpha} } {\mathbb G}       | c = Definition of $z$ }} {{eqn | r = g \sqbrk {\map {\paren {\complement_T \circ \alpha} } {\mathbb G} }       | c = as $\complement_S \circ \complement_S = I_{\powerset S}$ }} {{end-eqn}} At this stage, a diagram can be helpful: :500px Let $h: S \to T$ be the mapping defined as: :$\forall x \in S: \map h x = \begin {cases} \map f x & : x \in \mathbb G \\ \map {g^{-1} } x & : x \in \relcomp S {\mathbb G} \end {cases}$ From the above, we have that: :$\relcomp S {\mathbb G} \subseteq g \sqbrk T$ Therefore, as $g$ is an injection, it follows that the preimage $\map {g^{-1} } x$ is a singleton. So $h$ is a bijection by dint of the injective nature of both $f$ and $g^{-1}$. {{qed}}
From the facts that $T \sim S_1$ and $S \sim T_1$, we can set up the two bijections: :$f: S \to T_1$ :$g: T \to S_1$ Let: :$S_2 = g \sqbrk {f \sqbrk S} = g \sqbrk {T_1} \subseteq S_1$ and: :$T_2 = f \sqbrk {g \sqbrk T} = f \sqbrk {S_1} \subseteq T_1$ So $S_2 \subseteq S_1$ and $S_2 \sim S$, while $T_2 \subseteq T_1$, and $T_2 \sim T$. For each natural number $k$, let $S_{k + 2} \subseteq S$ be the image of $S_k$ under the mapping $g \circ f$. {{explain|Why do these superset relations hold?}} Then: :$S \supseteq S_1 \supseteq S_2 \supseteq \ldots \supseteq S_k \supseteq S_{k + 1} \ldots$ Let $\displaystyle D = \bigcap_{k \mathop = 1}^\infty S_k$. {{explain|Why can we represent it so?}} Now we can represent $S$ as: {{begin-eqn}} {{eqn | n = 1       | l = S       | r = \paren {S \setminus S_1} \cup \paren {S_1 \setminus S_2} \cup \ldots \cup \paren {S_k \setminus S_{k + 1} } \cup \ldots \cup D }} {{end-eqn}} where $S \setminus S_1$ denotes set difference. {{explain}} Similarly, we can represent $S_1$ as: {{begin-eqn}} {{eqn | n = 2       | l = S_1       | r = \paren {S_1 \setminus S_2} \cup \paren {S_2 \setminus S_3} \cup \ldots \cup \paren {S_k \setminus S_{k + 1} } \cup \ldots \cup D }} {{end-eqn}} Now let: {{begin-eqn}} {{eqn | l = M       | r = \paren {S_1 \setminus S_2} \cup \paren {S_3 \setminus S_4} \cup \paren {S_5 \setminus S_6} \cup \ldots       | c =  }} {{eqn | l = N       | r = \paren {S \setminus S_1} \cup \paren {S_2 \setminus S_3} \cup \paren {S_4 \setminus S_5} \cup \ldots       | c =  }} {{eqn | l = N_1       | r = \paren {S_2 \setminus S_3} \cup \paren {S_4 \setminus S_5} \cup \paren {S_6 \setminus S_7} \cup \ldots       | c =  }} {{end-eqn}} and rewrite $(1)$ and $(2)$ as: {{begin-eqn}} {{eqn | n = 3       | l = S       | r = D \cup M \cup N }} {{eqn | n = 4       | l = S_1       | r = D \cup M \cup N_1 }} {{end-eqn}} Now: {{begin-eqn}} {{eqn | l = g \circ f \sqbrk {S \setminus S_1} = \paren {S_2 \setminus S_3}       | o = \leadsto       | r = \paren {S \setminus S_1} \sim \paren {S_2 \setminus S_3}       | c =  }} {{eqn | l = g \circ f \sqbrk {S_2 \setminus S_3} = \paren {S_4 \setminus S_5}       | o = \leadsto       | r = \paren {S_2 \setminus S_3} \sim \paren {S_4 \setminus S_5}       | c =  }} {{end-eqn}} {{handwaving}} and so on. So $N \sim N_1$. It follows from $(3)$ and $(4)$ that a bijection can be set up between $S$ and $S_1$. But $S_1 \sim T$. Therefore $S \sim T$. {{qed}}
By Injection to Image is Bijection: :$g: T \to g \sqbrk T$ is a bijection. Thus $T$ is equivalent to $g \sqbrk T$. By Composite of Injections is Injection $g \circ f: S \to g \sqbrk T \subset S$ is also an injection (to a subset of the domain of $g \circ f$). Then by Cantor-Bernstein-Schröder Theorem: Lemma: :There exists a bijection $h: S \to g \sqbrk T$. Thus $S$ is equivalent to $g \sqbrk T$. We already know that $T$ is equivalent to $g \sqbrk T$. Thus by Set Equivalence is Equivalence Relation, $S$ is equivalent to $T$. By the definition of set equivalence: :There is a bijection $\phi: S \to T$. {{qed}}
To prove that the trivial ordering is in fact an ordering, we need to checking each of the criteria for an ordering: === Reflexivity === :$\forall a \in S: a \mathrel \RR a$: From its definition, we have $\forall a, b \in S: a = b \implies a \mathrel \RR b$. Thus, as $a = a$, we have $\forall a \in S: a \mathrel \RR a$. So reflexivity is proved. === Transitivity === :$\forall a, b, c \in S: a \mathrel \RR b \land b \mathrel \RR c \implies a \mathrel \RR c$: From the definition: :$a \mathrel \RR b \iff a = b$ :$b \mathrel \RR c \iff b = c$ So as $a = b \land b = c \implies a = c$ from transitivity of equals, we have that $a \mathrel \RR c$ and thus transitivity is proved. === Antisymmetry === :$\forall a, b \in S: a \mathrel \RR b \land b \mathrel \RR a \implies a = b$: From the definition: :$a \mathrel \RR b \iff a = b$ :$b \mathrel \RR a \iff b = a$ Antisymmetry follows from symmetry of equals. The trivial ordering is by definition the same as the diagonal relation, and is therefore universally compatible. {{qed}} Category:Order Theory fabrlm275a3wq81prjzwuimdfgrewj4
By definition: :$\forall x \in S: I_S \left({x}\right) = x$ So: :$x \preceq y \implies I_S \left({x}\right) \preceq I_S \left({y}\right)$ As $I_S$ is a bijection, we also have: :$I_S^{-1} \left({x}\right) = x$ So: :$x \preceq y \implies I_S^{-1} \left({x}\right) \preceq I_S^{-1} \left({y}\right)$ {{qed}}
An ordered set is a relational structure where order isomorphism is a special case of relation isomorphism. The result follows directly from Identity Mapping is Relation Isomorphism. {{Qed}}
Follows directly from the definition of order isomorphism. Let $\phi: \left({S, \preceq_1}\right) \to \left({T, \preceq_2}\right)$ be an order isomorphism. Then: :$\forall x, y, \in S: x \preceq_1 y \implies \phi \left({x}\right) \preceq_2 \phi \left({y}\right)$ and: :$\forall \phi \left({x}\right), \phi \left({y}\right) \in T: \phi \left({x}\right) \preceq_2 \phi \left({y}\right) \implies \phi^{-1} \left({\phi \left({x}\right)}\right) \preceq_1 \phi^{-1} \left({\phi \left({y}\right)}\right)$ That is: :$\forall \phi \left({x}\right), \phi \left({y}\right) \in T: \phi \left({x}\right) \preceq_2 \phi \left({y}\right) \implies x \preceq_1 y$ Hence the result. {{Qed}}