hcju/proofwikiqa huggingface dataset:
Let $m$ and $n$ be natural numbers such that $n^+ = m^+$. By construction: :$n \in n^+$ and: :$m \in m^+$ Thus as $n^+ = m^+$ we have: :$n \in m^+$ and: :$m \in n^+$ This gives us: :$n \in m \lor n = m$ and: :$m \in n \lor m = n$ {{AimForCont}} that $n \ne m$. Then from $n \in m \lor n = m$ we have: :$n \in m$ and from $m \in n \lor m = n$ we have: :$m \in n$ In summary, if $n \ne m$ we have  :$n \in m$ and $m \in n$ But from Natural Numbers cannot be Elements of Each Other, this is not possible. Hence by Proof by Contradiction: :$n^+ = m^+ \implies n = m$ and the result follows by definition of injection. {{qed}}
The proof proceeds by induction. For all $n \in \N$, let $\map P n$ be the proposition: :for all $m \in n$: $m$ is a natural number. === Basis for the Induction === $\map P 0$ is the case: :for all $m \in 0$: $m$ is a natural number. This is true vacuously, as $\O$ has no elements by definition of empty class. Thus $\map P 0$ is seen to hold. This is the basis for the induction. === Induction Hypothesis === Now it needs to be shown that if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k^+}$ is true. So this is the induction hypothesis: :for all $m \in k$: $m$ is a natural number. from which it is to be shown that: :for all $m \in k^+$: $m$ is a natural number. === Induction Step === This is the induction step: Let $m \in k^+$. Then either: :$m = k$ or:  :$m \in k$ In the first case, $m$ is the natural number $k$. In the second case, it follows by the induction hypothesis that $m$ is a natural number. In both cases $m$ is a natural number. So $\map P k \implies \map P {k^+}$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\forall n \in \N: \forall m \in n$: $m$ is a natural number. {{qed}}
First we note that from Union of Empty Set we have: :$\bigcup \O = \O$ leading to: :$\O \subseteq \bigcup \O$ thus disposing of the degenerate case. Let $n \in \N$ such that $n \ne \O$. By definition of the von Neumann construction: :$n = \set {0, 1, 2, \ldots, n - 1}$ Thus, by definition, $m \in n$ for $m = 0, 1, 2, \ldots, n - 1$. From Natural Numbers cannot be Elements of Each Other, it cannot therefore be the case that $n \in m$ for $m \in \n$. By definition of union of class: :$\bigcup n = \set {x: \exists X \in n: x \in X}$ But it has been established that there is no $X \in n$ such that $n \in X$. Hence the result. {{qed}}
Let $n \in \N$. From Natural Number is Transitive Set, $n$ is transitive. From Class is Transitive iff Union is Subset it follows directly that: :$\bigcup n \subseteq n$ {{qed}}
{{begin-eqn}} {{eqn | o =        | r = \map \bigcup {n^+}       | c =  }} {{eqn | r = \map \bigcup {\set n \cup n}       | c = {{Defof|Von Neumann Construction of Natural Numbers}} }} {{eqn | r = \bigcup \set n \cup \bigcup n       | c = Union Distributes over Union }} {{eqn | r = n \cup \bigcup n       | c = Union of Singleton }} {{end-eqn}} From Natural Number is Superset of its Union we have: :$\bigcup n \subseteq n$ Then from Union with Superset is Superset: :$\bigcup n \subseteq n \iff \paren {n \cup \bigcup n} = n$ and the result follows. {{qed}}
By hypothesis, $\omega \in V$, where $V$ is a Zermelo universe. By the Axiom of Infinity we have that $\omega$ is a set. By the Axiom of Transitivity, $\omega$ is transitive. Hence: :$\bigcup \omega \subseteq \omega$ {{qed|lemma}} Let $n \in \omega$. Then by definition of the von Neumann construction: :$n + 1 = n \cup \set n$ That is: :$\exists X \in \omega: n \in X$ where in this case $X = n + 1$. Thus we have that: :$n \in \set {x: \exists X \in \omega: x \in X}$ and so by definition of union of class: :$n \in \bigcup \omega$ Thus we have that: :$\omega \subseteq \bigcup \omega$ which, together with: :$\bigcup \omega \subseteq \omega$ gives us, by set equality: :$\bigcup \omega = \omega$ {{qed}}
We have that: :$\omega^+ = \omega \cup \set \omega$ and so: :$\omega \subseteq \omega^+$ {{qed|lemma}} By definition: :$\bigcup \omega^+ = \set {x: \exists y \in \omega^+: x \in y}$ Thus: :$x \in \bigcup \omega^+ \implies x \in \omega$ {{qed|lemma}} So by definition of set equality: :$\bigcup \omega^+ = \omega$ {{qed}}
=== $(1)$ implies $(2)$ === Let it be given that $A$ is inductive under $g$. Let $A$ be a minimally inductive class under $g$ by definition 1. Then by definition: :$A$ has no proper subclass $B$ such that $B$ is inductive under $g$. Let $A$ have a subclass $C$ which is inductive under $g$. Then by definition, $C$ is not a proper subclass of $A$. Thus by definition of proper subclass: :$C = A$ By definition of subclass: :$A \subseteq C$ and: :$C \subseteq A$ That is: :$\forall x \in A: x \in C$ and: :$\forall x \in C: x \in A$ That is: $C$ contains all elements of $A$. Thus $A$ is a minimally inductive class under $g$ by definition 2. {{qed|lemma}} === $(2)$ implies $(1)$ === Let it be given that $A$ is inductive under $g$. Let $A$ be a minimally inductive class under $g$ by definition 2. Then by definition: :Every subclass of $A$ which is inductive under $g$ contains all the elements of $A$. Let $C$ be a subclass of $A$ which is inductive under $g$. By definition of subclass: :$C \subseteq A$ But by hypothesis: :$\forall x \in A: x \in C$ That is: :$A \subseteq C$ By definition of equality of classes it follows that: :$A = C$ and so by definition $C$ cannot be a proper subclass of $A$. Thus $A$ is a minimally inductive class under $g$ by definition 1. {{qed|lemma}} === $(1)$ implies $(3)$ === Let $A$ be a minimally inductive class under $g$ by definition 1. Then by definition: {{begin-axiom}} {{axiom | n = 1         | t = $A$ is inductive under $g$ }} {{axiom | n = 2         | t = No proper subclass of $A$ is inductive under $g$. }} {{end-axiom}} By definition of inductive class: :$\O \in A$ :$A$ is closed under $g$. Putting these definitions together, we have that: :No proper subclass of $A$ such that $\O \in A$ is closed under $g$. That is,  $A$ is '''minimally closed under $g$ with respect to $\O$''' Thus $A$ is a minimally inductive class under $g$ by definition 3. {{qed|lemma}} === $(3)$ implies $(1)$ === Let $A$ be a minimally inductive class under $g$ by definition 3. That is, $A$ is minimally closed under $g$ with respect to $\O$. That is: :$(1): \quad \O \in A$ :$(2): \quad$ No proper subclass of $A$ such that $\O \in A$ is closed under $g$. Hence by definition of inductive class: {{begin-axiom}} {{axiom | n = 1         | t = $A$ is inductive under $g$ }} {{axiom | n = 2         | t = No proper subclass of $A$ is inductive under $g$. }} {{end-axiom}} Thus $A$ is a minimally inductive class under $g$ by definition 1. {{qed}} Category:Minimally Inductive Classes 3n8dtkvys9p3dy7r2sy6yqux9m07gad
We are given that $M$ is a minimally inductive class under $g$. That is, $M$ is an inductive class under $g$ with the extra property that $M$ has no proper class which is also inductive class under $g$. Let $P$ be a propositional function on $M$ which has the properties specified: :$(1): \quad \map P \O = \T$ :$(2): \quad \forall x \in M: \map P x = \T \implies \map P {\map g x} = \T$ Thus by definition, the class $S$ of all elements of $M$ such that $\map P x = \T$ is an inductive class under $g$. But because $M$ is minimally inductive under $g$, $S$ contains all elements of $M$. That is: :$\forall x \in M: \map P x = \T$ as we were to show. {{Qed}}
Consider Peano's axioms: {{:Axiom:Peano's Axioms}} From Inductive Construction of Natural Numbers fulfils Peano's Axioms, $\omega$ fulfils Peano's axioms. We note that from {{PeanoAxiom|1}}: :$\O \in \omega$ We acknowledge from {{PeanoAxiom|2}}: :the successor mapping defines that $n^+ := n \cup \set n$ and from {{PeanoAxiom|5}} the result follows. {{qed}}
Let $T_i \subseteq S_i$ for all $i \in I$. Then: {{begin-eqn}} {{eqn | l = \family {x_i}       | o = \in       | r = T       | c =  }} {{eqn | ll= \leadsto       | lo= \forall i \in I:       | l = x_i       | o = \in       | r = T_i       | c = {{Defof|Cartesian Product of Family}} }} {{eqn | ll= \leadsto       | lo= \forall i \in I:       | l = x_i       | o = \in       | r = S_i       | c = {{Defof|Subset}} }} {{eqn | ll= \leadsto       | l = \family {x_i}       | o = \in       | r = S       | c = {{Defof|Cartesian Product of Family}} }} {{end-eqn}} Thus $T \subseteq S$ by the definition of a subset. {{qed}} Category:Cartesian Product Category:Subsets 8cz2xt7hc8o1zc5kird65vnhhb0v6oo
The proof proceeds by general induction. Let an element $x$ of $M$ be defined as: :'''left normal''' with respect to $\RR$ {{iff}} $\map \RR {x, y}$ for all $y \in M$ :'''right normal''' with respect to $\RR$ {{iff}} $\map \RR {y, x}$ for all $y \in M$. Let the hypothesis be assumed. First we demonstrate a lemma: === Lemma === {{:Double Induction Principle/Lemma}}{{qed|lemma}} We now show by general induction that every $x \in M$ is right normal with respect to $\RR$. It then follows from the lemma that $\map \RR {x, y}$ for all $x, y \in M$. === Basis for the Induction === $\map P \O$ is the case: From condition $\text D_1$ of the definition of $\RR$, we have immediately that: :$\map \RR {x, \O}$ for all $x \in M$. That is, that $\O$ is right normal with respect to $\RR$. Thus $\map P \O$ is seen to hold. This is the basis for the induction. === Induction Hypothesis === Now it needs to be shown that if $\map P x$ is true, where $x \in M$, then it logically follows that $\map P {\map g x}$ is true. So this is the induction hypothesis: :$x$ is right normal with respect to $\RR$ from which it is to be shown that: :$\map g x$ is right normal with respect to $\RR$. === Induction Step === This is the induction step: Let $x \in M$ be right normal with respect to $\RR$: :$\forall y \in M: \map \RR {y, x}$ By the lemma we have that $x$ is left normal with respect to $\RR$. That is: :$\forall y \in M: \map \RR {x, y}$ Thus by condition $\text D_2$ of the definition of $\RR$: :$\forall y \in M: \map \RR {y, \map g x}$ That is, $\map g x$ is right normal with respect to $\RR$. So $\map P x \implies \map P {\map g x}$ and by the Principle of General Induction: :$\forall x \in M$: $x$ is right normal with respect to $\RR$. The result follows. {{qed}}
By definition, a minimally inductive class under $g$ is a minimally closed class under $g$ with respect to $\O$. Recall the Double Induction Principle for Minimally Closed Class: Let $\RR$ be a relation on $M$ which satisfies: {{begin-axiom}} {{axiom | n = \text D_1         | q = \forall x \in M         | m = \map \RR {x, b} }} {{axiom | n = \text D_2         | q = \forall x, y \in M         | m = \map \RR {x, y} \land \map \RR {y, x} \implies \map \RR {x, \map g y} }} {{end-axiom}} Then $\map \RR {x, y}$ holds for all $x, y \in M$. In this context: :$b = \O$ and the result follows. {{qed}}
The proof proceeds by general induction. Let $x \in M$ be right normal with respect to $\RR$ Let $\map P y$ be the proposition: :$\map \RR {x, y}$ holds. === Basis for the Induction === By condition $\text D_1$ of the definition of $\RR$: :$\map \RR {x, \O}$ for all $x \in M$. Thus $\map P \O$ is seen to hold. This is the basis for the induction. === Induction Hypothesis === Now it needs to be shown that if $\map P y$ is true, where $x \in M$, then it logically follows that $\map P {\map g y}$ is true. So this is the induction hypothesis: :$\map \RR {x, y}$ holds from which it is to be shown that: :$\map \RR {x, \map g y}$ holds === Induction Step === This is the induction step: Let $\map \RR {x, y}$ hold. As $x$ is right normal with respect to $\RR$: :$\map \RR {y, x}$ holds. Thus by condition $\text D_2$ of the definition of $\RR$: :$\map \RR {x, \map g y}$ holds. So $\map P x \implies \map P {\map g x}$ and the result follows by the Principle of General Induction. Therefore: :$\forall x \in M$: if $x$ is a right normal element of $M$ with respect to $\RR$, then $x$ is a left normal element of $M$ with respect to $\RR$ {{qed}}
By definition of the von Neumann construction: :$n^+ = n \cup \set n$ from which it follows that: :$n \subseteq n^+$ {{qed}}
From Cartesian Product of Family of Subsets: :$\paren {\forall i \in I: T_i \subseteq S_i} \implies T \subseteq S$ So it remains to show that: :$T \subseteq S \implies \forall i \in I : T_i \subseteq S_i$. Let $T \subseteq S$. Let $x_j \in T_j$ for some $j \in I$. Let $\map x j = x_j$ Suppose $k \in I: k \ne j$. As $T_k \ne \O$ it is possible to use the axiom of choice to choose $\map x k \in T_k$. Then: :$x \in T \subseteq S$ By definition of the Cartesian product $S$: :$\forall i \in I: \map x i \in S_i$ In particular: :$\map x j = x_j \in S_j$ As $x_j$ is an arbitrary element of $T_j$: :$T_j \subseteq S_j$ As $j$ is an arbitrary element of $I$: :$\forall i \in I: T_i \subseteq S_i$ The result follows.  {{qed}} {{AoC}} Category:Cartesian Product Category:Subsets plcdb1han305sgi1wm0m79k3pqzm33p
=== Sufficient Condition === {{:Projection is Injection iff Factor is Singleton/Family of Sets/Sufficient Condition}}{{qed|lemma}} === Necessary Condition === {{:Projection is Injection iff Factor is Singleton/Family of Sets/Necessary Condition}}{{qed}} Category:Projections Category:Injections Category:Singletons dlpp7532yqnwmzllerktj7c29d703pt
In the following, let $x$ and $y$ be arbitrary elements of $A$ in the domain of $g$. From Empty Set is Subset of All Sets: :$\O \subseteq y$ Thus by the Rule of Addition: :$\map g y \subseteq \O \lor \O \subseteq y$ and so it is seen that $\map \RR {y, \O}$. That is, $(1)$ holds. {{qed|lemma}} Let $\map \RR {x, y}$ and $\map \RR {y, x}$ both hold. Thus by definition of $\RR$, both of the following hold: :$(\text a): \quad \map g x \subseteq y$ or $y \subseteq x$ :$(\text b): \quad \map g y \subseteq x$ or $x \subseteq y$ We are to show that $\map \RR {x, \map g y}$ holds. That is: :$\map g x \subseteq \map g y$ or $\map g y \subseteq x$ Let $\map g x \subseteq y$. Because $g$ is a progressing mapping: :$y \subseteq \map g y$ and so: :$\map g x \subseteq \map g y$ Thus $\map \RR {x, \map g y}$ holds. {{qed|lemma}} Let $\map g y \subseteq x$. Then by definition $\map \RR {x, \map g y}$ holds. {{qed|lemma}} If neither $\map g x \subseteq y$ nor $\map g y \subseteq x$, then by $(\text a)$ and $(\text b)$: :$y \subseteq x$ and $x \subseteq y$ and so by definition of set equality: :$x = y$ Hence: :$\map g x = \map g y$ Thus, again by definition of set equality: :$\map g x \subseteq \map g y$ and so again, by definition, $\map \RR {x, \map g y}$ holds. {{qed|lemma}} Thus it has been shown that in all cases, if $\map \RR {x, y}$ and $\map \RR {y, x}$ both hold, then $\map \RR {x, \map g y}$ holds. Thus condition $(2)$ has been satisfied. Hence the result. {{qed}}
Let $\RR$ be the relation on $M$ defined as: :$\forall x, y \in M: \map \RR {x, y} \iff \map g x \subseteq y \lor y \subseteq x$ We are given that $g$ is a progressing mapping. From the Progressing Function Lemma, we have that: {{begin-axiom}} {{axiom | n = 1         | q = \forall y \in \Dom g         | ml= \map \RR {y, \O} }} {{axiom | n = 2         | q = \forall x, y \in \Dom g         | ml= \map \RR {x, y} \land \map \RR {y, x}         | mo= \implies         | mr= \map \RR {x, \map g y} }} {{end-axiom}} From the Double Induction Principle, if $\RR$ is a relation on $M$ which satisfies: {{begin-axiom}} {{axiom | n = \text D_1         | q = \forall x \in M         | m = \map \RR {x, \O} }} {{axiom | n = \text D_2         | q = \forall x, y \in M         | m = \map \RR {x, y} \land \map \RR {y, x} \implies \map \RR {x, \map g y} }} {{end-axiom}} then $\map \RR {x, y}$ holds for all $x, y \in M$. Thus: :$\forall x, y \in M: \map g x \subseteq y \lor y \subseteq x$ follows directly. {{qed|lemma}} Then the fact that: :$\map g x \subseteq y \lor y \subseteq x$ implies that: :$x \subseteq y \lor y \subseteq x$ because: :$x \subseteq \map g x$ Hence the result by definition of a nest. {{qed}}
A minimally inductive class under $g$ is the same thing as a minimally closed class under $g$ with respect to $\O$. The result then follows by a direct application of Minimally Closed Class under Progressing Mapping induces Nest. {{qed}}
We are given that: :for all $x, y \in A$, either $\map g x \subseteq y$ or $y \subseteq x$. Let $x, y \in A$ such that: :$x \subseteq y \subseteq \map g x$ {{AimForCont}} both $x \subset y$ and $y \subset \map g x$ From $x \subset y$, it follows by definition of proper subset that: :$\exists a \in y: a \notin x$ and so it is not the case that $y \subseteq x$. From $y \subset \map g x$, it follows by definition of proper subset that: :$\exists b \in \map g x: b \notin y$ and so it is not the case that $\map g x \subseteq y$. Hence neither $y \subseteq x$ nor $\map g x \subseteq y$. This contradicts the property of $g$. Hence by Proof by Contradiction it follows that $y \subset \map g x$ and $x \subset y$ cannot both be true. Therefore either $x = y$ or $y = \map g x$. Hence the result. {{Qed}}
We are given that: :for all $x, y \in A$, either $\map g x \subseteq y$ or $y \subseteq x$. Let $x, y \in A$ such that: :$x \subseteq y \subseteq \map g x$ Then either we have: :$\map g x \subseteq y$ and $y \subseteq \map g x$ in which case, by definition of set equality: :$y = \map g x$ or we have that: :$x \subseteq y$ and $y \subseteq x$ in which case, by definition of set equality: :$x = y$ Thus either $y = \map g x$ or $x = y$ and the result follows. {{Qed}}
Let $x \subseteq y$. Suppose $x = y$. Then $\map g x \subseteq \map g y$ and the result holds. {{qed|lemma}} Suppose that $x \ne y$. Then $x \subset y$ It follows from Corollary 1 that: :$\map g x \subseteq y$ As $g$ is a progressing mapping on $A$: :$y \subseteq \map g y$ Hence by Subset Relation is Transitive: :$\map g x \subseteq \map g y$ {{qed|lemma}} So in either case: :$\map g x \subseteq \map g y$ {{Qed}}
By definition of progressing mapping: :$\forall x \in A: x \subseteq \map g x$ Thus by Subset Relation is Transitive: :$\map g x \subseteq y \implies x \subseteq y$ and it follows that: :$\forall x, y \in A: x \subseteq y \lor y \subseteq x$ Hence the result by definition of nest. {{qed}}
From Minimally Inductive Class under Progressing Mapping induces Nest, we have that $M$ is a nest in which: :$\forall x, y \in M: \map g x \subseteq y \lor y \subseteq x$ Thus the corollary 1 of the Sandwich Principle applies directly. {{qed}}
From Minimally Inductive Class under Progressing Mapping induces Nest, we have that $M$ is a nest in which: :$\forall x, y \in M: \map g x \subseteq y \lor y \subseteq x$ Thus the corollary 2 of the Sandwich Principle applies directly. {{qed}}
By definition, every element of $B$ is a subset of $x$. Then every element of $B$ is an element of the power set $\powerset x$ of $x$. Thus $B$ is a subclass of $\powerset x$. By the Axiom of Powers/Class Theory, $\powerset x$ is a set. That is, $\powerset x$ is an element of $V$. As $V$ is a swelled class, then by definition, then every subclass of $\powerset x$ is an element of $V$. That includes $B$. That is, $B$ is an element of $V$. Thus, by definition of $V$, $B$ is a set. {{qed}}
Let the hypothesis be assumed. The proof proceeds by general induction. For all $x \in M$, let $\map P x$ be the proposition: :Every non-empty subset of $M$ which is bounded by $x$ has a greatest element. === Basis for the Induction === Let $x = \O$. The only non-empty subset of $M$ which is bounded by $\O$ is $\set \O$. Then it is seen that $\set \O$ has a greatest element, that is $\O$. Thus $\map P \O$ is seen to hold. This is the basis for the induction. === Induction Hypothesis === Now it needs to be shown that if $\map P x$ is true, where $x \in M$, then it logically follows that $\map P {\map g x}$ is true. So this is the induction hypothesis: :Every non-empty subset of $M$ which is bounded by $x$ has a greatest element. from which it is to be shown that: :Every non-empty subset of $M$ which is bounded by $\map g x$ has a greatest element. === Induction Step === This is the induction step: Let the induction hypothesis be assumed. Let $B$ be a non-empty subset of $M$ which is bounded by $\map g x$. Thus, by definition, every element of $B$ is a subset of $\map g x$. It is to be shown that $B$ has a greatest element $m$. That is, that $m \in B$ and every element of $B$ is a subset of $m$ Let $\map g x \in B$. By definition of subset bounded by $\map g x$, every element of $B$ is a subset of $\map g x$. That is, $\map g x$ is itself the greatest element of $B$. Suppose that $\map g x \notin B$. Then every element of $B$ is a proper subset of $\map g x$. Let $y \in B$ be arbitrary. Because $y$ is a proper subset of $\map g x$, there exists $z \in \map g x$ such that $z \notin y$. Thus $\map g x$ is not a subset of $y$. From Minimally Inductive Class under Progressing Mapping induces Nest: :$\forall x, y \in M: \map g x \subseteq y \lor y \subseteq x$ Because it is not the case that $\map g x \subseteq y$, it must be that: :$y \subseteq x$ Thus, as $y$ is arbitrary, it follows that every element of $B$ is a proper subset of $x$. Thus $B$ is bounded by $x$. It follows by the induction hypothesis that $B$ has a greatest element. So $\map P x \implies \map P {\map g x}$ and the result follows by the Principle of General Induction. Therefore: :For all $x \in M$, every non-empty subset of $M$ which is bounded by $x$ has a greatest element. That is: :Every non-empty bounded subset of $M$ has a greatest element. {{qed}}
Let $\pr_j$ be an injection. Then: :$\forall x, y \in S: \map {\pr_j} x = \map {\pr_j} y \implies x = y$ We have that $\family {S_i}_{i \mathop \in I}$ is a non-empty family of non-empty sets Hence, by the axiom of choice (formulation $2$), $S$ is non-empty. Let $z \in S$. By the definition of Cartesian product $S$: :$\forall i \in I: \map z i \in S_i$. Let $x_k \in S_k$ for some $k \in I \setminus \set j$ Let $z' \in S$ be defined by: :$\map {z'} i = \begin{cases} \map z i & : i \in I \setminus \set k \\ x_k & : i = k  \end{cases}$ Then: {{begin-eqn}} {{eqn | l = \map {\pr_j} {z'}       | r = \map {z'} j        | c = {{Defof|Projection}} }} {{eqn | r = \map z j       | c = as $j \ne k$  }} {{eqn | r = \map {\pr_j} z       | c = {{Defof|Projection}} }} {{end-eqn}} Thus $z' = z$. In particular: :$x_k = \map {z'} k = \map z k = z_k$ It follows that: :$S_k = \set {z_k}$ Since $k$ was arbitrary: :$\forall k \in I \setminus \set j: S_k = \set {z_k}$ The result follows.
Let $S_i = \set {s_i}$ for all $i \in I \setminus \set {j}$. Let $\map {\pr_j} x = \map {\pr_j} y = s_j$ for $x, y \in S$. By definition of $j$th projection: :$\map x j = \map {\pr_j} x = s_j$ :$\map y j = \map {\pr_j} y = s_j$ and so $\map x j = \map y j$. By the definition of Cartesian product, for all $i \in I \setminus \set j$: :$\map x i, \map y i \in S_i = \set {s_i}$ and so $\map x i = \map y i$ for all $i \in I \setminus \set j$. Thus $x = y$. Thus $\pr_j$ is an injection by definition.