hcju/proofwikiqa huggingface dataset:
Let $\displaystyle \bigcup A \subseteq A$. Let $x \in \displaystyle \bigcup A$. Then by definition: :$\exists y \in A: x \in y$ By definition of subclass: :$x \in A$ Thus we have that: :$x \in y \land y \in A \implies x \in A$ It follows by definition that $A$ is a transitive class.
Let $A$ be transitive. Let $x \in \displaystyle \bigcup A$. By Class is Transitive iff Union is Subset we have that  :$\displaystyle \bigcup A \subseteq A$ Thus by definition of subclass: :$x \in A$ As $A$ is transitive: :$x \subseteq A$ Let $z \in x$. As $x \subseteq A$, it follows by definition of subclass that: :$z \in A$ Thus we have that: :$\exists x \in A: z \in x$ and so by definition of union of class: :$z \in \displaystyle \bigcup A$ Thus we have that: :$z \in x \implies z \in \displaystyle \bigcup A$ and so by definition of subclass: :$x \subseteq \displaystyle \bigcup A$ Thus we have that: :$x \in \displaystyle \bigcup A \implies x \subseteq \displaystyle \bigcup A$ Hence $\displaystyle \bigcup A$ is a transitive class by definition. {{qed}}
Let $A$ be transitive. By Class is Transitive iff Union is Subset: :$\displaystyle \bigcup A \subseteq A$ By Union of Subclass is Subset of Union of Class: :$\displaystyle \map \bigcup {\bigcup A} \subseteq \bigcup A$ Then by Class is Transitive iff Union is Subset: :$\bigcup A$ is transitive. {{qed}}
Let $A$ be such that every $y \in A$ is transitive. Let $x \in \bigcup A$. Then $x$ is an element of some element $y$ of $A$. By hypothesis, $y$ is transitive. Hence, by definition of transitive, $x \subseteq y$. Because $y \in A$, by definition of union of class, $y \subseteq \bigcup A$. So $x \subseteq \bigcup A$. As this is true for all $x \in A$, it follows by definition that $\bigcup A$ is transitive. {{qed}}
From Set Difference with Set Difference is Union of Set Difference with Intersection: :$R \setminus \paren {S \setminus T} = \paren {R \setminus S} \cup \paren {R \cap T}$ where $R, S, T$ are sets. Hence: {{begin-eqn}} {{eqn | l = T \setminus \paren {S \setminus T}       | r = \paren {T \setminus S} \cup \paren {T \cap T}       | c = Set Difference with Set Difference is Union of Set Difference with Intersection }} {{eqn | r = \paren {T \setminus S} \cup T       | c = Intersection is Idempotent }} {{eqn | r = T       | c = Set Difference Union First Set is First Set }} {{end-eqn}} {{qed}}
By Axiom of Powers, $\powerset x$ exists and is a set. Let $\powerset x$, $\map \QQ x$ both be power sets of $x$. From definition of power sets, $\forall T$: :$T \in \powerset x \iff T \subseteq x$ :$T \in \map \QQ x \iff T \subseteq x$ Since Biconditional is Commutative and Transitive: :$T \in \powerset x \iff T \in \map \QQ x$ By Axiom of Extension, $\powerset x = \map \QQ x$. Hence the power set is unique. {{qed}}
Let $x \in A$. By definition of class, $x$ is a set. Let $y \in x$. By definition of union of $A$: :$\displaystyle \bigcup A := \set {y: \exists x \in A: y \in x}$ It follows directly from that definition that: :$y \in \displaystyle \bigcup A$ The result follows by definition of subset. {{qed}} Category:Class Union 6s5jafp39xqpc0pkr80gxortwgipbda
Let $z \in x$. By Element of Class is Subset of Union of Class: :$z \subseteq \displaystyle \bigcup x$ By definition of power set: :$z \in \powerset {\displaystyle \bigcup x}$ The result follows by definition of subset. {{qed}}
From Set is Element of its Power Set: :$x \in \powerset x$ From Element of Class is Subset of Union of Class it follows that: :$x \subseteq \displaystyle \map \bigcup {\powerset x}$ {{qed|lemma}} Let $z \in \displaystyle \map \bigcup {\powerset x}$ Then by definition of union: :$\exists y \in \powerset x: z \in y$ By definition of $\powerset x$ :$\exists y \subseteq x: z \in y$ But by definition of subset, that means: :$z \in x$ Thus, again by definition of subset: :$\displaystyle \map \bigcup {\powerset x} \subseteq x$ {{qed|lemma}} We have that: :$x \subseteq \displaystyle \map \bigcup {\powerset x}$ and: :$\displaystyle \map \bigcup {\powerset x} \subseteq x$ Hence by definition of set equality: :$x = \displaystyle \map \bigcup {\powerset x}$ {{qed}}
Let $A$ and $B$ be sets in $V$. Because $V$ is a basic universe, the basic universe axioms apply. Hence by the axiom of pairing $\set {A, B}$ is a set. Then by the axiom of unions $\displaystyle \bigcup \set {A, B}$ is also a set. We have that $A \cup B = \displaystyle \bigcup \set {A, B}$. By the axiom of powers $\powerset {A \cup B}$ is a set. Therefore, also by the axiom of powers, so is $\powerset {\powerset {A \cup B} }$. It remains to be shown that $A \times B$ is a subclass of $\powerset {\powerset {A \cup B} }$. Let $x \in A \times B$. Then $x = \tuple {a, b}$ for some $a \in A$ and $b \in B$. By definition of ordered pair: :$x = \set {\set a, \set {a, b} }$ From Set is Subset of Union, we have that $A \subseteq A \cup B$ and $B \subseteq A \cup B$. Hence by definition of subset: :$a \in A \cup B$ and: :$b \in A \cup B$ Thus: :$\set a \subseteq A \cup B$ and: :$\set {a, b} \subseteq A \cup B$ Thus by definition of power set: :$\set a \in \powerset {A \cup B}$ and: :$\set {a, b} \in \powerset {A \cup B}$ Hence: :$\set {\set a, \set {a, b} } \subseteq \powerset {A \cup B}$ That is: :$\tuple {a, b} \subseteq \powerset {A \cup B}$ and so: :$x \subseteq \powerset {A \cup B}$ which means: :$x \in \powerset {\powerset {A \cup B} }$ Hence by definition of subclass: :$A \times B \subseteq \powerset {\powerset {A \cup B} }$ We have that $\powerset {\powerset {A \cup B} }$ is a set. By the axiom of swelledness it follows that $A \times B$ is also a set. {{Qed}}
As the codomain of $f$ is $\closedint a b$, it follows that the image of $f$ is a subset of $\closedint a b$. Thus $\map f a \ge a$ and $\map f b \le b$. Let us define the real function $g: \closedint a b \to \R$ by $\map g x = \map f x - x$. Then by the Combined Sum Rule for Continuous Functions, $\map g x$ is continuous on $\closedint a b$. But $\map g a \ge 0$ and $\map g b \le 0$. By the Intermediate Value Theorem, $\exists \xi: \map g \xi = 0$. Thus $\map f \xi = \xi$. {{qed}}
Let $c, d \in \closedint a b$ such that $c < d$. Then $f$ satisfies the conditions of the Mean Value Theorem on $\closedint c d$. Hence: :$\exists \xi \in \openint c d: \map {f'} \xi = \dfrac {\map f d - \map f c} {d - c}$ Let $f$ be such that: :$\forall x \in \openint a b: \map {f'} x \ge 0$ Then: :$\map {f'} \xi \ge 0$ and hence: :$\map f d \ge \map f c$ Thus $f$ is increasing on $\closedint a b$. {{qed}}
Let $c, d \in \closedint a b$ such that $c < d$. Then $f$ satisfies the conditions of the Mean Value Theorem on $\closedint c d$. Hence: :$\exists \xi \in \openint c d: \map {f'} \xi = \dfrac {\map f d - \map f c} {d - c}$ Let $f$ be such that: :$\forall x \in \openint a b: \map {f'} x > 0$ Then: :$\map {f'} \xi > 0$ and hence: :$\map f d > \map f c$ Thus $f$ is strictly increasing on $\closedint a b$. {{qed}}
Let $c, d \in \closedint a b$ such that $c < d$. Then $f$ satisfies the conditions of the Mean Value Theorem on $\closedint c d$. Hence: :$\exists \xi \in \openint c d: \map {f'} \xi = \dfrac {\map f d - \map f c} {d - c}$ Let $f$ be such that: :$\forall x \in \openint a b: \map {f'} x < 0$ Then: :$\map {f'} \xi < 0$ and hence: :$\map f d < \map f c$ Thus $f$ is strictly decreasing on $\closedint a b$. {{qed}}
Let $c, d \in \closedint a b$ such that $c < d$. Then $f$ satisfies the conditions of the Mean Value Theorem on $\closedint c d$. Hence: :$\exists \xi \in \openint c d: \map {f'} \xi = \dfrac {\map f d - \map f c} {d - c}$ Let $f$ be such that: :$\forall x \in \openint a b: \map {f'} x \le 0$ Then: :$\map {f'} \xi \le 0$ and hence: :$\map f d \le \map f c$ Thus $f$ is decreasing on $\closedint a b$. {{qed}}
Recall the definition of a limit of a real function: :$\forall \epsilon \in \R_{>0}: \exists \delta \in \R_{>0}: \forall x \in \R: 0 < \size {x - c} < \delta \implies \size {\map f x - L} < \epsilon$ Denote by $\map P {\epsilon,\delta}$ the proposition that the above statement holds. Suppose that $\map P {\epsilon,\delta}$ holds for all rational $\epsilon$ in the interval of interest: :$\forall \epsilon \in \Q_{>0}: \exists \delta \in \Q_{>0}: \forall x \in \R: 0 < \size {x - c} < \delta \implies \size {\map f x - L} < \epsilon$ Let $\epsilon_0 > 0$ be a real number. From Between two Real Numbers exists Rational Number, there is some rational $\epsilon_0'$ such that $0 < \epsilon_0' < \epsilon_0$. Then $\size {\map f x - L} < \epsilon_0'$ implies $\size {\map f x - L} < \epsilon_0$. Thus $\map P {\epsilon,\delta}$ holding for all rational $\epsilon$ implies $\map P {\epsilon,\delta}$ for all real $\epsilon$. Now suppose $\map P {\epsilon,\delta}$ has been established for some rational $\epsilon$ and some real $\delta_0$. From Between two Real Numbers exists Rational Number, there is some rational $\delta_0'$ such that $\size {x-c} < \delta_0' < \delta_0$. Then $\size {x-c} < \delta_0$ is implied by $\size {x-c} < \delta_0'$ This means $\map P {\epsilon,\delta_0'}$ implies $\map P {\epsilon,\delta_0}$ holds. We conclude: :$\map P {\epsilon,\delta}$ holds for all rational $\epsilon$ and $\delta$ in the interval of interest is a stronger statement than: :$\map P {\epsilon,\delta}$ holds for all real $\epsilon$ and $\delta$ in the interval of interest. {{qed}}
Suppose it is given that $B$ has occurred. Then the probability of $A$ having occurred may not be $\map \Pr A$ after all. In fact, we ''can'' say that $A$ has occurred {{iff}} $A \cap B$ has occurred. So, if we ''know'' that $B$ has occurred, the conditional probability of $A$ given $B$ is $\map \Pr {A \cap B}$. It follows then, that if we ''don't'' actually know whether $B$ has occurred or not, but we know its probability $\map \Pr B$, we can say that: :The probability that $A$ and $B$ have both occurred is the conditional probability of $A$ given $B$ multiplied by the probability that $B$ has occurred. Hence: :$\map \Pr {A \mid B} = \dfrac {\map \Pr {A \cap B} } {\map \Pr B}$ {{qed}}
Denote by $\map P {2^{-n},\delta}$ the proposition considered in the theorem exposition: :$\forall \in \N: \exists \delta \in \R_{>0}: \forall x \in \R: 0 < \size {x - c} < \delta \implies \size {\map f x - L} < 2^{-n} $ Let the limit of $f$ as $x \to c$ exist and equal $L$, as described in the definition of limit: :$\forall \epsilon \in \R_{>0}: \exists \delta \in \R_{>0}: \forall x \in \R: 0 < \size {x - c} < \delta \implies \size {\map f x - L} < \epsilon$. Denote by $\map P {\epsilon,\delta}$ the proposition that the above statement holds. From the Archimedean Principle: :$\forall \epsilon > 0: \exists n \in \N: n > \dfrac 1 \epsilon$. As $2^n > n$ for all $n$, we have that $2^n > \dfrac 1 \epsilon$. By Ordering of Reciprocals, $2^{-n} < \epsilon$. Therefore $\size {\map f x - L} < \epsilon$ is implied by $\size {\map f x - L} < 2^{-n}$. Thus $\map P {\epsilon,\delta}$ is a weaker statement $\map P {2^{-n},\delta}$  {{qed}}
Trivially, the reverse implication holds, as $\set{B \in \BB: B \subseteq U} \subseteq \BB$. We now show that the forward implication holds. Suppose that $U \in \tau$. Then, by definition: :$\exists \AA \subseteq \BB: U = \bigcup \AA$ By Union is Smallest Superset: General Result, we have: :$\forall B \in \AA: B \subseteq U$ By the definition of a subset, it follows that: :$\AA \subseteq \set{B \in \BB: B \subseteq U}$ From Union of Subset of Family is Subset of Union of Family: :$U = \bigcup \AA \subseteq \bigcup \set{B \in \BB: B \subseteq U}$ By Union is Smallest Superset: General Result: :$\bigcup \set{B \in \BB: B \subseteq U} \subseteq U$ By definition of set equality: :$U = \bigcup \set{B \in \BB: B \subseteq U}$
From Set is Subset of Union: General Result, the forward implication directly follows. We now show that the reverse implication holds. By hypothesis, we have that: :$U \subseteq \bigcup \left\{{B \in \mathcal B: B \subseteq U}\right\}$ By Union is Smallest Superset: General Result: :$\bigcup \left\{{B \in \mathcal B: B \subseteq U}\right\} \subseteq U$ By definition of set equality: :$U = \bigcup \left\{{B \in \mathcal B: B \subseteq U}\right\}$ The result follows.
=== Projection is Continuous === {{:Projection from Product Topology is Continuous}}{{qed|lemma}} === Projection is Open === {{:Projection from Product Topology is Open}}{{qed}} Category:Tychonoff Topology Category:Continuous Mappings Category:Open Mappings Category:Projections Category:Projection from Product Topology is Open and Continuous 84kyj91lyyr81b0ezz855q4h76nf27e
=== Projection is Continuous === {{:Projection from Product Topology is Continuous/General Result/Proof}}{{qed|lemma}} === Projection is Open === {{:Projection from Product Topology is Open/General Result/Proof}}{{qed}} Category:Projection from Product Topology is Open and Continuous ns6a93h20bpf8vlqd937bjl5h861lon
Let $x = \family {x_\alpha}_{\alpha \mathop \in I} \in S$. Let $W \in \tau$ such that $x \in W$. From Natural Basis of Tychonoff Topology, the set $\BB'$ of cartesian products of the form $\displaystyle \prod_{\alpha \mathop \in I} V_\alpha$ where: :for all $\alpha \in I : V_\alpha \in \tau_\alpha$ :for all but finitely many indices $\alpha : V_\alpha = X_\alpha$ is a basis for $\tau$. Therefore, there exists $V = \displaystyle \prod_{\alpha \mathop \in I} V_\alpha \in \BB'$: :$x \in V \subseteq W$ Let $J = \set{\alpha \in I : V_\alpha \ne S_\alpha}$. By definition of $\BB'$, $J$ is finite. Now for all $\alpha \in J$, $x_\alpha \in V_\alpha$ and $V_\alpha \in \tau_\alpha$. Since $\BB_\alpha$ is a basis for $\tau_\alpha$ then for all $\alpha \in J$ there exists $N_\alpha \in \BB_\alpha$: :$x_\alpha \in N_\alpha \subseteq V_\alpha$ For all $\alpha \in I$, let: :$U_\alpha = \begin {cases} N_\alpha & : \alpha \in J \\ S_\alpha & : \alpha \notin J \end {cases}$ Let $U = \displaystyle \prod_{\alpha \mathop \in I} U_\alpha$. For $\alpha \in J$: :$U_\alpha = N_\alpha \subseteq V_\alpha$ :$x_\alpha \in N_\alpha = U_\alpha$ :$U_\alpha = N_\alpha \in \BB_\alpha$ For $\alpha \notin J$: :$U_\alpha = S_\alpha = V_\alpha$ :$x_\alpha \in S_\alpha = U_\alpha$ Thus: :$x \in U$ :$U \subseteq V \subseteq W$ By the definition of $\BB$ and because $J$ is finite: :$U \in \BB$. The result follows. {{qed}} Category:Topological Bases Category:Product Spaces 0dogy2ti2wzkj349waodn19f4a2e0h1
From Domain of Relation is Subset of Union of Union of Relation: :$\Dom \RR \subseteq \map \bigcup {\bigcup \RR}$ From Image of Relation is Subset of Union of Union of Relation: :$\Img \RR \subseteq \map \bigcup {\bigcup \RR}$ We are given that $\RR$ is a set. From the Axiom of Unions: :$\bigcup \RR$ is a set. Applying the Axiom of Unions a second time: :$\map \bigcup {\bigcup \RR}$ is a set. We have that $\RR \subseteq V$, by definition of basic universe From the Axiom of Swelledness, it follows that every subclass of $V$ is a set. The result follows. {{qed}}
=== Necessary Condition === Let $S$ be transitive. Let $s \in S$. By definition of transitive set: :$s \subseteq S$ Then by definition of power set: :$s \in \powerset S$ Hence, by definition of subset: :$S \subseteq \powerset S$ {{qed|lemma}} === Sufficient Condition === Let $S \subseteq \powerset S$. Let $s \in S$. Then by definition of subset: :$s \in \powerset S$ By definition of power set: :$s \subseteq S$ As this is true for all $s \in S$, it follows by definition that $S$ is transitive. {{qed}}
Let $S$ be transitive. By Set is Transitive iff Subset of Power Set: :$S \subseteq \powerset S$ Then by Power Set of Subset: :$\powerset S \subseteq \powerset {\powerset S}$ Thus by Set is Transitive iff Subset of Power Set: :$\powerset S$ is a transitive set. {{qed}}
By definition, $V$ is the class of all sets. As $a \in V$, by definition of $V$ it follows that $a$ is a set. Consider the power set $\powerset a$ of $a$. From the axiom of powers: :$\powerset a$ is also a set and: :$\powerset {\powerset a}$ is also a set. By definition: :$a \in \powerset a$ and so: :$\set a \in \powerset {\powerset a}$ as $\powerset {\powerset a}$ is a set, it follows that: :$\powerset {\powerset a} \in V$ But we have by definition of class difference that $\set a \notin V$. Hence we have an element of an element of $V$ which is not itself in $V$. Hence, by definition, $V \setminus \set a$ is not a transitive class. {{qed}}
By definition of basic universe, $V$ is a class containing all sets as elements. By the axiom of the empty set: :the empty class $\O$ is a set. Hence $\O$ is an element of $V$. By the Axiom of Powers, if $x$ is a set, then $\powerset x$ is a set. By definition of power set: :$\set x \subseteq \powerset x$ By the Axiom of Swelledness, $\set x$ is a set. By the Axiom of Unions, if $x$ is a set, then $\bigcup x$ is a set. Thus $x \bigcup \set x$ is a set. The result follows. {{qed}}
Let the axiom of infinity, in each of its formulations, be applied to a basic universe $V$ separately, as follows. === Formulation $1$ implies Formulation $2$ === Let formulation 1 be taken as an axiom. Then the class $\omega$ as so defined is a set. But, by definition, $\omega$ is itself an inductive set. Hence an inductive set exists. Thus formulation 2 holds. {{qed|lemma}} === Formulation $2$ implies Formulation $1$ === Let formulation 2 be taken as an axiom. That is, there exists an inductive set, which we shall call $b$. By the inductive set definition of the natural numbers: :$\forall n \in \omega: n \in b$ where $\omega$ denotes the set of natural numbers. That is: :$\omega \subseteq b$ We have that: :$b$ is a set and :$\omega$ is a subclass of $b$. Because the axiom of infinity be applied to a basic universe $V$. By the Axiom of Swelledness, $\omega$ is a set. Thus formulation 1 holds. {{qed|lemma}} === Formulation $1$ implies Formulation $3$ === Let formulation 1 be taken as an axiom. In the basic universe $V$, we are able to create the set: :$S = \set {\O, \set \O}$ which is the natural number $2$. By the Axiom of Powers, its power set is also a set: :$\powerset S = \set {\O, \set \O, \set {\set \O}, \set {\set \O, \set \O} }$ But $\powerset S$ is not a natural number. Thus formulation 3 holds. {{qed|lemma}} === Formulation $3$ implies Formulation $2$ === Let formulation 3 be taken as an axiom. {{AimForCont}} formulation 2 did not hold. Then there would be no inductive sets. But then vacuously every set is an element of all inductive sets. Thus, by the inductive set definition of the natural numbers: :Every set is a natural number. {{qed|lemma}} Thus it is seen that each of these formulations of the axiom of infinity directly or indirectly implies each of the other two. Hence the result. {{qed}}
By definition of inductive set: :$\O \in P$ By definition of the natural numbers, $\O$ is identified with $0$ (zero). Thus {{PeanoAxiom|1}} holds. {{qed|lemma}} Let $x$ be a natural number. By definition, $x$ is an element of every inductive set. Thus if $x \in P$ it follows that $x^+ \in P$. Thus {{PeanoAxiom|2}} holds. {{qed|lemma}} By Inductive Construction of Natural Numbers fulfils Peano's Axiom of Injectivity, {{PeanoAxiom|3}} holds. {{qed|lemma}} For all $n$, $n$ is an element of $n^+$. But $0$ is identified with the empty set $\O$. By definition, $\O$ has no elements. Therefore it is not possible for $\O$ to equal $n^+$ for any $n$. Thus {{PeanoAxiom|4}} holds. {{qed|lemma}} {{PeanoAxiom|5}} holds immediately by definition of inductive set. {{qed|lemma}}
The proof proceeds by induction. For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition: :$n$ is a transitive set. === Basis for the Induction === From Empty Class is Supercomplete, we have that $\O$ is supercomplete. Thus by definition of supercomplete, $\O$ is a transitive class. By the axiom of the empty set, $\O$ is a set. Hence $\O$ is a transitive set. 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 $k \ge 0$, then it logically follows that $\map P {k + 1}$ is true. So this is the induction hypothesis: :$k$ is a transitive set. from which it is to be shown that: :$k^+$ is a transitive set. === Induction Step === This is the induction step: {{begin-eqn}} {{eqn | l = x       | o = \in       | r = k^+       | c =  }} {{eqn | ll= \leadsto       | l = x       | o = \in       | r = k \cup \set k       | c =  }} {{eqn | ll= \leadsto       | l = x       | r = k       | c =  }} {{eqn | lo= \lor       | l = x       | o = \in       | r = k       | c = {{Defof|Transitive Set}}: from Induction Hypothesis }} {{end-eqn}} If $x \in k$ then, because $k$ is transitive, $x \subseteq k$. If $x = k$, then, because $k^+ = k \cup \set k$, it follows that $x \subseteq k$ by definition of union of class. In either case: :$x \subseteq k$ But we have $k \subseteq k^+$ by definition of union of class. Hence: :$x \subseteq k^+$ So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\forall n \in \N: n$ is a transitive set. {{qed}}
The proof proceeds by induction. For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition: :$n$ is an ordinary set. === Basis for the Induction === By the axiom of the empty set, $\O$ is a set. By definition, $\O$ has no elements. Hence: :$\O \notin \O$ Thus by definition $\O$ is an ordinary set. 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 $k \ge 0$, then it logically follows that $\map P {k + 1}$ is true. So this is the induction hypothesis: :$k$ is an ordinary set. from which it is to be shown that: :$k^+$ is an ordinary set. === Induction Step === This is the induction step: By the Induction Hypothesis: :$k \notin k$ {{AimForCont}} $k^+ \in k^+$. {{begin-eqn}} {{eqn | l = k^+       | r = k       | c =  }} {{eqn | lo= \lor       | l = k^+       | o = \in       | r = k       | c =  }} {{end-eqn}} If $k^+ \in k$ then, because $k$ is transitive, $k^+ \subseteq k$. If $k^+ = k$, then, because $k^+ = k \cup \set k$, it follows that $k^+ \subseteq k$ by definition of union of class. In either case: :$k^+ \subseteq k$ But we have: :$k \in k^+$ and because $k^+ \subseteq k$ it follows that: :$$ \in $$ But this contradicts our supposition that $k \notin k$. Hence by Proof by Counterexample: :$k^+ \notin k^+$ So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\forall n \in \N: n$ is an ordinary set. {{qed}}
Let $W \in \tau$ such that $x \in W$. From Natural Basis of Tychonoff Topology, the set $\BB$ of cartesian products of the form $\displaystyle \prod_{\alpha \mathop \in I} V_\alpha$ where: :for all $\alpha \in I : V_\alpha \in \tau_\alpha$ :for all but finitely many indices $\alpha : V_\alpha = X_\alpha$ is a basis for $\tau$. Therefore, there exists $V = \displaystyle \prod_{\alpha \mathop \in I} V_\alpha \in \BB$: :$x \in V \subseteq W$ Let $J = \set {\alpha \in I : V_\alpha \ne S_\alpha}$. By definition of $\BB$, $J$ is finite. Now for all $\alpha \in J$, $x_\alpha \in V_\alpha$ and $V_\alpha \in \tau_\alpha$. Since $\BB_\alpha$ is a local basis for $\tau_\alpha$ then for all $\alpha \in J$ there exists $N_\alpha \in \BB_\alpha$: :$x_\alpha \in N_\alpha \subseteq V_\alpha$ For all $\alpha \in I$, let: :$U_\alpha = \begin {cases} N_\alpha & : \alpha \in J \\ S_\alpha & : \alpha \notin J \end {cases}$ Let $U = \displaystyle \prod_{\alpha \mathop \in I} U_\alpha$. For $\alpha \in J$: :$U_\alpha = N_\alpha \subseteq V_\alpha$ :$x_\alpha \in N_\alpha = U_\alpha$ :$U_\alpha = N_\alpha \in \BB_\alpha$ For $\alpha \notin J$: :$U_\alpha = S_\alpha = V_\alpha$ :$x_\alpha \in S_\alpha = U_\alpha$ Thus: :$x \in U$ :$U \subseteq V \subseteq W$ By the definition of $\BB_x$ and since $J$ is finite: :$U \in \BB_x$. The result follows. {{qed}} Category:Topological Bases Category:Product Spaces Category:Local Bases pyxtzu4rg5pqw6d6sc2nkd8s9v1qzr2
{{AimForCont}} both $m \in n$ and $n \in m$. We have $m \in n$ From Natural Number is Transitive Set: :$m \subseteq n$ by definition of transitive. Thus: :$n \in m \subseteq n$ and so: :$n \in n$ But from Natural Number is Ordinary Set: :$n \notin n$ The result follows by Proof by Contradiction. {{qed}}