hcju/proofwikiqa huggingface dataset:
From Cartesian Product of Family of Subsets, $Y \subseteq X$. Thus the topological subspace $T_Y$ is well-defined. From Natural Basis of Tychonoff Topology, a (synthetic) basis for $T$ is: :$\displaystyle \BB_T = \set {\prod_{i \mathop \in I} U_i : U_i \in \tau_i, U_i = X_i \text{ for all but finitely many } i \in I}$ From Basis for Topological Subspace a (synthetic) basis for $T_Y$ is: :$\BB_Y = \set {U \cap Y: U \in \BB_T}$ That is: {{begin-eqn}} {{eqn | l = \BB_Y       | r = \set {\paren{ \prod_{i \mathop \in I} U_i} \bigcap \paren{\prod_{i \mathop \in I} Y_i}: U_i \in \tau_i, U_i = X_i \text{ for all but finitely many } i \in I} }} {{eqn | r = \set {\prod_{i \mathop \in I} \paren {U_i \cap Y_i} : U_i \in \tau_i, U_i = X_i \text{ for all but finitely many } i \in I}       | c = General Case of Cartesian Product of Intersections }} {{end-eqn}} By definition of a subspace topology, :$\forall i \in I: \upsilon_i = \set{U \cap S_i : U \in \tau_i}$ From Natural Basis of Tychonoff Topology a (synthetic) basis for $S$ is: :$\displaystyle \BB_S = \set {\prod_{i \mathop \in I} V_i : V_i \in \upsilon_i, V_i = Y_i \text{ for all but finitely many } i \in I}$ That is: {{begin-eqn}} {{eqn | l = \BB_S       | r = \set {\prod_{i \mathop \in I} \paren{U_i \cap Y_i} : U_i \in \tau_i, U_i \cap Y_i = Y_i \text{ for all but finitely many } i \in I} }} {{eqn | r = \set {\prod_{i \mathop \in I} \paren{U_i \cap Y_i} : U_i \in \tau_i, U_i = X_i \text{ for all but finitely many } i \in I}       | c = as $Y_i \subseteq X_i$ for all $i \in I$ }} {{end-eqn}} Thus $\BB_S = \BB_Y$ and therefore $S = T_Y$. {{qed}} Category:Cartesian Product Category:Subsets pjd3vizfi2u53n3jmkppkkb6bvhmgqb
Let $w = -z_2$. Then: {{begin-eqn}} {{eqn | l = \overline {z_1 - z_2}       | r = \overline {z_1 + \paren {-z_2} }       | c = {{Defof|Complex Subtraction}} }} {{eqn | r = \overline {z_1 + w}       | c = Definition of $w$ }} {{eqn | r = \overline {z_1} + \overline w       | c = Sum of Complex Conjugates }} {{eqn | r = \overline {z_1} + \overline {-z_2}       | c = Definition of $w$ }} {{eqn | r = \overline {z_1} - \overline {z_2}       | c = {{Defof|Complex Subtraction}} }} {{end-eqn}} {{qed}}
The proof will proceed by the Principle of Finite Induction on $\N \setminus \set 0$ === Basis for the Induction === :$\map s 0 = 1$ So 1 is the successor of 0. This is our basis for the induction === Induction Hypothesis === This is our induction hypothesis: :$\exists m \in \N: \map s m = k$ Then we need to show: :$\exists m' \in \N: \map s m' = k+1$ === Induction Step === This is our induction step: {{begin-eqn}} {{eqn | o =        | r = \exists m \in \N: \map s m = k       | c =  }} {{eqn | o = \implies       | r = k \in \N       | c = {{PeanoAxiom|2}} }} {{eqn | o = \implies       | r = \map s k = k+1       | c =  }} {{eqn | o = \implies       | r = \exists m' \in \N: \map s m' = k+1 }} {{end-eqn}} {{qed}} Category:Proofs by Induction Category:Natural Numbers irytocb7177a83bgadx53xcg59gkyxy
Let $\mathbf A = \sqbrk a_n$ be a square matrix of order $n$. Let $\map \det {\mathbf A}$ be the determinant of $\mathbf A$. Let $1 \le r < s \le n$. Let $\mathbf B$ be $\mathbf A$ with columns $r$ and $s$ transposed. Consider: :the transpose $\mathbf A^\intercal$ of $\mathbf A$ :the transpose $\mathbf B^\intercal$ of $\mathbf B$. hence $\mathbf B^\intercal$ is $\mathbf A^\intercal$ with rows $r$ and $s$ transposed. From Determinant with Rows Transposed: :$\map \det {\mathbf B^\intercal} = -\map \det {\mathbf A^\intercal}$ From from Determinant of Transpose: :$\map \det {\mathbf B^\intercal} = \map \det {\mathbf B}$ :$\map \det {\mathbf A^\intercal} = \map \det {\mathbf A}$ and the result follows. {{qed}}
For all $y \in Y_i$: {{begin-eqn}} {{eqn | l = \map {\pr_i {\restriction_{Y_i} } } y       | r = \map {\pr_i} y       | c = {{Defof|Restriction of Mapping}}: $\pr_i {\restriction_{Y_i} } : Y_i \to X_i$ }} {{eqn | r = y_i       | c = {{Defof|Projection}}: $\pr_i: X \to X_i$ }} {{eqn | r = \map {p_i} y       | c = {{Defof|Projection}}: $p_i: Y_i \to X_i$ }} {{end-eqn}} By Equality of Mappings: :$\pr_i {\restriction_{Y_i} } = p_i$ {{qed}} Category:Subspace of Product Space Homeomorphic to Factor Space 8p5jidq5acmd8cke22y88j5prb6mnl9
{{begin-eqn}} {{eqn | r = x \in Y_i       | o = }} {{eqn | ll = \leadstoandfrom       | r = \forall j \in I : x_j = \begin{cases} z_j & j \ne i \\ x_i \in X_i & i = j \end{cases}       | o =       | c = Definition of $Y_i$ }} {{eqn | ll = \leadstoandfrom       | r = \forall j \in I : x_j \in Z_j       | o =       | c = Definition of $Z_j$ for all $j \in I$ }} {{eqn | ll = \leadstoandfrom       | r = x \in \prod_{j \mathop \in I} Z_j       | o =       | c = {{Defof|Cartesian Product}} }} {{end-eqn}} The result follows by definition of set equality. {{qed}} Category:Subspace of Product Space Homeomorphic to Factor Space igffiro4fexfs4m4dwkepn6z6uoobie
Let the hypothesis be assumed. Let $A$ be an arbitrary non-empty subclass of $N$. Let $L$ be the class of all elements $y$ of $N$ such that $y$ is a proper subset of all elements of $A$. Let $x$ be the greatest element of $L$. It is to be shown that $\map g x$ is the smallest element of $A$. We have that $x$ is a proper subset of all elements of $A$. Then by Corollary 1 to the Sandwich Principle: :$\map g x$ is a subset of all elements of $A$. It remains to be shown that $\map g x$ is itself an element of $A$. We have that $x$ is a proper subset of all elements of $A$. We also have that $x$ contains at least one element. Therefore $x$ cannot be the greatest element of $N$. Therefore, by hypothesis: :$x \ne \map g x$ Thus $x$ is a proper subset of $\map g x$. Because $x$ is the greatest element of $L$, $\map g x$ is not therefore an element of $L$. Therefore there exists at least $1$ element $y$ of $A$ such that $\map g x$ is not a proper subset of $y$. But we have shown that $\map g x \subseteq y$. Hence: :$\map g x = y$ and so: :$\map g x \in A$ Hence the result. {{qed}}
Let $\RR$ be the relation on $N$ defined as: :$\forall x, y \in N: \map \RR {x, y} \iff \map g x \subseteq y \lor y \subseteq x$ 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}} In what follows, $(1)$ is not needed. From the Double Induction Principle for Minimally Closed Class, if $\RR$ is a relation on $N$ which satisfies: {{begin-axiom}} {{axiom | n = \text D_1         | q = \forall x \in N         | m = \map \RR {x, b} }} {{axiom | n = \text D_2         | q = \forall x, y \in N         | 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 $x, y \in N$. As $\Dom g = N$ it is seen that $(\text D 2)$ is fulfilled by $(2)$ directly. It remains to be demonstrated that $(\text D 1)$ holds. Indeed, from Smallest Element of Minimally Closed Class under Progressing Mapping, we have that: :$\forall x \in N: b \subseteq x$ By the Rule of Addition: :$\forall x \in N: \map g x \subseteq b \lor b \subseteq x$ That is: :$\forall x \in N: \map \RR {x, b}$ Thus by the Double Induction Principle for Minimally Closed Class: :$\forall x, y \in N: \map \RR {x, y}$ That is: :$\forall x, y \in N: \map g x \subseteq y \lor y \subseteq x$ {{qed|lemma}} As $g$ is a progressing mapping, we have: :$x \subseteq \map g x$ Hence from: :$\map g x \subseteq y \lor y \subseteq x$ we have: :$x \subseteq y \lor y \subseteq x$ Hence the result by definition of a nest. {{qed}}
According to hypothesis, let $M$ be a minimally closed under $g$ with respect to $b$. By Minimally Closed Class under Progressing Mapping induces Nest: :$\forall x, y \in M: \map g x \subseteq y \lor y \subseteq x$ By Fixed Point of Progressing Mapping on Minimally Closed Class is Greatest Element: :if $x$ is a fixed point of $g$, then $x$ is the greatest element of $M$. Thus the conditions for Closed Class under Progressing Mapping Lemma hold. In order to show that $M$ is well-ordered under the inclusion relation, it is to be shown that every non-empty subclass $A$ of $M$ has a smallest element. Let $A$ be an arbitrary non-empty subclass of $N$. Let $L$ be the class of all elements $y$ of $N$ such that $y$ is a proper subset of all elements of $A$. If $b \in A$, then from Smallest Element of Minimally Closed Class under Progressing Mapping, $b$ is the smallest element of $A$. So, consider the case where $b \notin A$. Then $b$ is a proper subset of every element of $A$. Thus $b \in L$ and so $L$ is non-empty. By definition, $L$ is bounded by all elements of $A$. Since $A$ is non-empty, $L$ is bounded by at least one element of $A$. Hence by Non-Empty Bounded Subset of Minimally Inductive Class under Progressing Mapping has Greatest Element: :$L$ has a greatest element $x$. Therefore, by Closed Class under Progressing Mapping Lemma, $\map g x$ is the smallest element of $A$. {{qed}}
{{AimForCont}} $b$ is not the smallest element of $N$. Then there exists $m \in N$ such that $b \nsubseteq m$. In particular: :$m \ne b$ Let $B$ be the subclass of $A$ defined as: :$B = \set {x \in A: \paren {x = b} \lor \paren {\exists y \in B: x = \map g y} }$ This is a subclass of $A$ containing $b$ which is closed under $g$. Let $z \in B$. We have that: :$b \nsubseteq m$ Suppose that: :$z \nsubseteq m$ As $g$ is a Progressing Mapping, we have that: :$z \subseteq \map g z$ It follows that: :$\map g z \nsubseteq m$ By the Principle of General Induction for Minimally Closed Class it follows that: :$\forall z \in B: m \nsubseteq \map g z$ and in particular: :$\forall z \in B: m \ne \map g z$ In summary: :$m \ne b$ and: :$\not \exists z \in B: m = \map g z$ It follows that: :$m \notin B$ Thus we have created a proper subclass $B$ of $A$ containing $b$ which is closed under $g$. This contradicts the assumption that $A$ is a minimally closed class under $g$ with respect to $b$ Hence, by Proof by Contradiction, our supposition that $b$ is not the smallest element of $N$ was false. The result follows. {{qed}}
Let the hypothesis be assumed. The proof proceeds by general induction. For all $x \in N$, let $\map P x$ be the proposition: :Every subset of $N$ which is bounded by $x$ has a greatest element. === Basis for the Induction === Let $x = b$. From Smallest Element of Minimally Closed Class under Progressing Mapping, $b$ is the smallest element of $N$. That is: :$\forall x \in N: b \subseteq x$ So the only subset $B$ of $N$ such that $B$ is bounded by $b$ is $B = \set b$. By definition, $\set b$ trivially has a greatest element, which is $b$. Thus $\map P b$ 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 subset of $N$ which is bounded by $x$ has a greatest element. from which it is to be shown that: :Every subset of $N$ 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 subset of $N$ 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 Closed 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 $x \in N$, every subset of $N$ which is bounded by $x$ has a greatest element. That is: :Every bounded subset of $N$ has a greatest element. {{qed}}
Suppose $g$ has a fixed point. Let $x$ be an element of $M$ such that $\map g x = x$. We have from Smallest Element of Minimally Closed Class under Progressing Mapping that: :$b \subseteq x$ Suppose that $y \subseteq x$. Then by Image of Proper Subset under Progressing Mapping on Minimally Closed Class: :$\map g y \subseteq \map g x$ But we have that $\map g x = x$. Thus: :$\map g y \subseteq x$ So we have: :$b \subseteq x$ and: :$y \subseteq x \implies \map g y \subseteq x$ and the result follows by the Principle of General Induction for Minimally Closed Class. {{qed}}
Let it be given that $A$ is closed under $g$. === $(1)$ implies $(2)$ === Let $A$ be a minimally closed class under $g$ by definition 1. Then by definition: :there exists $b \in A$ such that $A$ has no proper subclass $B$ such that: :$b \in B$ :$B$ is closed under $g$. Let $A$ have a subclass $C$ which is closed under $g$ such that $b \in C$. 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 closed class under $g$ by definition 2. {{qed|lemma}} === $(2)$ implies $(1)$ === Let $A$ be a minimally closed class under $g$ by definition 2. Then by definition: :Every subclass of $A$ containing $b$ which is closed under $g$ contains all the elements of $A$. Let $C$ be a subclass of $A$ which is closed under $g$ such that $b \in C$. 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 closed class under $g$ by definition 1. {{qed}} Category:Minimally Closed Classes r383ti5xco1rd5dzza256x3v20kvqo3
We are given that $M$ is minimally closed under $g$ with respect to $b$. That is, $M$ is closed under $g$ with the extra property that $M$ has no proper class containing $b$ which is also closed under $g$. Let $P$ be a propositional function on $M$ which has the properties specified: :$(1): \quad \map P b = \T$ :$(2): \quad \forall x \in M: \map P x = \T \implies \map P {\map g x} = \T$ Let $S \subseteq M$ be the subclass of $M$ defined as: :$S = \set {x \in M: \map P x}$ We have: {{begin-eqn}} {{eqn | l = \map P b       | r = \T       | c =  }} {{eqn | l = b       | o = \in       | r = M       | c =  }} {{eqn | ll= \leadsto       | l = b       | o = \in       | r = S       | c =  }} {{end-eqn}} By definition, the class $S$ of all elements of $M$ such that $\map P x = \T$ is closed under $g$. But because $M$ is minimally closed under $g$ with respect to $b$, if $b \in S$ then $S$ contains all elements of $M$. That is: :$S = M$ and so: :$\forall x \in M: \map P x = \T$ as we were to show. {{Qed}}
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/Minimally Closed Class/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 === $\map P b$ is the case: From condition $\text D_1$ of the definition of $\RR$, we have immediately that: :$\map \RR {x, b}$ for all $x \in M$. That is, that $b$ is right normal with respect to $\RR$. Thus $\map P b$ 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 for Minimally Closed Class: :$\forall x \in M$: $x$ is right normal with respect to $\RR$. The result follows. {{qed}} Category:Double Induction Principle sgs8i1n37qyxovpj6s2h76yrccn4xtn
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 $\RR$: :$\map \RR {x, b}$ 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}} Category:Double Induction Principle dh3hayz3877232k9wavgxn0bn5rqq4k
From Minimally Closed Class under Progressing Mapping induces Nest, we have that $N$ is a nest in which: :$\forall x, y \in N: \map g x \subseteq y \lor y \subseteq x$ Thus the Sandwich Principle applies directly. {{qed}} Category:Sandwich Principle Category:Minimally Closed Classes Category:Inflationary Mappings 0ivlry033cq7r7sj382ql93ru90jcxk
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 Sandwich Principle applies directly. {{qed}}
By definition of minimally inductive class, $M$ is minimally closed under $g$ with respect to $\O$. The result is then seen to be a direct application of Sandwich Principle for Minimally Closed Class. {{qed}}
From Minimally Closed Class under Progressing Mapping induces Nest, we have that $N$ is a nest in which: :$\forall x, y \in N: \map g x \subseteq y \lor y \subseteq x$ Thus the corollary 1 of the Sandwich Principle applies directly. {{qed}} Category:Minimally Closed Classes under Progressing Mapping 22xzgeswg06469but0wp4i9whdq75qp
{{begin-eqn}} {{eqn | l = \map g t       | r = \map f {-t}       | c = Repeated Fourier Transform of Real Function }} {{eqn | r = \map f t       | c = {{Defof|Even Function}} }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = \map g t       | r = \map f {-t}       | c = Repeated Fourier Transform of Real Function }} {{eqn | r = -\map f t       | c = {{Defof|Odd Function}} }} {{end-eqn}} {{qed}}
At a point of continuity we have: {{begin-eqn}} {{eqn | l = \dfrac {\map f {t^+} + \map f {t^-} } 2       | r = \dfrac {\map f t + \map f t} 2       | c = as $\map f t = \map f {t^+} = \map f {t^-}$ at a point of continuity }} {{eqn | r = \dfrac {2 \map f t} 2       | c =  }} {{eqn | r = \map f t       | c =  }} {{end-eqn}} The result follows from Fourier's Theorem: Integral Form. {{qed}} {{Namedfor|Joseph Fourier}}
We have that $p_i$ is a bijection from the lemmas: :$p_i$ is an injection  :$p_i$ is a surjection  Let $x \in X_i$. Then: {{begin-eqn}} {{eqn | r = x \in \map {p_i^\to} {\map{\pr_k^\gets } {V_k} \cap Y_i}       | o = }} {{eqn | ll = \leadstoandfrom       | o =       | r = \map {p_i^{-1} } x \in \map{\pr_k^\gets } {V_k} \cap Y_i        | c = {{Defof|Direct Image Mapping of Mapping}} }} {{eqn | ll = \leadstoandfrom       | o =       | r = \map {p_i^{-1} } x \in \map{\pr_k^\gets } {V_k}        | c = As $\map {p_i^{-1} } x \in Y_i$  }} {{eqn | ll = \leadstoandfrom       | o =       | r = \map {\pr_k} {\map {p_i^{-1} } x} \in V_k       | c = {{Defof|Direct Image Mapping of Mapping}} }} {{end-eqn}} By definition of $p_i$: :$\map {p_i^{-1} } x = y$ where: :$\forall j \in I : y_j = \begin{cases} z_j & j \ne i \\ x & j = i \end{cases}$ ==== Case 1: $k = i$ ==== Let $k = i$. Then: {{begin-eqn}} {{eqn | l = \map {\pr_i} {\map {p_i^{-1} } x}       | r = \map {\pr_i} {y} }} {{eqn | r = y_i }} {{eqn | r = x }} {{end-eqn}} So: :$x \in \map {p_i} {\map{\pr_i^\gets } {V_i} \cap Y_i}$ {{iff}} $x \in V_i$ That is: :$\map {p_i} {\map{\pr_i^\gets } {V_i} \cap Y_i} = V_i \in \tau_i$ {{qed|lemma}} ==== Case 2: $k \neq i$ ==== Let $k \ne i$. Then: {{begin-eqn}} {{eqn | l = \map {\pr_k} {\map {p_i^{-1} } x}       | r = \map {\pr_k} {y} }} {{eqn | r = y_k }} {{eqn | r = z_k }} {{end-eqn}} So: :$x \in \map {p_i} {\map{\pr_i^\gets } {V_i} \cap Y_i}$ {{iff}} $z_k \in V_k$ Since $x \in X_i$ was arbitrary, then: :$\map {p_i} {\map{\pr_i^\gets } {V_i} \cap Y_i} = X_i$ {{iff}} $z_k \in V_k$ Let $w \in \map{\pr_k^\gets } {V_k} \cap Y_i$ which is guaranteed since $\map{\pr_k^\gets } {V_k} \cap Y_i \ne \O$ By the definition of inverse image mapping then: :$w_k = \map {\pr_k} w \in V_k$ By the definition of $Y_i$ then: :$w_k = z_k$ So $z_k \in V_k$. It follows that: :$\map {p_i} {\map{\pr_i^\gets } {V_i} \cap Y_i} = X_i \in \tau_i$ {{qed|lemma}} In either case: :$\map {p_i^\to} {\map{\pr_k^\gets } {V_k} \cap Y_i}$ is open in $\struct {X_i, \tau_i}$. {{qed}} Category:Subspace of Product Space Homeomorphic to Factor Space drrf0dqxcvm94uq6aki3qlgcsipf2a1
Let $x, y \in Y_i$. Then for all $j \in I \setminus \set i$: :$x_j = z_j = y_j$ Let $\map {p_i} x = \map {p_i} y$. Then: :$x_i = y_i$ Thus: :$x = y$ It follows that $p_i$ is an injection by definition.
Let $x \in X_i$. Let $y \in Y_i$ be defined by: :$\forall j \in I: y_j = \begin{cases} z_j & j \ne i \\ x & j = i \end{cases}$ Then: :$\map {p_i} y = y_i = x$ It follows that $p_i$ is an surjection by definition.
Let $V \in \tau_i$. Let $\displaystyle U = \prod_{i \mathop \in I} U_i$ where: :$U_j = \begin{cases} X_j & j \ne i \\ V & j = i \end{cases}$ From Natural Basis of Tychonoff Topology, $U$ is an element of the the natural basis. By definition of the Tychonoff topology $\tau$ on the product space $\struct {X, \tau}$ the natural basis is a basis for the Tychonoff topology. It follows that: :$U$ is open in $\struct {X, \tau}$ Let $x \in Y_i$. Now: {{begin-eqn}} {{eqn | r = x \in \map {p_i^\gets} V       | o = }} {{eqn | ll = \leadstoandfrom       | o =       | r = \map {p_i} x \in V       | c = {{Defof|Inverse Image Mapping of Mapping}}: $p_i^\gets$ }} {{eqn | ll = \leadstoandfrom       | o =       | r = \map {\pr_i} x \in V       | c = {{Defof|Restriction of Mapping}} $p_i$ }} {{eqn | ll = \leadstoandfrom       | o =       | r = x \in U       | c = Definition of $U$ }} {{eqn | ll = \leadstoandfrom       | o =       | r = x \in U \cap Y_i       | c = As $x \in Y_i$ }} {{end-eqn}} By set equality: :$ \map {p_i^\gets} V = U \cap Y_i$ By definition of the subspace topology on $Y_i$: :$\map {p_i^\gets} V \in \upsilon_i$ It follows that $p_i$ is continuous by definition.