hcju/proofwikiqa huggingface dataset:
By definition, we have that: :$\alpha \beta := \begin {cases} \size \alpha \, \size \beta & : \alpha \ge 0^*, \beta \ge 0^* \\ -\paren {\size \alpha \, \size \beta} & : \alpha < 0^*, \beta \ge 0^* \\ -\paren {\size \alpha \, \size \beta} & : \alpha \ge 0^*, \beta < 0^* \\ \size \alpha \, \size \beta & : \alpha < 0^*, \beta < 0^* \end {cases}$ where: :$\size \alpha$ denotes the absolute value of $\alpha$ :$\size \alpha \, \size \beta$ is defined as in Multiplication of Positive Cuts :$0^*$ denotes the rational cut associated with the (rational) number $0$. :$\ge$ denotes the ordering on cuts. We have that: :$0 < \dfrac 1 2 < 1$ and so: :$\dfrac 1 2 \notin 0^*$ but: :$\dfrac 1 2 \in 1^*$ Thus by definition of the strict ordering of cuts: :$1^* > 0^*$ Thus by definition of absolute value of $\alpha$: :$\size {1^*} = 1^*$ Thus: :$\alpha 1^* := \begin {cases} \size \alpha \, 1^* & : \alpha \ge 0^* \\ -\paren {\size \alpha \, 1^*} & : \alpha < 0^* \end {cases}$ Let $\alpha > 0^*$. Let $r \in \alpha 1^*$. Then by definition of Multiplication of Positive Cuts, either: :$r < 0$ or :$\exists p \in \beta, q \in 1^*: r = p q$ where $p \ge 0$ and $q \ge 0$. Because $0 \le q < 1$ it follows that $p q < p$. Thus if $r > 0$ it follows that $r < p$. That is: :$r \in \alpha$ If $r < 0$ then $r \in \alpha$ because $0^* < \alpha$. Thus in either case $r \in \alpha$. That is: :$\alpha 1^* \le \alpha$ {{finish|tedious}}
From Sum of Cuts is Cut, $p^* + q^*$ is a cut. Let $r \in p^* + q^*$. Then: :$r = s + t$ where $s < p$ and $t < q$ Thus: :$r < p + q$ and so: :$r \in \paren {p + q}^*$ Hence: :$p^* + q^* \subseteq \paren {p + q}^*$ {{qed|lemma}} Let $r \in \paren {p + q}^*$. Then: :$r < p + q$ Let: :$h = p + q - r$ :$s = p - \dfrac h 2$ :$t = q - \dfrac h 2$ Then: :$s \in p^*$ :$t \in q^*$ Hence: :$r = s + t$ and so: :$r \in p^* + q^*$ So: :$\paren {p + q}^* \subseteq p^* + q^*$ {{qed|lemma}} Hence the result by definition of set equality. {{Qed}}
From Product of Cuts is Cut, $p^* q^*$ is a cut. {{ProofWanted}}
Let $p < q$. Then $p \notin p^*$ but $q \in q^*$. Thus $p^* < q^*$ by definition of the strict ordering on cuts . Let $p^* < q^*$. Then: :$\exists r \in \Q: r \notin p^*, r \in q^*$ Hence: :$p \le r < q$ and so: :$p < q$ {{qed}}
Let $\alpha < \beta$. Then there exists a rational number $p$ such that $p \in \beta$ and $p \notin \alpha$. Let $r \in \Q$ such that $r > p$ and $r \in \beta$. Because $r \in \beta$ and $r \notin r^*$, we have that $r^* < \beta$. Because $p \in r^*$ and $p \notin \alpha$, we have that $\alpha < r^*$. {{qed}}
Let $p$ be a rational number such that $p \in \alpha$. Then by definition of rational cut: :$p \notin p^*$ Thus: :$p \in \alpha \implies p^* < \alpha$ {{qed|lemma}} Let $p^* < \alpha$. Then there exists a rational number $q$ such that $q \in \alpha$ and $q \notin p$. Thus $q \ge p$. But as $q \in \alpha$ it follows that $p \in \alpha$. {{Qed}}
Suppose $f$ is continuous. From Projection from Product Topology is Continuous $\pr_i$ is continuous for each $i \in I$. By Continuity of Composite Mapping it follows that $\pr_i \circ f$ is continuous for each $i \in I$. Conversely, suppose that each $\pr_i \circ f$ is continuous.  Let $U = \pr_{i_1}^{-1} \sqbrk {U_{i_1} } \cap \cdots \cap \pr_{i_n}^{-1} \sqbrk {U_{i_n} }$ be an open set in the natural basis of the product topology for $Y$. Note that by the definition of natural basis, $U$ is '''constrained''' to be the intersection of a '''finite number''' of sets of $\mathbb S$. Then by Preimage of Intersection under Mapping we find: :$f^{-1} \sqbrk U = f^{-1} \sqbrk {\pr_{i_1}^{-1} \sqbrk {U_{i_1} } } \cap \cdots \cap f^{-1} \sqbrk {\pr_{i_n}^{-1} \sqbrk {U_{i_n} } }$ Since each $\pr_{i_j} \circ f$ is continuous, each: :$\paren {\pr_{i_j} \circ f}^{-1} \sqbrk {U_{i_j} } = f^{-1} \sqbrk {\pr_{i_j}^{-1} \sqbrk {U_{i_j} } }$  is open. Therefore $f^{-1} \sqbrk U$ is open. By the definition of basis, each open set of $Y$ can be written as a union of sets in the natural basis of the product topology for $Y$. From Preimage of Union under Mapping, the union of the preimages of open sets of $Y$ equals the preimages of the union of those sets. Thus the preimage of an open set of $Y$ is open in $X$. Hence $f$ is continuous. {{qed}} Category:Product Spaces Category:Continuous Mappings b16lw2u9dopx269azdonuy2zqap4pdn
First we show that $\struct {\CC, +, \times}$ is a field by demomstrating that it fulfils the field axioms: {{:Definition:Field Axioms}} It has been established from Set of Cuts under Addition forms Abelian Group that $\struct {\CC, +}$ forms an abelian group. Thus $\text A 0$ through to $\text A 4$ are fulfilled. It remains to verify that $\struct {\CC, +, \times}$ fulfils the remaining field axioms. === $\text M 0$: Closure === From Product of Cuts is Cut: :$\forall \alpha, \beta \in \CC: \alpha \beta \in \CC$ Thus $\struct {\CC, \times}$ is closed. {{qed|lemma}} ===  $\text M 1$: Associativity === From Multiplication of Cuts is Associative: :$\paren {\alpha \beta} \gamma = \alpha \paren {\beta \gamma}$ Thus the operation $\times$ of multiplication of cuts is associative on $\struct {\CC, \times}$. {{qed|lemma}} === $\text M 2$: Commutativity === From Multiplication of Cuts is Commutative: :$\alpha \beta = \beta \alpha$ Thus the operation $\times$ of multiplication of cuts is commutative on $\struct {\CC, \times}$. {{qed|lemma}} === $\text M 3$: Identity === Consider the rational cut $1^*$ associated with the (rational) number $1$: :$1^* = \set {r \in \Q: r < 1}$ From Cut Associated with 1 is Identity for Multiplication of Cuts: :$\alpha \times 1^* = \alpha$ for all $\alpha \in \CC$. From Multiplication of Cuts is Commutative it follows that: :$1^* \times \alpha = \alpha$ That is, $1^*$ is the identity element of $\struct {\CC, \times}$. {{qed|lemma}} === $\text M 4$: Inverses === We have that $1^*$ is the identity element of $\struct {\CC, \times}$. Let $\alpha \ne 0^*$. From Existence of Unique Inverse Element for Multiplication of Cuts, there exists a unique cut $\dfrac 1 \alpha$ such that: :$\alpha \times \dfrac 1 \alpha = 1^*$ From Multiplication of Cuts is Commutative it follows that: :$\dfrac 1 \alpha \times \alpha = 1^*$ Thus every element $\alpha$ of $\struct {\CC, \times}$ such that $\alpha \ne 0^*$ has an inverse $\dfrac 1 \alpha$. {{qed|lemma}} === $\text D$: Distributivity === From Multiplication of Cuts Distributes over Addition: :$\alpha \paren {\beta + \gamma} = \alpha \beta + \alpha \gamma$ for all $\alpha$, $\beta$ and $\gamma$ in $\CC$. {{qed|lemma}} All the field axioms are thus seen to be fulfilled, and so $\struct {\CC, +, \times}$ is a field. {{qed|lemma}} Finally it is shown that $<$ is a total ordering on $\struct {\CC, +, \times}$: This is demonstrated in Ordering on Cuts is Total. From Properties of Ordered Field, this is all that is required to be shown. {{qed}}
We demonstrate that $\struct {\RR, +, \times}$ is a field by showing it is a subfield of the structure $\struct {\CC, +, \times}$, where $\CC$ denotes the set of all cuts. We do this by establishing that all $4$ criteria of the Subfield Test are satisfied. We note that $0^* \in \RR$, where $0^*$ is the rational cut associated with the (rational) number $0$: :$0^* = \set {r \in \Q: r < 0}$ So $\RR \ne \O$. Thus criterion $(1)$ is satisfied. {{qed|lemma}} {{finish}} {{qed}}
From Natural Basis of Tychonoff Topology and Basis for Box Topology, it follows immediately that the natural basis $\BB$ for the Tychonoff topology is contained in the basis $\BB'$ for the box topology. From Corollary of Basis Condition for Coarser Topology, it follows that $\tau \subseteq \tau'$. {{qed}} Category:Tychonoff Topology Category:Box Topology osdpble5zm28utlpmiq9jg14i0c3vv9
Consider the subset $S$ of the real numbers $\R$ defined as: :$S = \set {\dfrac 1 n: n \in \Z_{>0} }$ It is seen that: :$S = \set {1, \dfrac 1 2, \dfrac 1 3, \ldots}$ and hence $\sup S = 1$. Thus $\sup S \in S$. Consider the subset $T$ of the real numbers $\R$ defined as: :$T = \set {-\dfrac 1 n: n \in \Z_{>0} }$ It is seen that: :$T = \set {-1, -\dfrac 1 2, -\dfrac 1 3, \ldots}$ and hence $\sup T = 0$. Thus $\sup T \notin T$. {{Qed}}
Consider the subset $S$ of the real numbers $\R$ defined as: :$S = \set {\dfrac 1 n: n \in \Z_{>0} }$ It is seen that: :$S = \set {1, \dfrac 1 2, \dfrac 1 3, \ldots}$ and hence $\inf S = 0$. Thus $\inf S \notin S$. Consider the subset $T$ of the real numbers $\R$ defined as: :$T = \set {-\dfrac 1 n: n \in \Z_{>0} }$ It is seen that: :$T = \set {-1, -\dfrac 1 2, -\dfrac 1 3, \ldots}$ and hence $\inf T = 1$. Thus $\inf T \in T$. {{Qed}}
Let $f$ be the real function defined on the unbounded closed interval $\hointr 0 \to$ defined by $\map f y = y^n$. Consider first the case of $n > 0$. By Strictly Positive Integer Power Function is Unbounded Above: :$\exists q \in \R_{>0}: \map f q \ge x$ Since $x \ge 0$: :$\map f 0 \le x$ By the Intermediate Value Theorem: :$\exists y \in \R: 0 \le y \le q, \map f y = x$ Hence the result has been shown to hold for $n > 0$. {{qed}}
Let $S$ be the set consisting of the positive real numbers $t$ such that $t^n < x$. Let $t = \dfrac x {1 - x}$. Then $0 < t < 1$ and so: :$t^n < t < x$ demonstrating that $S \ne \O$. Let $t_0 = 1 + x$. Then $t > t_0$ implies $t^n \ge t > x$. So $t \notin S$. Hence $t_0$ is an upper bound of $S$. By the Continuum Property, $S$ has a supremum. Let $y = \sup S$. {{AimForCont}} $y_n < x$. Choose $h \in \R$ such that $0 < h < 1$ and such that: :$h < \dfrac {x - y^n} {\paren {1 + y}^n - y^n}$ Let $\dbinom n m$ denote a binomial coefficient. By the Binomial Theorem, the coefficient of $z^m$ in the expansion of $\paren {1 + z}^n$ is We have: {{begin-eqn}} {{eqn | l = \paren {y + h}^n       | r = y^n + \dbinom n 1 y^{n - 1} h + \dbinom n 2 y^{n - 2} h^2 + \dotsb + \dbinom n n h^n       | c = Binomial Theorem }} {{eqn | o = \le       | r = y^n + h \paren {\dbinom n 1 y^{n - 1} + \dbinom n 2 y^{n - 2} + \dotsb + \dbinom n n}       | c = as $0 < h < 1$ }} {{eqn | r = y^n + h \paren {\paren {1 + y}^n - y^n}       | c = Binomial Theorem }} {{eqn | o = <       | r = y^n + \paren {x - y^n}       | c = Definition of $h$ }} {{eqn | r = x       | c =  }} {{end-eqn}} Thus we have $y + h \in S$, which contradicts the fact that $y$ is an upper bound of $S$. {{AimForCont}} $y^n > x$. Let $k \in \R$ be chosen such that: :$0 < k < 1$ :$k < y$ :$k < \dfrac {y^n - x} {\paren {1 + y}^n - y^n}$ Let $t \ge y - k$. Then: {{begin-eqn}} {{eqn | l = t^n       | o = \ge       | r = \paren {y - k}^n       | c =  }} {{eqn | r = y^n - \dbinom n 1 y^{n - 1} h + \dbinom n 2 y^{n - 2} h^2 - \dotsb + \paren {-1} \dbinom n n h^n       | c = Binomial Theorem }} {{eqn | r = y^n - k \paren {\dbinom n 1 y^{n - 1} + \dbinom n 2 y^{n - 2} k - \dotsb + \paren {-1} \dbinom n n k^{n - 1} }       | c =  }} {{eqn | o = \ge       | r = y^n - k \paren {\dbinom n 1 y^{n - 1} + \dbinom n 2 y^{n - 2} + \dotsb + \dbinom n n}       | c = as $0 < k < 1$ }} {{eqn | r = y^n - k \paren {\paren {1 + y}^n - y^n}       | c = Binomial Theorem }} {{eqn | o = >       | r = y^n - \paren {y^n - x}       | c = Definition of $k$ }} {{eqn | r = x       | c =  }} {{end-eqn}} Thus we have $y - k$ is an upper bound of $S$ This contradicts the fact that $y$ is the supremum of $S$. It has been shown that both $y^n < x$ and $y^n > x$ lead to a contradiction. Hence it must be the case that $y^n = x$. Hence the result. {{qed}}
Let $\BB_1$ and $\BB_2$ satisfy: :$\forall U \in \mathcal B_1 : \exists \AA \subseteq \BB_2: U = \bigcup \AA$ Let $U \in \BB_1$. Then there exists $\AA_2 \subseteq \BB_2$ such that $U = \bigcup \AA_2$. By definition of the topology generated by $\BB_2$, it follows that $U \in \tau_2$. Since $U$ was arbitrary, it follows that $\BB_1 \subseteq \tau_2$. Let $W \in \tau_1$. By definition of the topology generated by $\BB_1$: :there exists $\AA_1 \subseteq \BB_1$ such that $W = \bigcup \AA_1$. Since $\BB_1 \subseteq \tau_2$, it follows that $\AA_1 \subseteq \tau_2$. By the open set axiom $(O1)$, then $W \in \tau_2$. Since $W$ was arbitrary, it follows that $\tau_1 \subseteq \tau_2$. The result follows. {{qed}} Category:Topology 1p6fcu1389e64hpn6s5q9ruzum2nrrf
Let $\BB_1$ and $\BB_2$ satisfy: :$\forall U \in \BB_1: \forall x \in U: \exists V \in \BB_2: x \in V \subseteq U$ Let $U \in \BB_1$. Let $\AA = \set {V \in \BB_2 : V \subseteq U}$ From Union of Family of Sets is Smallest Superset: :$\bigcup \AA \subseteq V$ Let $x \in U$. Then: :$\exists V_x \in \BB_2 : x \in V_x \subseteq U$ Thus: :$V_x \in \AA$ and :$x \in V_x \subseteq \bigcup \AA$ Since $x$ was arbitrary, it follows that: :$U \subseteq \bigcup \AA$ By the definition of set equality: :$U = \bigcup \AA$ Since $U$ was arbitrary, it follows that $\BB_1$ and $\BB_2$ satisfy: :$\forall U \in \BB_1 : \exists \AA \subseteq \BB_2: U = \bigcup \AA$ From Basis Condition for Coarser Topology, $\tau_1$ is coarser than $\tau_2$. {{qed}} Category:Topology 1zhb26tgjsjt1fduu2m276b0j0lvc7r
Let $\BB_1 \subseteq \BB_2$. Let $U \in \BB_1$. Let $\AA = \set U$. Then: :$\AA \subseteq \BB_2$ and :$U = \bigcup \AA$ So: :$\forall U \in \mathcal B_1 : \exists \AA \subseteq \BB_2: U = \bigcup \AA$ From Basis Condition for Coarser Topology: :$\tau_1$ is coarser than $\tau_2$ {{qed}} Category:Topology 7m31w4bckmhmiw8tm0n6q8vomlv2gjs
Let $U \in \tau'_2$. Since $\tau'_2$ is a coarser topology than $\tau_2$: :$U \in \tau_2$ By definition of continuity: :$\map {f^{-1}} U \in \tau_1$ Since $\tau'_1$ is a finer topology than $\tau_1$: :$\map {f^{-1}} U \in \tau'_1$ Since $U$ was arbitrary, by definition of continuity: :$f: \struct {X, \tau'_1} \to \struct {Y, \tau'_2}$ is a continuous mapping {{qed}} Category:Topology e1qsjh48v81uxnduhrf29bekhv9qwob
Let $\tau'$ be the Tychonoff topology on $X$. From Projection from Product Topology is Continuous: :$\forall i \in I : \pr_i : \struct{X, \tau'} \to \struct{X_i, \tau_i}$ is continuous From Box Topology contains Tychonoff Topology, $\tau$ is a finer topology than $\tau'$ on $X$. From Continuous Mapping on Finer Domain and Coarser Codomain Topologies is Continuous,  :$\forall i \in I : \pr_i : \struct{X, \tau} \to \struct{X_i, \tau_i}$ is continuous {{qed}} Category:Box Topology 769ivowebzsonorcducgfyokrzrffn6
=== Necessary Condition === Let $\tau' \subseteq \tau$. From Equivalence of Definitions of Initial Topology: :for each $i \in I$, $f_i: \struct{Y, \tau'} \to \struct{X_i, \tau_i}$ is $\tuple{\tau', \tau_i}$-continuous From Continuous Mapping on Finer Domain and Coarser Codomain Topologies is Continuous: :for each $i \in I$, $f_i: \struct{Y, \tau} \to \struct{X_i, \tau_i}$ is $\tuple{\tau, \tau_i}$-continuous {{qed|lemma}} === Sufficient Condition === For all $i \in I$, let $f_i: \struct{Y, \tau} \to \struct{X_i, \tau_i}$ be $\tuple{\tau, \tau_i}$-continuous. From Equivalence of Definitions of Initial Topology: :$\tau'$ is the coarsest topology on $Y$ such that each $f_i: Y \to X_i$ is $\tuple{\tau', \tau_i}$-continuous. Then: :$\tau' \subseteq \tau$ {{qed}} Category:Initial Topology s6vlsuxh0qtyb82z4rn9lxezbjapgt2
=== Necessary Condition === Let $\tau \subseteq \tau'$. From Equivalence of Definitions of Final Topology: :for each $i \in I$, $f_i: \struct{X_i, \tau_i} \to \struct{Y, \tau'}$ is $\tuple{\tau_i, \tau'}$-continuous From Continuous Mapping on Finer Domain and Coarser Codomain Topologies is Continuous: :for each $i \in I$, $f_i: \struct{X_i, \tau_i} \to \struct{Y, \tau}$ is $\tuple{\tau_i, \tau}$-continuous {{qed|lemma}} === Sufficient Condition === For all $i \in I$, let $f_i: \struct{X_i, \tau_i} \to \struct{Y, \tau}$ be $\tuple{\tau, \tau_i}$-continuous. From Equivalence of Definitions of Final Topology: :$\tau'$ is the finest topology on $Y$ such that each $f_i: \struct{X_i, \tau_i} \to \struct{Y, \tau'}$ is $\tuple{\tau_i, \tau'}$-continuous. Then: :$\tau \subseteq \tau'$ {{qed}} Category:Final Topology kgw38qii4f0p0f2y7cn0y2qk7rv1ts9
=== Definition 1 Implies Definition 2 === {{:Equivalence of Definitions of Final Topology/Definition 1 Implies Definition 2}}{{qed|lemma}} === Definition 2 Implies Definition 1 === {{:Equivalence of Definitions of Final Topology/Definition 2 Implies Definition 1}}{{qed}} Category:Final Topology k9a6qti8xt063sfig9ichbv24l3gf69
By definition, the trivial relation in $S$ to $T$ is the set: :$\RR = \set {\tuple {s, t}: s \in S, t \in T}$ Proof by Cases: :$(1)$: Let $\card S = 0$. That is, $S = \O$. By definition of the trivial relation, in this case $\RR = \O$. This is the empty mapping. From Empty Mapping is Mapping, this is a mapping. :$(2)$: Let $\card S \ne 0$. That is, $S \ne \O$. ::$(2) \, \text (a)$: Let $\card T = 0$. That is, $T = \O$. From Relation to Empty Set is Mapping iff Domain is Empty, $\RR$ cannot be a mapping for $S \ne \O$. ::$(2) \, \text (b)$: Let $\card T = 1$. So, let $T = \set t$. Then the trivial relation in $S$ to $T$ is: :$\RR = \set {\tuple {s, t}: s \in S}$ We have that each element of $S$ is associated with exactly one element of $T$. Hence $\RR$ is a mapping by definition. ::$(2) \, \text (c)$: Let $\card T > 1$. Then: :$\exists t_1, t_2 \in T: t_1 \ne t_2$ We have that $S \ne \O$, so $\exists s \in S$. Thus by definition of trivial relation: :$\tuple {s, t_1} \in \RR$ and: :$\tuple {s, t_2} \in \RR$ but $t_1 \ne t_2$. Thus $\RR$ is not a mapping. Hence the result. {{qed}}
Let $S \ne \O$. Then $\exists s \in S$. But there exists no $t \in \O$. Hence there is no $\tuple {s, t} \in \RR$. So $\RR$ is not a mapping by definition. Let $S = \O$. Then $\RR$ is the empty mapping by definition. From Empty Mapping is Mapping, it is demonstrated that $\RR$ is indeed a mapping. Hence the result. {{qed}} Category:Empty Set Category:Mapping Theory px0y5wob8d8d79a384wro6umcyjdlqx
==== Mappings are Continuous ==== Let $i \in I$. Let $U \in \tau_i$. Then $\map {f_i^{-1} } U$ is an element of the subbase $\SS$ of $X$, and is therefore trivially in $\tau$. {{qed|lemma}} ==== $\tau$ is the Coarsest such Topology ==== Let $\struct {X, \vartheta}$ be a topological space. Let the mappings $\family {f_i: X \to Y_i}_{i \mathop \in I}$ be $\tuple {\vartheta, \tau_i}$-continuous. Let $U \in \SS$. Then for some $i \in I$ and some $V \in \tau_i$: :$U = \map {f_i^{-1} } V$ By definition of the continuity of $f_i$: :$U \in \vartheta$ From Equivalence of Definitions of Topology Generated by Synthetic Sub-Basis: :$\tau \in \vartheta$ That is, $\tau$ is coarser than $\vartheta$.
Let $\map \tau \SS$ be the topology on $X$ generated  by the subbase $\SS$. ==== $\tau$ contains Topology Generated by $\SS$ ==== Let $i \in I$ and $U \in \tau_i$. By definition of $\tuple{\tau, \tau_i}$-continuity: :$\map {f_i^{-1} } U \in \tau$ Since $i \in I$ and $U \in \tau_i$ were arbitrary, then $\SS \subseteq \tau$. From Equivalence of Definitions of Topology Generated by Synthetic Sub-Basis: :$\map \tau \SS \subseteq \tau$ {{qed|lemma}} ==== Topology Generated by $\SS$ Contains $\tau$ ==== Let $i \in I$. Let $U \in \tau_i$. Then $\map {f_i^{-1} } U$ is an element of the subbase $\SS$ of $X$, and is therefore trivially in $\map \tau \SS$. By definition of $\tuple{\map \tau \SS, \tau_i}$-continuity: :each $f_i: X \to Y_i$ is $\tuple{\map \tau \SS, \tau_i}$-continuous Because $\tau$ is the coarsest topology on $X$ such that each $f_i: X \to Y_i$ is $\tuple{\tau, \tau_i}$-continuous: :$\tau \subseteq \map \tau \SS$ By definition of set equality: :$\tau = \map \tau \SS$
Let $E$ denote the set of even natural numbers. {{AimForCont}} $E$ is finite. Then there exists $n \in \N$ such that $E$ has $n$ elements. Let $m$ be the greatest element of $E$. But then $m + 2$ is an even natural number. But $m + 2 > m$, and $m$ is the greatest element of $E$. Therefore $m + 2$ is an even natural number that is not an element of $E$. So $E$ does not contain all the even natural numbers. From that contradiction it follows by Proof by Contradiction that $E$ is not finite. {{Qed}}
Let $S$ denote the set of points on a line segment. {{AimForCont}} $S$ is finite. Then there exists $n \in \N$ such that $S$ has $n$ elements. Let $s_1$ and $s_2$ be two arbitrary adjacent points in $S$. That is, such that there are no points in $S$ between $s_1$ and $s_2$. But there exists (at least) one point on the line segment between $s_1$ and $s_2$ which is not in $S$. Hence there must be more than $n$ elements of $S$. From that contradiction it follows by Proof by Contradiction that $S$ is not finite. {{Qed}}
We can write the elements of $S \times T$ in the form of an infinite table: :$\begin{array} {*{4}c}  \tuple {1, 0} &  &  &  \\  \tuple {2, 0} & \tuple {2, 1} &  & \\  \tuple {3, 0} & \tuple {3, 1} & \tuple {3, 2} & \\  \vdots  & \vdots  & \vdots & \ddots \\ \end{array}$ This table clearly contains all the elements of $S$. Let $\set {n_1, n_2} \in S$ be written such that $n_1 > n_2$. Then $\set {n_1, n_2}$ will be found in row $n_1$ and column $n_2 + 1$ of the table. The index of the $n$th row is the $n$th triangular number. Let $f: S \to \N$ be defined as: :$\forall \set {n_1, n_2} \in S: \map f {\set {n_1, n_2} } = \dfrac {n_1 \paren {n_1 + 1} } 2 + n_2$ It is seen that $f$ is an injection. With a little more work it can be shown to be a bijection as well, but that condition  is not necessary. The result follows. {{qed}}
From Final Topology is Topology, $\tau$ is a topology. ==== Mappings are continuous ==== Let $U \in \tau$. Let $i \in I$. Then $\map {f_i^{-1}} {U} \in \tau_i$ by definition of $\tau$. It follows that for each $i \in I$, $f_i: Y_i \to X$ is $\tuple {\tau_i, \tau}$-continuous. {{qed|lemma}} ==== $\tau$ is the finest such topology ==== Let $\struct{X, \vartheta}$ be a topological space. Let the mappings $\family {f_i: Y_i \to X}_{i \mathop \in I}$ be $\tuple {\tau_i, \vartheta}$-continuous. Let $U \in \vartheta$. By definition of continuity, for every $i \in I$: :$\map {f_i^{-1} } U \in \tau_i$ Then by definition of $\tau$: :$U \in \tau$ It follows that: :$\vartheta \subseteq \tau$ That is, $\tau$ is finer than $\vartheta$.
Let $\tau' =  \set{U \subseteq X: \forall i \in I: \map {f_i^{-1} } U \in \tau_i}$. ==== $\tau'$ contains $\tau$ ==== Let $U \in \tau$. By definition of  $\tuple {\tau_i, \tau}$-continuity for each $i \in I$: :$\forall i \in I : \map {f_i^{-1} } U \in \tau_i$ So: :$U \in \tau'$ Since $U$ was arbitrary: :$\tau \subseteq \tau'$ {{qed|lemma}} ==== $\tau$ contains $\tau'$ ==== From Final Topology is Topology then $\tau'$ is a topology. Let $U \in \tau'$. Let $i \in I$. Then $\map {f_i^{-1}} {U} \in \tau_i$ by definition of $\tau'$. It follows that for each $i \in I$, $f_i: Y_i \to X$ is $\tuple{\tau_i, \tau'}$-continuous. So $\tau'$ is a topology on $X$ such that each $f_i: Y_i \to X$ is $\tuple {\tau_i, \tau'}$-continuous. Since $\tau$ is the finest topology on $X$ such that each $f_i: Y_i \to X$ is $\tuple {\tau_i, \tau}$-continuous then: :$\tau' \subseteq \tau$ By definition of set equality: :$\tau = \tau'$
Consider the mapping $f: \N \to \Z_{>0}$ defined as: :$\forall x \in \N: \map f x = x + 1$ Then $f$ is trivially seen to be a bijection. The result follows by definition of cardinality.
=== Outine === $\powerset \N$ is demonstrated to have the same cardinality as the set of real numbers. This is done by identifying a real number with its basis expansion in binary notation. Such a basis expansion is a sequence of $0$s and $1$s. Each $1$ of a real number $x$ expressed in such a way can be identified with the integer which identifies the power of $2$ representing the $1$ in that place. Hence $x$ corresponds to the subset of $\N$ whose elements are the positions in $x$ which hold a $1$. Thus $\R$ and $\powerset \N$ are equivalent. From the Cantor-Dedekind Hypothesis, the real numbers are in one-to-one correspondence with the points on a line segment. The points on a line segment form a continuum. {{ProofWanted|Formalise the above}}
Let $P$ be the property defined as: :$\forall x: \map P x := \neg \paren {x = x}$ Hence we form the set: :$\O := \set {x: \neg \paren {x = x} }$ Since we have that: :$\forall x: x = x$ it is seen that $\O$ as defined here has no elements. Hence the result by definition of the '''empty set'''. {{qed}}
Let $P$ be the property defined as: :$\forall x: \map P x := \paren {x = a \lor x = b}$ where $\lor$ is the disjunction operator. Hence we form the set: :$\set {a, b} := \set {x: x = a \lor x = b}$ {{qed}}