hcju/proofwikiqa huggingface dataset:
Let $\struct {\CC, <}$ denote the relational structure defined from the above. From Ordering on Cuts is Transitive, $<$ is a transitive relation on $\CC$. From Ordering on Cuts satisfies Trichotomy Law, we have that: :$\alpha < \beta \implies \lnot \paren {\beta < \alpha}$ demonstrating that $<$ is asymmetric. Hence, by definition, $<$ is a strict ordering. Also by Ordering on Cuts satisfies Trichotomy Law, $\struct {\CC, <}$ has no non-comparable pairs: :$\forall \alpha, \beta \in \CC: \alpha \ne \beta \implies \alpha < \beta \lor \beta < \alpha$ The result follows by definition of strict total ordering. {{Qed}}
By definition of cut, neither $\alpha$ nor $\beta$ are empty. Hence there exist $p \in \alpha$ and $q \in \beta$. Hence there exists $r = p + q$ and so $\gamma$ is likewise not empty. Let $s, t \in \Q$ such that $s \notin \alpha$ and $t \notin \beta$, where $\Q$ denotes the set of rational numbers. Such $s$ and $t$ are bound to exist because by definition of cut, neither $\alpha$ nor $\beta$ equal $\Q$. We have: {{begin-eqn}} {{eqn | l = p       | o = <       | r = s       | c = {{Defof|Cut (Analysis)|Cut}}: $s \notin \alpha$ }} {{eqn | l = q       | o = <       | r = t       | c = {{Defof|Cut (Analysis)|Cut}}: $t \notin \beta$ }} {{eqn | ll= \leadsto       | lo= \forall p \in \alpha, q \in \beta:       | l = p + q       | o = <       | r = s + t       | c = Rational Numbers form Ordered Field }} {{eqn | ll= \leadsto       | l = s + t       | o = \notin       | r = \gamma       | c = {{Defof|Cut (Analysis)|Cut}} }} {{end-eqn}} Thus it is demonstrated that $\gamma$ does not contain every rational number. Thus condition $(1)$ of the definition of a cut is fulfilled. {{qed|lemma}} Let $r \in \gamma$. Let $s \in \Q$ such that $s < r$. Then $r = p + q$ for some $p \in \alpha, q \in \beta$. Let $t \in \Q$ such that $s = t + q$. Then $t < p$. Hence $t \in \alpha$. Hence by definition of $\gamma$, $t + q = s \in \gamma$. Thus we have that $r \in \gamma$ and $s < r$ implies that $s \in \gamma$. Thus condition $(2)$ of the definition of a cut is fulfilled. {{qed|lemma}} {{AimForCont}} $r \in \gamma$ is the greatest element of $\gamma$. Then $r = p + q$ for some $p \in \alpha, q \in \beta$. By definition of a cut, $\alpha$ has no greatest element: Hence: :$\exists s \in \Q: s > p: s \in \alpha$ But then $s + q \in \gamma$ while $s + q > r$. This contradicts the supposition that $r$ is the greatest element of $\gamma$. Hence $\gamma$ itself can have no greatest element. Thus condition $(3)$ of the definition of a cut is fulfilled. {{qed|lemma}} Thus it is seen that all the conditions are fulfilled for $\gamma$ to be a cut. {{qed}}
$\alpha + \beta$ is the set of all rational numbers of the form $p + q$ such that $p \in \alpha$ and $q \in \beta$. Similarly, $\beta + \alpha$ is the set of all rational numbers of the form $q + p$ such that $p \in \alpha$ and $q \in \beta$. From Rational Addition is Commutative we have that: :$p + q = q + p$ The result follows. {{qed}}
$\paren {\alpha + \beta} + \gamma$ is the set of all rational numbers of the form $\paren {p + q} + r$ such that $p \in \alpha$, $q \in \beta$ and $r \in \gamma$. Similarly, $\alpha + \paren {\beta + \gamma}$ is the set of all rational numbers of the form $p + \paren {q + r}$ such that $p \in \alpha$, $q \in \beta$ and $r \in \gamma$. From Rational Addition is Associative we have that: :$\paren {p + q} + r = p + \paren {q + r}$ The result follows. {{qed}}
Let $r \in \alpha + 0^*$. $\alpha + 0^*$ is the set of all rational numbers of the form $p + q$ such that $p \in \alpha$, $q \in 0^*$, that is, $q < 0$. It follows that: :$p + q < p$ and so: :$p + q \in \alpha$ that is: :$r \in \alpha$ Let $r \in \alpha$. Let $s \in \Q$ be a rational number such that $s > r$ and $s \in \alpha$. This is possible because $r$ is not the greatest element of $\alpha$ by definition of a cut. Let $q = r - s$. Then $q < 0$ and so $q \in 0^*$. Thus we have: :$r = s + q$ and so: :$r \in \alpha + 0^*$ We have that: :$r \in \alpha \implies r \in \alpha + 0^*$ and: :$r \in \alpha + 0^* \implies r \in \alpha$ and so by definition of set equality: :$\alpha + 0^* = \alpha$ {{qed}}
Let $s \in \alpha$ be a rational number. For $n = 0, 1, 2, \ldots$ let $s_n = s + n r$. Then there exists a unique integer $m$ such that: :$s_m \in \alpha$ and: :$s_{m + 1} \notin \alpha$ If $s_{m + 1}$ is not the smallest upper number of $\alpha$, take: :$p = s_m$ :$q = s_{m + 1}$ If $s_{m + 1}$ ''is'' the smallest upper number of $\alpha$, take: :$p = s_m + \dfrac r 2$ :$q = s_{m + 1} + \dfrac r 2$ The result follows. {{qed}}
=== Proof of Uniqueness === Suppose $\alpha + \beta_1 = \alpha + \beta_2 = 0^*$. We have: {{begin-eqn}} {{eqn | l = \beta_2       | r = 0^* + \beta_2       | c = Identity Element for Addition of Cuts }} {{eqn | r = \paren {\alpha + \beta_1} + \beta_2       | c = Definition of $\beta_1$ }} {{eqn | r = \paren {\alpha + \beta_2} + \beta_1       | c = Addition of Cuts is Commutative and Addition of Cuts is Associative }} {{eqn | r = 0^* + \beta_1       | c = Definition of $\beta_2$ }} {{eqn | r = \beta_1       | c = Identity Element for Addition of Cuts }} {{end-eqn}} So if such a $\beta$ exists, it is indeed unique. {{qed|lemma}} === Proof of Existence === Let $\beta$ be the set of all rational numbers $p$ such that $-p$ is an upper number of $\alpha$, but not the smallest upper number, were it to exist. We verify that the conditions are satisfied for $\beta$ to be a cut. It follows directly from the definition that $\beta \ne \O$ and $\beta \ne \Q$. Thus condition $(1)$ is satisfied. {{qed|lemma}} Let $p \in \beta$. Let $q \in \Q$ be a rational number such that $q < p$. Then: :$-p \notin \alpha$ and: :$-q > -p$ Thus $-q$ is an upper number of $\alpha$, but not the smallest. Thus $q \in \beta$. Thus condition $(2)$ is satisfied. {{qed|lemma}} Let $p \in \beta$. Then by definition $-p$ is an upper number of $\alpha$, but not the smallest. Thus there exists a rational number $q$ such that $-q < -p$ and such that $-q \notin \alpha$. Let: :$r = \dfrac {p + q} 2$ By Mediant is Between: :$-q < -r < -p$ so $-r$ is also an upper number of $\alpha$, but not the smallest. So we have found a rational number $r$ such that $r > p$ where $r \in \beta$. Thus condition $(3)$ is satisfied. {{qed|lemma}} We have demonstrated that $\beta$ is a cut. We now verify that $\alpha + \beta = 0^*$. Suppose $p \in \alpha + \beta$. Then: :$p = q + r$ for some $q \in \alpha, r \in \beta$. Hence: :$-r \notin \alpha$ :$-r < q$ :$q + r < 0$ and: :$p \in 0^*$ Suppose $p \in 0^*$. Then: :$p < 0$ By Existence of Upper and Lower Numbers of Cut whose Difference equal Given Rational, there exist $q \in \alpha, r \notin \alpha$ such that: :$r$ is not the smallest upper number of $\alpha$ $r - q = -p$ Because $-r \in \beta$: {{begin-eqn}} {{eqn | l = p       | r = q - r       | c =  }} {{eqn | r = q + \paren {-r}       | c =  }} {{eqn | o = \in       | r = \alpha + \beta       | c =  }} {{end-eqn}} Thus by definition of set equality: :$\alpha + \beta = 0^*$ and the proof is complete. {{qed}}
By definition of the strict ordering on cuts and addition of cuts: :$\alpha + \beta \le \alpha + \gamma$ {{explain|Details needed for the above}} Suppose $\alpha + \beta = \alpha + \gamma$. Then: {{begin-eqn}} {{eqn | l = \beta       | r = 0^* + \beta       | c = Identity Element for Addition of Cuts }} {{eqn | r = \paren {-\alpha} + \paren {\alpha + \beta}       | c = {{Defof|Negative of Cut}} }} {{eqn | r = \paren {-\alpha} + \paren {\alpha + \gamma}       | c =  }} {{eqn | r = 0^* + \gamma       | c =  }} {{eqn | r = \gamma       | c =  }} {{end-eqn}} Thus: :$\beta = \gamma$ and the result follows. {{qed}}
From Ordering on Cuts is Compatible with Addition of Cuts :$0^* + 0^* < 0^* + \alpha$ :$\alpha + 0^* < \alpha + \gamma$ The result follows from Ordering on Cuts is Transitive. {{qed}}
From Ordering on Cuts is Compatible with Addition of Cuts: :$\gamma_1 \ne \gamma_2 \implies \alpha + \gamma_1 \ne \alpha + \gamma_2$ That demonstrates uniqueness. Let: :$\gamma = \beta + \paren {-\alpha}$ where $-\alpha$ is the negative of $\alpha$. Then by Identity Element for Addition of Cuts: {{begin-eqn}} {{eqn | l = \alpha + \gamma       | r = \alpha + \paren {\beta + \paren {-\alpha} }       | c = Definition of $\gamma$ }} {{eqn | r = \alpha + \paren {\paren {-\alpha} + \beta}       | c = Addition of Cuts is Commutative }} {{eqn | r = \paren {\alpha + \paren {-\alpha} } + \beta       | c = Addition of Cuts is Associative }} {{eqn | r = 0^* + \beta       | c = Existence of Unique Inverse Element for Addition of Cuts }} {{eqn | r = \beta       | c = Identity Element for Addition of Cuts }} {{end-eqn}} The result follows. {{qed}}
In the below, $\alpha$, $\beta$ and $\gamma$ denote arbitrary cuts. Taking the abelian group axioms in turn: === $\text G 0$: Closure === From Sum of Cuts is Cut: :$\forall \alpha, \beta \in \CC: \alpha + \beta \in \CC$ Thus $\struct {\CC, +}$ is closed. {{qed|lemma}} ===  $\text G 1$: Associativity === From Addition of Cuts is Associative: :$\paren {\alpha + \beta} + \gamma = \alpha + \paren {\beta + \gamma}$ Thus the operation $+$ of addition of cuts is associative on $\struct {\CC, +}$. {{qed|lemma}} === $\text C$: Commutativity === From Addition of Cuts is Commutative: :$\alpha + \beta = \beta + \alpha$ Thus the operation $+$ of addition of cuts is commutative on $\struct {\CC, +}$. {{qed|lemma}} === $\text G 2$: Identity === Consider the rational cut $0^*$ associated with the (rational) number $0$: :$0^* = \set {r \in \Q: r < 0}$ From Identity Element for Addition of Cuts: :$\alpha + 0^* = \alpha$ for all $\alpha \in \CC$. From Addition of Cuts is Commutative it follows that: :$0^* + \alpha = \alpha$ That is, $0^*$ is the identity element of $\struct {\CC, +}$. {{qed|lemma}} === $\text G 3$: Inverses === We have that $0^*$ is the identity element of $\struct {\CC, +}$. From Existence of Unique Inverse Element for Addition of Cuts, there exists a unique cut $-\alpha$ such that: :$\alpha + \paren {-\alpha} = 0^*$ From Addition of Cuts is Commutative it follows that: :$\paren {-\alpha} + \alpha = 0^*$ Thus every element of $\struct {\CC, +}$ has an inverse $-\alpha$. {{qed|lemma}} All the abelian group axioms are thus seen to be fulfilled, and so $\struct {\CC, +}$ is a group. {{qed}}
By definition of $\gamma$, we have that $r < 0 \implies r \in \gamma$. Hence $\gamma$ is not empty. First suppose that either $\alpha = 0^*$ or $\beta = 0^*$. Then by definition of cut: :$p \in \alpha \implies p < 0$ :$q \in \beta \implies q < 0$ and so there exist no $r \in \gamma$ such that $r = p q$. Thus $\gamma$ consists entirely of rational numbers $r$ such that $r < 0$. That is: :$\gamma = \set {r \in \Q: r < 0}$ and so: :$\gamma = 0^*$ Hence $\gamma$ is a cut, as we needed to demonstrate. {{qed|lemma}} Suppose that neither $\alpha$ nor $\beta$ are equal to $0^*$ That is: :$\alpha > 0$ and: :$\beta > 0$ Hence by definition of cut: :$0 \in \alpha$ and: :$0 \in \beta$ As a cut has no greatest element: :$\exists p \in \alpha: p > 0$ :$\exists q \in \alpha: q > 0$ Let $s, t \in \Q$ such that $s \notin \alpha$ and $t \notin \beta$, where $\Q$ denotes the set of rational numbers. Such $s$ and $t$ are bound to exist because by definition of cut, neither $\alpha$ nor $\beta$ equal $\Q$. Note that as $\alpha \ge 0^*$ and $\beta \ge 0^*$, it follows by definition of cut that, both $s \ge 0$ and $t \ge 0$. We have: {{begin-eqn}} {{eqn | l = p       | o = <       | r = s       | c = {{Defof|Cut (Analysis)|Cut}}: $s \notin \alpha$ }} {{eqn | l = q       | o = <       | r = t       | c = {{Defof|Cut (Analysis)|Cut}}: $t \notin \beta$ }} {{eqn | ll= \leadsto       | lo= \forall p \in \alpha, q \in \beta, p > 0, q > 0:       | l = p q       | o = <       | r = s t       | c = Rational Numbers form Ordered Field }} {{eqn | ll= \leadsto       | l = s t       | o = \notin       | r = \gamma       | c = {{Defof|Cut (Analysis)|Cut}} }} {{end-eqn}} Thus it is demonstrated that $\gamma$ does not contain every rational number. Thus condition $(1)$ of the definition of a cut is fulfilled. {{qed|lemma}} Let $r \in \gamma$. Let $s \in \Q$ such that $s < r$. If $r < 0$, then $s < 0$ and so $s \in \gamma$ by definition of $\gamma$. If $r > 0$ and $s < 0$, then $s \in \gamma$ by definition of $\gamma$. Otherwise, we have the case where $r > 0$ and $s > 0$. By definition of $\gamma$: :$r = p q$ for some $p \in \alpha, q \in \beta$ such that $p > 0$ and $q > 0$. Let $t \in \Q$ such that $s = t q$. Then $0 < t < p$. Hence $t \in \alpha$. Hence by definition of $\gamma$, $t q = s \in \gamma$. Thus we have that $r \in \gamma$ and $s < r$ implies that $s \in \gamma$. Thus condition $(2)$ of the definition of a cut is fulfilled. {{qed|lemma}} {{AimForCont}} $r \in \gamma$ is the greatest element of $\gamma$. Then $r = p q$ for some $p \in \alpha, q \in \beta$ such that $p > 0$ and $q > 0$. By definition of a cut, $\alpha$ has no greatest element: Hence: :$\exists s \in \Q: s > p: s \in \alpha$ But then $s q \in \gamma$ while $s q > r$. This contradicts the supposition that $r$ is the greatest element of $\gamma$. Hence $\gamma$ itself can have no greatest element. Thus condition $(3)$ of the definition of a cut is fulfilled. {{qed|lemma}} Thus it is seen that all the conditions are fulfilled for $\gamma$ to be a cut. {{qed}}
From $z^n + a = 0$ we have that: :$z^n = - a$ Let $b = a^{1 / n}$, hence $b^n = a$, with $a, b \in \C$. From Roots of Complex Number: {{begin-eqn}} {{eqn | l = z^{1 / n}       | r = \set {a^{1 / n} e^{i \paren {\theta + 2 k \pi} / n}: k \in \set {0, 1, 2, \ldots, n - 1}, \theta = \arg -a}       | c = $z^n = -a$ so we need $\theta = \arg -a$ }} {{eqn | r = \set {a^{1 / n} e^{i \paren {2 k + 1} \pi / n}: k \in \set {0, 1, 2, \ldots, n - 1} }       | c = $\theta = \pi$ }} {{eqn | r = \set {b e^{i \paren {2 k + 1} \pi / n}: k \in \set {0, 1, 2, \ldots, n - 1} }       | c = $b = a^{1 / n}$ }} {{end-eqn}} and so each of $\alpha_k b = b e^{i \paren {2 k + 1} \pi / n}$ is a unique root of $z^n + a$ with $n \in \Z_{>0}$. From the corollary to the Polynomial Factor Theorem: If $\zeta_1, \zeta_2, \ldots, \zeta_n \in \C$ such that all are different, and $\map P {\zeta_1} = \map P {\zeta_2} = \cdots = \map P {\zeta_n} = 0$, then: :$\displaystyle \map P z = k \prod_{j \mathop = 1}^n \paren {z - \zeta_j}$ where $k \in \C$. $z^n + a$ is a monic polynomial, hence $k = 1$ in the above product. Choose $\zeta_j = \alpha_{j - 1} b$ and we have the desired result. {{qed}} Category:Algebra Category:Complex Roots rw5q3we51ls3ujjufu78nkmszgdco20
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 $\alpha$ :$0^*$ denotes the rational cut associated with the (rational) number $0$ :$\ge$ denotes the ordering on cuts. Let $\alpha \ge 0^*$ and $\beta \ge 0^*$. :$\gamma := \alpha \beta$ where $\gamma$ is the set of all rational numbers $r$ such that: :$\exists p \in \alpha, q \in \beta: r = p q$ $\alpha \beta$ is the set of all rational numbers $r$ either of the form $r < 0$ of the form $r = p q$ such that $p \in \alpha$ and $q \in \beta$. Similarly, $\beta \alpha$ is the set of all rational numbers $r$ either of the form $r < 0$ of the form $r = q p$ such that $p \in \alpha$ and $q \in \beta$. From Rational Multiplication is Commutative we have that: :$p q = q p$ Thus we have that: :$\size \alpha \, \size \beta = \size \beta \, \size \alpha$ and the result follows. {{qed}}
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 $\alpha$ :$0^*$ denotes the rational cut associated with the (rational) number $0$ :$\ge$ denotes the ordering on cuts. Let $\alpha \ge 0^*$, $\beta \ge 0^*$ and $\gamma \ge 0^*$. $\paren {\alpha \beta} \gamma$ is the set of all rational numbers $s$ of the form $s = \paren {p q} r$ such that $s < 0$ or $p \in \alpha$, $q \in \beta$ and $r \in \gamma$. Similarly, $\alpha \paren {\beta \gamma}$ is the set of all rational numbers $s$ of the form $s = p \paren {q r}$ such that $s < 0$ or $p \in \alpha$, $q \in \beta$ and $r \in \gamma$. From Rational Multiplication is Associative we have that: ::$\paren {p q} r = p \paren {q r}$ Thus we have that: :$\size \alpha \paren {\size \beta \, \size \gamma} = \paren {\size \alpha \, \size \beta} \size \gamma$ and the result follows. {{qed}}
Let $B_1, B_2 \in \BB$. Let $B_1 = \displaystyle \prod_{i \mathop \in I} U_i$ where: :for all $i \in I : U_i \in \tau_i$ :for all but finitely many indices $i : U_i = X_i$ Let $J_1$ be the finite set of indices such that: :$J_1 = \set {j \in I : U_i \neq X_i}$ Let $B_2 = \displaystyle \prod_{i \mathop \in I} V_i$ where: :for all $i \in I : V_i \in \tau_i$ :for all but finitely many indices $i : V_i = X_i$ Let $J_2$ be the finite set of indices such that: :$J_2 = \set{j \in I : V_i \neq X_i}$ Then: {{begin-eqn}} {{eqn | l = B_1 \cap B_2       | r = \paren{\prod_{i \mathop \in I} U_i } \cap \paren{\prod_{i \mathop \in I} V_i} }} {{eqn | r =\prod_{i \mathop \in I} \paren{ U_i \cap V_i}       | c = General Case of Cartesian Product of Intersections }} {{end-eqn}} Now: {{begin-eqn}} {{eqn | lo= \forall i \in I :       | l = U_i, V_i \in \tau_i       | o = \leadsto       | r = U_i \cap V_i \in \tau_i       | c = {{Defof|Topology|topology}} $\tau_i$ }} {{end-eqn}} and  {{begin-eqn}} {{eqn | lo = \forall i \in I \setminus \paren {J_1 \cup J_2}:       | l = U_i \cap V_i       | r = X_i \cap X_i       | c = Definition of $J_1$ and $J_2$ }} {{eqn | r = X_i       | c = Intersection is Idempotent }} {{end-eqn}} From Finite Union of Finite Sets is Finite, $J_1 \cup J_2$ is finite. Then: :for all but finitely many indices $i : U_i \cap V_i = X_i$ Thus $B_1 \cap B_2 \in \BB$ {{qed}} Category:Natural Basis of Tychonoff Topology nhbbp5qovgau0pr14avlwsarsnaera8
Let $B \in \BB$. Let $B = \displaystyle \prod_{i \mathop \in I} U_i$ where :for all $i \in I : U_i \in \tau_i$ :for all but finitely many indices $i : U_i = X_i$ Let $J = \set{j \in I : U_i \ne X_i}$. Then $J$ is a finite set and: :$\forall i \in I \setminus J : U_i = X_i$ For all $j \in J$, let: :$\pr_j^{-1} \sqbrk {U_j} = \displaystyle \prod_{i \mathop \in I} V^j_i$ where: :$V^j_j = U_j$ :$\forall i \ne j : V^j_i = X_i$ Then: {{begin-eqn}} {{eqn | l = \bigcap_{j \mathop \in J} \pr_j^{-1} \sqbrk {U_j}       | r = \bigcap_{j \mathop \in J} \paren {\prod_{i \mathop \in I} V^j_i} }} {{eqn | r = \prod_{i \mathop \in I} \paren {\bigcap_{j \mathop \in J}  V^j_i}       | c = General Case of Cartesian Product of Intersections }} {{end-eqn}} To complete the proof it remains to show that: :$\forall i \in I : U_i = \bigcap_{j \mathop \in J}  V^j_i$ Let $i \in I$. === Case: $i \not \in J$ === Let $i \not \in J$. Then: {{begin-eqn}} {{eqn | l = \bigcap_{j \mathop \in J}  V^j_i       | r = \bigcap_{j \mathop \in J}  X_i       | c = as $\forall j \in J : i \ne j$ }} {{eqn | r = X_i       | c = Intersection is Idempotent }} {{eqn | r = U_i       | c = Definition of $J$ }} {{end-eqn}} {{qed|lemma}} === Case: $i \in J$ === Let $i \in J$. Then: {{begin-eqn}} {{eqn | l = \bigcap_{j \mathop \in J}  V^j_i       | r = V^i_i \cap \bigcap_{j \mathop \in J \mathop \setminus i}  V^j_i }} {{eqn | r = U_i \cap \bigcap_{j \mathop \in J \mathop \setminus i}  V^j_i       | c = Definition of $V^i_i$ }} {{eqn | r = U_i \cap \bigcap_{j \mathop \in J \mathop \setminus i}  X_i       | c = Definition of $V^j_i$ when $j \ne i$ }} {{eqn | r = U_i \cap X_i       | c = Intersection is Idempotent }} {{eqn | r = U_i       | c = Intersection with Subset is Subset }} {{end-eqn}} {{qed|lemma}} Thus: :$\displaystyle \forall i \in I : U_i = \bigcap_{j \mathop \in J}  V^j_i$ The result follows. {{qed}} Category:Natural Basis of Tychonoff Topology l2q2m8zinjmitjg1kws5ghnp2tiqvy4
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 $\alpha$ :$0^*$ denotes the rational cut associated with the (rational) number $0$ :$\ge$ denotes the ordering on cuts. Let $\alpha \ge 0^*$, $\beta \ge 0^*$ and $\gamma \ge 0^*$. $\alpha \paren {\beta + \gamma}$ is the set of all rational numbers $s$ of the form: :$s = p \paren {q + r}$ such that: :$s < 0$ or: :$p \in \alpha$, $q \in \beta$ and $r \in \gamma$. $\alpha \beta + \alpha \gamma$ is the set of all rational numbers $s$ of the form: :$s = p q + p r$ such that: :$s < 0$ or: :$p \in \alpha$, $q \in \beta$ and $r \in \gamma$. From Rational Multiplication Distributes over Addition: $p \paren {q + r} = p q + p r$ and the result follows. {{qed}}
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 :$\ge$ denotes the ordering on cuts. From Absolute Value of Cut is Zero iff Cut is Zero: :$\size {0^*} = 0^*$ Thus: :$\alpha 0^* := \begin {cases} \size \alpha \, 0^* & : \alpha \ge 0^* \\ -\paren {\size \alpha \, 0^*} & : \alpha < 0^* \end {cases}$ From Absolute Value of Cut is Greater Than or Equal To Zero Cut, we have that: :$\size \alpha = \beta$ where $\beta \ge 0^*$. By definition of Multiplication of Positive Cuts: $\beta 0^*$ is the set of all rational numbers $r$ such that either: :$r < 0$ or :$\exists p \in \beta, q \in 0^*: r = p q$ where $p \ge 0$ and $q \ge 0$ By definition of $0^*$, it is not possible for $q \ge 0$. Thus: :$\beta 0^* = \set {r \in \Q: r < 0}$ Hence by definition of a cut: :$\beta 0^* = 0^*$ By Identity Element for Addition of Cuts and definition of negative of cut: :$-0^* = 0^*$ and the result follows. {{qed}}
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 :$\ge$ denotes the ordering on cuts. First we note that from Absolute Value of Cut is Zero iff Cut is Zero: :$\size {0^*} = 0^*$ === Necessary Condition === Let $\alpha = 0^*$ or $\beta = 0^*$. Then from Product of Cut with Zero Cut equals Zero Cut it follows directly that: :$\alpha \beta = 0^*$ {{qed|lemma}} === Sufficient Condition === Let $\alpha \beta = 0^*$. Suppose $\alpha > 0^*$ and $\beta > 0^*$. By definition of multiplication of positive cuts: $\alpha \beta$ is the set of all rational numbers $r$ such that either: :$r < 0$ or :$\exists p \in \alpha, q \in \beta: r = p q$ where $p \ge 0$ and $q \ge 0$. We have that $\alpha \beta = 0^*$. {{AimForCont}} $r = p q$ where $p \ge 0$ and $q \ge 0$. Then $r \ge 0$. But then $r \notin 0^*$ by definition of $0^*$. Thus there is no $p \in \alpha, q \in \beta$ such that $r = p q$ where $p \ge 0$ and $q \ge 0$. It follows that $\alpha > 0^*$ and $\beta > 0^*$ cannot both be satisfied. It follows further that for $\size \alpha \, \size \beta = 0^*$ it is not possible for $\size \alpha > 0^*$ and $\size \beta > 0^*$. The result follows. {{qed}}