hcju/proofwikiqa huggingface dataset:
It can be seen that $(1)$ is a nonhomogeneous linear second order ODE in the form: :$y'' + p y' + q y = \map R x$ where: :$p = 0$ :$q = 1$ :$\map R x = e^{-x} \cos x$ First we establish the solution of the corresponding constant coefficient homogeneous linear second order ODE: :$y'' + y = 0$ From Linear Second Order ODE: $y'' + y = 0$, this has the general solution: :$y_g = C_1 \sin x + C_2 \cos x$ It remains to find a particular solution $y_p$ to $(1)$. We have that: :$\map R x = e^{-x} \cos x$ From the Method of Undetermined Coefficients for Exponential of Sine and Cosine: :$y_p = e^{-x} \paren {A \cos x + B \sin x}$ where $A$ and $B$ are to be determined. Hence: {{begin-eqn}} {{eqn | l = y_p       | r = e^{-x} \paren {A \cos x + B \sin x}       | c =  }} {{eqn | ll= \leadsto       | l = {y_p}'       | r = -e^{-x} \paren {A \cos x + B \sin x} + e^{-x} \paren {-A \sin x + B \cos x}       | c = Product Rule for Derivatives etc. }} {{eqn | r = e^{-x} \paren {\paren {B - A} \cos x - \paren {A + B} \sin x}       | c =  }} {{eqn | ll= \leadsto       | l = {y_p}''       | r = -e^{-x} \paren {\paren {B - A} \cos x - \paren {A + B} \sin x} + e^{-x} \paren {-\paren {B - A} \sin x - \paren {A + B} \cos x}       | c = Product Rule for Derivatives etc. }} {{eqn | r = e^{-x} \paren {-2 B \cos x + 2 A \sin x}       | c =  }} {{end-eqn}} Substituting into $(1)$: {{begin-eqn}} {{eqn | l = e^{-x} \paren {-2 B \cos x + 2 A \sin x} + e^{-x} \paren {A \cos x + B \sin x}       | r = e^{-x} \cos x       | c =  }} {{eqn | ll= \leadsto       | l = -2 B \cos x + A \cos x       | r = \cos x       | c = equating coefficients }} {{eqn | l = 2 A \sin x + B \sin x       | r = 0       | c =  }} {{eqn | ll= \leadsto       | l = A - 2 B       | r = 1       | c =  }} {{eqn | l = 2 A + B       | r = 0       | c =  }} {{eqn | ll= \leadsto       | l = 5 A       | r = 1       | c =  }} {{eqn | l = 5 B       | r = -2       | c =  }} {{eqn | ll= \leadsto       | l = A       | r = \dfrac 1 5       | c =  }} {{eqn | l = B       | r = -\dfrac 2 5       | c =  }} {{end-eqn}} Hence the result: :$y_p = \dfrac {e^{-x} } 5 \paren {\cos x - 2 \sin x}$ So from General Solution of Linear 2nd Order ODE from Homogeneous 2nd Order ODE and Particular Solution: :$y = y_g + y_p = \dfrac {e^{-x} } 5 \paren {\cos x - 2 \sin x} + C_1 \sin x + C_2 \cos x$ is the general solution to $(1)$. {{qed}}
It can be seen that $(1)$ is a constant coefficient homogeneous linear second order ODE. Its auxiliary equation is: :$(2): \quad: m^2 - 5 m + 6 = 0$ From Solution to Quadratic Equation with Real Coefficients, the roots of $(2)$ are: :$m_1 = 2$ :$m_2 = 3$ These are real and unequal. So from Solution of Constant Coefficient Homogeneous LSOODE, the general solution of $(1)$ is: :$y = C_1 e^{2 x} + C_2 e^{3 x}$ {{qed}}
It can be seen that $(1)$ is a nonhomogeneous linear second order ODE in the form: :$y'' + p y' + q y = \map R x$ where: :$p = -5$ :$q = 5$ :$\map R x = \cos x + \sin x$ First we establish the solution of the corresponding constant coefficient homogeneous linear second order ODE: :$y'' - 5 y' + 6 y = 0$ From Linear Second Order ODE: $y'' - 5 y' + 6 y = 0$, this has the general solution: :$y_g = C_1 e^{2 x} + C_2 e^{3 x}$ It remains to find a particular solution $y_p$ to $(1)$. We have that: :$\map R x = e^{-x} \cos x$ From the Method of Undetermined Coefficients for Sine and Cosine: :$y_p = A \cos x + B \sin x$ where $A$ and $B$ are to be determined. Hence: {{begin-eqn}} {{eqn | l = y_p       | r = A \cos x + B \sin x       | c =  }} {{eqn | ll= \leadsto       | l = {y_p}'       | r = -A \sin x + B \cos x       | c = Derivative of Sine Function, Derivative of Cosine Function }} {{eqn | ll= \leadsto       | l = {y_p}''       | r = -A \cos x - B \sin x       | c = Derivative of Sine Function, Derivative of Cosine Function }} {{end-eqn}} Substituting into $(1)$: {{begin-eqn}} {{eqn | l = \paren {-A \cos x - B \sin x} - 5 \paren {-A \sin x + B \cos x} + 6 \paren {A \cos x + B \sin x}       | r = \cos x + \sin x       | c =  }} {{eqn | ll= \leadsto       | l = \paren {5 A - 5 B} \cos x       | r = \cos x       | c = equating coefficients }} {{eqn | l = \paren {5 B + 5 A} \sin x       | r = \sin x       | c =  }} {{eqn | ll= \leadsto       | l = 10 A       | r = 2       | c =  }} {{eqn | l = 10 B       | r = 0       | c =  }} {{eqn | ll= \leadsto       | l = A       | r = \dfrac 1 5       | c =  }} {{eqn | l = B       | r = 0       | c =  }} {{end-eqn}} Hence the result: :$y_p = \dfrac {\cos x} 5$ So from General Solution of Linear 2nd Order ODE from Homogeneous 2nd Order ODE and Particular Solution: :$y = y_g + y_p = C_1 e^{2 x} + C_2 e^{3 x} + \dfrac {\cos x} 5$ is the general solution to $(1)$. {{qed}}
It can be seen that $(1)$ is a nonhomogeneous linear second order ODE with constant coefficients in the form: :$y'' + p y' + q y = \map R x$ where: :$p = 0$ :$q = -4$ :$\map R x = x^2 - 3 x - 4$ First we establish the solution of the corresponding constant coefficient homogeneous linear second order ODE: :$y'' - 4 y = 0$ From Linear Second Order ODE: $y'' - 4 y = 0$, this has the general solution: :$y_g = C_1 e^{2 x} + C_2 e^{-2 x}$ We have that: :$\map R x = x^2 - 3 x - 4$ and so from the Method of Undetermined Coefficients for Polynomial function: :$y_p = A_1 x^2 + A_2 x + A_3$ Hence: {{begin-eqn}} {{eqn | l = y_p       | r = A_1 x^2 + A_2 x + A_3       | c =  }} {{eqn | ll= \leadsto       | l = {y_p}'       | r = 2 A_1 x + A_2       | c =  }} {{eqn | ll= \leadsto       | l = {y_p}''       | r = 2 A_1       | c =  }} {{end-eqn}} Substituting in $(1)$: {{begin-eqn}} {{eqn | l = \paren {2 A_1} - 4 \paren {A_1 x^2 + A_2 x + A_3}       | r = x^2 - 3 x - 4       | c =  }} {{eqn | ll= \leadsto       | l = 2 A_1 - 4 A_3       | r = -4       | c = equating coefficients }} {{eqn | l = -4 A_2       | r = -3       | c =  }} {{eqn | l = -4 A_1       | r = 1       | c =  }} {{eqn | ll= \leadsto       | l = A_1       | r = -\dfrac 1 4       | c =  }} {{eqn | l = A_2       | r = \dfrac 3 4       | c =  }} {{eqn | l = A_3       | r = \dfrac 7 8       | c =  }} {{end-eqn}} So from General Solution of Linear 2nd Order ODE from Homogeneous 2nd Order ODE and Particular Solution: :$y = y_g + y_p = C_1 e^{2 x} + C_2 e^{-2 x} - \dfrac {x^2} 4 + \dfrac {3 x} 4 + \dfrac 7 8$ {{qed}}
For all $n \ge m$, let: :$\alpha_n = \displaystyle \sum_{i \mathop = m}^n d_i p^i$ By assumption: :$\sequence{\alpha_n}$ is a representative of $\mathbf a$ By definition of the induced norm: :$\norm{\mathbf a}_p = \displaystyle \lim_{n \mathop \to \infty} \norm{\alpha_n}_p$ From Eventually Constant Sequence Converges to Constant it is sufficient to show: :$\forall n \ge l + 1 : \norm{\alpha_n}_p = p^{-l}$ Let $n \ge l + 1$. Then: {{begin-eqn}} {{eqn | l = \norm {\sum_{i \mathop = m}^n d_i p^i}_p       | r = \norm {\paren {\sum_{i \mathop = m}^{l - 1} d_i p^i} + d_l p^l + \paren {\sum_{i \mathop = l + 1}^n d_i p^i} }_p }} {{eqn | r = \norm {\paren {\sum_{i \mathop = m}^{l - 1} 0 \cdot p^i } + d_l p^l + \paren{\sum_{i \mathop = l + 1}^n d_i p^i} }_p       | c = by choice of $l$, for all $i : m \le i < l \implies d_i = 0$ }} {{eqn | r = \norm {d_l p^l + \paren {\sum_{i \mathop = l + 1}^n d_i p^i} }_p }} {{eqn | r = \max \set {\norm {d_l p^l }_p, \norm {\sum_{i \mathop = l + 1}^n d_i p^i}_p}       | o = \le       | c = {{NormAxiom|4}} }} {{end-eqn}} The sum $\displaystyle \sum_{i \mathop = l + 1}^n d_i p^i$ can be rewritten: :$\displaystyle \sum_{i \mathop = l + 1}^n d_i p^i = p^{l + 1} \sum_{i \mathop = l + 1}^n d_i p^{i - \paren {l + 1} }$ By definition of divisor: :$\displaystyle p^{l + 1} \divides \sum_{i \mathop = l + 1}^n d_i p^i$ Then: {{begin-eqn}} {{eqn | l = \norm {\sum_{i \mathop = l + 1}^n d_i p^i}_p       | o = \le       | r = p^{-\paren {l + 1} }       | c = {{Defof|P-adic Norm|$p$-adic norm}} }} {{eqn | o = <       | r = p^{-l}       | c = Power Function on Base between Zero and One is Strictly Decreasing }} {{end-eqn}} By definition of a $p$-adic expansion: :$0 < d_l < p$ Then: :$p^l \divides d_l p^l$ :$p^{l + 1} \nmid d_l p^l$ By definition of $p$-adic norm: :$\norm {d_l p^l}_p = p^{-l}$ Thus: :$\displaystyle \norm {\sum_{i \mathop = l + 1}^n d_i p^i}_p < p^{-l} = \norm {d_l p^l}_p$ Finally: {{begin-eqn}} {{eqn | l = \norm {\sum_{i \mathop = m}^n d_i p^i}_p       | r = \max \set {\norm {d_l p^l}_p, \norm {\sum_{i \mathop = l + 1}^n d_i p^i}_p}       | c = Corollary of Three Points in Ultrametric Space have Two Equal Distances }} {{eqn | r = \norm {d_l p^l}_p       | c = as $\norm {\sum_{i \mathop = l + 1}^n d_i p^i}_p < \norm {d_l p^l}_p$ }} {{eqn | r = p^{-l}       | c = derived earlier }} {{end-eqn}} The result follows. {{qed}}
Let $\sequence {y_n}$ be the subsequence of $\sequence {\norm {x_n}}$ defined by: :$\forall n: y_n =x_{N + n}$ The $\sequence {y_n}$ is the constant sequence $\tuple {\lambda, \lambda, \lambda, \dotsc}$. Then: {{begin-eqn}} {{eqn | l = \displaystyle \lim_{n \mathop \to \infty} x_n       | r = \displaystyle \lim_{n \mathop \to \infty} y_n       | c = Limit of Subsequence equals Limit of Sequence in Normed Division Ring }} {{eqn | r = \lambda       | c = Constant Sequence Converges to Constant in Normed Division Ring }} {{end-eqn}} {{qed}} Category:Sequences Category:Normed Division Rings ffxsfudtlibs04pt7ljl9sfyahebkvj
Taking the group axioms in turn: === $\text G 0$: Closure === Non-Zero Complex Numbers Closed under Multiplication. {{qed|lemma}} === $\text G 1$: Associativity === Complex Multiplication is Associative. {{qed|lemma}} === $\text G 2$: Identity === From Complex Multiplication Identity is One, the identity element of $\struct {\C_{\ne 0}, \times}$ is the complex number $1 + 0 i$. {{qed|lemma}} === $\text G 3$: Inverses === From Inverse for Complex Multiplication, the inverse of $x + i y \in \struct {\C_{\ne 0}, \times}$ is: :$\dfrac 1 z = \dfrac {x - i y} {x^2 + y^2} = \dfrac {\overline z} {z \overline z}$ where $\overline z$ is the complex conjugate of $z$. {{qed}}
Taking the group axioms in turn: === $\text G 0$: Closure === Complex Addition is Closed. {{qed|lemma}} === $\text G 1$: Associativity === Complex Addition is Associative. {{qed|lemma}} === $\text G 2$: Identity === From Complex Addition Identity is Zero, we have that the identity element of $\struct {\C, +}$ is the complex number $0 + 0 i$: :$\paren {x + i y} + \paren {0 + 0 i} = \paren {x + 0} + i \paren {y + 0} = x + i y$ and similarly for $\paren {0 + 0 i} + \paren {x + i y}$. {{qed|lemma}} === $\text G 3$: Inverses === From Inverse for Complex Addition, the inverse of $x + i y \in \struct {\C, +}$ is $-x - i y$. {{qed}}
From Special Linear Group is Subgroup of General Linear Group we have that the special linear group $\SL {n, K}$ is a subgroup of $\GL {n, K}$. From Special Linear Group is not Abelian, $\SL {n, K}$ is not abelian. From Subgroup of Abelian Group is Abelian it follows by the Rule of Transposition that $\GL {n, K}$ is not abelian. {{qed}}
{{begin-eqn}} {{eqn | l = \map {\cl_H} {W \cap H}       | r = \map {\cl_T} {W \cap H} \cap H       | c = Closure in Subspace }} {{eqn | o = \subseteq       | r = \map {\cl_T} W \cap H       | c = Topological Closure of Subset is Subset of Topological Closure }} {{end-eqn}} {{qed}} Category:Set Closures Category:Topological Subspaces 045nymh5qlj7kr9tg6clxl5vkm9qo9m
Let the field of complex numbers be denoted $\struct {\C, +, \times}$. By Complex Numbers under Addition form Abelian Group, $\struct {\C, +}$ is an abelian group. From Complex Multiplication Distributes over Addition: {{begin-eqn}} {{eqn | lo= \forall x, y, z \in \C:       | l = x \times \paren {y + z}       | r = x \times y + x \times z }} {{eqn | l = \paren {y + z} \times x       | r = y \times x + z \times x }} {{end-eqn}} From Complex Multiplication is Associative: :$\forall x, y, z \in \C: x \times \paren {y \times z} = \paren {x \times y} \times z$ From Complex Multiplication Identity is One: :$\forall x \in \C: 1 \times x = x$ Therefore $\struct {\C, +, \times}$ forms a vector space. {{qed}}
Let the set of quaternions be denoted $\struct {\H, +, \times}$. From Quaternions form Skew Field, the algebraic structure $\struct {\H, +, \times}$ is a skew field. By definition, a skew field is a division ring. {{ProofWanted}}
By definition of simple event: {{begin-eqn}} {{eqn | l = e_1       | r = \set {s_1} }} {{eqn | l = e_2       | r = \set {s_2} }} {{end-eqn}} for some elementary events $s_1$ and $s_2$ of $\EE$ such that $s_1 \ne s_2$. It follows that: {{begin-eqn}} {{eqn | l = e_1 \cap e_2       | r = \set {s_1} \cap \set {s_2}       | c = Definition of $e_1$ and $e_2$ }} {{eqn | r = \O       | c = {{Defof|Set Intersection}} }} {{end-eqn}} The result follows by definition of mutually exclusive events. {{qed}}
By definition of event, $e$ is a subset of the sample space $\Omega$ of $\EE$. By hypothesis: :$e \ne \O$ and so: :$\exists s \in \Omega: s \in e$ Let $S$ be the set defined as: :$S = \set {\set s: s \in e}$ By Union is Smallest Superset: Set of Sets it follows that: :$\displaystyle \bigcup S \subseteq e$ Let $x \in e$. Then by Singleton of Element is Subset: :$\set x \subseteq e$ and by definition of $S$ it follows that: :$\set x \in S$ and so by definition of set union: :$x \in \displaystyle \bigcup S$ Thus we have: :$e \subseteq \displaystyle \bigcup S$ The result follows by definition of set equality. {{qed}}
By Set is Subset of Itself: :$\Omega \subseteq \Omega$ That is, $\Omega$ is itself an event in $\EE$. The result as an application of Non-Trivial Event is Union of Simple Events. {{qed}}
{{begin-eqn}} {{eqn | l = \map \complement {\map \complement A \cup \map \complement B}       | r = \map \complement {\map \complement A} \cap \map \complement {\map \complement B}       | c = De Morgan's Laws: Complement of Union }} {{eqn | r = A \cap B       | c = Complement of Complement }} {{eqn | ll= \leadstoandfrom       | l = \map \complement {\map \complement {\map \complement A \cup \map \complement B} }       | r = \map \complement {A \cap B}       | c = taking complements of both sides }} {{eqn | ll= \leadstoandfrom       | l = \map \complement A \cup \map \complement B       | r = \map \complement {A \cap B}       | c = Complement of Complement }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | o =        | r = x \in \overline {T_1 \cup T_2} }} {{eqn | o = \leadstoandfrom       | r = x \notin \paren {T_1 \cup T_2}       | c = {{Defof|Set Complement}} }} {{eqn | o = \leadstoandfrom       | r = \neg \paren {x \in T_1 \lor x \in T_2}       | c = {{Defof|Set Union}} }} {{eqn | o = \leadstoandfrom       | r = \neg \paren {x \in T_1} \land \neg \paren {x \in T_2}       | c = De Morgan's Laws (Logic): Conjunction of Negations }} {{eqn | o = \leadstoandfrom       | r = x \in \overline {T_1} \land x \in \overline {T_2}       | c = {{Defof|Set Complement}} }} {{eqn | o = \leadstoandfrom       | r = x \in \overline {T_1} \cap \overline {T_2}       | c =  }} {{end-eqn}} By definition of set equality: :$\overline {T_1 \cup T_2} = \overline {T_1} \cap \overline {T_2}$ {{qed}}
Let $a$ and $b$ be expressed in canonical form: :$a = \dfrac {p_1} {q_1}, b = \dfrac {p_2} {q_2}$ where $p_1, p_2 \in \Z$ and $q_1, q_2 \in \Z_{>0}$. As $\forall a, b \in \Q_{>0}$ it follows that $p_1, p_2 \in \Z_{>0}$. By definition of rational addition: :$\dfrac {p_1} {q_1} + \dfrac {p_2} {q_2} = \dfrac {p_1 q_2 + p_2 q_1} {q_1 q_2}$ From Integers form Ordered Integral Domain, it follows that: {{begin-eqn}} {{eqn | l = p_1 q_2       | o = >       | r = 0 }} {{eqn | l = p_2 q_1       | o = >       | r = 0 }} {{eqn | l = q_1 q_2       | o = >       | r = 0 }} {{eqn | ll= \leadsto       | l = p_1 q_2 + p_2 q_1       | o = >       | r = 0 }} {{eqn | ll= \leadsto       | l = \dfrac {p_1 q_2 + p_2 q_1} {q_1 q_2}       | o = >       | r = 0 }} {{end-eqn}} {{qed}}
Let $a$ and $b$ be expressed in canonical form: :$a = \dfrac {p_1} {q_1}, b = \dfrac {p_2} {q_2}$ where $p_1, p_2 \in \Z$ and $q_1, q_2 \in \Z_{>0}$. As $\forall a, b \in \Q_{>0}$ it follows that $p_1, p_2 \in \Z_{>0}$. By definition of rational multiplication: :$\dfrac {p_1} {q_1} \times \dfrac {p_2} {q_2} = \dfrac {p_1 \times p_2} {q_1 \times q_2}$ From Integers form Ordered Integral Domain, it follows that: {{begin-eqn}} {{eqn | l = p_1 \times p_2       | o = >       | r = 0 }} {{eqn | l = q_1 \times q_2       | o = >       | r = 0 }} {{eqn | ll= \leadsto       | l = \dfrac {p_1} {q_1} \times \dfrac {p_2} {q_2}       | o = >       | r = 0 }} {{end-eqn}} {{qed}}
{{AimForCont}} $p \in A$ is the greatest element of $A$. Then by definition of $A$: :$p^2 < 2$ Let $h \in \Q$ be a rational number such that $0 < h < 1$ such that: :$h < \dfrac {2 - p^2} {2 p + 1}$ This is always possible, because by definition $2 - p^2 > 0$ and $2 p + 1 > 0$. Let $q = p + h$. Then $q > p$, and: {{begin-eqn}} {{eqn | l = q^2       | r = p^2 + \paren {2 p + h} h       | c =  }} {{eqn | o = <       | r = p^2 + \paren {2 p + 1} h       | c =  }} {{eqn | o = <       | r = p^2 + \paren {2 - p^2}       | c =  }} {{eqn | r = 2       | c =  }} {{end-eqn}} That means $q \in A$. This contradicts our assertion that $p$ is the greatest element of $A$. It follows that $A$ can have no greatest element. {{qed}}
{{AimForCont}} $p \in B$ is the smallest element of $B$. Then by definition of $B$: :$p^2 > 2$ Let $q = p - \dfrac {p_2 - 2} {2 p}$. Then $q > p$, and: {{begin-eqn}} {{eqn | l = q       | r = p - \dfrac {p_2 - 2} {2 p}       | c =  }} {{eqn | r = \dfrac p 2 + \dfrac 1 p       | c =  }} {{end-eqn}} Hence: :$0 < q < p$ and so: {{begin-eqn}} {{eqn | l = q^2       | r = p^2 - \dfrac {p^2 - 2} + \paren {\dfrac {p^2 - 2} {2 p} }^2       | c =  }} {{eqn | o = >       | r = p^2 - \paren {p^2 - 2}       | c =  }} {{eqn | r = 2       | c =  }} {{end-eqn}} That means $q \in B$. This contradicts our assertion that $p$ is the smallest element of $B$. It follows that $B$ can have no smallest element. {{qed}}
We have been given that $p \in \alpha$ while $q \notin \alpha$. {{AimForCont}} $q \le p$. If $q = p$ then $q \in \alpha$ immediately. If $q < p$ then $q \in \alpha$ by property $(2)$ of the definition of cut. In either case $q \in \alpha$ which contradicts $q \notin \alpha$. Hence by Proof by Contradiction it follows that $q > p$. {{Qed}}
That $\alpha$ fulfils conditions $(1)$ and $(2)$ of the definition of a cut follows directly from that definition. Then it is noted that if $p \in \alpha$ then from Mediant is Between: :$p < \dfrac {p + r} 2 < r$ and so $p$ cannot be the greatest element of $\alpha$. Hence it is seen that $\alpha$ fulfils all conditions to be a cut. Because it is not the case that $r < r$, it follows that $r \notin \alpha$. Thus $r$ is an upper number of $\alpha$. {{AimForCont}} $q \in \Q$ is an upper number of $\alpha$ such that $q < r$. But then by definition of a cut, $q \in \alpha$. That is, $q$ is not an upper number of $\alpha$. Hence by Proof by Contradiction it follows that there is no upper number of $\alpha$ which is smaller than $r$. That is, $r$ is the smallest upper number of $\alpha$. {{Qed}}
Let $\NN$ denote the natural basis on $X$. By definition of the natural basis, $\NN$ is the basis generated by: :$\SS = \set {\pr_i^{-1} \sqbrk U: i \in I, \, U \in \tau_i}$ where for each $i \in I$, $\pr_i: X \to X_i$ denotes the $i$th projection on $X$: :$\forall \family {x_j}_{j \mathop \in I} \in X: \map {\pr_i} {\family {x_j}_{j \mathop \in I} } = x_i$ === Lemma 1 === {{:Natural Basis of Tychonoff Topology/Lemma 1}}{{qed|lemma}} === Lemma 2 === {{:Natural Basis of Tychonoff Topology/Lemma 2}}{{qed|lemma}} From Synthetic Basis formed from Synthetic Sub-Basis: :$\NN = \set {\bigcap \FF: \FF \subseteq \SS, \, \FF \text{ is finite} }$ From Lemma 1 and Lemma 2 it follows that: :$\NN \subseteq \BB$. === Lemma 3 === {{:Natural Basis of Tychonoff Topology/Lemma 3}}{{qed|lemma}} From Lemma 3 and Synthetic Basis formed from Synthetic Sub-Basis: :$\BB \subseteq \NN$ The result follows from the definition of set equality. {{qed}} Category:Tychonoff Topology Category:Natural Basis of Tychonoff Topology 8s28fvzjpjdhhly4irrx2itgc9mg2mm
From Natural Basis of Tychonoff Topology, the natural basis on $X$ is the set $\BB$ of cartesian products of the form $\displaystyle \prod_{i \mathop \in I} U_i$ where: :for all $k = 1, \dots, n: U_k \in \tau_i$ :for all but finitely many indices $k : U_k = X_k$ Let $\BB' = \set {\displaystyle \prod_{k \mathop = 1}^n U_k : \forall k : U_k \in \tau_k}$ By definition of $\BB$: :$\BB \subseteq \BB'$ Let $B \in \BB'$, that is:  :$B = \displaystyle \prod_{k \mathop = 1}^n U_k$ where $\forall k : U_k \in \tau_k$. Let $J = \set {k : U_k \neq X_k}$. Since $J \subseteq \set{ 1, \dots, n}$ is finite then: :for all but finitely many indices $k : U_k = X_k$ Thus $B \in \BB$. Since $B$ was arbitrary, then $\BB' \subseteq \BB$. It follows that $\BB = \BB'$ from the definition of set equality. {{qed}} Category:Natural Basis of Tychonoff Topology pmu495fnwvu31h14w0j93xoyrcpcgco
From Projection from Box Topology is Continuous, the projections $\family{\pr_i} _{i \mathop \in I}$ are continuous. From Tychonoff Topology is Coarsest Topology such that Projections are Continuous, the Tychonoff topology is the coarsest topology such that the projections $\family{\pr_i} _{i \mathop \in I}$ are continuous. Therefore, it is sufficient to show that in some cases the Tychonoff topology is strictly coarser than the box topology. That is, it is sufficient to show that in some cases the Tychonoff topology is not equal to the box topology. === Lemma === {{:Box Topology may not be Coarsest Topology such that Projections are Continuous/Lemma}}{{qed|lemma}} Let $\struct{\R, \tau}$ denote the real numbers with Euclidean topology. Let $\epsilon \in \R_{>0}$. Let $U$ be the open interval $\openint {-\epsilon} {\epsilon}$. From the Lemma: :$U^\N$ is open in the box topology on $\R^\N$ :$U^\N$ is not open in the Tychonoff topology on $\R^\N$ where $\R^\N$ is the countable-dimensional real Cartesian space. {{qed}} Category:Box Topology nwkp2jvm1uqrsqt9zc0nrqtfo2h75ah
Let $\alpha = \beta$. By definition of equality of cuts: :$p \in \alpha \iff p \in \beta$ By definition of strict ordering of cuts it follows that neither $\alpha < \beta$ or $\alpha > \beta$. {{AimForCont}} both $\alpha < \beta$ and $\alpha > \beta$. Because $\alpha < \beta$ there exists $p \in \Q$ such that: :$p \in \beta, p \notin \alpha$ Because $\alpha > \beta$ there exists $q \in \Q$ such that: :$q \in \alpha, q \notin \beta$ By Rational Number Not in Cut is Greater than Element of Cut: :$p \in \beta$ and $q \notin \beta$ implies $p < q$. :$q \in \alpha$ and $p \notin \alpha$ implies $p > q$. But from Rational Numbers form Ordered Field, $p < q$ and $p > q$ is not possible. Hence by Proof by Contradiction, it is not possible for both $\alpha < \beta$ and $\alpha > \beta$. Suppose either $\alpha < \beta$ or $\alpha > \beta$. Then by definition of strict ordering of cuts it follows that $\alpha \ne \beta$. Thus mutual exclusivity of the three conditions has been demonstrated. That is, at most one of the three conditions holds. {{qed|lemma}} It remains to be shown that at least one of the three conditions holds. Suppose $\alpha \ne \beta$. Then by definition of equality of cuts, either: :$\exists p \in \alpha: p \notin \beta$ in which case $\alpha > \beta$, or: :$\exists q \in \beta: q \notin \alpha$ in which case $\alpha < \beta$. Hence the result. {{Qed}}
We have that: :$\alpha < \beta$ By definition of strict ordering of cuts: :$\exists p \in \Q: p \in \beta, p \notin \alpha$ Similarly, we have that: :$\beta < \gamma$ and so: :$\exists q \in \Q: q \in \gamma, q \notin \beta$ We have by definition of a cut that: :$p \in \beta$ and $q \notin \beta$ implies that $p < q$ Together with $p \notin \alpha$, this implies that $q \notin \alpha$. Thus we have: :$q \in \gamma$ and $q \notin \alpha$ from which by definition of strict ordering of cuts: :$\alpha < \gamma$ {{Qed}}