hcju/proofwikiqa huggingface dataset:
{{NotZFC}} By the definition of a transitive class: :$A \in B \implies A \subseteq B$ But $A \ne B$ because $\paren {A = B \land A \in B} \implies A \in A$, which by No Membership Loops is a contradiction. Therefore $A \subsetneq B$. {{qed}}
The proof is carried out in the following three steps: === LSC implies Closed Epigraph === Let $f: S \to \overline \R$ be LSC on $S$. Let $\operatorname{epi} \left({f}\right)$ denote the epigraph of $f$. Take a sequence $\sequence{\tuple{x_n, a_n}}_{n \mathop \in \N} \in \map {\operatorname{epi}} f$ such that: :$\tuple{x_n, a_n} \to \tuple{\bar x, \bar a} \in S \times \R$ as $n \to \infty$. This implies that $x_n \to \bar x$ and $a_n \to \bar a$ by definition of the product topology on $S \times \R$ and because a Continuous Mapping is Sequentially Continuous. Consequently: {{begin-eqn}} {{eqn | l =  \map f {\bar x}       | r = \liminf_{x \mathop \to \bar x} \map f x       | c = Definition of Lower Semicontinuous }} {{eqn | o = \le       | r = \liminf_{n \mathop \to \infty} \map f {x_n}       | c = Relationship between Limit Inferior and Lower Limit }} {{eqn | o = \le       | r = \liminf_{n \mathop \to \infty} a_n       | c = because $\tuple{x_n, a_n} \in \map {\operatorname{epi} } f$ }} {{eqn | r = \bar a       | c = because $a_n \to \bar a$ }} {{end-eqn}} Thus: :$\tuple{\bar x, \bar a} \in \map {\operatorname{epi}} f$ {{proofread}} {{questionable|This only proves sequential closedness.}} {{qed|lemma}} === Closed Epigraph implies Closed Level Sets === Let $\map {\operatorname{epi}} f$ be a closed set in $S \times \R$. Let $a \in \R$. Then the $\alpha$-lower level set: :$\displaystyle \operatorname{lev} \limits_{\mathop \le a} = \map {\operatorname{epi}} f \cap S \times \set{a}$ {{questionable|The identity claimed here is not true}} is closed in $S$ because: : closedness is preserved under Intersectionintersection and: : $S \times \set{a}$ is a closed set with respect with the product topology of $S \times \R$. {{qed|lemma}} === Closed Level Sets implies LSC === Let all lower level sets of $f$ be closed. Let $x \in S$. Let $\displaystyle a: = \liminf_{t \mathop \to x} \map f t$. In order to prove that $f$ is LSC it suffices to show that $a = \map f x$.  We know already that $a \le \map f x$. Thus it suffices to show that $\map f x \le a$. Let $a = \infty$. As $\infty \le \infty$ the proof is complete. Let $a < \infty$. Then we can find a sequence $\sequence{x_n}_{n \mathop \in \N}$ such that: : $x_n \to x$ and $\map f {x_n} \to a$ as $n \to \infty$. For any $b > a$: :$\map f {x_n} \le b$ or equivalently: :$x_n \in \displaystyle \operatorname{lev} \limits_{\mathop \le b} f$ which by hypothesis is closed. Therefore $\displaystyle \operatorname{lev} \limits_{\mathop \le b} f$ contains the limit point $x$. We have that for all $b > a$: :$\map f x \le b$ Therefore: :$\map f x \le a$ Hence the result. {{qed|lemma}} So: : $(1) \implies (2) \implies (3)$ and the proof is complete. {{qed}}
By definition of foundational relation, every nonempty subset of $S$ has a minimal element. By Subset Relation is Transitive, every subset of $T$ is also a subset of $S$. Therefore every nonempty subset of $T$ has a minimal element. Thus by definition, $\RR$ is a foundational relation on $T$. {{qed}}
By Restriction of Foundational Relation is Foundational, $R$ is a foundational relation on $B$. By Restriction of Total Ordering is Total Ordering, $R$ is a total ordering on $B$. By the above two statements, $R$ is a strict well-ordering of $B$. {{explain|Indicate the specific definition which identifies that a foundation relation which is a total ordering is a well-ordering.}} {{qed}}
=== Necessary Condition === Suppose that $S$ is an ordinal. Then $S$ is transitive. By definition, the strict well-ordering on $S$ is given by the $\in$-relation. Hence, the necessary condition is satisfied. {{qed|lemma}} === Sufficient Condition === Suppose that $S$ is a transitive set that is strictly well-ordered by the $\in$-relation. Let $a \in S$. Then: {{begin-eqn}} {{eqn | l = a       | r = S \cap a       | c = $S$ is transitive, $a \subseteq S$; then apply Intersection with Subset is Subset }} {{eqn | r = \left\{ {x \in S: x \in a} \right\} }} {{eqn | r = S_a       | c = Definition of Initial Segment }} {{end-eqn}} That is, $S$ is an ordinal. {{qed}}
{{questionable|This whole proof probably needs to be rewritten from scratch.}} {{NotZFC}} The epsilon relation is equivalent to the strict subset relation when restricted to ordinals by Transitive Set is Proper Subset of Ordinal iff Element of Ordinal. It follows that: :$\forall x \in \On: x \subset \On$ {{explain|We have a problem here since $\On$ is a proper class. The notation $\in $ an $\subseteq$ can naturally be extended to classes, but one must be careful with their use.}} The initial segment of the class of ordinals is: :$\set {x \in \On : x \subset \On}$ {{explain|What is meant by $\subset$ here: $\subseteq$ or $\subsetneqq$? 
Initial segment makes no sense here because $\On$ is not a set.
<br/>Sorry, but "makes no sense" is a bit strong: "initial segment" has been defined on {{ProofWiki}} in the context of a well-ordered set, but the term makes perfect sense if defined in the context of a class.}} This class is equal to $\On$. Therefore, by the definition of ordinal, $\On$ is an ordinal. {{qed}}
=== Necessary Condition === Suppose that $B \in A$. From Ordinal is Transitive, it follows that $B \subseteq A$. Also, $B \ne A$ by Ordinal is not Element of Itself. Therefore, $B \subsetneq A$, as desired. {{qed|lemma}} === Sufficient Condition === Suppose that $B \subsetneq A$. By the definition of set equality, the set difference $A \setminus B$ is non-empty. By the definition of a strict well-ordering, there exists a minimal element $x$ of $A \setminus B$. As $x \in A$, it follows from Ordinal is Transitive that $x \subseteq A$. Next, since the strict well-ordering on $A$ is given by the epsilon restriction $\Epsilon \! \restriction_A$, it follows by the definition of a minimal element that: :$\forall y \in A \setminus B: y \notin x$ Therefore: {{begin-eqn}} {{eqn | l = \O       | r = \paren {A \setminus B} \cap x }} {{eqn | r = \paren {A \cap x} \setminus B       | c = Intersection with Set Difference is Set Difference with Intersection }} {{eqn | r = x \setminus B       | c = Intersection with Subset is Subset }} {{eqn | ll= \leadsto       | l = x       | o = \subseteq       | r = B       | c = by Set Difference with Superset is Empty Set }} {{end-eqn}} Suppose that $z \in B$. Since $B \subset A$, it follows that $z \in A$. Recall that the strict well-ordering on $A$ is given by the epsilon restriction $\Epsilon \! \restriction_A$. From the Trichotomy Law (Ordering), it follows that $z \in x$ or $x = z$ or $x \in z$. If $x = z$ or $x \in z$, then it follows by the transitivity of $B$ that $x \in B$. This contradicts the definition of $x$. Hence, $z \in x$. That is, $B \subseteq x$. We have shown that $x \subseteq B$ and $B \subseteq x$. By definition of set equality: :$B = x \in A$ {{qed}}
{{begin-eqn}} {{eqn | l = \int \arcsin x \rd x       | r = \int 1 \cdot \arcsin x \rd x }} {{eqn | r = x \arcsin x - \int x \paren {\frac \rd {\rd x} \arcsin x} \rd x       | c = Integration by Parts }} {{eqn | r = x \arcsin x - \int \frac x {\sqrt {1 - x^2} } \rd x       | c = Derivative of Arcsine Function }} {{end-eqn}} Substitute: {{begin-eqn}} {{eqn | l = u        | r = 1 - x^2 }} {{eqn | l = \frac {\d u} {\d x}       | r = -2 x       | c = differentiating {{WRT|Differentiation}} $x$ }} {{eqn | ll = \leadsto       | l = -\frac 1 2 \frac {\d u} {\d x}       | r = x }} {{eqn | l = \int \arcsin x \rd x        | r = x \arcsin x - \int \frac {- \frac 1 2} {\sqrt u} \frac {\d u} {\d x} \rd x }} {{eqn | l = \int \arcsin x \rd x        | r = x \arcsin x + \frac 1 2 \int u^{-1/2} \rd u       | c = Integration by Substitution }} {{eqn | r = x \arcsin x + u^{1/2} + C       | c = Integral of Power }} {{eqn | r = x \arcsin x + \sqrt {1 - x^2} + C       | c = from $u = 1 - x^2$ }} {{end-eqn}} {{qed}}
From Real Function is Convex iff Derivative is Increasing, $f$ is convex {{iff}} $f'$ is increasing. From Derivative of Monotone Function, $f'$ is increasing {{iff}} its second derivative $f'' \ge 0$. {{qed}}
By Relation between Two Ordinals, it follows that: :$\left({A = B}\right) \lor \left({A \subset B}\right) \lor \left({B \subset A}\right)$ By Transitive Set is Proper Subset of Ordinal iff Element of Ordinal, the result follows. {{qed}}
From Real Function is Concave iff Derivative is Decreasing, $f$ is concave {{iff}} $f'$ is decreasing. From Derivative of Monotone Function, $f'$ is decreasing {{iff}} its second derivative $f'' \le 0$. {{qed}}
By hypothesis $A$ is an ordinal From Ordinal Class is Ordinal and Ordinal Membership is Trichotomy: :$A \in \On \lor A = \On \lor \On \in A$ But by the Burali-Forti Paradox $\On$ is a proper class. Therefore: :$A \in \On \lor A = \On$ {{qed}} {{explain|please say or link to why being a proper class means that $\On \notin A$}}
{{NotZFC}} By Ordinal is Member of Ordinal Class: :$A \in \On \lor A = \On$. In either case: :$A \subseteq \On$ since $\On$ is transitive.
We check all the Kolmogorov axioms in turn: === First Axiom === From Empty Set is Subset of All Sets and from the definitions of the event space and sample space: :$\varnothing \subseteq \Sigma \subseteq \Omega $ From Cardinality of Empty Set and Cardinality of Subset of Finite Set: :$0 \le \card \Sigma \le \card \Omega$ Dividing all terms by $\card \Omega$: :$0 \le \dfrac {\card \Sigma} {\card \Omega} \le 1$ The middle term is the asserted definition of $\map \Pr {\, \cdot \,}$. {{explain|Link to a page expanding on that statement -- perhaps better to say that it's the definition of the image of Pr?}} {{qed|lemma}} === Second Axiom === By hypothesis: :$\map \Pr \Omega = \dfrac {\card \Omega} {\card \Omega} = 1$ {{qed|lemma}} === Third Axiom === Follows from Cardinality is Additive Function and the corollary to the Inclusion-Exclusion Principle. {{qed}} {{proofread}} Category:Probability Theory o3l502qow7dfw5nt0ovn2rjryjdeq5x
From Image of Set Difference under Relation: Corollary 1 we have: :$\relcomp {\mathcal R \sqbrk {S_2} } {\mathcal R \sqbrk {S_1} } \subseteq \mathcal R \sqbrk {\relcomp {S_2} {S_1} }$ where $\mathcal R \subseteq S \times T$ is a relation on $S \times T$. As $f$, being a mapping, is also a relation, it follows directly that: :$\relcomp {f \sqbrk {S_2} } {f \sqbrk {S_1} } \subseteq f \sqbrk {\relcomp {S_2} {S_1} }$ {{qed}} Category:Image of Set Difference under Mapping 6vparbmf8toamnwm6qzwws9aegj6j36
From Image of Set Difference under Relation: Corollary 2 we have: :$\relcomp {\Img {\mathcal R} } {\mathcal R \sqbrk X} \subseteq \mathcal R \sqbrk {\relcomp S X}$ where $\mathcal R \subseteq S \times T$ is a relation on $S \times T$. As $f$, being a mapping, is also a relation, it follows directly that: :$\relcomp {\Img f} {f \sqbrk X} \subseteq f \sqbrk {\relcomp S X}$ {{Qed}}
From One-to-Many Image of Set Difference: Corollary 1 we have: :$\complement_{\mathcal R \left[{T_2}\right]} \left({\mathcal R \left[{T_1}\right]}\right) = \mathcal R \left[{\complement_{T_2} \left({T_1}\right)}\right]$ where $\mathcal R \subseteq T \times S$ is a one-to-many relation on $T \times S$. Hence as $f^{-1}: T \to S$ is a one-to-many relation: :$\complement_{f^{-1} \left[{T_2}\right]} \left({f^{-1} \left[{T_1}\right]}\right) = f^{-1} \left[{\complement_{T_2} \left({T_1}\right)}\right]$ {{qed}} Category:Preimages under Mappings Category:Set Difference Category:Set Complement kzmeqg8okrg6o116vd5hzaexl3zv0cn
From One-to-Many Image of Set Difference: Corollary 2 we have: :$\relcomp {\Img \RR} {\RR \sqbrk {S_1} } = \RR \sqbrk {\relcomp S {S_1} }$ where: :$S_1 \subseteq S$ :$\RR \subseteq T \times S$ is a one-to-many relation on $T \times S$. Hence as $f^{-1}: T \to S$ is a one-to-many relation: :$\relcomp {\Preimg f} {f^{-1} \sqbrk {T_1} } = f^{-1} \sqbrk {\relcomp T {T_1} }$ But from Preimage of Mapping equals Domain, we have that: :$\Preimg f = S$ Hence: :$\relcomp S {f^{-1} \sqbrk {T_1} } = f^{-1} \sqbrk {\relcomp T {T_1} }$ {{Qed}}
We have that $A \subseteq B$. Then by definition of relative complement: :$\relcomp B A = B \setminus A$ :$\relcomp {\RR \sqbrk B} {\RR \sqbrk A} = \RR \sqbrk B \setminus \RR \sqbrk A$ Hence, when $A \subseteq B$: :$\relcomp {\RR \sqbrk B} {\RR \sqbrk A} = \RR \sqbrk {\relcomp B A}$ means exactly the same thing as: :$\RR \sqbrk B \setminus \RR \sqbrk A = \RR \sqbrk {B \setminus A}$ Hence the result from One-to-Many Image of Set Difference. {{qed}} Category:Set Difference Category:Relation Theory kidx68wz3t76abdkb6vnfv5eq6tkopc
By definition of the image of $\mathcal R$: :$\Img {\mathcal R} = \mathcal R \sqbrk S$ So, when $B = S$ in One-to-Many Image of Set Difference: Corollary 1: :$\relcomp {\Img {\mathcal R} } {\mathcal R \sqbrk A} = \relcomp {\mathcal R \sqbrk S} {\mathcal R \sqbrk A}$ Hence: :$\relcomp {\Img {\mathcal R} } {\mathcal R \sqbrk A} = \mathcal R \sqbrk {\relcomp S A}$ means exactly the same thing as: :$\relcomp {\mathcal R \sqbrk S} {\mathcal R \sqbrk A} = \mathcal R \sqbrk {\relcomp S A}$ that is: :$\mathcal R \sqbrk S \setminus \mathcal R \sqbrk A = \mathcal R \sqbrk {S \setminus A}$ Hence the result from One-to-Many Image of Set Difference. {{Qed}} Category:Set Difference Category:Relation Theory 9adga48l5jrmaq0u5x1l3ojd7d37hzr
=== Even Index === Suppose $n$ is even. Let $x \ne 0$. Then $1^n = \paren {-1}^n$ by Power of Ring Negative, so $f_n$ is not injective. Also, by Even Power is Non-Negative, $f_n$ is not surjective. By definition, a bijection is both injective and surjective. It follows that for even $n$, $f_n$ is not bijective. {{qed|lemma}} === Odd Index === Now suppose $n$ is odd. From Odd Power Function is Strictly Increasing, $f_n$ is injective. From Odd Power Function is Surjective, $f^n$ is surjective. So when $n$ is odd, $f_n$ is both injective and surjective, and so by definition bijective. Hence the result. {{qed}}
As $S$ has more than one element, it must have at least two. So, let $a, b \in S$ be those two elements. Let $Z$ be the set of all sequences from $\set {a, b}$. Suppose $S^\infty$ were countable. From Subset of Countably Infinite Set is Countable, $Z$ is likewise countable. So by definition, it would be possible to set up a bijection $\phi: Z \leftrightarrow \N$ between $Z$ and the set $\N$ of natural numbers: {{begin-eqn}} {{eqn | l = 0       | o = \leftrightarrow       | r = \sequence {z_0} = \tuple {z_{00}, z_{01}, z_{02}, \ldots, z_{0n}, \ldots} }} {{eqn | l = 1       | o = \leftrightarrow       | r = \sequence {z_1} = \tuple {z_{10}, z_{11}, z_{12}, \ldots, z_{1n}, \ldots} }} {{eqn | l =        | o = \vdots       | r =  }} {{eqn | l = n       | o = \leftrightarrow       | r = \sequence {z_n} = \tuple {z_{n0}, z_{n1}, z_{n2}, \ldots, z_{nn}, \ldots} }} {{eqn | l =        | o = \vdots       | r =   }} {{end-eqn}} where each of $z_{rs}$ is: : an element of $\set {a, b}$ : the $s$'th element of $\sequence {z_r}$, the particular sequence which is in correspondence with $r \in \N$. Now we create the sequence $Q = \sequence {q_{dd} }$ where: :$q_{dd} = \begin{cases} a & : z_{dd} = b \\ b & : z_{dd} = a \end{cases}$ Thus: :$\forall n \in \N: q_{nn} \notin \sequence {z_n}$ and so: :$\forall n \in \N: Q \ne \sequence {z_n}$ So $Q \notin Z$. That is, we have constructed a sequence which has not been placed into correspondence with an element of $\N$. Thus by definition $Z$ is uncountable. Hence, by the Rule of Transposition, $S^\infty$ is likewise uncountable. {{qed}}
Let $S := \left\{{0, 1, 2}\right\}$ be a set. Let: : $\tau_1 := \left\{{\varnothing, \left\{{0}\right\}, \left\{{1}\right\}, \left\{{0, 1}\right\}, S}\right\}$  : $\tau_2 := \left\{{\varnothing, \left\{{0}\right\}, \left\{{2}\right\}, \left\{{0, 2}\right\}, S}\right\}$ be topologies on $S$.  Then: : $\tau := \tau_1 \cup \tau_2 = \left\{{\varnothing, \left\{{0}\right\}, \left\{{1}\right\}, \left\{{2}\right\}, \left\{{0, 1}\right\}, \left\{{0, 2}\right\}, S}\right\}$ For $\tau$ to be a topology the union of any number of elements of $\tau$ should also be in $\tau$. But: :$\left\{{1}\right\} \cup \left\{{2}\right\} = \left\{{1, 2}\right\} \not \in \tau$  Therefore $\tau$ is not a topology on $S$. Hence the result.  {{qed}}
We will prove that if both $\dfrac \theta \pi$ and $\cos \theta$ are rational then: :$\theta \in \left\{ {0, \dfrac \pi 3, \dfrac \pi 2}\right\}$ === Lemma === {{:Niven's Theorem/Lemma}} {{refactor|Extract this lemma into its own page.}} For any integer $n \ge 1$, there exists a polynomial $F_n \left({x}\right)$ such that: :$F_n \left({2 \cos t}\right) = 2 \cos n t$ In addition: :$\deg F_n = n$ and $F_n$ is a monic polynomial with integer coefficients. === Proof of Lemma === The proof proceeds by induction. For $n = 1$, it is seen that: :$F_1 \left({x}\right) = x$ fulfils the propositions. For $n = 2$: :$F_2 \left({x}\right) = x^2 - 2$ For $n > 2$: {{begin-eqn}} {{eqn | l = 2 \cos \left({n - 1}\right) t \cos t       | r = \cos n t + \cos \left({n - 2}\right) t       | c =  }} {{eqn | ll= \implies       | l = 2 \cos n t       | r = \left({2 \cos \left({n - 1}\right) t}\right) \left({2 \cos t}\right) - 2 \cos \left({n - 2}\right) t       | c =  }} {{eqn | r = 2 \cos t F_{n - 1} \left({2 \cos t}\right) - F_{n - 2} \left({2 \cos t}\right)       | c =  }} {{end-eqn}} so: : $F_n \left({x}\right) = x F_{n - 1} \left({x}\right) - F_{n - 2} \left({x}\right) \in \Z \left[{x}\right]$ will fulfil: : $F_n \left({2 \cos t}\right) = 2 \cos n t$ Because $\deg F_{n - 1} = n - 1$ and $\deg F_{n - 2} = n - 2$, we can conclude that: :$\deg F_n = \deg \left({x F_{n - 1} \left({x}\right) - F_{n - 2} \left({x}\right)}\right) = n$ In addition, the leading coefficient of $F_n$ is equal to the leading coefficient of $F_{n - 1}$, which is $1$. Hence the lemma. {{qed|lemma}} Suppose that $\dfrac \theta \pi$ is rational, meaning: :$\theta = \dfrac {2 \pi k} n$ where $k, n \in \Z$ and $n \ge 1$. Suppose also that $\cos \theta \in \Q$. Denoting $c = 2 \cos \theta \in \Q$, we get: :$F_n \left({c}\right) = F_n \left({2 \cos \dfrac {2 \pi k} n}\right) = 2 \cos \left({2 \pi k}\right) = 2$ So $c$ is a rational root of $F_n \left({x}\right) - 2$, which is a monic polynomial with integer coefficients. By Rational Root Theorem, $c$ must be an integer. But: :$\left\vert{c}\right\vert = \left\vert{2 \cos \theta}\right\vert \le 2$ so: :$c \in \left\{ {-2, -1, 0, 1, 2}\right\}$ Assuming that $0 \le \theta \le \dfrac \pi 2$, we get that: :$\theta \in \left\{ {0, \dfrac \pi 3, \dfrac \pi 2}\right\}$ Thus for any $\theta$ in the range $0 \le \theta \le \dfrac \pi 2$ such that both $\dfrac \theta \pi$ and $\cos \theta$ are rational, then: :$\theta \in \left\{ {0, \dfrac \pi 3, \dfrac \pi 2}\right\}$ Instead of the above, suppose that: :$0 \le \alpha \le \dfrac \pi 2$ and both of $\dfrac \alpha \pi$ and $\sin \alpha$ are rational. Then we can denote $\theta = \dfrac \pi 2 - \alpha$ and get that: :$0 \le \theta \le \dfrac \pi 2$ :$\dfrac \theta \pi \in Q$ :$\cos \theta \in \Q$ So: :$\dfrac \pi 2 - \alpha = \theta \in \left\{ {0, \dfrac \pi 3, \dfrac \pi 2}\right\}$ therefore: :$\alpha \in \left\{ {0, \dfrac \pi 6, \dfrac \pi 2}\right\}$ {{qed}} {{Namedfor|Ivan Morton Niven|cat = Niven}} It is suspected that this result is considerably older, and may date back as far as {{AuthorRef|Charles Hermite}}. Category:Trigonometry m1u0lyspoe8aleud92enyws7wdojzv7
We check all the Kolmogorov axioms in turn: === First Axiom === Let $n$ be the number of times a certain event $\omega$ is observed to happen. Let $n'$ be the number of times $\omega$ is observed not to happen. By Law of Excluded Middle and Principle of Non-Contradiction, $\omega$ either happened or did not, and not both. Therefore $n + n'$ is the total number of observations. It is supposed that at least one observation is actually made, so that $n + n' \ne 0$. The relative frequency model says that the probability of $\omega$ occurring can be defined as: :$\map \Pr \omega = \dfrac n {n + n'}$ the numerator and denominator of which are positive integers such that $n + n' \ge n$. If $\omega$ is observed never to happen, then $n = 0$ and $\map \Pr \omega = \dfrac 0 {0 + n'} = 0$.  If $\omega$ is observed to always happen, then $n' = 0$ and $\map \Pr \omega = \dfrac n {n + 0} = 1$. Otherwise, if $n, n' \ne 0$, from Mediant is Between: :$0 = \dfrac 0 {0 + n'} <  \dfrac n {n + n'} < \dfrac n {n + 0} = 1$ Thus $\Pr$ is bounded: :$0 \le \map \Pr {\cdot} \le 1$ {{qed|lemma}} === Second Axiom === By hypothesis: :$\map \Pr \Omega = \dfrac {n + n'} {n + n'} = 1$ {{qed|lemma}} === Third Axiom === This is a proof by induction. ==== Basis for the Induction ==== The case $j = 2$ is verified as follows: Let $A$ and $B$ be two pairwise disjoint events. Let $p$ and $q$ be the number of times $A$ and $B$ have been observed, respectively. Let $n$ be the total number of trials observed.  By the definition of pairwise disjoint, $A$ and $B$ never happened at the same time. By the same reasoning as the proof for the first axiom, in all $n$ observations: :$A$ happened $p$ times :$B$ happened $q$ times :$A \lor B$ happened $p + q$ times. {{MissingLinks|There's a result that can be linked to so as to replace all these words. In fact, now I think of it, there's a result which allows us to replace this entire inductive argument, we just need to find it.}} By hypothesis: {{begin-eqn}} {{eqn | l = \map \Pr {A \cup B}       | r = \dfrac {p + q} n }} {{eqn | l =       | r = \dfrac p n + \dfrac q n }} {{eqn | l =        | r = \map \Pr A + \map \Pr B }} {{end-eqn}} This is the basis for the induction. ==== Induction Hypothesis ==== Let $A_1, A_2, \ldots, A_j$ be $j$ pairwise disjoint events. By the definition of the relative frequency model, $j$ is finite. Assume: :$\displaystyle \map \Pr {\bigcup_{i \mathop = 1}^j A_i} = \map \Pr {A_1} + \map \Pr {A_2} + \cdots + \map \Pr {A_j}$ This is our induction hypothesis. ==== Induction Step ==== This is our induction step: Let $A_1, A_2, \ldots, A_j, A_{j + 1}$ be $j+1$ pairwise disjoint events. Define $C = A_1 \lor A_2 \lor A_3 \lor \cdots \lor A_j$. Then $C$ and $A_{j+1}$ are also pairwise disjoint. By the base case:  :$\map \Pr {C \cup A_{j + 1} } = \map \Pr C + \map \Pr {A_{j + 1} }$ By the definition of $C$, this equation is logically equivalent to: :$\displaystyle \map \Pr {\bigcup_{i \mathop = 1}^{j + 1} A_i} = \sum_{i \mathop = 1}^{j + 1} \map \Pr {A_i}$ By the definition of the relative frequency model, $j + 1$ is finite. The result follows by the Principle of Mathematical Induction. {{qed}} {{LEM}}
We have to check that $\BB = \set {\hointr a b: a, b \in \R}$ fulfills the axioms of being a basis for a topology. From Equivalent Definitions of Synthetic Basis we only have to check that: :$(1): \quad \bigcup \BB = \R$ :$(2): \quad \forall B_1, B_2 \in \BB: \exists V \in \BB: V \subseteq B_1 \cap B_2$ We have that: :$\forall n \in \Z: \hointr n {n + 1} \in \BB$ :$\R = \displaystyle \bigcup_{n \mathop \in \Z} \hointr n {n + 1} \subseteq \bigcup \BB$ Hence $\R = \bigcup \BB$ and condition $(1)$ is fulfilled. Now take $ B_1, B_2 \in \BB$ where: :$B_1 = \hointr {a_1} {b_1}$ :$B_2 = \hointr {a_2} {b_2}$ Let $B_3$ be constructed as: :$B_3 := \hointr {\max \set {a_1, a_2} } {\min \set {b_1, b_2} } \in \BB$ From the method of construction, it is clear that $B_3 = B_1 \cap B_2$. Thus taking $V = B_3$, condition $(2)$ is fulfilled. {{qed}}
Take $x, y \in \R$ such that $x \ne y$. {{WLOG}}, assume that $x < y$. From Real Numbers are Close Packed: :$\exists t \in \R: x < t < y$ Then: :$\hointr x t \cap \hointr t {y + 1} = \O$ We also have that $x \in \hointr x t$ by definition of half-open interval. Also, as $t < y < y+1$ it is clear that $y \in \hointr t {y + 1}$. By definition of the Sorgenfrey line, both are open in $T$. Thus we have found two disjoint subsets of $\R$ which are open in $T$, such that one contains $x$ and the other contains $y$. Hence the Sorgenfrey Line is Hausdorff by definition. {{qed}}