hcju/proofwikiqa huggingface dataset:
From Probability Generating Function of Negative Binomial Distribution (Second Form), we have: :$\displaystyle \Pi_X \left({s}\right) = \left({\frac {ps} {1 - qs}}\right)^n$ where $q = 1 - p$. From Expectation of Discrete Random Variable from PGF, we have: :$E \left({X}\right) = \Pi'_X \left({1}\right)$ We have: {{begin-eqn}} {{eqn | l=\Pi'_X \left({s}\right)       | r=\frac {\mathrm d} {\mathrm ds} \left({\frac {ps} {1 - qs} }\right)^n       | c= }} {{eqn | r=n p \left({\frac {\left({ps}\right)^{n-1} } {\left({1 - qs}\right)^{n+1} } }\right)       | c=First Derivative of PGF of Negative Binomial Distribution/Second Form }} {{end-eqn}} Plugging in $s = 1$: {{begin-eqn}} {{eqn | l=\Pi'_X \left({1}\right)       | r=n p \left({\frac {p^{n-1} } {\left({1 - q}\right)^{n+1} } }\right)       | c= }} {{eqn | r=n p \left({\frac {p^{n-1} } {p^{n+1} } }\right)       | c=as $p = 1 - q$ }} {{eqn | ll=\implies       | l=E \left({X}\right)       | r=\frac n p       | c=simplifying }} {{end-eqn}} {{qed}}
The Probability Generating Function of Negative Binomial Distribution (Second Form) is: :$\Pi_X \left({s}\right) = \left({\dfrac {p s} {1 - q s}}\right)^n$ We have that for a given negative binomial distribution , $n, p$ and $q$ are constant. {{ProofWanted}} Category:Negative Binomial Distribution Category:Derivatives of PGFs bzrcxgr6g5ckj57pyj8u84gh18l9xut
The Probability Generating Function of Negative Binomial Distribution (Second Form) is: :$\displaystyle \map {\Pi_X} s = \paren {\frac {p s} {1 - q s} }^n$ We have that for a given negative binomial distribution, $n, p$ and $q$ are constant. Thus we have: {{begin-eqn}} {{eqn | l = \frac \d {\d s} \map {\Pi_X} s       | r = \frac \d {\d s} \paren {\frac {p s} {1 - q s} }^n       | c =  }} {{eqn | r = n \paren {\frac {p s} {1 - q s} }^{n - 1} \frac \d {\d s} \paren {\frac {p s} {1 - q s} }       | c = Power Rule for Derivatives and Chain Rule for Derivatives }} {{eqn | r = n \paren {\frac {p s} {1 - q s} }^{n - 1} \paren {\frac {\paren {1 - q s} \frac \d {\d s} \paren {p s} - p s \frac \d {\d s} \paren {1 - q s} } {\paren {1 - q s}^2} }       | c = Quotient Rule for Derivatives and Chain Rule for Derivatives }} {{eqn | r = n \paren {\frac {p s} {1 - q s} }^{n - 1} \paren {\frac {\paren {1 - q s} p - p s \paren {-q} } {\paren {1 - q s}^2} }       | c = Power Rule for Derivatives where $n = 1$ }} {{eqn | r = n \paren {\frac {p s} {1 - q s} }^{n - 1} \paren {\frac {p - p q s + p q s} {\paren {1 - q s}^2} }       | c =  }} {{eqn | r = n \paren {\frac {p s} {1 - q s} }^{n - 1} \paren {\frac p {\paren {1 - q s}^2} }       | c =  }} {{eqn | r = n p \paren {\frac {\paren {p s}^{n - 1} } {\paren {1 - q s}^{n + 1} } }       | c =  }} {{end-eqn}} Hence the result. {{qed}} Category:Negative Binomial Distribution Category:Derivatives of PGFs 21kn3n9l8swrscycjgr0lgfq6da9z08
Follows directly from Nth Derivative of Reciprocal of Mth Power by putting $m = 1$. {{Qed}} Category:Derivatives Category:Reciprocals 5esfog7mo3vmt4mb8i0vdfikz5koxty
The Probability Generating Function of Negative Binomial Distribution (First Form) is: :$\map {\Pi_X} s = \paren {\dfrac q {1 - p s} }^n$ We have that for a given negative binomial distribution, $n, p$ and $q$ are constant. Thus we have: {{begin-eqn}} {{eqn | l = \frac \d {\d s} \map {\Pi_X} s       | r = \map {\frac \d {\d s} } {\paren {\dfrac q {1 - p s} }^n}       | c =  }} {{eqn | r = n \paren {\frac q {1 - p s} }^{n - 1} \map {\frac \d {\d s} } {\frac q {1 - p s} }       | c = Power Rule for Derivatives and Chain Rule for Derivatives }} {{eqn | r = n \paren {\frac q {1 - p s} }^{n - 1} \paren {-p} \paren {\frac {-q} {\paren {1 - p s}^2} }       | c = Power Rule for Derivatives and Chain Rule for Derivatives }} {{eqn | r = \frac {n p} q \paren {\frac q {1 - p s} }^{n + 1}       | c = simplifying }} {{end-eqn}} Hence the result. {{qed}} Category:Negative Binomial Distribution Category:Derivatives of PGFs 6816adzsbsp3olo0xuds1lsoellzqhi
From Probability Generating Function of Negative Binomial Distribution (First Form), we have: :$\Pi_X \left({s}\right) = \left({\dfrac q {1 - ps}}\right)^n$ where $q = 1 - p$. From Expectation of Discrete Random Variable from PGF, we have: :$E \left({X}\right) = \Pi'_X \left({1}\right)$ We have: {{begin-eqn}} {{eqn | l=\Pi'_X \left({s}\right)       | r=\frac {\mathrm d} {\mathrm ds} \left({\dfrac q {1 - ps} }\right)^n       | c= }} {{eqn | r=\dfrac {n p} q \left({\dfrac q {1 - ps} }\right)^{n+1}       | c=First Derivative of PGF of Negative Binomial Distribution/First Form }} {{end-eqn}} Plugging in $s = 1$: {{begin-eqn}} {{eqn | l=\Pi'_X \left({1}\right)       | r=\frac {n p} q \left({\frac q {1 - p} }\right)^{n+1}       | c= }} {{eqn | r=\frac {n p} q \left({\frac q q }\right)^{n+1}       | c=as $p = 1 - q$ }} {{eqn | ll=\implies       | l=E \left({X}\right)       | r=\frac {n p} q       | c=simplifying }} {{end-eqn}} {{qed}} Category:Expectation Category:Negative Binomial Distribution 375rbxm4qyxdbbwwfgzwke8lcw5gu8f
=== Necessary Condition === Let $\struct {X, \tau}$ be a Hausdorff space. Let $x, y \in X$ with $x \ne y$. Then by the definition of Hausdorff space there exist open sets $U, V \subseteq X$ such that: :$x \in U$ :$y \in V$ :$U \cap V = \O$ By Empty Intersection iff Subset of Complement, $U \subseteq X \setminus V$. By definition, $X \setminus V$ is closed. Thus by definition of closure: :$U^- \subseteq X \setminus V$ Since $y \in V$, by definition of relative complement it follows that $y \notin X \setminus V$. By definition of subset, $y \notin U^-$. {{qed|lemma}} === Sufficient Condition === Suppose that for each $x, y \in X$ such that $x \ne y$ there exists an open set $U$ such that $x \in U$ and $y \notin U^-$. Let $x, y \in X$ with $x \ne y$. Let $U$ be as described. Let $V = X \setminus U^-$. Then $y \in V$ by the definition of set difference. Since $U \subseteq U^-$, it follows that $U \cap V = \O$. As such $U$ and $V$ exist for all such $x$ and $y$, it follows that $\struct {X, \tau}$ is a Hausdorff space. {{qed}} Category:Hausdorff Spaces 3ymtzz86sy7g52016qat6i3jx6vo45i
The Probability Generating Function of Negative Binomial Distribution (Second Form) is: :$\displaystyle \Pi_X \left({s}\right) = \left({\frac {ps} {1 - qs}}\right)^n$ We have that for a given negative binomial distribution, $n, p$ and $q$ are constant. From First Derivative of PGF of Negative Binomial Distribution/Second Form: {{begin-eqn}} {{eqn | l = \frac {\mathrm d} {\mathrm d s} \Pi_X \left({s}\right)       | r = n p \left({\dfrac {\left({ps}\right)^{n-1} } {\left({1 - qs}\right)^{n+1} } }\right)       | c =  }} {{eqn | r = \frac n {p s^2} \left({\frac {ps} {1 - qs} }\right)^{n+1}       | c =  }} {{end-eqn}} Thus we have: {{begin-eqn}} {{eqn | l=\frac {\mathrm d^2} {\mathrm d s^2} \Pi_X \left({s}\right)       | r=\frac {\mathrm d} {\mathrm d s} \left({\frac n {p s^2} \left({\frac {ps} {1 - qs} }\right)^{n+1} }\right)       | c= }} {{eqn | r=\frac n {p s^2} \frac {\mathrm d} {\mathrm d s} \left({\left({\frac {ps} {1 - qs} }\right)^{n+1} }\right) + \frac {\mathrm d} {\mathrm d s} \left({\frac n {p s^2} }\right) \left({\frac {ps} {1 - qs} }\right)^{n+1}       | c=Product Rule for Derivatives }} {{eqn | r=\frac n {p s^2} \left({\frac {n+1}{p s^2} \left({\frac {ps} {1 - qs} }\right)^{n+2} }\right) + \frac {\mathrm d} {\mathrm d s} \left({\frac n {p s^2} }\right) \left({\frac {ps} {1 - qs} }\right)^{n+1}       | c=First Derivative of PGF of Negative Binomial Distribution/Second Form }} {{eqn | r=\frac n {p s^2} \left({\frac {n+1}{p s^2} \left({\frac {ps} {1 - qs} }\right)^{n+2} }\right) + \left({\frac {- 2 n} {p s^3} }\right) \left({\frac {ps} {1 - qs} }\right)^{n+1}       | c=Power Rule for Derivatives where $n = -2$ }} {{eqn | r=\frac {n \left({n+1}\right)} {p^2 s^4} \left({\frac {ps} {1 - qs} }\right)^{n+2} + \left({\frac {- 2 n} {p s^3} }\right) \left({\frac {ps} {1 - qs} }\right)^{n+1}       | c=dismayingly messy algebra }} {{eqn | r=\left({\frac {ps} {1 - qs} }\right)^{n+1} \left({\frac {n \left({n+1}\right)} {p s^3} \left({\frac 1 {1 - qs} }\right) + \left({\frac {- 2 n} {p s^3} }\right) }\right)       | c= }} {{eqn | r=\left({\frac {ps} {1 - qs} }\right)^{n+1} \left({\frac {n \left({n+1}\right) - 2 n \left({1 - qs}\right)} {p s^3 \left({1 - qs}\right)} }\right)       | c= }} {{eqn | r=\left({\frac {ps} {1 - qs} }\right)^{n+1} \left({\frac {n^2 + n - 2 n + 2 n q s} {p s^3 \left({1 - qs}\right)} }\right)       | c= }} {{eqn | r=\left({\frac {ps} {1 - qs} }\right)^{n+1} \left({\frac {n^2 - n + 2 n q s} {p s^3 \left({1 - qs}\right)} }\right)       | c= }} {{eqn | r=\left({\frac {ps} {1 - qs} }\right)^{n+1} \left({\frac {n \left({n-1}\right) + 2 n q s} {p s^3 \left({1 - qs}\right)} }\right)       | c= }} {{eqn | r=\left({\frac {ps} {1 - qs} }\right)^{n+2} \left({\frac {n \left({n-1}\right) + 2 n q s} {p^2 s^4 } }\right)       | c=multiplying top and bottom by $p s$ and gathering terms }} {{eqn | r=\left({\frac {ps} {1 - qs} }\right)^{n+2} \left({\frac {n \left({n-1}\right) + 2 n q s} {\left({p s^2}\right)^2} }\right)       | c=final tidy up }} {{end-eqn}} {{qed}} {{proofread}} Category:Negative Binomial Distribution Category:Derivatives of PGFs r7awsuh5jw57kmtticmwq9u2bfcwz05
Proof by induction: The Probability Generating Function of Negative Binomial Distribution (First Form) is: :$\displaystyle \Pi_X \left({s}\right) = \left({\frac q {1 - p s}}\right)^n$ For all $k \in \N_{> 0}$, let $P \left({k}\right)$ be the proposition: :$\dfrac {\mathrm d^k} {\mathrm d s^k} \Pi_X \left({s}\right) = \dfrac {n^{\overline k} p^k} {q^k} \left({\dfrac q {1 - p s} }\right)^{n + k}$ === Basis for the Induction === $P \left({1}\right)$ is the case: :$\dfrac {\mathrm d} {\mathrm d s} \Pi_X \left({s}\right) = \dfrac {n^{\overline 1} p} q \left({\dfrac q {1 - p s} }\right)^{n + 1}$ which is proved in First Derivative of PGF of Negative Binomial Distribution: First Form. Note that from Number to Power of One Rising is Itself: :$n^\overline 1 = n$ This is our basis for the induction. === Induction Hypothesis === Now we need to show that, if $P \left({j}\right)$ is true, where $j \ge 1$, then it logically follows that $P \left({j+1}\right)$ is true. So this is our induction hypothesis: :$\dfrac {\mathrm d^j} {\mathrm d s^j} \Pi_X \left({s}\right) = \dfrac {n^{\overline j} p^j} {q^j} \left({\dfrac q {1 - p s} }\right)^{n + j}$ Then we need to show: :$\dfrac {\mathrm d^{j + 1} } {\mathrm d s^{j + 1} } \Pi_X \left({s}\right) = \dfrac {n^{\overline {j + 1}} p^{j + 1} } {q^{j + 1} } \left({\dfrac q {1 - p s} }\right)^{n + j + 1}$ === Induction Step === This is our induction step: {{begin-eqn}} {{eqn | l = \dfrac {\mathrm d} {\mathrm d s} \left({\dfrac {\mathrm d^j} {\mathrm d s^j} \Pi_X \left({s}\right)}\right)       | r = \dfrac {\mathrm d} {\mathrm d s} \left({\dfrac {n^{\overline j} p^j} {q^j} \left({\dfrac q {1 - p s} }\right)^{n + j} }\right)       | c = Induction Hypothesis }} {{eqn | r = \dfrac {n^{\overline j} p^j} {q^j} \dfrac {\mathrm d} {\mathrm d s} \left({\left({\dfrac q {1 - p s} }\right)^{n + j} }\right)       | c = Derivative of Constant Multiple }} {{eqn | r = \dfrac {n^{\overline j} p^j} {q^j} \left({\dfrac {\left({n + j}\right) p} q \left({\dfrac q {1 - p s} }\right)^{n + j + 1} }\right)       | c = First Derivative of PGF of Negative Binomial Distribution/First Form }} {{eqn | r = \dfrac {n^{\overline {j + 1} } p^{j + 1} } {q^{j + 1} } \left({\dfrac q {1 - p s} }\right)^{n + j + 1}       | c = Simplification, and definition of Rising Factorial }} {{end-eqn}} So $P \left({j}\right) \implies P \left({j + 1}\right)$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\forall k \in \N_{> 0}: \dfrac {\mathrm d^k} {\mathrm d s^k} \Pi_X \left({s}\right) = \dfrac {n^{\overline k} p^k} {q^k} \left({\dfrac q {1 - p s} }\right)^{n + k}$ {{qed}} Category:Negative Binomial Distribution Category:Derivatives of PGFs onouvm3q9tc6295trx77k2l523rmhjz
The Probability Generating Function of Negative Binomial Distribution (First Form) is: :$\Pi_X \left({s}\right) = \left({\dfrac q {1 - ps}}\right)^n$ From Derivatives of PGF of Negative Binomial Distribution:First Form: :$(1): \quad \dfrac {\mathrm d^k} {\mathrm d s^k} \Pi_X \left({s}\right) = \dfrac {n^{\overline k} p^k} {q^k} \left({\dfrac q {1 - p s} }\right)^{n+k}$ where: : $n^{\overline k}$ is the rising factorial: $n^{\overline k} = n \left({n + 1}\right) \left({n + 2}\right) \cdots \left({n + k - 1}\right)$ : $q = 1 - p$ Putting $k = 2$ in $(1)$ above yields the required solution. {{qed}} Category:Negative Binomial Distribution Category:Derivatives of PGFs 0597n2lys8a14q6llwkyyn2olx5ewwb
From Variance of Discrete Random Variable from PGF, we have: :$\operatorname{var} \left({X}\right) = \Pi''_X \left({1}\right) + \mu - \mu^2$ where $\mu = E \left({x}\right)$ is the expectation of $X$. From the Probability Generating Function of Negative Binomial Distribution (Second Form), we have: :$\Pi_X \left({s}\right) = \dfrac {ps} {1 - qs}$ where $q = 1 - p$. From Expectation of Negative Binomial Distribution/Second Form, we have: :$\mu = \dfrac n p$ From Second Derivative of PGF of Negative Binomial Distribution/Second Form, we have: :$\dfrac {\mathrm d^2} {\mathrm d s^2} \Pi_X \left({s}\right) = \left({\dfrac {ps} {1 - qs} }\right)^{n+2} \left({\dfrac {n \left({n-1}\right) + 2 n q s} {\left({p s^2}\right)^2} }\right)$ Putting $s = 1$ and using the formula $\Pi''_X \left({1}\right) + \mu - \mu^2$: {{begin-eqn}} {{eqn | l = \operatorname{var} \left({X}\right)       | r = \left({\frac p {1 - q} }\right)^{n+2} \left({\frac {n \left({n-1}\right) + 2 n q} {p^2} }\right) + \frac n p - \frac {n^2} {p^2}       | c =  }} {{eqn | r = \left({\frac p p}\right)^{n+2} \left({\frac {n \left({n-1}\right) + 2 n q} {p^2} }\right) + \frac {n \left({1 - q}\right)} {p^2} - \frac {n^2} {p^2}       | c = as $q = 1 - p$ and so $p = 1 - q$ }} {{eqn | r = \frac {n^2 - n + 2 n q + n - n q - n^2} {p^2}       | c = gathering terms and multiplying out }} {{eqn | ll = \implies       | l = \operatorname{var} \left({X}\right)       | r = \frac {n q} {p^2}       | c = simplification }} {{end-eqn}} {{qed}}
Proof by induction: For all $n \in \N_{> 0}$, let $\map P n$ be the proposition: :$\displaystyle \map {\Pi_Z} s = \prod_{j \mathop = 1}^m \map {\Pi_{X_j} } s$ whenever $m \le n$ $\map P 1$ is true, as this just says $\map {\Pi_{X_1} } s = \map {\Pi_{X_1} } s$. === Basis for the Induction === $\map P 2$ is the case: :$\map {\Pi_{X + Y} } s = \map {\Pi_X} s \, \map {\Pi_Y} s$ which is proved in PGF of Sum of Independent Discrete Random Variables. This is our basis for the induction. === Induction Hypothesis === Now we need to show that, if $\map P j$ is true, where $j \ge 2$, then it logically follows that $\map P {k + 1}$ is true. So this is our induction hypothesis: :$\displaystyle \map {\Pi_Z} s = \prod_{j \mathop = 1}^m \map {\Pi_{X_j} } s$ whenever $m \le k$. Then we need to show: :$\displaystyle \map {\Pi_Z} s = \prod_{j \mathop = 1}^m \map {\Pi_{X_j} } s$ whenever $m \le {k + 1}$. === Induction Step === This is our induction step: Let $Z = X_1 + X_2 + \cdots + X_k + X_{k + 1}$ {{begin-eqn}} {{eqn | l = \map {\Pi_Z} s       | r = \map {\Pi_{X_1 + X_2 + \cdots + X_k + X_{k + 1} } } s       | c = by definition of $Z$ }} {{eqn | r = \map {\Pi_{\paren {X_1 + X_2 + \cdots + X_k} + X_{k + 1} } } s       | c =  }} {{eqn | r = \map {\Pi_{\paren {X_1 + X_2 + \cdots + X_k} } } s \, \map {\Pi_{X_{k + 1} } } s       | c = Basis for the Induction }} {{eqn | r = \prod_{j \mathop = 1}^k \map {\Pi_{X_j} } s \, \map {\Pi_{X_{k + 1} } } s       | c = Induction Hypothesis }} {{eqn | r = \prod_{j \mathop = 1}^{k + 1} \map {\Pi_{X_j} } s       | c =  }} {{end-eqn}} So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction. Therefore: :$\forall n \in \N: \map {\Pi_Z} s = \displaystyle \prod_{j \mathop = 1}^n \map {\Pi_{X_j} } s$ {{qed}}
Let $\mathcal F$ be the set of all ordered pairs $\left({A, B}\right)$ such that $A$ and $B$ are open, $x \in A$, and $A \cap B = \varnothing$. As a set of ordered pairs, $\mathcal F$ constitutes a relation on $\tau$: :$\mathcal F \subseteq \tau \times \tau$ By the definition of Hausdorff space, for each $y \in C$ there exists an element $\left({A, B}\right) \in \mathcal F$ such that $y \in B$. Thus the image of $\mathcal F$ covers $C$. By the definition of compactness, there exists a finite subset $\mathcal G \subseteq \mathcal F$ such that $\displaystyle C \subseteq V = \bigcup \operatorname{Im} \left({G}\right)$. {{explain|This does not follow directly. See the parent page for an outline of how the principle of finite choice leads to it.}} Then $\operatorname{Im}^{-1} \left({G}\right)$ is also finite, so by the definition of a topology: : $\displaystyle U = \bigcap \operatorname{Im}^{-1} \left({G}\right)$ is open. Then $x \in U$, $C \subseteq V$, and $U \cap V = \varnothing$. {{qed}} Category:Compact Spaces Category:Hausdorff Spaces 5oy62q1poz8qhuu0i3mqt5hcwccezv4
From Probability Generating Function of Poisson Distribution, we have that the probability generating functions of $X$ and $Y$ are given by: : $\Pi_X \left({s}\right) = \left({q + ps}\right)^m$ : $\Pi_Y \left({s}\right) = \left({q + ps}\right)^n$ respectively. Now because of their independence, we have: {{begin-eqn}} {{eqn | l = \Pi_{X + Y} \left({s}\right)       | r = \Pi_X \left({s}\right) \Pi_Y \left({s}\right)       | c = PGF of Sum of Independent Discrete Random Variables }} {{eqn | r = \left({q + ps}\right)^m \left({q + ps}\right)^n       | c = by definition of $X$ and $Y$ }} {{eqn | r = \left({q + ps}\right)^{m + n}       | c = Exponent Combination Laws: Product of Powers }} {{end-eqn}} This is the probability generating function for a discrete random variable with a binomial distribution: :$\operatorname{B} \left({\left({m + n}\right), p}\right)$ Therefore: :$Z = X + Y \sim \operatorname{B} \left({\left({m + n}\right), p}\right)$ {{qed}}
Let $F$ be a field. Let $\FF$ be the collection of all extensions of $F$.  {{explain|Is $\FF$ a set or a proper class? It is not obvious that this collection is "small enough" to be a set, and I was unable to find a proof of this on ProofWiki.<br/>(On the other hand, does it matter which it is? // Yes, because Zorn's lemma only covers the case of it being a set.)}} Define an ordering on $\FF$ thus: :$\forall K, L \in \FF: K \preceq L \iff L$ is an extension of $K$. 	 Let $C$ be a chain in $\FF$. Let $E = \bigcup_{K \mathop \in C} K$. $E$ satisfies all field axioms, so $E \in \FF$. {{explain|Why does $E$ satisfy all field axioms?}} By Set is Subset of Union, $E$ is an upper bound for $C$.  	 By Zorn's Lemma, $C$ has a maximal element $m$. Then $m$ is an algebraic closure of $F$. {{explain|Why is this an algebraic closure of $F$?}} {{qed}} {{BPI}} Category:Field Theory 6ravysqwxem9dk60vu9am8rajmj2oum
Since $\struct {X, \tau}$ is separable, $X$ has a countable everywhere dense subset $S$. {{WLOG}}, assume that $\O \notin \mathcal F$. By the definition of point finite, $\set {V \in \mathcal F: x \in V}$ is finite for each $x \in S$. From Open Set Characterization of Denseness, each element of $\mathcal F$ contains an element of $S$. From Union of Set of Sets when a Set Intersects All: :$\mathcal F = \displaystyle \bigcup_{x \mathop \in S} \set {V \in \mathcal F: x \in V}$ Thus by Countable Union of Finite Sets is Countable, $\mathcal F$ is countable. {{qed}} Category:Topology peav638k41ru8v0pgjpbwib9lvlbmu4
Let $d$ denote the metric induced by $\norm {\, \cdot \,}$, that is: :$\map d {x, y} = \norm {x - y}$ From Metric Induced by Norm is Metric we have that $d$ is indeed a  metric. Then, from the Reverse Triangle Inequality as applied to metric spaces: :$\forall x, y, z \in X: \size {\norm {x - z} - \norm {y - z} } \le \norm {x - y}$ Then: :$\forall x, y \in X: \size {\norm x - \norm y} = \size {\norm {x - 0} - \norm {y - 0} } \le \norm {x - y}$. {{qed}}
=== Form 1 implies Form 2 === Suppose that Axiom of Countable Choice: Form 1 holds. Let $S$ be a countable set of non-empty sets. Then $S$ is either finite or countably infinite. If $S$ is finite, then it has a choice function by the Principle of Finite Choice. Suppose instead that $S$ is countably infinite. Then there exists a bijection $f: \N \to S$. Thus by Form 1, there exists a mapping $g: \N \to \bigcup S$ such that: :$\forall n \in \N: g \left({n}\right) \in f \left({n}\right)$ Then $g \circ f^{-1}: S \to \bigcup S$ is a choice function for $S$. As every countable set has a choice function, Axiom of Countable Choice: Form 2 holds. {{qed|lemma}} === Form 2 implies Form 1 === Suppose that Axiom of Countable Choice: Form 2 holds. Let $\langle x_n \rangle_{n \mathop \in \N}$ be a sequence of non-empty sets. By the Axiom of Replacement: :$I = \left\{{x_n: n \in \N}\right\}$ is a set. By Surjection from Natural Numbers iff Countable, $I$ is countable. By Axiom of Countable Choice: Form 2, $I$ has a choice function $h: I \to \bigcup I$. Define a sequence in $\bigcup I$ by letting $y_n = h \left({x_n}\right)$. Then by the definition of choice function: :$\forall n \in \N: y_n \in x_n$ As this holds for all such sequences, Axiom of Countable Choice: Form 1 holds. {{qed}} Category:Axiom of Countable Choice 28oe4z2qid06czulkvz9gxdwl9o0148
First we show that such an element $d$ exists. By Ring of Integers is Principal Ideal Domain there exists a generator $e$ of $I$. If $e < 0$, then since the units of $\Z$ are $\set {\pm 1}$, we have by definition that $-e$ is an associate of $e$. Therefore by $(3)$ of Principal Ideals in Integral Domain $-e > 0$ is also a generator of $I$. In particular setting $d = \max \set {e, -e}$, $d$ is a non-negative generator of $I$. By Bézout's Lemma, we are required to show that $d$ is the smallest positive integer combination of $m$ and $n$. By definition: :$I = \set {a m + b n: a, b \in \Z}$ Thus we are required to show that $d$ is the smallest positive element of $I$. Suppose that $d' \le d$ is a positive element of $I$, not larger than $d$. By hypothesis $d$ generates $I$, so there exists $a \in \Z$ such that $d' = ad$. Since $d > 0$, we can therefore write $a = \dfrac {d'} d \in \Q$. Moreover, because $d' > 0$, by $(6)$ of Properties of Ordered Ring we have: :$0 = \dfrac 0 d < \dfrac {d'} d$ Using the hypothesis that $d' \le d$, we have the pair of inequalities: :$0 < a = \dfrac {d'} d \le 1$ By the definition we have $a \in \Z$, so this shows that $\dfrac {d'} d = 1$. It follows that $d' = d$. Therefore there is no positive element of $I$ smaller than $d$. {{Qed}} Category:Greatest Common Divisor 3q9ivkfsvkoj3r0bbf43mu1jbo0a7q8
=== Necessary Condition === Let $S$ be everywhere dense in $X$. Let $U$ be open and non-empty. Then $U$ has an element $x$. Since $S$ is everywhere dense in $X$, $x \in S^-$, the closure of $S$. By Equivalence of Definitions of Adherent Point, every open neighborhood of $x$ contains an element of $S$. Thus in particular, $U$ contains an element of $S$. {{qed|lemma}} === Sufficient Condition === Suppose that every non-empty open set in $X$ contains an element of $S$. Let $x \in X$. Let $U$ be an open neighborhood of $x$. Then $U$ contains an element $s$ of $S$. As this holds for all open neighborhoods of $x$, Equivalence of Definitions of Adherent Point shows that $x \in S^-$. As this holds for all $x \in X$, $S$ is everywhere dense in $X$. {{qed}} Category:Denseness h61ats4ilaanie2ojggvvq9i0te8jtv
Suppose that $B \in F$. Then $B \cap S$ has an element $x_B$. Thus $B \in \set {A \in F: x_B \in A}$. By the definition of union: :$\ds B \in \bigcup_{x \mathop \in S} \set {A \in F: x \in A}$ Suppose instead that $\ds B \in \bigcup_{x \mathop \in S} \set {A \in F: x \in A}$. Then by the definition of union, there exists an $x_B \in S$ such that $B \in \set {A \in F: x_B \in A} \subseteq F$. Thus $B \in F$. We have shown that $\ds \forall B: \paren {B \in F \iff B \in \bigcup_{x \mathop \in S} \set {A \in F: x \in A} }$. Therefore $\ds F = \bigcup_{x \mathop \in S} \set {A \in F: x \in A}$ by the Axiom of Extension. {{qed}} Category:Set Union 7jyioquf6sl4x14g7cxhhckygid8ad3
By Reduced Residue System under Multiplication forms Abelian Group, $U$ is equal to the set of integers modulo $n$ which are coprime to $n$. It follows by Cardinality of Reduced Residue System: :$\order U = \map \phi n$ {{qed}}