hcju/proofwikiqa huggingface dataset:
Let $x \in S$ and $x \in U \in \tau$. Let $\mathcal B_x = \set{ W \in \tau : x \in W, W \text{ is connected in } T}$.   By definition of local basis, we have to show that there exists a connected open set $V \in \mathcal B_x$ with $x \in V \subset U$. Let $V = \map {\operatorname{Comp}_x} U$ denote the component of $x$ in $U$. From Open Set in Open Subspace, it suffices to show that $V$ is open in $U$. By Set is Open iff Neighborhood of all its Points, we may do this by showing that $V$ is a neighborhood in $U$ of all of its points. Let $y \in V$. By assumption, there exists a connected neighborhood $W$ of $y$ in $T$ with $W \subset U$. By Neighborhood in Topological Subspace, $W$ is a neighborhood of $y$ in $U$. By definition of component: :$W \subseteq \map {\operatorname{Comp}_y} U = \map {\operatorname{Comp}_x} U = V$ By Neighborhood iff Contains Neighborhood, $V$ is a neighborhood of $y$ in $U$. Because $y$ was arbitrary, $V$ is open in $U$. From Open Set in Open Subspace, $V$ is open in $T$. Hence $V \in \mathcal B_x$. Because $U$ was arbitrary, $\mathcal B_x$ is a local basis of $x$ consisting of (open) connected sets. Since $x$ was arbitrary, then each point of $T$ has a local basis consisting entirely of connected sets in $T$.
By definition of local basis, each of these connected sets in $\mathcal D_x$ is open in $T$. Consider the set $\displaystyle \mathcal D = \bigcup_{x \mathop \in S} \mathcal D_x$. From Union of Local Bases is Basis, $\mathcal D$ is a basis for the topology $\tau$. Since each $\mathcal D_x$ consists entirely of connected sets, $\mathcal D$ also consists entirely of connected sets.
For each $x \in S$ we define: : $\mathcal B_x = \set {B \in \mathcal B: x \in B}$ From Basis induces Local Basis, $\mathcal B_x$ is a local basis. As each element of $\mathcal B_x$ is also an element of $\mathcal B$, it follows that $\mathcal B_x$ is also formed of connected sets. Thus, for each point $x \in S$, there is a local basis which consists entirely of connected sets.
Let $U$ be an open subset of $T$. From Open Set is Union of Elements of Basis, $U$ is a union of open connected sets in $T$. From Open Set in Open Subspace and Connected Set in Subspace, $U$ is a union of open connected sets in $U$. From Components are Open iff Union of Open Connected Sets, the components of $U$ are open in $U$. From Open Set in Open Subspace then the components of $U$ are open in $T$.
Let $\mathcal B = \set {U \in \tau : U \text{ is connected in } T}$. Let $U$ be open in $T$. By assumption, the components of $U$ are open in $T$. From Connected Set in Subspace, the components of $U$ are connected in $T$. By the definition of the components of a topological space, $U$ is the union of its components. Hence $U$ is the union of open connected sets in $T$. By definition, $\mathcal B$ is an basis for $T$.
Let $H$ be a connected component of $T$. By the definition of connected component, $H$ is connected. From Closure of Connected Set is Connected then the closure $H^-$ is connected. By the definition of the closure, $H \subseteq H^-$. By the definition of connected component, $H$ is a maximal connected set. Hence $H = H^-$. From Set is Closed iff Equals Topological Closure, $H$ is closed. {{qed}} Category:Connected Sets 7wrc491ds5yv7byhfpne2zqogz84jjh
First note that if $S$ is a singleton, then there exists only one constant mapping on $S$. In such a circumstance, $f = g$ and so $f \circ g \ne g \circ f$. {{qed|lemma}} So, let $\card S > 1$. Then there exist at least $2$ distinct elements $a$ and $b$ of $S$. Thus, let $f$ and $g$ be defined as: :$\forall x \in S: \map f x = a$ :$\forall x \in S: \map g x = b$ for some $a, b \in S$ such that $a \ne b$. We have that: {{begin-eqn}} {{eqn | l = \map {\paren {f \circ g} } x       | r = \map f {\map g x}       | c =  }} {{eqn | r = \map f b       | c = Definition of $\map g x$ }} {{eqn | r = a       | c = Definition of $\map f x$ }} {{eqn | l = \map {\paren {g \circ f} } x       | r = \map g {\map f x}       | c =  }} {{eqn | r = \map g a       | c = Definition of $\map f x$ }} {{eqn | r = b       | c = Definition of $\map g x$ }} {{end-eqn}} So $f \circ g \ne g \circ f$ as required. {{qed}}
Let $\eqclass x {\mathcal R}, \eqclass y {\mathcal R} \in S / \mathcal R$ === Sufficient Condition === Let $q_{\mathcal R}: S \to S / \mathcal R$ be an injection. Then: {{begin-eqn}} {{eqn | l = x       | o = \mathcal R       | r = y       | c =  }} {{eqn | l = \eqclass x {\mathcal R}       | r = \eqclass y {\mathcal R}       | c = {{Defof|Equivalence Class}} }} {{eqn | ll= \leadsto       | l = \map {q_{\mathcal R} } x       | r = \map {q_{\mathcal R} } y       | c = {{Defof|Quotient Mapping}} }} {{eqn | ll= \leadsto       | l = x       | r = y       | c = {{Defof|Injection}} }} {{end-eqn}} That is: :$\mathcal R$ is the equality relation. {{qed|lemma}} === Necessary Condition === Let $q_{\mathcal R}: S \to S / \mathcal R$ be a mapping which is specifically not an injection. Then: {{begin-eqn}} {{eqn | lo= \exists a, b \in S, a \ne b:       | l = \map {q_{\mathcal R} } a       | r = \map {q_{\mathcal R} } b       | c = $q_{\mathcal R}$ is not an injection }} {{eqn | ll= \leadsto       | l = \eqclass a {\mathcal R}       | r = \eqclass b {\mathcal R}       | c = {{Defof|Quotient Mapping}} }} {{eqn | ll= \leadsto       | l = a       | o = \mathcal R       | r = b       | c = {{Defof|Equivalence Class}} }} {{end-eqn}} That is: :$a \ne b$ but: :$a \mathrel {\mathcal R} b$ and so $\mathcal R$ is not the equality relation. From Rule of Transposition it follows that: :if $\mathcal R$ is the equality relation then $q_{\mathcal R}$ is an injection. {{qed}}
{{begin-eqn}} {{eqn | l = \map {\cl_H} A       | r = \bigcap \set {K \subseteq H: A \subseteq K, K \text{ is closed in } T_H}       | c = {{Defof|Closure (Topology)}} }} {{eqn | r = \bigcap \set {N \cap H: A \subseteq N, N \text{ is closed in } T}       | c = Closed Set in Topological Subspace }} {{eqn | r = H \cap \bigcap \set {N: A \subseteq N, N \text{ is closed in } T}       | c = Intersection Distributes over Intersection of Family of Sets }} {{eqn | r = H \cap \map \cl A       | c = {{Defof|Closure (Topology)}} }} {{end-eqn}} {{qed}} Category:Topology r45jlaeftv0bf5esgeqlyof0gfrs5ra
From Local Basis for Open Sets Implies Neighborhood Basis of Open Sets, it follows directly that: :each point of $T$ has a neighborhood basis  consisting entirely of path-connected sets in $T$.
Let $x \in S$ and $x \in U \in \tau$. Let $\mathcal B_x = \set{ W \in \tau : x \in W, W \text{ is path-connected in } T}$.   By definition of local basis, we have to show that there exists a path-connected open set $V\in\tau$ with $x\in V\subset U$. Let $V = \map {\operatorname{PC}_x} U$ denote the path component of $x$ in $U$. From Open Set in Open Subspace, it suffices to show that $V$ is open in $U$. By Set is Open iff Neighborhood of all its Points, we may do this by showing that $V$ is a neighborhood in $U$ of all of its points. Let $y \in V$. By assumption, there exists a path-connected neighborhood $W$ of $y$ in $T$ with $W\subset U$. By Neighborhood in Topological Subspace, $W$ is a neighborhood of $y$ in $U$. By definition of path component: :$W \subset \map {\operatorname{PC}_y} U = \map {\operatorname{PC}_x} U = V$ By Neighborhood iff Contains Neighborhood, $V$ is a neighborhood of $y$ in $U$. Because $y$ was arbitrary, $V$ is open in $U$. From Open Set in Open Subspace, $V$ is open in $T$. Hence $V \in \mathcal B_x$. Because $U$ was arbitrary, $\mathcal B_x$ is a local basis of $x$ consisting of (open) path-connected sets. Since $x$ was arbitrary, then each point of $T$ has a local basis consisting entirely of path-connected sets in $T$.
By definition of local basis, each of these path-connected sets in $\mathcal D_x$ is open in $T$. Consider the set $\displaystyle \mathcal D = \bigcup_{x \mathop \in S} \mathcal D_x$. From Union of Local Bases is Basis, $\mathcal D$ is a basis for the topology $\tau$. Since each $\mathcal D_x$ consists entirely of path-connected sets, $\mathcal D$ also consists entirely of path-connected sets.
For each $x \in S$ we define: : $\mathcal B_x = \set{B \in \mathcal B: x \in B}$ From Basis induces Local Basis, $\mathcal B_x$ is a local basis. As each element of $\mathcal B_x$ is also an element of $\mathcal B$, it follows that $\mathcal B_x$ is also formed of path-connected sets. Thus, for each point $x \in S$, there is a local basis which consists entirely of path-connected sets.
Let $U$ be an open subset of $T$. From Open Set is Union of Elements of Basis, $U$ is a union of open path-connected sets in $T$. From Open Set in Open Subspace and Path-Connected Set in Subspace, $U$ is a union of open path-connected sets in $U$. From Path Components are Open iff Union of Open Path-Connected Sets, the path components of $U$ are open in $U$. From Open Set in Open Subspace then the path components of $U$ are open in $T$.
Let $\mathcal B = \set{U \in \tau : U \text{ is path-connected in } T}$. Let $U$ be open in $T$. By assumption, the path components of $U$ are open in $T$. From Path-Connected Set in Subspace, the path components of $U$ are path-connected in $T$. By the definition of the path components of a topological space, $U$ is the union of its path components. Hence $U$ is the union of open path-connected sets in $T$. By definition, $\mathcal B$ is an basis for $T$.
By definition of binomial coefficient: {{begin-eqn}} {{eqn | l = \dbinom {2 n} n       | r = \dfrac {\paren {2 n}!} {\paren {n!}^2}       | c = where $n!$ denotes the factorial of $n$ }} {{eqn | ll= \leadsto       | l = \dbinom {2 n} n \paren {n!}^2       | r = \paren {2 n}!       | c =  }} {{end-eqn}} Let $\mathbb P$ denote the set of prime numbers. Let $p \in \mathbb P$ such that $n < p < 2 n$. By definition of factorial: :$p \divides \paren {2 n}!$ That is: :$p \divides \dbinom {2 n} n \paren {n!}^2$ From Prime iff Coprime to all Smaller Positive Integers: :$p \nmid n!$ where $\nmid$ denotes non-divisibility. and so: :$p \nmid \paren {n!}^2$ But from Euclid's Lemma for Prime Divisors, $p$ is a divisor of either $\dbinom {2 n} n$ or $\paren {n!}^2$. Hence it must be the case that: :$p \divides \dbinom {2 n} n$ {{qed}}
Let $\tau_A$ be the subspace topology on $A$ induced by $\tau$. Let $\tau'_A$ be the subspace topology on $A$ induced by $\tau_B$. By the definition of a path-connected set, $A$ is path-connected in $T$ {{iff}} every two points in $A$ are path-connected in $\struct {A, \tau_A}$. Similarly, $A$ is path-connected in $T_B$ {{iff}} every two points in $A$ are path-connected in $\struct {A, \tau'_A}$. From Subspace of Subspace is Subspace, we have $\tau_A = \tau'_A$. Hence every two points in $A$ are path-connected in $\struct {A, \tau_A}$ {{iff}} every two points in $A$ are path-connected in $\struct {A, \tau'_A}$. {{qed}} Category:Path-Connected Spaces fjn398nft5bqigl7msccdtvfw6sjie6
We apply the Method of Truth Tables. As can be seen by inspection, the truth values under the main connectives match for all boolean interpretations. $\begin{array}{|cccc||ccccc|} \hline P & (A & B & C) & (((A & \land & \neg & B) & \land & \neg & C) & \lor & ((\neg & A & \land & B) & \land & \neg & C)) & \lor & ((\neg & A & \land & \neg & B) & \land & C) \\ \hline F & F & F & F & F & F & T & F & F & T & F & F & T & F & F & F & F & T & F & F & T & F & T & T & F & F & F \\ T & F & F & T & F & F & T & F & F & F & T & F & T & F & F & F & F & F & T & T & T & F & T & T & F & T & T \\ T & F & T & F & F & F & F & T & F & T & F & T & T & F & T & T & T & T & F & T & T & F & F & F & T & F & F \\ F & F & T & T & F & F & F & T & F & F & T & F & T & F & T & T & F & F & T & F & T & F & F & F & T & F & T \\ T & T & F & F & T & T & T & F & T & T & F & T & F & T & F & F & F & T & F & T & F & T & F & T & F & F & F \\ F & T & F & T & T & T & T & F & F & F & T & F & F & T & F & F & F & F & T & F & F & T & F & T & F & F & T \\ F & T & T & F & T & F & F & T & F & T & F & F & F & T & F & T & F & T & F & F & F & T & F & F & T & F & F \\ F & T & T & T & T & F & F & T & F & F & T & F & F & T & F & T & F & F & T & F & F & T & F & F & T & F & T \\ \hline \end{array}$ {{qed}}
First note that from Composition of Mappings is Associative: :$\paren {h \circ g} \circ f = h \circ \paren {g \circ f}$ and so on. However, while there is no need to use parenthesis to establish the order of composition of mappings, in the following the technique will be used in order to clarify what is being done. We have that: :$\paren {h \circ g} \circ f = I_A$ It follows from Injection iff Left Inverse that $f$ is an injection. We also have that: :$f \circ \paren {h \circ g} = I_B$ It follows from Surjection iff Right Inverse that $f$ is a surjection. So $f$ is a bijection. It follows from the corollary to Bijection iff Left and Right Inverse that $h \circ g$ is also a bijection. Thus we have that $h \circ g$ is both a left inverse and a right inverse of $f$. It follows by definition that $h \circ g$ is the inverse of $f$: :$f^{-1} = h \circ g$ The same argument, mutatis mutandis, can be used to show that: :$g$ and $h$ are bijections :the inverse of $g$ is $f \circ h$ :the inverse of $h$ is $g \circ f$. {{qed}}
Let:  :$\displaystyle Z = \sum_{i \mathop = 1}^n \alpha_i X_i$  Let $M_Z$ be the moment generating function of $Z$.  We aim to show that:  :$\displaystyle Z \sim \Gaussian {\sum_{i \mathop = 1}^n \alpha_i \mu_i} {\sum_{i \mathop = 1}^n \alpha^2_i \sigma^2_i}$ By Moment Generating Function of Gaussian Distribution and Moment Generating Function is Unique, it is therefore sufficient to show that:  :$\displaystyle \map {M_Z} t = \map \exp {\paren {\sum_{i \mathop = 1}^n \alpha_i \mu_i} t + \frac 1 2 \paren {\sum_{i \mathop = 1}^n \alpha^2_i \sigma^2_i} t^2}$ We also have, by Moment Generating Function of Gaussian Distribution, that the moment generating function of $X_i$, $M_{X_i}$, is given by:  :$\displaystyle \map {M_{X_i}} t = \map \exp {\mu_i t + \frac 1 2 \sigma^2_i t^2}$ We have: {{begin-eqn}} {{eqn	| l = \map {M_Z} t 	| r = \prod_{i \mathop = 1}^n \map {M_{X_i} } {\alpha_i t} 	| c = Moment Generating Function of Linear Combination of Independent Random Variables }} {{eqn	| r = \prod_{i \mathop = 1}^n \map \exp {\mu_i \alpha_i t + \frac 1 2 \sigma^2_i \alpha^2_i t^2} }} {{eqn	| r = \map \exp {\sum_{i \mathop = 1}^n \paren {\mu_i \alpha_i t + \frac 1 2 \sigma^2_i \alpha^2_i t^2} } 	| c = Exponential of Sum }} {{eqn	| r = \map \exp {\paren {\sum_{i \mathop = 1}^n \alpha_i \mu_i} t + \frac 1 2 \paren {\sum_{i \mathop = 1}^n \alpha^2_i \sigma^2_i} t^2} }} {{end-eqn}} {{qed}} Category:Gaussian Distribution ndnwg2zith4ibmqisppio6m3l0izyir
{{begin-eqn}} {{eqn	| l = \map {M_Z} t 	| r = \expect {\map \exp {t Z} } 	| c = {{Defof|Moment Generating Function}} }} {{eqn	| r = \expect {\map \exp {t \paren {\alpha X + \beta} } } }} {{eqn	| r = \expect {\map \exp {\paren {\alpha t} X} \map \exp {\beta t} } 	| c = Exponential of Sum }} {{eqn	| r = \map \exp {\beta t} \expect {\map \exp {\paren {\alpha t} X} } 	| c = Linearity of Expectation Function }} {{eqn	| r = e^{\beta t} \map {M_X} {\alpha t} 	| c = {{Defof|Moment Generating Function}} }} {{end-eqn}} {{qed}} Category:Moment Generating Functions gqrh8x8a1pdm6gzokzlek4tb3nrekl2
{{begin-eqn}} {{eqn	| l = \frac {X - \mu} \sigma 	| r = \frac 1 \sigma X - \frac \mu \sigma }} {{eqn	| o = \sim 	| r = \Gaussian {\frac \mu \sigma - \frac \mu \sigma} {\paren {\frac 1 \sigma}^2 \sigma^2} 	| c = Linear Transformation of Gaussian Random Variable }} {{eqn	| r = \Gaussian 0 1 }} {{end-eqn}} {{qed}} Category:Gaussian Distribution 9y2s9b0zahlr8acby6a1jbxsakimu87
Let $Y \sim \chi^2_n$ where $\chi^2_n$ is the chi-squared distribution with $n$ degrees of freedom.  Let $Z \sim \chi^2_m$ where $\chi^2_m$ is the chi-squared distribution with $m$ degrees of freedom. Then, by Quotient of Independent Random Variables with Chi-Squared Distribution Divided by Degrees of Freedom has F-Distribution: :$X = \dfrac {Y / n} {Z / m} \sim F_{n, m}$ Applying Quotient of Independent Random Variables with Chi-Squared Distribution Divided by Degrees of Freedom has F-Distribution again, we have:  :$\dfrac 1 X = \dfrac {Z / m} {Y / n} \sim F_{m, n}$ {{qed}} Category:F-Distribution 8a4cedwfk7a2anrt2pzbonmkfd0bdfc
Let $Y \sim F_{1, k}$.  We aim to show that:  :$\map \Pr {Y < x^2} = \map \Pr {|X| < x}$ for all $x \ge 0$.  That is:  :$\map \Pr {Y < x^2} = \map \Pr {-x < X < x}$ for all $x \ge 0$.  We have:  {{begin-eqn}} {{eqn	| l = \map \Pr {Y < x^2} 	| r = \int_0^{x^2} \frac {k^{k/2} 1^{1/2} u^{\paren {1/2} - 1} } {\paren {k + u}^{\paren {1 + k}/2} \map \Beta {1/2, k/2} } \rd u 	| c = {{Defof|F-Distribution}} }} {{eqn	| r = \frac {k^{k/2} } {k^{k/2} \sqrt k \map \Beta {1/2, k/2} } \int_0^{x^2} \frac 1 {\sqrt u} \paren {1 + \frac u k}^{-\paren {1 + k}/2} \rd u 	| c = extracting a factor of $k^{-\paren {1 + k}/2}$, rewriting }} {{eqn	| r = \frac {\map \Gamma {\frac {k + 1} 2} } {\sqrt k \map \Gamma {\frac 1 2} \map \Gamma {\frac k 2} } \int_0^{x^2} \frac 1 {\sqrt u} \paren {1 + \frac u k}^{-\paren {1 + k}/2} \rd u 	| c = {{Defof|Beta Function}} }} {{eqn	| r = \frac {\map \Gamma {\frac {k + 1} 2} } {\sqrt {\pi k} \map \Gamma {\frac k 2} } \int_0^{x^2} \frac 1 {\sqrt u} \paren {1 + \frac u k}^{-\paren {1 + k}/2} \rd u 	| c = Gamma Function of One Half }} {{eqn	| r = \frac {\map \Gamma {\frac {k + 1} 2} } {\sqrt {\pi k} \map \Gamma {\frac k 2} } \int_0^x \frac {2 v} {\sqrt {v^2} } \paren {1 + \frac {v^2} k}^{-\paren {1 + k}/2} \rd v 	| c = substituting $u = v^2$ }} {{eqn	| r = \frac {\map \Gamma {\frac {k + 1} 2} } {\sqrt {\pi k} \map \Gamma {\frac k 2} } \int_0^x 2 \paren {1 + \frac {v^2} k}^{-\paren {1 + k}/2} \rd v }} {{eqn	| r = \frac {\map \Gamma {\frac {k + 1} 2} } {\sqrt {\pi k} \map \Gamma {\frac k 2} } \int_{-x}^x \paren {1 + \frac {v^2} k}^{-\paren {1 + k}/2} \rd v 	| c = Definite Integral of Even Function }} {{eqn	| r = \map \Pr {-x < X < x} 	| c = {{Defof|Student's t-Distribution}} }} {{end-eqn}} {{qed}} Category:Student's t-Distribution Category:F-Distribution lv3r6ec9fpyvl7gzbwxg9o7i5n4ivnd
Let $B = \displaystyle \bigcup_{\alpha \mathop \in A} B_\alpha$. From definition 3 of a connected set, $B$ is connected in $T$ {{iff}} the subspace $\struct {B, \tau_B}$ is a connected space. From definition 4 of a connected space, $\struct {B, \tau_B}$ is connected {{iff}} the only clopen sets in $\struct {B, \tau_B}$ are $B$ and $\O$. Let $U$ be any clopen set of the subspace of $\struct {B, \tau_B}$. Let $V = B \setminus U$. From Complement of Clopen Set is Clopen, $V$ is also clopen. Hence $U, V$ are disjoint clopen sets such that $B = U \cup V$.  {{WLOG}} assume $x \in U$. From Set is Subset of Union: :$\displaystyle \forall \alpha \in A : B_\alpha \subseteq B = U \cup V$ From Connected Subset of Union of Disjoint Open Sets: :$\forall \alpha \in A : B_\alpha \subseteq U$ From Union is Smallest Superset: :$B \subseteq U$ Since $U \subseteq B$ then $U = B$. From Set Difference with Self is Empty Set, :$V = B \setminus U = B \setminus B = \O$ Hence the only clopen sets in $\struct {B, \tau_B}$ are $B$ and $\O$. The result follows.  {{qed}}
Follows immediately from Union of Connected Sets with Non-Empty Intersections is Connected. {{qed}}
Let $U' = A \cap U$ and $V' = A \cap V$. By definition $U'$ and $V'$ are open sets in the subspace $\struct{A, \tau_A}$.  From Intersection is Empty Implies Intersection of Subsets is Empty $U'$ and $V'$ are disjoint. Hence $U'$ and $V'$ are separated sets by definition. Now {{begin-eqn}} {{eqn | l = A       | r = A \cap \paren {U \cup V}       | c = Intersection with Subset is Subset }} {{eqn | r = \paren {A \cap U} \cup \paren {A \cap V}       | c = Intersection Distributes over Union }} {{eqn | r = U' \cup V' }} {{end-eqn}} Since $A$ is connected then one of $U'$ or $V'$ is empty. {{WLOG}} assume that $V' = \O$. Then {{begin-eqn}} {{eqn | l = A       | r = U' \cup V' }} {{eqn | r = U' \cup \O }} {{eqn | r = U'       | c = Union with Empty Set }} {{eqn | r = A \cap U }} {{eqn | ll= \leadsto       | l = A       | o = \subseteq       | r = U       | c = Intersection with Subset is Subset }} {{end-eqn}} {{qed}}
Taking Laplace transforms, we have:  :$\laptrans {y'' + y} = \laptrans 0$ From Laplace Transform of Constant Mapping, we have:  :$\laptrans 0 = 0$ We also have:  {{begin-eqn}} {{eqn	| l = \laptrans {y'' + y} 	| r = \laptrans {y''} + \laptrans y 	| c = Linear Combination of Laplace Transforms }} {{eqn	| r = s^2 \laptrans y - s \map y 0 - \map {y'} 0 + \laptrans y 	| c = Laplace Transform of Second Derivative }} {{end-eqn}} So:  :$\paren {s^2 + 1} \laptrans y = s \map y 0 + \map {y'} 0$ Giving:  :$\laptrans y = \map y 0 \dfrac s {s^2 + 1} + \map {y'} 0 \dfrac 1 {s^2 + 1}$ So:  {{begin-eqn}} {{eqn	| l = y 	| r = \map { {\mathcal L}^{-1} } {\map y 0 \frac s {s^2 + 1} + \map {y'} 0 \frac 1 {s^2 + 1} } 	| c = {{Defof|Inverse Laplace Transform}} }} {{eqn	| r = \map y 0 \map { {\mathcal L}^{-1} } {\frac s {s^2 + 1} } + \map {y'} 0 \map { {\mathcal L}^{-1} } {\frac 1 {s^2 + 1} } 	| c = Linear Combination of Laplace Transforms }} {{eqn	| r = \map y 0 \map { {\mathcal L}^{-1} } {\laptrans {\cos x} } + \map {y'} 0 \map { {\mathcal L}^{-1} } {\laptrans {\sin x} } 	| c = Laplace Transform of Cosine, Laplace Transform of Sine }} {{eqn	| r = \map y 0 \cos x + \map {y'} 0 \sin x 	| c = {{Defof|Inverse Laplace Transform}} }} {{end-eqn}} Setting $C_1 = \map {y'} 0$ and $C_2 = \map y 0$ gives the result. {{qed}}
{{AimForCont}} there exists $x, y \in \N$ such that $n x < y$ for every natural number $n$. Consider the set $S$, defined as: :$S := \set {y - n x: n \in \N}$ By hypothesis, $S$ contains only natural numbers. By the Well-Ordering Principle, $S$ contains a smallest element, $y - m x$ for example. But $y - \paren {m + 1} x$ is also in $S$, as $S$ contains all natural numbers of this form. Furthermore, we have: :$y - \paren {m + 1} x = \paren {y - m x} - x < y - m x$ But this contradicts our assertion that $y - m x$ is the smallest element of $S$. This assertion originated from the assumption that there exists $x, y \in \N$ such that $n x < y$ for every natural number $n$. Hence that assumption cannot hold. Hence the result by proof by Contradiction. {{qed}} {{Namedfor|Archimedes of Syracuse|cat = Archimedes}}
Consider $\N$ defined as a naturally ordered semigroup. The result follows directly from Principle of Mathematical Induction for Naturally Ordered Semigroup: General Result. {{qed}}
Let $T$ be the set of all $1$-based natural numbers not in $S$: :$T = \N_{>0} \setminus S$ {{AimForCont}} $T$ is non-empty. From the Well-Ordering Principle, $T$ has a smallest element. Let this smallest element be denoted $a$. We have been given that $1 \in S$. So: :$a > 1$ and so: :$0 < a - 1 < a$ As $a$ is the smallest element of $T$, it follows that: :$a - 1 \notin T$ That means $a - 1 \in S$. But then by hypothesis: :$\paren {a - 1} + 1 \in S$ But: :$\paren {a - 1} + 1 = a$ and so $a \notin T$. This contradicts our assumption that $a \in T$. It follows by Proof by Contradiction that $T$ has no such smallest element. Hence it follows that $T$ can have no elements at all. That is: :$\N_{>0} \setminus S = \O$ That is: :$S = \N_{>0}$
Let $\displaystyle \bigcap_{\alpha \mathop \in I} B_\alpha = \O$. From Set Intersection Preserves Subsets/Families of Sets: :$\displaystyle \bigcap_{\alpha \mathop \in I} A_\alpha \subseteq \bigcap_{\alpha \mathop \in I} B_\alpha = \O$ From Subset of Empty Set: :$\displaystyle \bigcap_{\alpha \mathop \in I} A_\alpha = \O$ {{qed}} Category:Set Intersection Preserves Subsets Category:Indexed Families rt2bjcc1j08lpm15m8r0yhqjler403o
Let $B = \displaystyle \bigcup \family {B_\alpha}_{\alpha \mathop \in A}$. Let $a, b \in B$. Then  :$\exists \alpha, \beta \in A: a \in B_\alpha \land b \in B_\beta$. As $B_\alpha$ is a path-connected set in $T$ then $a$ and $x$ are path-connected points. Similarly, $x$ and $b$ are path-connected points. From Joining Paths makes Another Path, $a$ and $b$ are path-connected points. Since $a$ and $b$ were arbitrary points then $B$ is a path-connected set of $T$. {{qed}}
From Sum of Geometric Sequence: :$\displaystyle \sum_{j \mathop = 0}^{n - 1} x^j = \frac {x^n - 1} {x - 1}$ The result follows by setting $x = 2$. {{Qed}}