hcju/proofwikiqa huggingface dataset:
A special case of Primitive of $\dfrac 1 {x^4 + a^4}$, setting $a = 1$. {{begin-eqn}} {{eqn | l = \int \frac {\d x} {x^4 + a^4}       | r = \frac 1 {4 a^3 \sqrt 2} \map \ln {\frac {x^2 + a x \sqrt 2 + a^2} {x^2 - a x \sqrt 2 + a^2} } - \frac 1 {2 a^3 \sqrt 2} \paren {\map \arctan {1 - \frac {x \sqrt 2} a} - \map \arctan {1 + \frac {x \sqrt 2} a} } + C       | c =  }} {{eqn | ll= \leadsto       | l = \int \frac {\d x} {1 + x^4}       | r = \frac 1 {4 \sqrt 2} \map \ln {\frac {x^2 + x \sqrt 2 + 1} {x^2 - x \sqrt 2 + 1} } - \frac 1 {2 \sqrt 2} \paren {\map \arctan {1 - x \sqrt 2} - \map \arctan {1 + x \sqrt 2} } + C       | c =  }} {{eqn | r = \frac 1 {2 \sqrt 2} \paren {\dfrac 1 2 \map \ln {\frac {x^2 + x \sqrt 2 + 1} {x^2 - x \sqrt 2 + 1} } - \paren {\map \arctan {1 - x \sqrt 2} - \map \arctan {1 + x \sqrt 2} } } + C       | c = extracting factor }} {{eqn | r = \frac 1 {2 \sqrt 2} \paren {\dfrac 1 2 \map \ln {\frac {x^2 + x \sqrt 2 + 1} {x^2 - x \sqrt 2 + 1} } - \map \arctan {\dfrac {\paren {1 - x \sqrt 2} - \paren {1 + x \sqrt 2} } {1 + \paren {1 - x \sqrt 2} \paren {1 + x \sqrt 2} } } } + C       | c = Difference of Arctangents }} {{eqn | r = \frac 1 {2 \sqrt 2} \paren {\dfrac 1 2 \map \ln {\frac {x^2 + x \sqrt 2 + 1} {x^2 - x \sqrt 2 + 1} } - \map \arctan {\dfrac {-2 x \sqrt 2} {1 + \paren {1 - 2 x^2} } } } + C       | c = Difference of Two Squares }} {{eqn | r = \frac 1 {2 \sqrt 2} \paren {\dfrac 1 2 \map \ln {\frac {x^2 + x \sqrt 2 + 1} {x^2 - x \sqrt 2 + 1} } + \map \arctan {\sqrt 2 \dfrac x {1 - x^2} } } + C       | c = Inverse Tangent is Odd Function }} {{eqn | r = \frac 1 {2 \sqrt 2} \paren {\dfrac 1 2 \map \ln {\frac {x^2 + x \sqrt 2 + 1} {x^2 - x \sqrt 2 + 1} } + \paren {\frac \pi 2 - \map \arctan {\frac 1 {\sqrt 2} \paren {\frac {1 - x^2} x} } } } + C       | c = Sum of Arctangent and Arccotangent }} {{eqn | r = \frac 1 {2 \sqrt 2} \paren {\frac 1 2 \ln \size {\frac {x^2 + \sqrt 2 x + 1} {x^2 - \sqrt 2 x + 1} } + \map \arctan {\frac 1 {\sqrt 2} \paren {x - \frac 1 x} } } + C'       | c = Arctangent is Odd Function, setting $C' = \dfrac \pi 2 + C$ }} {{end-eqn}} {{qed}}
We have: :$\dfrac \d {\d x} \paren {q \sin a x + p \cos a x} = a q \cos a x - a p \sin a x$ Thus: {{begin-eqn}} {{eqn | l = \int \frac {\d x} {p + q \tan a x}       | r = \int \frac {\d x} {p + q \dfrac {\sin a x} {\cos a x} }       | c = Tangent is Sine divided by Cosine }} {{eqn | r = \int \frac {\cos a x \rd x} {p \cos a x + q \sin a x}       | c =  }} {{eqn | r = \frac 1 {p^2 + q^2} \int \frac {\paren {p^2 + q^2} \cos a x \rd x} {p \cos a x + q \sin a x}       | c =  }} {{eqn | r = \frac 1 {p^2 + q^2} \int \frac {p^2 \cos a x + p q \sin a x - p q \sin a x + q^2 \cos a x} {p \cos a x + q \sin a x} \rd x       | c =  }} {{eqn | r = \frac 1 {p^2 + q^2} \paren {\int \frac {p^2 \cos a x + p q \sin a x} {p \cos a x + q \sin a x} \ \d x + \int \frac {- p q \sin a x + q^2 \cos a x} {p \cos a x + q \sin a x} \rd x}       | c =  }} {{eqn | r = \frac 1 {p^2 + q^2} \paren {\int p \rd x + \frac q a \int \frac {\map \d {p \cos a x + q \sin a x} } {p \cos a x + q \sin a x} }       | c =  }} {{eqn | r = \frac {p x} {p^2 + q^2} + \frac q {a \paren {p^2 + q^2} } \ln \size {p \cos a x + q \sin a x} + C       | c = Primitive of Constant and Primitive of Reciprocal }} {{end-eqn}} {{qed}}
First, let $\arctan \dfrac p q = \phi$. Let $z = a x + \phi$. {{begin-eqn}} {{eqn | l = z       | r = \map \sin {a x + \phi}       | c =  }} {{eqn | ll= \leadsto       | l = \frac {\d z} {\d x}       | r = a \map \cos {a x + \phi}       | c = Derivative of $\sin a x$ etc. }} {{eqn | r = a \cos z       | c =  }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | l = \int \frac {\d x} {p + q \tan a x}       | r = \int \frac {\d x} {p + q \dfrac {\sin a x} {\cos a x} }       | c = Tangent is Sine divided by Cosine }} {{eqn | r = \int \frac {\cos a x \rd x} {p \cos a x + q \sin a x}       | c = multiplying top and bottom by $\cos a x$ }} {{eqn | r = \int \frac {\cos a x \rd x} {\sqrt {p^2 + q^2} \map \sin {a x + \phi} }       | c = Multiple of Sine plus Multiple of Cosine: Sine Form }} {{eqn | r = \frac 1 {\sqrt {p^2 + q^2} } \int \frac {\cos a x \rd x} {\map \sin {a x + \phi} }       | c = Primitive of Constant Multiple of Function }} {{eqn | r = \frac 1 {\sqrt {p^2 + q^2} } \paren {\frac {\ln \size {\map \sin {a x + \phi} } } {a \cos \phi} + \tan \phi \int \frac {\sin a x \rd x} {\map \sin {a x + \phi} } + C}       | c = Primitive of $\dfrac {\cos a x} {\map \sin {a x + \phi} }$ }} {{end-eqn}} {{finish}}
The proof proceeds by the Principle of Mathematical Induction: Let $\map P n$ be the proposition: :$\paren {\circ^n a} \circ b = b \circ \paren {\circ^n a}$ === Basis of the Induction === {{begin-eqn}} {{eqn | l = a \circ b       | r = b \circ a       | c =  }} {{eqn | ll= \leadsto       | l = \paren {\circ^1 a} \circ b       | r = b \circ \paren {\circ^1 a}       | c = Definition of $\circ^1$ }} {{end-eqn}} demonstrating that $\map P 1$ holds. This is the basis for the induction. === Induction Hypothesis === Suppose that $\map P k$ holds: :$\paren {\circ^k a} \circ b = b \circ \paren {\circ^k a}$ This is the induction hypothesis. It remains to be shown that: :$\map P k \implies \map P {k + 1}$ That is, that: :$\paren {\circ^{k + 1} a} \circ b = b \circ \paren {\circ^{k + 1} a}$ === Induction Step === This is the induction step: Thus: {{begin-eqn}} {{eqn | l = \paren {\circ^{k + 1} a} \circ b       | r = \paren {\paren {\circ^k a} \circ a} \circ b       | c = Definition of $\circ^n a$ }} {{eqn | r = \paren {\circ^k a} \circ \paren {a \circ b}       | c = $\circ$ is associative }} {{eqn | r = \paren {\circ^k a} \circ \paren {b \circ a}       | c = $b$ commutes with $a$ under $\circ$ }} {{eqn | r = \paren {\paren {\circ^k a} \circ b} \circ a       | c = $\circ$ is associative }} {{eqn | r = \paren {b \circ \paren {\circ^k a} } \circ a       | c = Induction Hypothesis }} {{eqn | r = b \circ \paren {\paren {\circ^k a} \circ a}       | c = $\circ$ is associative }} {{eqn | r = b \circ \paren {\circ^{k + 1} a}       | c = Definition of $\circ^n a$ }} {{end-eqn}} So $\map P {k + 1}$ is true. Thus: :$\forall m \in \N_{>0}: \paren {\circ^m a} \circ b = b \circ \paren {\circ^m a}$ {{qed|lemma}} By repeating the argument above, replacing $a$ with $b$ and $b$ with $\circ^m a$, we have: {{begin-eqn}} {{eqn | l = b \circ \paren {\circ^m a}       | r = \paren {\circ^m a} \circ b }} {{eqn | ll= \leadsto       | l = \paren {\circ^n b} \circ \paren {\circ^m a}       | r = \paren {\circ^m a} \circ \paren {\circ^n b} }} {{end-eqn}} Hence the result: :$\forall m, n \in \N_{>0}: \paren {\circ^m a} \circ \paren {\circ^n b} = \paren {\circ^n b} \circ \paren {\circ^m a}$ That is: :$\forall m, n \in \N_{>0}: a^m \circ b^n = b^n \circ a^m$ {{qed}}
It is established in Element under Right Operation is Left Identity that $\struct {S, \rightarrow}$ is a semigroup. From the definition of right operation: :$\forall x, y \in S: x \rightarrow y = y$ from which it can immediately be seen that all elements of $S$ are indeed right zeroes. {{qed}} From More than One Right Zero then No Left Zero, it also follows that there is no left zero.
It is established in Element under Left Operation is Right Identity that $\struct {S, \leftarrow}$ is a semigroup. From the definition of left operation: :$\forall x, y \in S: x \leftarrow y = x$ from which it can immediately be seen that all elements of $S$ are indeed left zeroes. {{qed}} From More than One Right Zero then No Left Zero, it also follows that there is no right zero.
Let $\struct {S, \circ}$ be an algebraic structure with more than one left zero. Take any two of these, and call them $z_{L_1}$ and $z_{L_2}$, where $z_{L_1} \ne z_{L_2}$. Suppose $\struct {S, \circ}$ has a right zero. Call it $z_R$. Then, by the behaviour of $z_R$, $z_{L_1}$ and $z_{L_2}$: :$z_{L_1} = z_{L_1} \circ z_R = z_R$ :$z_{L_2} = z_{L_2} \circ z_R = z_R$ So $z_{L_1} = z_R = z_{L_2}$, which contradicts the supposition that $z_{L_1}$ and $z_{L_2}$ are different. Therefore, in an algebraic structure with more than one left zero, there can be no right zero. {{qed}} The same argument can be applied to an algebraic structure with more than one right zero. Category:Zero Elements 48vfjejkwqkxzeq6sxzvhdtg9doj2qd
Let $\struct {S, \circ}$ be an algebraic structure such that: :$\exists z_L \in S: \forall x \in S: z_L \circ x = z_L$ :$\exists z_R \in S: \forall x \in S: x \circ z_R = z_R$ Then $z_L = z_L \circ z_R = z_R$ by both the above, hence the result. The uniqueness of the left and right zero is a direct result of Zero Element is Unique. {{Qed}}
{{begin-eqn}} {{eqn | l = \int \cot x \rd x       | r = \int \frac {\cos x} {\sin x} \rd x       | c = {{Defof|Real Cotangent Function}} }} {{eqn | r = \int \frac {\paren {\sin x}'} {\sin x} \rd x       | c = Derivative of Sine Function }} {{eqn | r = \ln \size {\sin x} + C       | c = Primitive of Function under its Derivative }} {{end-eqn}} {{qed}}
If $x \in \liminf_n C_n$ then there exist a sequence $\left \langle{x_k}\right \rangle_{n \in \N}$ such that $x_k \to x$ while: :$x_k \in C_{n_k}$  and  :$\left \langle{n_k}\right \rangle_{k \in \N} \subseteq \N$ is a strictly increasing sequence of indices. For any $V \in \mho \left({x}\right)$ there exists $N_0 \in\N$ such that for all $i \ge N_0$:  : $x_i \in V$ and: : $x_i \in C_{n_i}$ Thus: :$C_{n_i} \cap V \ne \varnothing$ Therefore $x \in \left\{{x \in \mathcal X: \forall V \in \mho \left({x}\right): \exists N_0 \in \N: \forall n \ge N_0: C_n \cap V \ne \varnothing}\right\}$. {{qed|lemma}} Let $x \in \left\{ {x \in \mathcal X: \forall V \in \mho \left({x}\right): \exists N \in \mathcal N_\infty: \forall n \in N: C_n \cap V \ne \varnothing}\right\}$. Thus: : $\forall V \in \mho \left({x}\right): \exists N \in \mathcal N_\infty: \forall n \in N: C_n \cap V \ne \varnothing$ Then there exists a strictly increasing sequence: :$\left \langle{n_k}\right \rangle_{k \in \N} \subseteq \N$ such that for every $V \in \mho \left({x}\right)$: :$\exists x_k \in C_{n_k}\cap V$. Hence $x_k \to x$ in the topology $\tau$. {{qed}} {{explain|A general increase in clarity needed.}}
This proof is similar to the proof of Taylor's theorem with the remainder in the Lagrange form, and is also based on Rolle's Theorem. Observe that: :$R \left({x_i}\right) = 0$ for $i = 0, \dotsc, n$ and that: :$R^{\left({n + 1}\right)} = f^{\left({n + 1}\right)}$ {{WLOG}}, assume that $x$ is different from all $x_i$ for $i = 0, \dotsc, n$. Let the function $g$ be defined by: :$g \left({t}\right) = R \left({t}\right) - \dfrac{\left({t - x_0}\right) \left({t - x_1}\right) \dotsm \left({t - x_n}\right) R \left({x}\right)} {\left({x - x_0}\right) \left({x - x_1}\right) \dotsm \left({x - x_n}\right)}$ Then $g \left({x_i}\right) = 0$ for $i = 0, \dotsc, n$, and $g \left({x}\right) = 0$. Denote by $J$ the interval spanned by $x$ and $x_i$, $i = 0, \dotsc, n$. Thus $g$ has at least $n + 2$ zeros in $J$. {{Proofread}} The Extended Rolle's Theorem is applied in $J$ successively to $g$, $g'$, $g''$ and so on until $g^{\left({n + 1}\right)}$, which thus has at least one zero $\xi$ located between the two known zeros of $g^{\left({n}\right)}$ in $J$. Then: :$0 = g^{\left({n + 1}\right)} \left({\xi}\right) = f^{\left({n + 1}\right)} \left({\xi}\right) - \dfrac{\left({n + 1}\right)! R \left({x}\right)} {\left({x - x_0}\right) \left({x - x_1}\right) \dotsm \left({x - x_n}\right)}$ and the formula for $R \left({x}\right)$ follows. {{qed}}
Let $p: H \to V$ be the orthogonal projection of $H$ onto $V$. Then the orthogonal projection of $H$ onto $V^\perp$ is $\mathbf 1 - p$, where $\mathbf 1$ is the identity map of $H$. The fact that $t$ stabilizes $V$ can be expressed as: :$\left({\mathbf 1 - p}\right) t p = 0$ or: :$p t p = t p$ The goal is to show that: :$p t \left({\mathbf 1 - p}\right) = 0$ We have that $\left({a, b}\right) \mapsto \operatorname{tr} \left({a b^*}\right)$ is an inner product on the space of endomorphisms of $H$. Here, $b^*$ denotes the adjoint operator of $b$. Thus it will suffice to show that $\operatorname{tr} \left({x x^*}\right) = 0$ for $x = p t \left({\mathbf 1 - p}\right)$. This follows from a direct computation, using properties of the trace and orthogonal projections: {{begin-eqn}} {{eqn | l = x x^*       | r = p t \left({\mathbf 1 - p}\right)^2 t^* p       | c =  }} {{eqn | r = p t \left({\mathbf 1 - p}\right) t^* p       | c =  }} {{eqn | r = p t t^* p - p t p t^* p       | c =  }} {{eqn | ll= \implies       | l = \operatorname{tr} \left({x x^*}\right)       | r = \operatorname{tr} \left({p t t^* p}\right) - \operatorname{tr} \left({p t p t^* p}\right)       | c =  }} {{eqn | r = \operatorname{tr} \left({p^2 t t^*}\right) - \operatorname{tr} \left({p^2 t p t^*}\right)       | c =  }} {{eqn | r = \operatorname{tr} \left({p t t^*}\right) - \operatorname{tr} \left({\left({p t p}\right) t^*}\right)       | c =  }} {{eqn | r = \operatorname{tr} \left({p t t^*}\right) - \operatorname{tr} \left({t p t^*}\right)       | c =  }} {{eqn | r = \operatorname{tr} \left({p t t^*}\right) - \operatorname{tr} \left({p t^* t}\right)       | c =  }} {{eqn | r = \operatorname{tr} \left({p \left({t t^* - t^*t}\right)}\right)       | c =  }} {{eqn | r = \operatorname{tr} \left({0}\right)       | c =  }} {{eqn | r = 0       | c =  }} {{end-eqn}} {{qed}} Category:Hilbert Spaces ln3fp79opuvp5z9ckxy0ptr7tuexxoc
Let: {{begin-eqn}} {{eqn | l = u       | r = \cot x + \csc x       | c =  }} {{eqn | ll= \leadsto       | l = \frac {\d u} {\d x}       | r = \frac \d {\d x} \cot x + \frac \d {\d x} \csc x        | c = Linear Combination of Derivatives }} {{eqn | r = -\csc^2 x + \frac \d {\d x} \csc x        | c = Derivative of Cotangent Function }} {{eqn | r = -\csc^2 x - \csc x \cot x        | c = Derivative of Cosecant Function }} {{eqn | r = -\csc x \paren {\csc x + \cot x}       | c = factorising }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | l = \int \csc x \rd x       | r = \int \frac {\csc x \paren {\csc x + \cot x} } {\csc x + \cot x} \rd x       | c = multiplying top and bottom by $\csc x + \cot x$ }} {{eqn | r = -\int \frac {-\csc x \paren {\csc x + \cot x} } {\csc x + \cot x} \rd x       | c = Primitive of Constant Multiple of Function }} {{eqn | r = -\ln \size {\csc x + \cot x} + C       | c = Primitive of Function under its Derivative }} {{end-eqn}} {{qed}}
First, let: {{begin-eqn}} {{eqn | l = \mathbf A       | r = \sum_{n \mathop = 0}^\infty \frac {\cos n} {K^n} }} {{eqn | l = \mathbf B       | r = \sum_{n \mathop = 0}^\infty \frac {\sin n} {K^n} }} {{end-eqn}} Now, consider: :$\mathbf A + i \, \mathbf B$ where $i$ is the imaginary unit: {{begin-eqn}} {{eqn | l = \mathbf{A} + i \, \mathbf B       | r = \sum_{n \mathop = 0}^\infty \frac {\cos n} {K^n} + \sum_{n \mathop = 0}^\infty \frac {i \sin n} {K^n} }} {{eqn | r = \sum_{n \mathop = 0}^\infty \frac {\cos n + i \sin n} {K^n} }} {{eqn | r = \sum_{n \mathop = 0}^\infty \frac {e^{i n} } {K^n}       | c = Euler's Formula }} {{eqn | r = \sum_{n \mathop = 0}^\infty \paren {\frac {e^i} K}^n }} {{eqn | r = \frac K {K - e^i}       | c = Sum of Infinite Geometric Sequence }} {{eqn-intertext | The plan now is to equate real and imaginary parts of this to re-obtain $\mathbf A$ and $\mathbf B$:}} {{eqn | r = \frac K {K - \cos 1 - i \sin 1}       | c = Euler's Formula }} {{eqn | r = \frac K {K - \cos 1 - i \sin 1} \cdot \frac {K - \cos 1 + i \sin 1} {K - \cos 1 + i \sin 1}       | c = multiplying by complex conjugate of denominator }} {{eqn | r = \frac {K \paren {K - \cos 1 + i \sin 1} } {K^2 - 2 K \cos 1 + \cos^2 1 + \sin^2 1} }} {{eqn | r = \frac {K \paren {K - \cos 1 + i \sin 1} } {K^2 - 2 K \cos 1 + 1}       | c = Sum of Squares of Sine and Cosine }} {{eqn | r = \paren {\frac {K \paren {K - \cos 1} } {K^2 - 2 K \cos 1 + 1} } + i \paren {\frac {K \sin 1} {K^2 - 2 K \cos 1 + 1} } }} {{end-eqn}} So equating real and imaginary parts: {{begin-eqn}} {{eqn | l = \mathbf A = \sum_{n \mathop = 0}^\infty \frac {\cos n} {K^n}       | r = \frac {K \paren {K - \cos 1} } {K^2 - 2 K \cos 1 + 1}  }} {{eqn | l = \mathbf B = \sum_{n \mathop = 0}^\infty \frac {\sin n} {K^n}       | r = \frac {K \sin 1} {K^2 - 2 K \cos 1 + 1} }} {{end-eqn}} This proves $(1)$ and $(2)$ respectively. $(3)$ is proved by finding the quotient of the two results: {{begin-eqn}} {{eqn | l = \frac {\sum_{n \mathop = 0}^\infty \frac {\sin n} {K^n} } {\sum_{n \mathop = 0}^\infty \frac {\cos n} {K^n} }       | r = \frac {K \sin 1} {K^2 - 2 K \cos 1 + 1} \cdot \frac {K^2 - 2 K \cos 1 + 1} {K \paren {K - \cos 1} } = \frac {\sin 1} {K - \cos 1} }} {{eqn | ll= \leadsto       | l = \sum_{n \mathop = 0}^\infty \frac {\sin n} {K^n}       | r = \paren {\frac {\sin 1} {K - \cos 1} } \sum_{n \mathop = 0}^\infty \frac {\cos n} {K^n} }} {{end-eqn}} {{qed}} Category:Trigonometric Identities bp153htcf489ze6b95cu3ob3ubm3kru
From Rule of Transposition, we may replace the ''only if'' statement by its contrapositive. Therefore, the following suffices: === Implication === Suppose $S$ is infinite. Let $n \in \N$ be an arbitrary natural number. We have that $\N_n = \left\{{m \in \N: m \le n}\right\}$ is a finite set. By Subset of Finite Set is Finite, it is not the case that $S \subseteq \N_n$. Hence, as $\le$ is a total ordering, there exists a $s \in S$ such that $n \le s$. Hence, as $n$ was arbitrary, $S$ is cofinal. {{qed|lemma}} === Contrapositive Implication === Suppose now that $S$ is finite. As $\le$ is a total ordering, $S$ has a maximal element. Let this maximal element of $S$ be $N$. Then as $N + 1 \not \le N$, from transitivity of $\le$ and maximality of $N$: :$\forall s \in S: N + 1 \not \le s$ Hence $S$ is not cofinal. {{qed}} Category:Order Theory Category:Natural Numbers tee3nlxha5twk3id7x003gh9762jr5l
$(1)$: Let: :$\displaystyle x \in \liminf_n \ C_n$ Let: :$\Sigma \in \NN_\infty^\#$ Let $W$ be a neighborhood of $x$.  Then there exists $N_0 \in \N$ such that for all $n \ge N_0$ such that $n \in \Sigma$: :$W \cap C_n \ne \O$ Thus: :$\displaystyle x \in \map \cl {\bigcup_{n \mathop \in \Sigma} C_n}$ $(2)$: Let: :$\displaystyle x \notin \liminf_n C_n$ Then there exists an open neighborhood of $x$. Let $\map \mho x := \set {V \in \tau: x \in V}$ denote the set of open neighborhoods of $x$. Let $W \in \map \mho x$ such that: :$\Sigma_0 := \set {n \in \N: W \cap C_n = \O}$ is cofinal. Then: :$\displaystyle x \notin \map \cl {\bigcup_{n \mathop \in \Sigma_0} C_n}$ This completes the proof. {{qed}}
Let $S$ be an ordinal by Definition 3. By definition: :$\forall a \in S: a = S_a \subseteq S$ where $S_a$ denotes the initial segment of $S$ determined by $a$. That is, $S$ is a transitive set. {{qed}} Category:Ordinals Category:Transitive Classes 7pw4myves5i5yz0bituyrotteg1v09k
Since $x_1, x_2, \ldots, x_n \in A$, there exists a $y$ such that $y = \left\{{x_1, x_2, \ldots, x_n}\right\}$. Then $y$ is a nonempty subset of $A$. So, by the definition of a foundational relation: : $\exists w \in y: \forall z \in y: \neg w \prec z$ Now, suppose $x_1 \prec x_2 \land x_2 \prec x_3 \cdots \land x_n \prec x_1$. But since the elements of $y$ are $x_1, x_2, \ldots, x_n$, then this contradicts the previous statement, since: : $\forall w \in y: \exists z \in y: w \prec z$ Thus a founded relation has no relational loops. {{explain|is a "founded relation" the same as a "foundational relation"? If so, use the same terminology throughout the entire page; if not, provide a definition for the former.}} {{qed}}
{{NotZFC}} By the axiom of foundation: :$\forall S: \paren {\exists x: x \in S \implies \exists y \in S: \forall x \in S: \neg x \in y}$ That is, by Nonempty Class has Members: :$\forall S: \paren {S \ne \O \implies \exists y \in S: \forall x \in S: \neg x \in y}$  This holds for all sets $S$ whose construction is based on the Zermelo-Fraenkel axioms. We can weaken the antecedent of the above statement with this statement: :$\forall S: \paren {\paren {S \ne \O \land S \subseteq A} \implies \exists y \in S: \forall x \in S: \neg x \in y}$ Note that this step does not require that $A$ be a set: it can be any class, even a proper class. By definition, it follows that $\Epsilon$ is a foundational relation on every class $A$. Category:Class Theory Category:Axiom of Foundation 695zkccvjz04mph4p8ung9e7zxklqcn
The proof is an immediate result of Inner Limit in Hausdorff Space by Open Neighborhoods since the arbitrary open sets can be here replaced by open balls. {{qed}}
According to Inner Limit in Hausdorff Space by Set Closures, the inner limit is given by an arbitrary intersection of closed sets which is closed in the topology $\tau$. {{qed}} Category:Limits of Sequence of Sets olfc9bqbfl4j70ofagsfnhu9jetdors
The proof will be carried out in various steps. We will construct a collection of star-shaped neighborhoods of $0_\XX$. Then we will show that it is indeed a local basis with the required properties. Firstly we define the following set: :$\BB_0 := \set {W \in \tau: 0 \in W, W \text{ is star-shaped} }$ === $\BB_0$ is a local basis for $0_\XX$ === Let $U \ni 0_\XX$ be an open set. Notice that: :$0 \cdot 0_\XX = 0_\XX$ This way we have proved that $\BB_0$ is a local basis for $0_\XX$. Since $\cdot: K \times \XX \to \XX$ is a continuous mapping, there is a neighborhood of $\tuple {0, 0_\XX} \in K \times \XX$ in the form: :$\openint {-\epsilon} \epsilon \times W$ such that: :$\map \cdot {\openint {-\epsilon} \epsilon \times W} \subseteq U$ which means that :$\openint {-\epsilon} \epsilon \cdot W \subseteq U$ Let: :$\displaystyle G := \bigcup_{\lambda \mathop \in \openint {-\epsilon} \epsilon, \lambda \mathop \ne 0} \lambda W \subseteq U$ We have that: :$(1): \quad$ For every $\lambda \ne 0$, the set $\lambda \cdot W$ is open. Hence $G$ is open as union of open sets. :$(2): \quad 0_\XX \in G$ since $0 \in W$. :$(3): \quad$ If $x \in G$ then $-x \in G$. Therefore, $G$ is open, star-shaped, contains $0_\XX$ and $G \subseteq U$. {{qed|lemma}} === Condition $(1)$ is satisfied by $\BB_0$ === Let $V \in \BB_0$. We have that $+\tuple {0_\XX, 0_\XX} = 0_\XX \in W$. Since $+$ is a continuous mapping and $\BB_0$ is a local basis of $0_\XX$, there exists a $V \in \BB_0$ such that: :$+\tuple {V, V} = V + V \subseteq W$ {{qed|lemma}} === $\BB_0$ consists of Absorbent Sets === We notice that: :$\tuple {\dfrac 1 n, x} \overset n \to \tuple {0, x}$ and: :$\map \cdot {\dfrac 1 n, x} \overset n \to \map \cdot {0, x} = 0_\XX$ Thus, there is a $n \in \N$ such that: :$\frac 1 n x \in V$ or what amounts to the same thing: :$x \in n V$ {{qed|lemma}} === Condition $(4)$ is satisfied by $\BB_0$ === It suffices that we find a $V \in \BB_0$ such that $x \notin V$. This is possible since $\tau$ is a Hausdorff topology. {{qed|lemma}} The proof is now complete. {{qed}} tse15mbihzw34nxocvkoaninbfxns2p
{{NotZFC}} Either $A_1, A_2, \ldots, A_n$ are all sets, or there exists a proper class $A_m$ such that $1 \le m \le n$. Suppose there exists a proper class $A_m$. Then, by the definition of a proper class, $\neg A_m \in A_{m+1}$, since it is not a member of any class. The result then follows directly. Otherwise it follows that all $A_1, A_2, \ldots, A_n$ are sets. Then, by the fact that Epsilon is Foundational and a Foundational Relation has no Relational Loops, it follows that:  :$\neg \paren {A_1 \in A_2 \land A_2 \in A_3 \land \cdots \land A_n \in A_1}$ {{qed}}
If $\phi$ and $\psi$ are both isomorphisms, then they are by definition: * homomorphisms * bijections. So: * From Composite of Homomorphisms on Algebraic Structure is Homomorphism we have that $\phi \circ \psi$ and $\psi \circ \phi$ are both homomorphisms * From Composite of Bijections is Bijection we have that $\phi \circ \psi$ and $\psi \circ \phi$ are both bijections and hence by definition also isomorphisms. {{qed}}
If $\phi$ and $\psi$ are both isomorphisms, then they are by definition: * homomorphisms * bijections. So: * From Composite of Homomorphisms for R-Algebraic Structures is Homomorphism we have that $\phi \circ \psi$ and $\psi \circ \phi$ are both homomorphisms * From Composite of Bijections is Bijection we have that $\phi \circ \psi$ and $\psi \circ \phi$ are both bijections; and hence by definition also isomorphisms. {{qed}} Category:Isomorphisms 5alihydv9np3orurmuk1lfl63nhxvnb