hcju/proofwikiqa huggingface dataset:
Let us establish that: :$n = m \implies \map f n = \map f m$ and so: :$n = m \implies \map f n \subseteq \map f m$ Hence it is sufficient to demonstrate that: :$\forall n < m: \map f n \subseteq \map f m$ The proof will proceed by the Principle of Finite Induction on $\N$. Let $S$ be the set defined as: :$S := \set {m \in \N: \forall n < m: \map f n \subseteq \map f m}$ That is, $S$ is to be the set of all $n$ such that: :$\forall n < m: \map f n \subseteq \map f m$ First we note that: :$\not \exists n \in \N: n < 0$ Thus $0 \in S$ vacuously. === Basis for the Induction === Consider the natural number $1$. There is only one $n \in \N$ such that $n < 1$, and that is: :$n = 0$ By definition of $f$: :$\map f 0 \subseteq \map f {0^+}$ But by definition of the set of natural numbers: :$0^+ = 1$ So $1 \in S$. This is the basis for the induction. === Induction Hypothesis === It is to be shown that if $k \in S$ where $k \ge 1$, then it follows that $k + 1 \in S$. This is the induction hypothesis: :$\forall n < k: \map f n \subseteq \map f k$ It is to be demonstrated that it follows that: :$\forall n < k^+: \map f n \subseteq \map f {k^+}$ === Induction Step === This is the induction step: Let $k \in S$. From the induction hypothesis: :$\forall n < k: \map f n \subseteq \map f k$ Then from the definition of $f$: :$\map f k \subseteq \map f {k^+}$ That is: :$\forall n \le k: \map f n \subseteq \map f {k^+}$ That is: :$\forall n < k^+: \map f n \subseteq \map f {k^+}$ So $k \in S \implies k^+ \in S$. The result follows by the Principle of Finite Induction: :$\forall n, m \in N: n \le m \implies \map f n \subseteq \map f m$ {{qed}}
The proof will proceed by the Principle of Finite Induction on $\N$. Let $S$ be the set defined as: :$S := \set {m \in \N: \forall n < m: \map f n \subsetneq \map f m}$ That is, $S$ is to be the set of all $n$ such that: :$\forall n < m: \map f n \subsetneq \map f m$ First we note that: :$\not \exists n \in \N: n < 0$ Thus $0 \in S$ vacuously. === Basis for the Induction === Consider the natural number $1$. There is only one $n \in \N$ such that $n < 1$, and that is: :$n = 0$ By definition of $f$: :$\map f 0 \subseteq \map f {0^+}$ But by definition of the set of natural numbers: :$0^+ = 1$ So $1 \in S$. This is the basis for the induction. === Induction Hypothesis === It is to be shown that if $k \in S$ where $k \ge 1$, then it follows that $k + 1 \in S$. This is the induction hypothesis: :$\forall n < k: \map f n \subsetneq \map f k$ It is to be demonstrated that it follows that: :$\forall n < k^+: \map f n \subsetneq \map f {k^+}$ === Induction Step === This is the induction step: Let $k \in S$. From the induction hypothesis: :$\forall n < k: \map f n \subsetneq \map f k$ Then from the definition of $f$: :$\map f k \subsetneq \map f {k^+}$ That is: :$\forall n \le k: \map f n \subsetneq \map f {k^+}$ That is: :$\forall n < k^+: \map f n \subsetneq \map f {k^+}$ So $k \in S \implies k^+ \in S$. The result follows by the Principle of Finite Induction: :$\forall n, m \in N: n \le m \implies \map f n \subsetneq \map f m$ {{qed}}
Let $A_1$ be the class which has a bijection $\phi_1$ from $A_1$ to $n$. Let $A_2$ be the class which has a bijection $\phi_2$ from $A_2$ to $n^+ \setminus \set 0$. Consider the mapping $\phi: A_1 \to A_2$ defined as: :$\forall k \in n: \map {\phi_1} k = k^+$ $\phi$ is trivially a bijection. The result follows. {{qed}}
Let it be assumed that all classes are subclasses of a basic universe $V$. The proof proceeds by induction. For all $n \in \N$, let $\map P n$ be the proposition: :If $A$ is a finite class with $n$ elements, then $A$ is a set. The axiom of the empty set gives that the empty class $\O$ is a set. From Empty Set is Unique, $\O$ is the only set (and hence class) with $0$ elements. Thus $\map P 0$ is seen to hold. === Basis for the Induction === Let $A$ be a singleton class. Thus by definition it has $1$ element, which we will call $x$. {{AimForCont}} $x$ is a class containing $1$ or more elements, one of which we may call $y$. We have that $A$ is a subclass of $V$. As $V$ is a basic universe, the axiom of transitivity holds. Hence: :$y \in x \land x \in A \implies y \in A$ But then $x \in A$ and $y \in A$ where $x \ne y$. This contradicts our assertion that $A$ is a singleton class. Hence: :$x = \O$ It has been established that $\O$ is a set. Hence $A$ is a class whose $1$ element is a set. From Singleton Class of Set is Set, it follows that $A$ is a set. Thus $\map P 1$ is seen to hold. This is the basis for the induction. === Induction Hypothesis === Now it needs to be shown that if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true. So this is the induction hypothesis: :If $A$ is a finite class with $k$ elements, then $A$ is a set. from which it is to be shown that: :If $A$ is a finite class with $k + 1$ elements, then $A$ is a set. === Induction Step === This is the induction step: Let $A$ have $k + 1$ elements. Let each of those elements be assigned a label: :$a_1, a_2, \ldots, a_{k + 1}$ according to a bijection $\phi: A \to k$: :$\map \phi {a_k} = k - 1$ This bijection $\phi$ is guaranteed to exist by definition of finite class. Consider $A$ as the union of the classes: :$\set {a_1, a_2, \ldots, a_k} \cup \set {a_{k + 1} }$ By the basis for the induction: :$\set {a_{k + 1} }$ is a set. By the induction hypothesis: :$\set {a_1, a_2, \ldots, a_k}$ is a set. By the axiom of unions: :$\set {a_1, a_2, \ldots, a_k} \cup \set {a_{k + 1} }$ is a set. So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction. Therefore, for all $\forall n \in \N$: :If $A$ is a finite class with $n$ elements, then $A$ is a set. That is, all finite classes are sets. {{qed}}
The proof proceeds by induction. By definition, if $A$ is a non-empty finite set, then $A$ has $m$ elements for some $m \in \N$ such that $m > 0$. So, for all $n \in \N_{>0}$, let $\map P n$ be the proposition: :If $A$ has $n$ elements, then $A$ has a greatest element. === Basis for the Induction === Let $A$ have one element which we will call $x$ Trivially: :$\forall y \in A: y \le x$ and so $A$ has a greatest element $x$, by definition. Thus $\map P 1$ is seen to hold. This is the basis for the induction. === Induction Hypothesis === Now it needs to be shown that if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true. So this is the induction hypothesis: :If $A$ has $k$ elements, then $A$ has a greatest element from which it is to be shown that: :If $A$ has $k + 1$ elements, then $A$ has a greatest element. === Induction Step === This is the induction step: Let $A$ have $k + 1$ elements: :$A = \set {a_1, a_2, \ldots, a_k, a_{k + 1} }$ Then: :$A = \set {a_1, a_2, \ldots, a_k} \cup \set {a_{k + 1} }$ By the induction hypothesis, $\set {a_1, a_2, \ldots, a_k}$ has a greatest element, which we will call $m$. By definition of greatest element, $m$ is one of the elements of $\set {a_1, a_2, \ldots, a_k}$. As $A$ is a set, all elements of $A$ are distinct. Hence: :$\forall a \in \set {a_1, a_2, \ldots, a_k}: a \ne a_{k + 1}$ It follows that $m \ne a_{k + 1}$. Suppose $a_{k + 1}$ is such that $a_{k + 1} > m$. By definition of greatest element: :$\forall y \in \set {a_1, a_2, \ldots, a_k}: y \le m$ and so: :$\forall y \in \set {a_1, a_2, \ldots, a_k}: y < a_{k + 1}$ It follows that: :$\forall y \in A: y \le a_{k + 1}$ and so $A$ has a greatest element, that is: $a_{k + 1}$. Suppose $a_{k + 1}$ is such that $a_{k + 1} < m$. Then it follows immediately that: :$\forall y \in A: y \le m$ and so $A$ has a greatest element, that is: $m$. In both cases $A$ has been shown to have a greatest element. So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction. Therefore, for all $n \in \N_{>0}$: :If $A$ has $n$ elements, then $A$ has a greatest element. {{Qed}}
The proof proceeds by general induction. For all $x \in M$, let $\map P x$ be the proposition: :$\set {y \in M: y \subseteq x}$ is finite. === Basis for the Induction === We have that: :$\set {y \in M: y \subseteq \O} = \O$ From Cardinality of Empty Set, $\O$ is finite. Thus $\map P \O$ is seen to hold. This is the basis for the induction. === Induction Hypothesis === Now it needs to be shown that if $\map P x$ is true, where $x \in M$, then it logically follows that $\map P {\map g x}$ is true. So this is the induction hypothesis: :$\set {y \in M: y \subseteq x}$ is finite. from which it is to be shown that: :$\set {y \in M: y \subseteq \map g x}$ is finite. === Induction Step === This is the induction step: Let $S = \set {y \in M: y \subseteq x}$. By the induction hypothesis, $S$ is finite. Consider the set: :$S' = \set {y \in M: y \subseteq \map g x}$ By definition of progressing mapping: :$x \subseteq \map g x$ Hence it follows that: :$S \subseteq S'$ Let $y \in S' \setminus S$. Then: :$y \notin S$ and so: :$y \not \subseteq x$ From Minimally Inductive Class under Progressing Mapping induces Nest: :$\forall x, y \in M: \map g x \subseteq y \lor y \subseteq x$ As it is not the case that $y \subseteq x$, it must be that: :$\map g x \subseteq y$ From sandwich principle: :$x \subseteq y \subseteq \map g x \implies x = y \lor y = \map g x$ As $x \ne y$ (or then $y \subseteq x$), it must be the case that: :$y = \map g x$ So: :$S' \setminus S = \map g x$ and so is finite, containing $1$ element. We have that: :$S' = S \cup \paren {S' \setminus S}$ and it follows from Union of Finite Sets is Finite that $S'$ is finite. So $\map P x \implies \map P {\map g x}$ and the result follows by the Principle of General Induction. Therefore: :$\forall x \in M: \set {y \in M: y \subseteq x}$ is finite.
By Set of Subsets of Element of Minimally Inductive Class is Finite, the set: :$\set {y \in M: y \subseteq x}$ is finite. From Fixed Point of Progressing Mapping on Minimally Inductive Class is Greatest Element, $x$ is the greatest element of $M$. Thus: :$M = \set {y \in M: y \subseteq x}$ Hence the result. {{qed}}
Let $\omega$ denote the set of natural numbers as defined by the von Neumann construction. By the Well-Ordering Principle, $\omega$ is well-ordered by the $\le$ relation. Thus from the Well-Ordering Principle, $S$ has a smallest element. Let this smallest element of $S$ be denoted $s_0$. Also from the Well-Ordering Principle, every subset of $S$ also has a smallest element. {{ProofWanted|establish a sequence where $s_k$ is the smallest element of $S_k$ where $S_k {{=}} S_{k - 1} \setminus \set {s_{k - 1} }$, and then build a mapping $f$ such that $\map f {s_k} \to \set {s_0, s_1, \ldots, s_{k - 1} }$ and show it's progressing, or something, but it's tedious.}}
{{AimForCont}} $A$ is finite. From Subset of Naturals is Finite iff Bounded, it follows that $A$ is bounded. Hence $A$ has a greatest element. This contradicts the fact that $A$ has no greatest element. Hence by Proof by Contradiction it follows that $A$ is not finite. From Set of Natural Numbers is either Finite or Denumerable, $A$ is denumerable. {{Qed}}
By Coulomb's Law of Electrostatics: :$\mathbf F_{a b} \propto \dfrac {q_a q_b {\mathbf r_{a b} } } {r^3}$ where: :$\mathbf F_{a b}$ is the force exerted on $b$ by the electric charge on $a$ :$\mathbf r_{a b}$ is the displacement vector from $a$ to $b$ :$r$ is the distance between $a$ and $b$. Let $q_a$ and $q_b$ both be positive. Then $q_a q_b$ is a positive number multiplied by a positive number.  Thus $q_a q_b$ is a positive number. Similarly, let $q_a$ and $q_b$ both be negative. Then $q_a q_b$ is a negative number multiplied by a negative number. Thus $q_a q_b$ is similarly a positive number. So, when $q_a$ and $q_b$ are like charges, $\dfrac {q_a q_b} {r^3}$ is a positive number. Hence $\mathbf F_{a b}$ is in the same direction as the displacement vector from $a$ to $b$. That is, the force exerted on $b$ by the electric charge on $a$ is in the direction away from $a$. The same applies to the force exerted on $a$ by the electric charge on $b$. That is, the force exerted on $b$ by the electric charge on $a$ is in the direction away from $b$. The effect of these forces is to cause $a$ and $b$ to push apart, that is, to repel each other.
By Coulomb's Law of Electrostatics: :$\mathbf F_{a b} \propto \dfrac {q_a q_b {\mathbf r_{a b} } } {r^3}$ where: :$\mathbf F_{a b}$ is the force exerted on $b$ by the electric charge on $a$ :$\mathbf r_{a b}$ is the displacement vector from $a$ to $b$ :$r$ is the distance between $a$ and $b$. {{WLOG}}, let $q_a$ be positive and $q_b$ be negative. Then $q_a q_b$ is a positive number multiplied by a negative number.  Thus $q_a q_b$ is a negative number. Hence $\mathbf F_{a b}$ is in the opposite direction to the displacement vector from $a$ to $b$. That is, the force exerted on $b$ by the electric charge on $a$ is in the direction towards $a$. The same applies to the force exerted on $a$ by the electric charge on $b$. That is, the force exerted on $b$ by the electric charge on $a$ is in the direction towards $b$. The effect of these forces is to cause $a$ and $b$ to pull together, that is, to attract each other.
The '''vacuum permittivity''' is the physical constant denoted $\varepsilon_0$ defined as: :$\varepsilon_0 := \dfrac 1 {\mu_0 c^2}$ where: :$\mu_0$ is the vacuum permeability defined in $\mathrm H \, \mathrm m^{-1}$ (henries per metre) :$c$ is the speed of light defined in $\mathrm m \, \mathrm s^{-1}$ $\mu_0$ has the value determined experimentally as: :$\mu_0 \approx 1 \cdotp 25663 \, 70621 \, 2 (19) \times 10^{-6} \, \mathrm H \, \mathrm m^{-1}$ $c$ is defined precisely as: :$c = 299 \, 792 \, 458 \, \mathrm m \, \mathrm s^{-1}$ Hence $\varepsilon_0$ can be calculated as: {{begin-eqn}} {{eqn | l = \varepsilon_0       | r = \dfrac 1 {\mu_0 c^2}       | rr= \dfrac 1 {\mathrm H \, \mathrm m^{-1} \times \paren {\mathrm m \, \mathrm s^{-1} }^2}       | c =  }} {{eqn | r = \dfrac 1 {1 \cdotp 25663 \, 70621 \, 2 (19) \times 10^{-6} \times \paren {299 \, 792 \, 458}^2}       | rr= \dfrac 1 {\frac {\mathrm {kg} \times \mathrm m^2} {\mathrm s^2 \times \mathrm A^2} \, \mathrm m^{-1} \times \paren {\mathrm m \, \mathrm s^{-1} }^2}       | c = Base Units of Henry }} {{eqn | r = 8 \cdotp 85418 \, 78128 (13) \times 10^{-12}       | rr= \dfrac {\mathrm A^2 \times \mathrm s^4} {\mathrm {kg} \times \mathrm m^3}       | c =  }} {{eqn | r = 8 \cdotp 85418 \, 78128 (13) \times 10^{-12}       | rr= \dfrac {\mathrm F} {\mathrm m}       | c = Base Units of Farad }} {{end-eqn}} {{qed}}
The '''vacuum permittivity''' is the physical constant denoted $\varepsilon_0$ defined as: :$\varepsilon_0 := \dfrac {e^2} {2 \alpha h c}$ where: :$e$ is the elementary charge :$\alpha$ is the fine-structure constant :$h$ is Planck's constant :$c$ is the speed of light defined in $\mathrm m \, \mathrm s^{-1}$ $e$ is defined precisely as: :$e = 1 \cdotp 60217 \, 6634 \times 10^{-19} \, \mathrm C$ (coulombs) $\alpha$ has the value determined experimentally as: :$\alpha \approx 0 \cdotp 00729 \, 73525 \, 693 (11)$ $h$ is defined precisely as: :$h = 6 \cdotp 62607 \, 015 \times 10^{-34} \, \mathrm J \, \mathrm s$ (joule seconds) $c$ is defined precisely as: :$c = 299 \, 792 \, 458 \, \mathrm m \, \mathrm s^{-1}$ (metres per second) Hence $\varepsilon_0$ can be calculated as: {{begin-eqn}} {{eqn | l = \varepsilon_0       | r = \dfrac {e^2} {2 \alpha h c}       | rr= \dfrac {\mathrm C^2} {\mathrm J \, s \times \mathrm m \, \mathrm s^{-1} }       | c =  }} {{eqn | r = \dfrac {\paren {1 \cdotp 60217 \, 6634 \times 10^{-19} }^2} {0 \cdotp 00729 \, 73525 \, 693 (11) \times 6 \cdotp 62607 \, 015 \times 10^{-34} \times 299 \, 792 \, 458}       | rr= \dfrac {\paren {\mathrm A \, \mathrm s}^2} {\frac {\mathrm {kg} \times \mathrm m^2} {\mathrm s^2} \times \mathrm s \times \mathrm m \, \mathrm s^{-1} }       | c = Base Units of Coulomb, Base Units of Joule }} {{eqn | r = 8 \cdotp 85418 \, 78128 (13) \times 10^{-12}       | rr= \dfrac {\mathrm A^2 \times \mathrm s^4} {\mathrm {kg} \times \mathrm m^3}       | c =  }} {{eqn | r = 8 \cdotp 85418 \, 78128 (13) \times 10^{-12}       | rr= \dfrac {\mathrm F} {\mathrm m}       | c = Base Units of Farad }} {{end-eqn}} {{qed}}
Consider the set of natural numbers as defined by the von Neumann construction. From Von Neumann Construction of Natural Numbers is Minimally Inductive, $\N$ is a minimally inductive class under the successor mapping. Let $s: \N \to \N$ denote the successor mapping: :$\forall x \in \N: \map s x := x + 1$ {{AimForCont}} $\exists n \in \N: n = n + 1$ From Fixed Point of Progressing Mapping on Minimally Inductive Class is Greatest Element, $n$ is the greatest element of $\N$. From Minimally Inductive Class with Fixed Element is Finite it follows that $\N$ is a finite set. This contradicts the fact that the natural numbers are by definition countably infinite. {{qed}}
{{begin-eqn}} {{eqn | l = y       | o = \in       | r = \set {x \in \R: \size {\xi - x} \ge \delta}       | c = }} {{eqn | ll= \leadstoandfrom       | l = y       | o = \notin       | r = \set {x \in \R: \size {\xi - x} < \delta}       | c =  }} {{eqn | ll= \leadstoandfrom       | l = y       | o = \notin       | r = \openint {\xi - \delta} {\xi + \delta}       | c = Open Interval Defined by Absolute Value }} {{eqn | ll= \leadstoandfrom       | l = y       | o = \in       | r = \R \setminus \openint {\xi - \delta} {\xi + \delta}       | c = {{Defof|Set Difference}} }} {{end-eqn}} {{qed}}
{{begin-eqn}} {{eqn | l = y       | o = \in       | r = \set {x \in \R: \size {\xi - x} > \delta}       | c = }} {{eqn | ll= \leadstoandfrom       | l = y       | o = \notin       | r = \set {x \in \R: \size {\xi - x} \le \delta}       | c =  }} {{eqn | ll= \leadstoandfrom       | l = y       | o = \notin       | r = \closedint {\xi - \delta} {\xi + \delta}       | c = Closed Interval Defined by Absolute Value }} {{eqn | ll= \leadstoandfrom       | l = y       | o = \in       | r = \R \setminus \closedint {\xi - \delta} {\xi + \delta}       | c = {{Defof|Set Difference}} }} {{end-eqn}} {{qed}}
:500pxthumbright First note that: {{begin-eqn}} {{eqn | l = T_n       | r = \set {y: \size {y - 1} < n \land \size {y + 1} > \dfrac 1 n}       | c =  }} {{eqn | r = \set {y: \size {y - 1} < n} \cap \set {y: \size {y + 1} > \dfrac 1 n}       | c =  }} {{end-eqn}} We have: {{begin-eqn}} {{eqn | l = \set {y: \size {y - 1} < n}       | r = \openint {1 - n} {1 + n}       | c = Open Interval Defined by Absolute Value }} {{eqn | l = \set {y: \size {y + 1} > \dfrac 1 n}       | r = \R \setminus \closedint {-1 - \dfrac 1 n} {-1 + \dfrac 1 n}       | c = Complement of Closed Interval Defined by Absolute Value }} {{eqn | ll= \leadsto       | l = \set {y: \size {y - 1} < n} \cap \set {y: \size {y + 1} > \dfrac 1 n}       | r = \openint {1 - n} {-1 - \dfrac 1 n} \cup \openint {-1 + \dfrac 1 n} {1 + n}       | c =  }} {{end-eqn}} {{qed}} Category:Unions of Families Category:Intersections of Families tmrlmfh1k5doftvxlyvw1fpx34g0rab
=== Union Subset of Intersection === Let $\displaystyle x \in S \cup \bigcap \mathbb T$. We need to show that: :$\displaystyle x \in \bigcap_{X \mathop \in \mathbb T} \paren {S \cup X}$ and then by definition of subset we will have shown that: :$\displaystyle S \cup \bigcap \mathbb T \subseteq \bigcap_{X \mathop \in \mathbb T} \paren {S \cup X}$. So, we have that $\displaystyle x \in S \cup \bigcap \mathbb T$. By definition of set union, $x \in S$ or $\displaystyle x \in \bigcap \mathbb T$. So there are two cases to consider: $(1): \quad$ Suppose $x \in S$. Then by definition of set union, $\displaystyle \forall X \in \mathbb T: x \in S \cup X$. So: :$\displaystyle x \in \bigcap_{X \mathop \in \mathbb T} \paren {S \cup X}$ $(2): \quad$ Suppose $\displaystyle x \in \bigcap \mathbb T$. Then by definition of set intersection: : $\displaystyle \forall X \in \mathbb T: x \in X$ So by definition of set union: :$\displaystyle \forall X \in \mathbb T: x \in S \cup X$ So: :$\displaystyle x \in \bigcap_{X \mathop \in \mathbb T} \paren {S \cup X}$ In both cases we see that: :$\displaystyle x \in \bigcap_{X \mathop \in \mathbb T} \paren {S \cup X}$ so by Proof by Cases, we have that: :$\displaystyle S \cup \bigcap \mathbb T \subseteq \bigcap_{X \mathop \in \mathbb T} \paren {S \cup X}$ {{qed|lemma}} === Intersection Subset of Union === Let $\displaystyle x \in \bigcap_{X \mathop \in \mathbb T} \paren {S \cup X}$. We need to show that: :$\displaystyle x \in S \cup \bigcap \mathbb T$ and then by definition of subset we will have shown that: :$\displaystyle \bigcap_{X \mathop \in \mathbb T} \paren {S \cup X} \subseteq S \cup \bigcap \mathbb T$. So, we have that: :$\displaystyle x \in \bigcap_{X \mathop \in \mathbb T} \paren {S \cup X}$ By definition of set intersection: : $(A): \quad \forall X \in \mathbb T: x \in S \cup X$ There are two cases to consider: :$(1): \quad \forall X \in \mathbb T: x \in X$ Then  by definition of set intersection: :$\displaystyle x \in \bigcap_{X \mathop \in \mathbb T} X$ and so by definition of set union: :$\displaystyle x \in S \cup \bigcap \mathbb T$ :$(2): \quad \exists X \in \mathbb T: x \notin X$ From $(A)$ we have that $x \in S \cup X$. But as $x \notin X$ it follows that $x \in S$. Then by definition of set union: :$\displaystyle x \in S \cup \bigcap \mathbb T$ In both cases we see that: :$\displaystyle x \in S \cup \bigcap \mathbb T$ so by Proof by Cases, we have that: :$\displaystyle \bigcap_{X \mathop \in \mathbb T} \paren {S \cup X} \subseteq S \cup \bigcap \mathbb T$ {{qed|lemma}} So we have that: :$S \cup \displaystyle \bigcap \mathbb T \subseteq \bigcap_{X \mathop \in \mathbb T} \paren {S \cup X}$ and :$\displaystyle \bigcap_{X \mathop \in \mathbb T} \paren {S \cup X} \subseteq S \cup \bigcap \mathbb T$ and so by definition of set equality: :$S \cup \displaystyle \bigcap \mathbb T = \bigcap_{X \mathop \in \mathbb T} \paren {S \cup X}$ {{qed}}
Suppose: :$\displaystyle x \in S \setminus \bigcap \mathbb T$ Note that by Set Difference is Subset we have that $x \in S$ (we need this later). Then: {{begin-eqn}} {{eqn | l = x       | o = \in       | r = S \setminus \bigcap \mathbb T       | c =  }} {{eqn | ll= \leadstoandfrom       | l = x       | o = \notin       | r = \bigcap \mathbb T       | c = {{Defof|Set Difference}} }} {{eqn | ll= \leadstoandfrom       | l = \neg (\forall T' \in \mathbb T: (x       | o = \in       | r = T'))       | c = {{Defof|Intersection of Set of Sets}} }} {{eqn | ll= \leadstoandfrom       | l = \exists T' \in \mathbb T: \neg (x       | o = \in       | r = T')       | c = De Morgan's Laws (Predicate Logic) }} {{eqn | ll= \leadstoandfrom       | l = \exists T' \in \mathbb T: x       | o = \in       | r = S \setminus T'       | c = {{Defof|Set Difference}}: note $x \in S$ from above }} {{eqn | ll= \leadstoandfrom       | l = x       | o = \in       | r = \bigcup_{T' \mathop \in \mathbb T} \paren {S \setminus T'}       | c = {{Defof|Union of Set of Sets}} }} {{end-eqn}} Therefore: :$S \setminus \bigcap \mathbb T = \bigcup_{T' \mathop \in \mathbb T} \paren {S \setminus T'}$ {{qed}}
Let $x \in \R$ be a real number. Let $y \in \R$ be an arbitrary real number. Let $z \in \C$ be the complex number defined as: :$z = x + i y$ Then we have: :$\map \Re z = x$ That is: :$\exists z \in \C: \map f z = x$ The result follows by definition of surjection. {{qed}}
Let $y \in \R$ be a real number. Let $x \in \R$ be an arbitrary real number. Let $z \in \C$ be the complex number defined as: :$z = x + i y$ Then we have: :$\map \Im z = y$ That is: :$\exists z \in \C: \map f z = y$ The result follows by definition of surjection. {{qed}}
From Condition for Composite Mapping on Left, we have: :$\exists \phi: S / \RR \to T$ such that $\phi$ is a mapping and $\phi \circ q_\RR = f$ {{iff}}: :$\forall x, y \in S: \map {q_\RR} x = \map {q_\RR} y \implies \map f x = \map f y$ But by definition of the quotient mapping induced by $\RR$: :$\map {q_\RR} x = \map {q_\RR} y \iff \tuple {x, y} \in \RR$ Hence the result. {{qed}}
By definition of automorphism, $\phi$ is a homomorphism It follows that this result is a specific instance of Homomorphic Image of Cyclic Group is Cyclic Group. {{qed}} Category:Cyclic Groups Category:Group Automorphisms kwnpb54bt14so7j5unset7a5q7styla