hcju/proofwikiqa huggingface dataset:
Let $G$ be a finite group whose identity is $e$. Either $G$ has a proper non-trivial normal subgroup or it does not. If not, then: :$\set e \lhd G$ is the composition series for $G$. Otherwise, $G$ has one or more proper non-trivial normal subgroup. Of these, one or more will have a maximum order. Select one of these and call it $G_1$. Again, either $G_1$ has a proper non-trivial normal subgroup or it does not. If not, then: :$\set e \lhd G_1 \lhd G$ is a composition series for $G$. By the Jordan-Hölder Theorem, there can be no other composition series which is longer. As $G_1$ is a proper subgroup of $G$: :$\order {G_1} < \order G$ where $\order G$ denotes the order of $G$. Again, if $G_1$ has one or more proper non-trivial normal subgroup, one or more will have a maximum order. Select one of these and call it $G_2$. Thus we form a normal series: :$\set e \lhd G_2 \lhd G_1 \lhd G$ The process can be repeated, and at each stage a normal subgroup is added to the series of a smaller  order than the previous one. This process can not continue infinitely. Eventually a $G_n$ will be encountered which has no proper non-trivial normal subgroup. Thus a composition series: :$\set e \lhd G_n \cdots \lhd G_2 \lhd G_1 \lhd G$ will be the result. {{qed}} {{proofread}}
Let $G$ be a finite group whose identity is $e$. Let: :$(1): \quad \set e = G_0 \lhd G_1 \lhd \cdots \lhd G_{n - 1} \lhd G_n = G$ be a normal series for $G$. === Necessary Condition === Suppose there exists $k$ such that $G_{k + 1} / G_k$ is not a simple group. Then there exists a normal subgroup $G'$ such that: :$\set e \lhd G' \lhd G_{k + 1} / G_k$ It follows that: :$G_k \lhd G'' \lhd G_{k + 1}$ where: :$G''$ is a normal subgroup of $G_{k + 1}$ and: :$G' = G'' / G_{k + 1}$ {{explain|why?}} Thus $(1)$ has a proper refinement and so is not a composition series. By the Rule of Transposition it follows that if normal series is a composition series, every factor group of that normal series is a simple group. {{qed|lemma}} === Sufficient Condition === Suppose $(1)$ is not a composition series for $G$. Then a proper refinement of $(1)$ can be constructed by inserting a group $G''$ into the series somewhere, for example: :$G_k \lhd G'' \lhd G_{k + 1}$ It follows that $G'' / G_k$ is a normal subgroup of $G_{k + 1} / G''$. {{explain|There's a specific result demonstrating that fact.}} Thus, by definition, $G_{k + 1} / G_k$ is not a simple group. Thus it has been shown that if a normal series is not a composition series, then it contains at least one factor group which is not a simple group By the Rule of Transposition it follows that if every factor group of a normal series is a simple group, then that normal series is a composition series. {{qed|lemma}} Hence the result. {{qed}}
From Composition Series of Group of Prime Power Order, $G$ has a sequence of subgroups: :$\set e = G_0 \subset G_1 \subset \ldots \subset G_n = G$ such that $\order {G_k} = p^k$, $G_k \lhd G_{k + 1}$ and $G_{k + 1} / G_k$ is cyclic and of order $p$. From Prime Group is Simple it follows that $G_{k + 1} / G_k$ is a simple group for all $k$. From Prime Group is Cyclic it follows that $G_{k + 1} / G_k$ is cyclic. The result follows from Condition for Composition Series. {{qed}}
Let the proposition (1) holds; assume the closed linear subspace $M'$ contains the set $A$, then because $M' \in \Bbb M$, we have :$ \qquad \vee A \subseteq M' $ The intersection of arbitrary family of subspaces is a subspace. For suppose $\mathcal C$ is a family of subspaces. Denotes $\bigcap \mathcal C =\{f\in H|\text{ for any }V\in \mathcal C \text{, there }f\in V \}$ If $f\in \bigcap \mathcal C$, then for any $V\in \mathcal C $, $f\in V$ there $af\in V$ for $a\in \Bbb F$. If $f\text{ , }g\in \bigcap \mathcal C$, we have for any $V\in \mathcal C $, $f+g\in V$. Therefore, $\vee A$ is a subspace. It is closed, as intersection of arbitrary family of closed sets is closed. The choice of $M'$ is arbitrary. Hence, (2) holds. This is the proof (1)=>(2). Next if (2) holds. Since $A\subseteq \vee A$, $\vee A \in \Bbb M$.  $\vee A$ is the smallest one in $\Bbb M$; hence :$\qquad \displaystyle \vee A = \bigcap \Bbb M$ We have established the equivalence between (1) and (2). Finally we come to (3): We prove that $\cl(\text{span}(A))$ is a subspace. Let $f\in \cl(\text{span}(A))$.  $H$, the Hilbert space, is a metric space, which satisfies the first countability axiom. For reason then one shall consider the collection of  open balls which all are centred at a point and all have their length of radiuses be rational. We then need "the sequence lemma," which is: :Let $A$ be a subset of a topological space $X$. If there is a sequence of points of $A$ converging to $x$, then $x\in \cl(A)$; :the converse holds if $X$ is first-countable. So there is a sequence $\{f_i\}$ in $\text{span}(A)$ that its limit is $f$. By the continuity of the function of the multiplication of numbers of the field $\Bbb F$ and points in $H$, the sequence $\{af_i\}$ with $a\in \Bbb F$ converges to $af$. Because all the terms of the sequence $\{af_i\}$ are points in $\text{span}(A)$, we have $af\in \cl(\text{span}(A))$. The proof regards the addition of two vectors is by a similar manner. If $f\in \cl(\text{span}(A))$ and $g\in \cl(\text{span}(A))$,  then a sequence $\{f_i\}$, converges to $f$, and a sequence $\{g_i\}$, converges to $g$, of $\text{span}(A)$ are given. By the continuity of the function of the addtion of points in $H$, $\{f_i+g_i\}$ converges to $f+g$. Because each terms of the sequence $f_i+g_i$ is in $\text{span}(A)$, we have $f+g\in \cl(\text {span}(A))$. Notice that in general, if $H'$ is a subspace, then $\cl(H')$ is a subspace. Finally we come to the proof of equivalence of (2) and (3): $\cl(\text{span}(A))$, the closed linear subspace, contains $\text{span}(A)$ and thus contains $A$. For any closed linear subspace $M$ which contains $A$, $\text{span}(A)\subseteq M$, since the linear span of $A$ is the smallest subspace that contains $A$. Because $M$ is closed, $\cl(\text{span}(A))\subseteq M$. $\cl(\text{span}(A))$ is the the smallest closed linear subspace $M$ of $H$ with $A \subseteq M$. Because arbitrary intersection of closed sets is closed and arbitrary intersection of subspaces is a subspace, the smallestness is unique. Hence $\vee A=\cl(\text{span}(A))$.{{qed}}
We have: :$\displaystyle\left\{{ h \in H : \left\Vert{h}\right\Vert = 1 }\right\} \subseteq \left\{{ h \in H : \left\Vert{h}\right\Vert \le 1 }\right\} \subseteq H$ So it follows from the definition of the supremum that  :$\left\Vert{L}\right\Vert_2 \le \left\Vert{L}\right\Vert_1 \le \left\Vert{L}\right\Vert_3$ Next we show that $\left\Vert{L}\right\Vert_2 = \left\Vert{L}\right\Vert_3$. {{begin-eqn}} {{eqn|l= \left\Vert{L}\right\Vert_2      |o= =      |r= \sup \left\{ {\left\Vert{Lh}\right\Vert: \left\Vert{h}\right\Vert = 1}\right\} }} {{eqn|l=       |o= =      |r= \sup \left\{ {\left\Vert{Lh'}\right\Vert: h \in H\setminus\left\{\mathbf 0\right\},\ h' = h/\left\Vert{h}\right\Vert} \right\}      |c= as $\left\Vert h/\left\Vert{h}\right\Vert \right\Vert = \left\Vert{h}\right\Vert/\left\Vert{h}\right\Vert = 1$ for all $h \in H\setminus\left\{\mathbf 0 \right\}$ }} {{eqn|l=       |o= =      |r= \sup \left\{ {\left\Vert{\frac1{\left\Vert{h}\right\Vert}Lh}\right\Vert: h \in H\setminus \left\{ \mathbf 0  \right\} }\right\}      |c= as $L$ is linear }} {{eqn|l=       |o= =      |r= \sup \left\{ {\frac{\left\Vert{Lh}\right\Vert}{\left\Vert{h}\right\Vert}: h \in H\setminus \left\{ \mathbf 0  \right\} }\right\}      |c= by property $(N2)$ of a norm }} {{eqn|l=       |o= =      |r= \left\Vert{L}\right\Vert_3 }} {{end-eqn}} Therefore :$\left\Vert{L}\right\Vert_1 = \left\Vert{L}\right\Vert_2 = \left\Vert{L}\right\Vert_3$ Moreover, if $\left|{Lh}\right| \le c \left\|{h}\right\|$ for all $h \in H\setminus\left\{\mathbf 0 \right\}$, then we have: :$\displaystyle\left\|{L}\right\|_3 = \sup \left\{{\dfrac {\left|{Lh}\right|} {\left\|{h}\right\|}: h \in H \setminus \left\{\mathbf 0 \right\}}\right\} \le c$ Taking the infimum over all such $c$ this reads: :$\left\Vert{L}\right\Vert_3 \le \left\Vert{L}\right\Vert_4$ Suppose that $c_0 := \left\Vert{L}\right\Vert_4 > \left\Vert{L}\right\Vert_3$. Then by the definitions of these two norms, this means that there exists $\epsilon > 0$ such that for every $h \in H\setminus\left\{\mathbf 0 \right\}$: :$\displaystyle\frac{|Lh|}{\Vert h\Vert} + \epsilon \le c_0$ But this in turn implies that for every $h \in H\setminus\left\{\mathbf 0 \right\}$: :$|Lh| \leq c_0 \Vert h\Vert - \epsilon \Vert h\Vert = \left( c_0 - \epsilon \right) \Vert h\Vert$ This contradicts the fact that $c_0$ is the least such number satisfying this inequality. Therefore: : $\left\Vert{L}\right\Vert_3 = \left\Vert{L}\right\Vert_4$ and the proof is complete. {{qed}}
If $L \equiv 0$ identically, then $L h = 0 = \innerprod h 0$, and the theorem holds. Otherwise, set: :$M = \map \ker L = \map {L^{-1} } {\set 0}$ Then $M$ is a subspace. Because $L$ is bounded, it is continuous. Because $\set 0$ is closed, the continuity of $L$ implies that $M$ is closed. Then we can decompose $H$ as a direct sum: :$H \cong M \oplus M^\perp$ As $L \not \equiv 0$: :$M^\perp \ne \set 0$ Choose a $z \in M^\perp$ with norm $1$. By linearity of $L$, for any $h \in H$: {{begin-eqn}} {{eqn | l = L \paren {z L h - h L z}       | r = L z L h - L h L z }} {{eqn | r = 0 }} {{end-eqn}} So: :$z L h - h L z \in \ker L = M$ Then: {{begin-eqn}} {{eqn | l = L h       | r = L h \innerprod z z       | c = as $\norm z = 1$ }} {{eqn | r = \innerprod {z L h} z       | c = linearity in the first argument }} {{eqn | r = \innerprod {z L h - h L z + h L z} z       | c = adding and subtracting $h L z$ in the first argument }} {{eqn | r = \innerprod {z L h - h L z} z + \innerprod {h L z} z       | c = linearity in the first argument }} {{eqn | r = \innerprod {h L z} z       | c = $z L h - h L z \in M, z \in M^\perp$ }} {{eqn | r = \innerprod h {z \paren {L z}^*}       | c = conjugate symmetry }} {{end-eqn}} Thus $L h = \innerprod h {h_0}$ for $h_0 = z (Lz)^*$. To show uniqueness, assume $h_0$ and $h_1$ both satisfy the above equation for all $h \in H$: {{begin-eqn}} {{eqn | l = \innerprod h {h_0}       | r = \innerprod h {h_1} }} {{eqn | ll= \leadsto       | l = \innerprod h {h_0} - \innerprod h {h_1}       | r = 0 }} {{eqn | r = \innerprod h {h_0 - h_1}       | c = additivity in the second argument }} {{end-eqn}} The result follows from Setting $h = h_0 - h_1$ and invoking the positive definiteness of the inner product. {{qed}} {{Namedfor|Frigyes Riesz|cat = Riesz F}}
Let $d = b - a$. As $a, b \in \Q: a < b$ it follows from Rational Numbers form Ordered Integral Domain that $d \in \Q: d > 0$. From Square Root of 2 is Irrational, $\sqrt 2$ is not a rational number, so it is an element of $\R \setminus \Q$. From Square Number Less than One, for any given real number $x$: :$x^2 < 1 \implies x \in \openint {-1} 1$ Let $k = \dfrac {\sqrt 2} 2$. Then from Lemma 1: :$k \in \R \setminus \Q$ As $k^2 = \dfrac 1 2$, it follows that $0 < k < 1$. Let $\xi = a + k d$. Then, since $a, d \in \Q$ and $k \in \R \setminus \Q$, it follows from Lemma 1 and Lemma 2 that: :$\xi \in \R \setminus \Q$ $d > 0$ and $k > 0$, so: :$\xi = a + k d > a + 0 \cdot 0 = a$ $k < 1$, so: :$\xi = a + k d < a + 1 \cdot d < a + \paren {b - a} = b$ We thus have: :$\xi \in \R \setminus \Q: \xi \in \openint a b$ Hence the result. {{qed}}
From Between two Real Numbers exists Rational Number, there exists $x \in \Q$ such that: :$a - \sqrt2 < x < b - \sqrt2$ Since Square Root of 2 is Irrational, by the Lemma: :$x + \sqrt 2$ is irrational. But: :$a < x + \sqrt 2 < b$ which is what we wanted to show. {{qed}}
First we note that the trivial group is (trivially) simple, but not prime because $1$ is not prime. Hence the specification of $G$ as being non-trivial. === Necessary Condition === Let $G$ be a simple non-trivial abelian group. Suppose that $G$ is an infinite group. Then by Infinite Group has Infinite Number of Subgroups $G$ has an infinite number of distinct subgroups. But then by Subgroup of Abelian Group is Normal, $G$ cannot be simple. So by contradiction $G$ must be finite. Let $n$ be the order of $G$. By definition of simple, $G$ has no non-trivial proper normal subgroups. From Subgroup of Abelian Group is Normal, it follows that $G$ can have no non-trivial proper subgroups at all. From Cauchy's Group Theorem, if $p$ is a prime number which is a divisor of $n$, then $G$ has a subgroup of order $p$. It follows that if $G$ is simple, there can be no prime number less than $n$ which is a divisor of $n$. It follows that $n$ is prime. {{qed|lemma}} === Sufficient Condition === Suppose $G$ is a prime group. By Prime Group is Simple, it follows that $G$ is simple. Because $G$ is prime, its order is greater than $1$. Thus $G$ is non-trivial. {{qed}}
Let $G$ be a cyclic group. From Cyclic Group is Abelian it follows that $G$ is an abelian group. The result follows from Abelian Group is Simple iff Prime. {{qed}} Category:Simple Groups Category:Cyclic Groups ihiodpl0ircj9fl5b5ya623qbow3qh2
By definition, a composition series is a normal series whose factor groups are all simple. A solvable group, by definition, is one which has a composition series whose factor groups are all cyclic. From Cyclic Group is Simple iff Prime, it follows that all the factor groups of a composition series of a solvable group must all be prime. {{qed}}
Let $x, y \in L, t \in \closedint 0 1$ be arbitrary. Then, as $L$ is by definition closed under addition and scalar multiplication, it follows immediately that :$t x + \paren {1 - t} y \in L$ Hence $L$ is convex. {{qed}}
For any $x, y \in S$, we have $x = y = v$. It follows that: :$\forall t \in \left[{0 \,.\,.\, 1}\right]: t x + \left({1 - t}\right)y = v \in S$ Hence $S$ is a convex set. {{qed}}
As $H \lhd G$ we can construct the normal series: :$(A): \quad \set e \lhd H \lhd G$ By Finite Group has Composition Series, $(A)$ can be refined to a composition series for $G$: :$(B): \quad \set e = G_0 \lhd G_1 \lhd \cdots \lhd G_n = G$ Suppose $G_k = H$. Then we can construct the composition series: :$(C): \quad \set e = G_0 \lhd G_1 \lhd \cdots \lhd G_k = H$ and: :$(D): \quad \set e = G_k / H \lhd G_{k + 1} / H \lhd \cdots \lhd G_n / H = G / H$ Furthermore, by the Third Isomorphism Theorem: :$\dfrac {G_{i + 1} / H} {G_i / H} \cong \dfrac {G_{i + 1} } {G_i}$ for all $k \le i \le n$. So each factor of the composition series for $G$ is a factor of either: :the composition series for $H$ or: :the composition series for $G / H$. The result follows. {{qed}}
Proof by strong induction on the order of $G$: For all $n \in \N_{>0}$, let $\map P n$ be the proposition: :All abelian groups of order $n$ and below are solvable. === Basis for the Induction === $\map P 1$ is true, as the trivial group is trivially solvable. From Prime Group is Simple, both $\map P 2$ and $\map P 3$ hold. This is the basis for the induction. === Induction Hypothesis === It remains to be shown that, if $\map P j$ holds, for all $j$ such that $1 \le j \le k$, where $k \ge 3$, then it logically follows that $\map P {k + 1}$ holds. So this is our induction hypothesis: :All abelian groups of order $k$ and below are solvable. Then we need to show: :All abelian groups of order $k+1$ and below are solvable. === Induction Step === This is our induction step: Let $G$ be an abelian group such that: :$\order G = k + 1$ where $\order G$ denotes the order of $G$. Then from Positive Integer Greater than 1 has Prime Divisor, there exists some prime $p$ which divides $k + 1$. From Cauchy's Lemma (Group Theory), $G$ has an element of order $p$. Since $G$ is abelian, this element generates a subgroup of $G$ which is normal by Subgroup of Abelian Group is Normal. If $p = n$ and $H = G$ the proof is finished, from Prime Power Group is Solvable. If $p < n$, then $H$ and $G / H$ (which has order $n / p$) are both solvable from the induction hypothesis. It follows from Group is Solvable iff Normal Subgroup and Quotient are Solvable that $G$ is solvable. The result follows by the strong principle of mathematical induction. {{qed}}
Let $x, y \in \displaystyle \bigcap \mathcal C$. Then by definition of set intersection, $\forall C \in \mathcal C: x, y \in C$. The convexity of each $C$ yields: :$\forall t \in \left[{0 \,.\,.\, 1}\right]: t x + \left({1 - t}\right) y \in C$ Therefore, these elements are also in $\displaystyle \bigcap \mathcal C$, by definition of set intersection. Hence $\displaystyle \bigcap \mathcal C$ is also convex. {{qed}}
A direct consequence of Factors of Composition Series for Prime Power Group and the definition of solvable group. {{qed}} Category:P-Groups Category:Solvable Groups a3862dezf4buwaskcacaosw1234emt5
Let $\set e = G_0 \lhd G_1 \lhd \cdots \lhd G_n = G$ be a normal series for $G$ with Abelian quotients. For every $i = 1, 2, \ldots, n$ we have: :$\paren {H \cap G_i} \cap G_{i - 1} = H \cap G_{i - 1}$ From the Second Isomorphism Theorem for Groups: :$\dfrac {\paren {H \cap G_i} \cap G_{i - 1} } {G_{i - 1} } \cong \dfrac {H \cap G_i} {H \cap G_{i - 1} }$ where $\cong$ denotes group isomorphism. Specifically, $H \cap G_{i - 1}$ is a normal subgroup of $H \cap G_i$. We have that: :$\paren {H \cap G_i} \cap G_{i - 1} \subseteq G_i$ and so from the Correspondence Theorem: :$\dfrac {\paren {H \cap G_i} \cap G_{i - 1} } {G_{i - 1} } \le G_i / G_{i - 1}$ We have that $G_i / G_{i - 1}$ is abelian. Thus from Subgroup of Abelian Group is Abelian: :$\dfrac {\paren {H \cap G_i } \cap G_{i - 1} } {G_{i - 1} }$ is abelian. Hence $\dfrac {H \cap G_i} {H \cap G_{i - 1} }$ is abelian. Therefore, the series : :$\set e = H \cap G_0 \lhd H \cap G_1 \lhd \cdots \lhd H \cap G_n = G$ is a normal series with abelian factor groups for $H$. Therefore $H$ is solvable. {{qed}} {{Proofread}}
A direct consequence of Burnside's Theorem: the smallest number with $3$ distinct prime factors is $2 \times 3 \times 5 = 60$. {{qed}}
First note that from Square of Real Number is Non-Negative: :$x^2 \ge 0$ From Ordering of Squares in Reals: :$(1): \quad x > 1 \implies x^2 > 1$ :$(2): \quad x < 1 \implies x^2 < 1$ From Identity Element of Multiplication on Numbers: :$1^2 = 1$ so it is clear that the strict inequalities apply above. For clarity, therefore, the case where $x = \pm 1$ can be ignored. Suppose $x \notin \left({-1 \,.\,.\, 1}\right)$. Then either: :$x < -1$ or: :$x > 1$ If $x > 1$ then: :$x^2 > 1$ If $x < -1$ then: :$-x > 1$ From Square of Real Number is Non-Negative: :$\left({-x}\right)^2 = x^2$ So again $x^2 > 1$. Thus: :$x \notin \left({-1 \,.\,.\, 1}\right) \implies x^2 > 1$ So by the Rule of Transposition: :$x^2 < 1 \implies x \in \left({-1 \,.\,.\, 1}\right)$ {{qed}} Category:Real Analysis 73izy5ub74rmzmsrkm35tsizjduv10g
Let $a \in M$. We are given that $d_N$ is a metric on $N$. So: :$\forall x, y, z \in N: d_N \left({x, z}\right) \le d_N \left({x, y}\right) + d_N \left({y, z}\right)$ We apply this property twice to assert that $\forall n \in \N, \forall x \in M$, we have: {{begin-eqn}} {{eqn | n = 1       | l = d_N \left({f \left({x}\right), f \left({a}\right)}\right)       | o = \le       | r = d_N \left({f \left({x}\right), f_n \left({x}\right)}\right) }} {{eqn | o = +       | r = d_N \left({f_n \left({x}\right), f_n \left({a}\right)}\right) }} {{eqn | o = +       | r = d_N \left({f_n \left({a}\right), f \left({a}\right)}\right) }} {{end-eqn}} Let $\epsilon >0$. Since $f_n \to f$ uniformly: : $\exists \mathcal N \in \R: \forall n \ge \mathcal N: \forall x \in M$: {{begin-eqn}} {{eqn | n = 2 a       | l = d_N \left({f \left({a}\right), f_n \left({a}\right)}\right)       | o = <       | r = \frac \epsilon 3 }} {{eqn | n = 2 b       | l = d_N \left({f \left({x}\right), f_n \left({x}\right)}\right)       | o = <       | r = \frac \epsilon 3 }} {{end-eqn}} We are given that $\forall n \in \N: f_n$ is continuous. So, for any $n\in\N$, $\exists \delta > 0: \forall x \in M$: {{begin-eqn}} {{eqn | n = 3       | l = d_M \left({x, a}\right)       | o = <       | r = \delta }} {{eqn | ll = \implies       | l = d_N \left({f_n \left({x}\right), f_n \left({a}\right)}\right)       | o = <       | r = \frac \epsilon 3 }} {{end-eqn}} By combining $(1)$, $(2 a)$, $(2 b)$ and $(3)$: : $\exists \delta > 0$ and $\exists n$ sufficiently large such that $\forall x \in M$: {{begin-eqn}} {{eqn | l = d_M \left({x, a}\right)       | o = <       | r = \delta }} {{eqn | ll = \implies       | l = d_N \left({f \left({x}\right), f \left({a}\right)}\right)       | o = \le       | r = \frac \epsilon 3 + \frac \epsilon 3 + \frac \epsilon 3 }} {{eqn | r = \epsilon }} {{end-eqn}} As $a$ and $\epsilon$ are arbitrary, it follows that: : $\forall a \in M: \forall \epsilon > 0: \exists \delta > 0: \forall x \in M$: {{begin-eqn}} {{eqn | l = d_M \left({x, a}\right)       | o = <       | r = \delta }} {{eqn | ll = \implies       | l = d_N \left({f \left({x}\right), f \left({a}\right)}\right)       | o = \le       | r = \epsilon }} {{end-eqn}} Hence, $f$ is continuous at every point of $M$. {{qed}}
{{begin-eqn}} {{eqn | l = a       | r = b       | c =  }} {{eqn | ll= \vdash       | l = \map P a       | o = \iff       | r = \map P b       | c = Leibniz's law }} {{eqn | l = b       | r = c       | c =  }} {{eqn | ll= \vdash       | l = \map P b       | o = \iff       | r = \map P c       | c = Leibniz's law }} {{eqn | ll= \vdash       | l = \map P a       | o = \iff       | r = \map P c       | c = Biconditional is Transitive }} {{eqn | ll= \vdash       | l = a       | r = c       | c = Leibniz's law }} {{end-eqn}} {{qed}}
The set $2 = \set {\O, \set \O}$ is used with the Axiom of Replacement as the domain for a mapping whose image is $\set {A, B}$. A suitable mapping would be: :$\paren {y = \O \land z = A} \lor \paren {y = \set \O \land z = B}$ The set $2$ is shown to exist as a member of the infinite set whose existence is asserted by the axiom of infinity. {{qed}} Category:Axiom of Pairing Category:Doubletons pqr9ervif1gvsz3x2iy9mcbnl1fomgi
It is to be shown that $2 = \set {\O, \set \O}$ is a set. By the Axiom of the Empty Set, it is seen that $\O$ is a set. By the Axiom of Powers, it is seen that $\powerset \O$ is also a set. It is to be shown that $\powerset \O = \set \O$. {{refactor|The above may already exist as a theorem on {{ProofWiki}} }} To determine the elements of $\powerset \O$, we must find all of the sets $z$ such that: :$\forall x: x \in z \implies x \in \O$ From the definition of the empty set, the right hand side of this expression is invariably false. So by Rule of Transposition, we deduce that: :$\forall x: x \notin z$ But this is the definition of the empty set. So by Empty Set is Unique, it follows that $\powerset \O = \set \O$. {{qed|lemma}} It is now to be shown that $\powerset {\set \O} = \set {\O, \set \O}$. By Empty Set is Subset of All Sets, $\O \in \powerset {\set \O}$. By Set is Subset of Itself, $\set \O \in \powerset {\set \O}$. And so: :$(1) \quad \set {\O, \set \O} \subseteq \powerset {\set \O}$ It is now to be shown that $\set {\O, \set \O} \supseteq \powerset {\set \O}$. {{AimForCont}} that there exists a set $S \in \powerset {\set \O}$ such that $S \ne \O$ and $S \ne \set \O$. But then $\exists x \ne \O: x \in \set \O$, which is false. Therefore by contradiction: :$(2) \quad \powerset {\set \O} \subseteq \set {\O, \set \O}$. Putting $(1)$ and $(2)$ together yields: :$\powerset {\set \O} = \set {\O, \set \O}$ {{qed|lemma}} The set $2 = \set {\O, \set \O}$ is used with the Axiom of Replacement as the domain for a mapping whose image is $\set {A, B}$. A suitable mapping would be: :$\paren {y = \O \land z = A} \lor \paren {y = \set \O \land z = B}$ Usually this mapping would be written as: :$\map f x = \begin{cases} A & : x = \O \\ B & : x = \set \O \end{cases}$ {{qed}} Category:Axiom of Pairing Category:Doubletons 0xyy1geaqee8imnyewfu01np6f4o34q
Let $A$ be a set which contains at least two elements. Let $a$ and $b$ be any two elements of $A$. Let $\map P x$ be the propositional function: :$\map P x := \paren {x = a \lor x = b}$ Then we may use the Axiom of Specification to define $B$ as: :$x \in B \iff \set {x \in A: \map P x}$ Hence we can define: :$B := \set {a, b}$ for any two $a, b \in A$. {{qed}}