avewright/proofwiki-math huggingface dataset:
Theorem: Intersection of Nested Closed Subsets of Compact Space is Non-Empty

Let $\struct {T, \tau}$ be a compact topological space.
Let $\sequence {V_n}$ be a sequence of non-empty closed subsets of $T$ with:
:$V_{i + 1} \subseteq V_i$
for each $i$.
Then:
:$\ds \bigcap_{n \mathop = 1}^\infty V_n \ne \O$

Proof:
From Closed Subspace of Compact Space is Compact:
:$V_n$ is compact for each $n$. 
{{AimForCont}} that:
:$\ds \bigcap_{n \mathop = 1}^\infty V_n = \O$
Then:
{{begin-eqn}}
{{eqn	| l = V_1
	| r = V_1 \setminus \paren {\bigcap_{n \mathop = 1}^\infty V_n}
}}
{{eqn	| r = \bigcup_{n \mathop = 1}^\infty \paren {V_1 \setminus V_n}
	| c = De Morgan's Laws: Difference with Intersection
}}
{{end-eqn}}
Since each $V_n$ is closed in $T$, from Closed Set in Topological Subspace: Corollary we have:
:$V_n$ is closed in $V_1$. 
So:
:$V_1 \setminus V_n$ is open for each $n$.
So:
:$\set {V_1 \setminus V_n : n \in \N}$
is a open cover of $V_1$.
Since $V_1$ is compact, there exists a finite subcover:
:$\set {V_1 \setminus V_{n_1}, V_1 \setminus V_{n_2}, \cdots, V_1 \setminus V_{n_j} }$
with:
:$n_1 < n_2 < \cdots < n_j$
so that:
:$\ds \bigcup_{i \mathop = 1}^j \paren {V_1 \setminus V_{n_i} } = V_1$
We then have, by De Morgan's Laws: Difference with Intersection:
:$\ds V_1 \setminus \paren {\bigcap_{i \mathop = 1}^j V_{n_i} } = V_1$
By construction we have:
:$\ds \bigcap_{i \mathop = 1}^j V_{n_i} \subseteq V_{n_1} \subseteq V_1$
so:
:$\ds \bigcap_{i \mathop = 1}^j V_{n_i} = \O$
Since each $V_{n_i}$ is non-empty, for every $x \in V_{n_j}$, there exists some $1 \le k < j$ such that: 
:$x \notin V_{n_k}$
But this is impossible since $V_{n_j} \subseteq V_{n_k}$, and so we have reached a contradiction.
So, we must have: 
:$\ds \bigcap_{n \mathop = 1}^\infty V_n \ne \O$
{{qed}}
Category:Compact Topological Spaces
Category:Closed Sets
m2em4hco0srl7r4nqusridb70t55zf6
Theorem: Family of Lipschitz Continuous Functions with same Lipschitz Constant is Uniformly Equicontinuous

Let $\struct {X, d}$ and $\struct {Y, d'}$ be metric spaces. 
Let $\map \CC {X, Y}$ be the set of continuous functions $X \to Y$.
Let $\FF \subset \map \CC {X, Y}$ be a set of Lipschitz continuous functions all with Lipschitz constant $M \ge 0$.
Then $\FF$ is uniformly equicontinuous.

Proof:
Note that for each $f \in \FF$, we have: 
:$\map {d'} {\map f x, \map f y} \le M \map d {x, y}$
for each $x, y \in X$.
Note that if $M = 0$, we have: 
:$\map {d'} {\map f x, \map f y} = 0$
for each $f \in \FF$ and $x, y \in X$. 
That is: 
:$\map {d'} {\map f x, \map f y} < \epsilon$
for each $\epsilon > 0$, $f \in \FF$ and $x, y \in X$. 
So $\FF$ is uniformly equicontinuous in this case.
Suppose now that $M > 0$. 
Note that whenever: 
:$\ds \map d {x, y} < \frac \epsilon M$
we have: 
{{begin-eqn}}
{{eqn	| l = \map {d'} {\map f x, \map f y}
	| o = \le
	| r = M \map d {x, y}
}}
{{eqn	| o = <
	| r = M \times \frac \epsilon M
}}
{{eqn	| r = \epsilon
}}
{{end-eqn}}
for each $f \in \FF$.
So $\FF$ is uniformly equicontinuous.
{{qed}}
Category:Uniform Equicontinuity
cra8cn2xaj48whhpuzvtsc0rdxxe73k
Theorem: Continuous Function from Compact Hausdorff Space to Itself Fixes a Non-Empty Set

Let $\struct {X, \tau}$ be a compact Hausdorff space. 
Let $f : X \to X$ be a continuous function.
Then there exists a non-empty subset $A \subseteq X$ such that: 
:$f \sqbrk A = A$

Proof:
Define a sequence of sets $\sequence {X_i}_{i \mathop \in \N}$ by:
:$X_i = \begin{cases} X & : i = 1 \\ f \sqbrk {X_{i - 1} } & : i \ge 2 \end{cases}$
Since:
:$f \sqbrk X \subseteq X$
we have: 
:$X_i \subseteq X$
for each $i \in \N$. 
Define:
:$\ds A = \bigcap_{n \mathop = 1}^\infty X_n$
We have: 
{{begin-eqn}}
{{eqn | l = A
      | r = \bigcap_{n \mathop = 1}^\infty X_n
}}
{{eqn | r = X_1 \cap \paren {\bigcap_{n \mathop = 2}^\infty X_n}
}}
{{eqn | r = \bigcap_{n \mathop = 2}^\infty X_n
}}
{{end-eqn}}
We aim to show that:
:$f \sqbrk A = A$
Theorem: Fundamental Property of Norm on Bounded Linear Transformation

Let $\HH, \KK$ be Hilbert spaces.
Let $A: \HH \to \KK$ be a bounded linear transformation.
Let $\norm A$ denote the norm of $A$ defined by:
:$\norm A = \inf \set {c > 0: \forall h \in \HH: \norm {A h}_\KK \le c \norm h_\HH}$
Then:
:$\forall h \in \HH: \norm {A h}_\KK \le \norm A \norm h_\HH$

Proof:
From Norm on Bounded Linear Transformation is Finite:
:$\norm A = \inf \set {c > 0: \forall h \in \HH: \norm {A h}_\KK \le c \norm h_\HH}$ exists
and
:$\norm A < \infty$
Let $x \in \HH \setminus \set{0_\HH}$
Let $\lambda \in \set {c > 0: \forall h \in \HH: \norm {A h}_\KK \le c \norm h_\HH}$.
Then:
{{begin-eqn}}
{{eqn | l = \norm {A x}_K
      | o = \le
      | r = \lambda \norm x_\HH
}}
{{eqn | ll= \leadstoandfrom
      | l = \dfrac {\norm {A x}_\KK} {\norm x_\HH}
      | o = \le
      | r = \lambda
}}
{{end-eqn}}
As $c$ was arbitrary, then:
:$\forall \lambda \in \set {c > 0: \forall h \in \HH: \norm {A h}_\KK \le c \norm h_\HH}: \dfrac {\norm {A x}_\KK} {\norm x_\HH} \le \lambda$
By the definition of the infimum:
:$\dfrac {\norm {A x}_\KK} {\norm x_\HH} \le \norm A$
Hence:
:$\norm {A x}_\KK \le \norm A \norm x_\HH$
Since $x$ was arbitrary:
:$\forall h \in \HH \setminus \set {0_\HH}: \norm {A h}_\KK \le \norm A \norm h_\HH$
Lastly, we have:
{{begin-eqn}}
{{eqn | l = \norm {A 0_\HH}_\KK
      | r = \norm {0_\KK}_\KK
      | c = 
}}
{{eqn | r = 0
      | c = 
}}
{{eqn | r = \norm A \cdot 0
}}
{{eqn | r = \norm A \norm {0_\HH}
}}
{{end-eqn}}
It follows that:
:$\forall h \in \HH: \norm {A h}_\KK \le \norm A \norm h_\HH$
{{qed}}
Theorem: Equivalence of Definitions of Norm of Linear Transformation/Lemma

:$\forall \lambda > 0 : \norm{A 0_H}_K = \lambda \norm{0_H}_H$

Proof:
Let $\lambda > 0$.
We have:
{{begin-eqn}}
{{eqn | l = \norm{A 0_H}_K
      | r = \norm{0_K}_K
      | c = Linear Transformation Maps Zero Vector to Zero Vector
}}
{{eqn | r = 0
      | c = {{Norm-axiom-mult|1}}
}}
{{eqn | r = \lambda \cdot 0
}}
{{eqn | r = \lambda \norm{0_H}
      | c = {{Norm-axiom-mult|1}}
}}
{{end-eqn}}
{{qed}}
Category:Equivalence of Definitions of Norm of Linear Transformation
qjceq6h241lzd9e7i71cl6b11h99whn
Theorem: Quotient of Symmetric Group by Alternating Group is Parity Group

Let $n \ge 2$ be a natural number.
Let $S_n$ denote the symmetric group on $n$ letters.
Let $A_n$ be the alternating group on $n$ letters.
Then the quotient of $S_n$ by $A_n$ is the parity group $C_2$.

Proof:
From Alternating Group is Normal Subgroup of Symmetric Group, $A_n$ is a normal subgroup of $S_n$ whose index is $2$.
By definition of index, $\dfrac {S_n} {A_n}$ is a group of order $2$.
The result follows by definition of the parity group.
{{qed}}
Theorem: Identity Mapping on Symmetric Group is Even Permutation

Let $S_n$ denote the symmetric group on $n$ letters.
Let $e \in S_n$ be the identity permutation on $S_n$.
Then $e$ is an even permutation on $S_n$.

Proof:
By definition of the identity permutation:
:$\forall i \in \N_{>0}: \map e i = i$
Thus $e$ fixes all elements of $S_n$.
We have that a Transposition is of Odd Parity.
Hence a permutation is odd {{iff}} if is the composition of an odd number of transpositions.
The identity permutation is the composition of $0$ (zero) transpositions.
Hence the result.
{{qed}}
Theorem: Even Permutation is Product of 3-Cycles

Let $n \in \N$ such that $n \ge 3$.
Let $\pi: \N_n \to \N_n$ be a '''permutation on $n$ letters'''.
Let $\pi$ be an even permutation.
Then $\pi$ can be expressed as the composition of cyclic permutations of length $3$.

Proof:
Let $A_n$ denote the alternating group on $n$ letters.
By Alternating Group is Set of Even Permutations, $A_n$ is the set of all even permutations of $S_n$.
$\pi$ can therefore be analysed in the context of elements of $A_n$.
The proof proceeds by induction.
For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition:
:$\pi \in A_n$ can be expressed as the composition of cyclic permutations of length $3$
From Identity Mapping on Symmetric Group is Even Permutation, the identity permutation $e$ on $S_n$ is even.
But there are no cyclic permutations of length $3$ in $S_1$ or $S_2$.
So $\map P 1$ and $\map P 2$ are false.
Theorem: Alternating Group is Generated by 3-Cycles

Let $n \in \N$ such that $n \ge 3$.
Let $A_n$ denote the alternating group on $n$ letters.
Then $A_n$ can be generated by the set of $3$-cycles:
:$\set {\tuple {1, i, n}: i \in \set {2, 3, \ldots, n - 1} }$

Proof:
{{tidy|in due course}}
{{MissingLinks|throughout}}
Let $B_n$ be the group generated by $\set {\tuple {1, i, n}: i \in \set {2, 3, \ldots, n - 1} }$.
As every $3$-cycle is even, we have:
:$B_n \subseteq A_n$
Therefore, we want to prove $A_n \subseteq B_n$.
We shall prove the statement by induction on $n$.
Theorem: Alternating Group is Simple except on 4 Letters/Lemma 1

Let $\alpha \in A_n$ be a permutation on $\N_n$ such that $\map \alpha 1 = 2$.
Let $\beta$ be the $3$-cycle $\tuple {3, 4, 5}$.
Then the permutation $\beta^{-1} \alpha^{-1} \beta \alpha$ fixes $1$.

Proof:
We let $\beta^{-1} \alpha^{-1} \beta \alpha$ act on $1$.
By construction:
:$\map \alpha 1 = 2$
Thus:
{{begin-eqn}}
{{eqn | l = \map {\beta^{-1} \alpha^{-1} \beta \alpha} 1
      | r = \map {\paren {3, 5, 4} \alpha^{-1} \tuple {3, 4, 5} } {\map \alpha 1}
      | c = 
}}
{{eqn | r = \map {\paren {3, 5, 4} \alpha^{-1} \tuple {3, 4, 5} } 2
      | c = 
}}
{{eqn | r = \map {\paren {3, 5, 4} \alpha^{-1} } 2
      | c = as $2$ is fixed by $\tuple {3, 4, 5}$
}}
{{eqn | r = \map {\paren {3, 5, 4} } 1
      | c = as $\map \alpha 1 = 2$
}}
{{eqn | r = 1
      | c = as $1$ is fixed by $\tuple {3, 4, 5}$
}}
{{end-eqn}}
{{qed}}
Theorem: Alternating Group is Simple except on 4 Letters/Lemma 2

Let $\alpha \in A_n$ be the permutation on $\N_n$ in the form:
:$\alpha = \tuple {1, 2} \tuple {3, 4}$
Let $\beta$ be the $3$-cycle $\tuple {3, 4, 5}$.
Then:
:$\beta^{-1} \alpha^{-1} \beta \alpha = \beta$

Proof:
{{begin-eqn}}
{{eqn | l = \beta^{-1} \alpha^{-1} \beta \alpha
      | r = \paren {3, 5, 4} \tuple {1, 2} \tuple {3, 4} \tuple {3, 4, 5} \tuple {1, 2} \tuple {3, 4}
      | c = 
}}
{{eqn | r = \paren {3, 5, 4} \tuple {1, 2} \tuple {3, 4} \tuple {1, 2} \tuple {3, 5}
      | c = 
}}
{{eqn | r = \paren {3, 5, 4} \tuple {3, 5, 4}
      | c = 
}}
{{eqn | r = \paren {3, 4, 5}
      | c = 
}}
{{end-eqn}}
{{qed}}
Theorem: Alternating Group has no Subgroup of Order 6

Let $A_4$ denote the alternating group on $4$ letters.
$A_4$ has no subgroup of order $6$.

Proof:
We list the subgroups of $A_4$:
{{:Alternating Group on 4 Letters/Subgroups}}
So there is no subgroup of $A_4$ of order $6$.
{{qed}}
Theorem: Alternating Group on More than 3 Letters is not Abelian

Let $n$ be an integer such that $n > 3$.
Then the $n$th alternating group $A_n$ is not abelian.

Proof:
Let $\tuple {1, 2, 3}$ and $\tuple {2, 3, 4}$ be elements of $A_n$.
Then we have:
{{begin-eqn}}
{{eqn | l = \tuple {1, 2, 3} \tuple {2, 3, 4}
      | r = \tuple {1 2} \tuple {3 4}
      | c = 
}}
{{eqn | l = \tuple {2, 3, 4} \tuple {1, 2, 3}
      | r = \tuple {1 3} \tuple {2 4}
      | c = 
}}
{{end-eqn}}
So:
:$\tuple {1, 2, 3} \tuple {2, 3, 4} \ne \tuple {2, 3, 4} \tuple {1, 2, 3}$
and so $A_n$ is not abelian.
{{qed}}
Theorem: Symmetric Group on Greater than 4 Letters is Not Solvable

Let $n \in \N$ such that $n > 4$.
Let $S_n$ denote the symmetric group on $n$ letters.
Then $S_n$ is not a solvable group.

Proof:
As stated, let $n > 4$ in the below.
Recall the definition of solvable group:
:A finite group $G$ is a '''solvable group''' {{iff}} it has a composition series in which each factor is a cyclic group.
:A '''composition series for $G$''' is a normal series for $G$ which has no proper refinement.
:A '''normal series''' for $G$ is a sequence of (normal) subgroups of $G$:
::$\set e = G_0 \lhd G_1 \lhd \cdots \lhd G_n = G$
:where $G_{i - 1} \lhd G_i$ denotes that $G_{i - 1}$ is a proper normal subgroup of $G_i$.
Consider the alternating group on $n$ letters $A_n$.
From Alternating Group is Normal Subgroup of Symmetric Group, $A_n$ is a proper normal subgroup of $S_n$.
But from Normal Subgroup of Symmetric Group on More than 4 Letters is Alternating Group, the ''only'' normal subgroup of $S_n$ is $A_n$
From Alternating Group is Simple except on 4 Letters, $A_n$ is simple.
That is, $A_n$ has no proper normal subgroup.
It follows that the only composition series for $S_n$ is:
:$\set e \lhd A_n \lhd S_n$
From Alternating Group on More than 3 Letters is not Abelian, $A_n$ is not an abelian group.
From Cyclic Group is Abelian, $A_n$ is not cyclic.
Thus we have demonstrated that the only composition series for $S_n$ contains a factor which is not cyclic.
Hence the result.
{{qed}}
Theorem: Isomorphism of Finite Group with Permutations of Quotient with Subgroup

Let $G$ be a finite group.
Let $H$ be a subgroup of $G$.
Let $H$ contain no non-trivial normal subgroup of $G$.
Let $G / H$ denote the left coset space of $G$ by $H$.
Then $G$ is isomorphic to a subgroup of the group of permutations $\map \Gamma {G / H}$ of $G / H$.

Proof:
Let a homomorphism $\phi: G \to \map \Gamma {G / H}$ be defined as:
:$\forall x \in G: \map {\map \phi g} {x H} = \paren {g x} H$
By Kernel is Normal Subgroup of Domain, $\ker \phi$ is a normal subgroup of $G$.
So:
{{begin-eqn}}
{{eqn | l = g
      | o = \in
      | r = \ker \phi
      | c = 
}}
{{eqn | ll= \leadstoandfrom
      | q = \forall x \in G
      | l = \paren {g x} H
      | r = x H
      | c = 
}}
{{eqn | ll= \leadstoandfrom
      | q = \forall x \in G
      | l = x^{-1} g x
      | o = \in
      | r = H
      | c = 
}}
{{end-eqn}}
That is, $\ker \phi$ is the intersection of $H$ and all its conjugates.
Thus:
:$\ker \phi \subseteq H$
But $H$ contains no non-trivial normal subgroup of $G$.
Thus $\ker \phi$ is the trivial subgroup of $G$
Thus by Kernel is Trivial iff Group Monomorphism:
:$G$ is isomorphic to $\Img \phi$.
{{qed}}
Theorem: Equivalence of Definitions of Norm of Linear Transformation/Definition 4 Greater or Equal Definition 2

Let $\HH, \KK$ be Hilbert spaces.
Let $A: \HH \to \KK$ be a bounded linear transformation.
Let:
:$\lambda_2 = \sup \set {\dfrac {\norm {A h}_\KK} {\norm h_\HH}: h \in \HH, h \ne 0_\HH}$
and
:$\lambda_4 = \inf \set {c > 0: \forall h \in \HH: \norm {A h}_\KK \le c \norm h_\HH}$
Then:
:$\lambda_4 \ge \lambda_2$

Proof:
From Fundamental Property of Norm on Bounded Linear Transformation:
:$\forall h \in \HH : \norm{A h}_\KK \le \lambda_4 \norm h_\HH$
Hence:
:$\forall h \in H, h \ne 0_\HH : \dfrac {\norm{A h}_\KK} {\norm h_\HH} \le \lambda_4$
From Continuum Property:
:$\lambda_2 = \sup \set {\dfrac {\norm {A h}_\KK} {\norm h_\HH}: h \in \HH, h \ne 0_\HH}$ exists
By definition of the supremum:
:$\lambda_2 \le \lambda_4$
Theorem: Equivalence of Definitions of Norm of Linear Transformation/Definition 2 Greater or Equal Definition 1

Let $H, K$ be Hilbert spaces.
Let $A: H \to K$ be a bounded linear transformation.
Let: 
:$\lambda_1 = \sup \set {\norm {A h}_K: \norm h_H \le 1}$
and
:$\lambda_2 = \sup \set {\dfrac {\norm {A h}_K} {\norm h_H}: h \in H, h \ne 0_H}$
Then:
:$\lambda_2 \ge \lambda_1$

Proof:
By definition of the supremum:
:$\forall h \in H, h \ne \mathbf 0_H: \dfrac {\norm {A h}_K} {\norm h_H} \le \lambda_2$
Hence:
:$\forall h \in H, h \ne \mathbf 0_H: \norm {A h}_K \le \lambda_2 \norm h_H$
From Lemma:
:$\norm{A 0_H}_K = \lambda_2 \norm{0_H}_H$
Hence:
:$\forall h \in H: \norm {A h}_K \le \lambda_2 \norm h_H$
In particular:
{{begin-eqn}}
{{eqn | q = \forall h \in H, \norm h_H \le 1
      | l = \norm {A h}_K
      | o = \le
      | r = \lambda_2 \norm h_H
}}
{{eqn | o = \le
      | r = \lambda_2
}}
{{end-eqn}}
From Continuum Property:
:$\lambda_1 = \sup \set {\norm {A h}_K: \norm h_H \le 1}$ exists
By definition of the supremum:
:$\lambda_1 \le \lambda_2$
Theorem: Equivalence of Definitions of Norm of Linear Transformation/Definition 1 Greater or Equal Definition 3

Let $H, K$ be Hilbert spaces.
Let $A: H \to K$ be a bounded linear transformation.
Let:
:$\lambda_1 = \sup \set {\norm {A h}_K: \norm h_H \le 1}$
and
:$\lambda_3 = \sup \set {\norm {A h}_K: \norm h_H = 1}$
Then:
:$\lambda_1 \ge \lambda_3$

Proof:
By definition of the supremum:
:$\forall h \in H, \norm h_H \le 1 : \norm{A h}_K \le \lambda_1$
In particular:
:$\forall h \in H, \norm h_H = 1 : \norm{A h}_K \le \lambda_1$
From Continuum Property:
:$\lambda_3 = \sup \set {\norm {A h}_K: \norm h_H = 1}$ exists
By definition of the supremum:
:$\lambda_3 \le \lambda_1$
Theorem: Equivalence of Definitions of Norm of Linear Transformation/Definition 3 Greater or Equal Definition 4

Let $H, K$ be Hilbert spaces.
Let $A: H \to K$ be a bounded linear transformation.
Let:
:$\lambda_3 = \sup \set {\norm {A h}_K: \norm h_H = 1}$
and
:$\lambda_4 = \inf \set {c > 0: \forall h \in H: \norm {A h}_K \le c \norm h_H}$
Let:
:$\lambda_3 \ge \lambda_4$

Proof:
Let $h \in H: h \ne 0_h$.
We have:
{{begin-eqn}}
{{eqn | l = \norm {\dfrac 1 {\norm h_H} h }_H
      | r = \dfrac {\norm h_H}{\norm h_H}
      | c = {{Norm-axiom-mult|2}}
}}
{{eqn | r = 1
}}
{{end-eqn}}
and
{{begin-eqn}}
{{eqn | l = \dfrac {\norm{A h}_K} {\norm h_H}
      | r = \norm {\dfrac 1 {\norm h_H} A h}_K
      | c = {{Norm-axiom-mult|2}}
}}
{{eqn | r = \norm {A \paren{ \dfrac 1 {\norm h_H} h } }_K
      | c = Linear Transformation Maps Zero Vector to Zero Vector
}}
{{eqn | o = \le 
      | r = \lambda_3
      | c = {{Defof|Supremum of Set}}
}}
{{end-eqn}}
Hence:
:$\forall h \in H: h \ne 0_h: \norm{A h}_K \le \lambda_3 \norm h_H$
From Lemma:
:$\norm {A 0_H}_K = \lambda_3 \norm {0_H}_H$
Hence:
:$\forall h \in H: \norm{A h}_K \le \lambda_3 \norm h_H$
That is,
:$\lambda_3 \in \set {c > 0: \forall h \in H: \norm {A h}_K \le c \norm h_H}$
By definition of the infimum:
:$\lambda_4 \le \lambda_3$
Theorem: Index of Proper Subgroup of Symmetric Group

Let $n \in \N$ be a natural number such that $n > 4$.
Let $S_n$ denote the symmetric group on $n$ letters.
Let $A_n$ denote the alternating group on $n$ letters.
$A_n$ is the only proper subgroup of $S_n$ whose index is less than $n$.

Proof:
From Normal Subgroup of Symmetric Group on More than 4 Letters is Alternating Group:
:$A_n$ is the only proper non-trivial normal subgroup of $S_n$.
Suppose $H$ is a subgroup of $S_n$ whose index $\index {S_n} H$ is less than $n$.
If $\index {S_n} H = 2$ then from Subgroup of Index 2 is Normal $H$ is normal.
Hence $H = A_n$.
Suppose $\index {S_n} H > 2$.
Then $A_n \nsubseteq H$.
Thus the hypotheses of Isomorphism of Finite Group with Permutations of Quotient with Subgroup are fulfilled.
Thus $S_n$ is isomorphic to a subgroup of the group of permutations $\map \Gamma {S_n / H}$ of $S_n / H$.
However:
{{begin-eqn}}
{{eqn | l = \order {\map \Gamma {S_n / H} }
      | r = \index {S_n} H !
      | c = 
}}
{{eqn | o = <
      | r = n!
      | c = as $\index {S_n} H < n$ {{hypothesis}}
}}
{{eqn | r = \order {S_n}
      | c = 
}}
{{end-eqn}}
which is a contradiction.
{{qed}}
Theorem: Number of Distinct Functions on n Variables obtained by Permutation

Let $\map f {x_1, x_2, \ldots, x_n}$ be a function on $n$ independent variables where $n > 4$.
Let $\nu$ denote the number of distinct functions that can be obtained when $\tuple {x_1, x_2, \ldots, x_n}$ are permuted.
Then:
:$\nu > 2 \implies \nu \ge n$

Proof:
{{ProofWanted|Probably uses Index of Proper Subgroup of Symmetric Group, as this looks suggestive}}
Theorem: Fundamental Solution to 2D Laplace's Equation

Let $\delta_{\tuple {0, 0}} \in \map {\DD'} {\R^2}$ be the Dirac delta distribution.
Let $\Delta = \dfrac {\partial^2} {\partial x^2} + \dfrac {\partial^2} {\partial y^2}$ be the Laplacian in $2$-dimensional Euclidean space.
Then in the distributional sense:
:$\ds \map \Delta {\frac {\ln r} {2 \pi}} = \delta_{\tuple {0, 0}}$
where $r = \sqrt {x^2 + y^2}$.

Proof:
Let $\norm {\,\cdot\,}_2$ be the 2-norm.
Theorem: Transitive Subgroup of Prime containing Transposition

Let $p$ be a prime number.
Let $S_p$ denote the symmetric group on $p$ letters.
Let $H$ be a transitive subgroup of $S_p$.
If $H$ contains a transposition, then $H = S_p$.

Proof:
{{WLOG}}, let $\tuple {1, 2}$ be the transposition contained by $H$.
Let us define an equivalence relation $\sim$ on the set $\N_p = \set {1, 2, \ldots, p}$ as:
:$i \sim j \iff \tuple {i, j} \in H$
Because $H$ is a transitive subgroup it follows that each $\sim$-equivalence class has the same number of elements.
In fact, if $\phi \in H$ and $\phi_1 := \map \phi 1 = i$, then $\phi$ yields a bijection from the $\sim$-equivalence class of $1$ to that of $i$, because:
:$\tuple {1, k} \in H \iff \tuple {i, \phi_k} = \tuple {\phi_1, \phi_k} = \phi \circ \tuple {1, k} \circ \phi^{-1} \in H$
The number $s$ of elements of any given $\sim$-equivalence class must be a divisor of $p$.
Thus $s = 1$ or $s = p$.
However, the $\sim$-equivalence class of $1$ contains at least both $1$ and $2$.
So there can be only one $\sim$-equivalence class which then contains $p$ elements.
In other words, $H$ contains all the transpositions of $S_p$.
From Existence and Uniqueness of Cycle Decomposition, every permutation is a composition of transpositions.
Hence:
:$H = S_p$
{{qed}}
Theorem: Inverse of Field Product with Inverse

Let $\struct {F, +, \times}$ be a field whose zero is $0$ and whose unity is $1$.
Let $a, b \in F$ such that $a \ne 0$.
Then:
:$\paren {a \times b^{-1} }^{-1} = b \times a^{-1}$
This can also be expressed using the notation of division as:
:$1 / \paren {a / b} = b / a$

Proof:
By definition, a field is a non-trivial division ring whose ring product is commutative.
By definition, a division ring is a ring with unity such that every non-zero element is a unit. 
Hence we can use Inverse of Division Product:
:$\paren {\dfrac a b}^{-1} = \dfrac {1_R} {\paren {a / b}} = \dfrac b a$
which applies to the group of units of a comutative ring with unity.
{{qed}}
Theorem: Addition of Division Products in Field

Let $\struct {F, +, \times}$ be a field whose zero is $0$ and whose unity is $1$.
Let $a, b, c, d \in F$ such that $b \ne 0$ and $d \ne 0$.
Then:
:$\dfrac a b + \dfrac c d = \dfrac {a d + b c} {b d}$
where $\dfrac x z$ is defined as $x \paren {z^{-1} }$.

Proof:
By definition, a field is a non-trivial division ring whose ring product is commutative.
By definition, a division ring is a ring with unity such that every non-zero element is a unit. 
Hence we can use Addition of Division Products:
:$\dfrac a b + \dfrac c d = \dfrac {a \circ d + b \circ c} {b \circ d}$
which applies to the group of units of a comutative ring with unity.
{{qed}}
Theorem: Norm on Bounded Linear Functional is Finite

Let $V$ be a normed vector space.
Let $L$ be a bounded linear functional on $V$.
Let $\norm L$ denote the norm on $L$ defined as:
:$\norm L = \inf \set {c > 0: \forall v \in V: \size {L v} \le c \norm v_V}$
Then:
:$\norm L < \infty$

Proof:
By definition of a bounded linear functional:
:$\exists c \in \R_{> 0}: \forall v \in V: \size{L v} \le c \norm v_V$
Hence:
:$\set {\lambda > 0: \forall v \in V: \size {L v} \le \lambda \norm v_V} \ne \O$
By definition:
$\set {\lambda > 0: \forall v \in V: \size {L v} \le \lambda \norm v_V}$ is bounded below by $0$.
From the Greatest Lower Bound Property:
:$\norm L = \inf \set {\lambda > 0: \forall v \in V: \size {L v} \le \lambda \norm v_V}$ exists.
We have:
{{begin-eqn}}
{{eqn | l = \norm L
      | o = \le
      | r = c
      | c = {{Defof|Infimum}}
}}
{{eqn | o = <
      | r = \infty
      | c = because $c \in \R_{>0}$
}}
{{end-eqn}}
The result follows.
{{qed}}
Category:Bounded Linear Functionals
8x411n7n3jih7x65njn1t295bafkjj8