nyuuzyou/wikis huggingface dataset:
Groups of Order 6

== Theorem ==
There exist exactly $2$ groups of order $6$, up to isomorphism:
:$C_6$, the cyclic group of order $6$
:$S_3$, the symmetric group on $3$ letters.
== Proof ==
From Existence of Cyclic Group of Order n we have that one such group of order $6$ is $C_6$ the cyclic group of order $6$:
This is exemplified by the additive group of integers modulo $6$, whose Cayley table can be presented as:
Then we have the symmetric group on $3$ letters.
From Order of Symmetric Group, this has order $6$.
It can be exemplified by the symmetry group of the equilateral triangle, whose Cayley table can be presented as:
It remains to be shown that these are the only $2$ groups of order $6$.
Let $G$ be a group of order $6$ whose identity is $e$.
By the First Sylow Theorem, $G$ has at least one Sylow $3$-subgroup.
By the Third Sylow Theorem, the number of Sylow $3$-subgroups is a divisor of $6$.
By the Fourth Sylow Theorem, the number of Sylow $3$-subgroups is equivalent to $1 \pmod p$.
Combining these results, this number is therefore $1$.
Call this Sylow $3$-subgroup $P$.
By Sylow $p$-Subgroup is Unique iff Normal, $P$ is normal.
From Prime Group is Cyclic, $P = \gen x$ for some $x \in G$ for $x^3 = e$.
By the First Sylow Theorem, $G$ also has at least one Sylow $2$-subgroup of order $2$.
Thus $G$ has an element $y$ such that $y^2 = e$.
We have that $P$ is normal.
Therefore:
:$y^{-1} x y \in P$
Therefore one of the following applies:
If $y^{-1} x y = e$ then it follows that $x = 1$, which is contrary to our deduction that $x^3 = e$.
Hence there remain two possibilities.
First suppose $y^{-1} x y = x$.
Then:
Hence we can calculate the powers of $x y$ in turn:
Thus the order of $x y$ is $6$, and so $G$ is cyclic of order $6$.
Now suppose $y^{-1} x y = x^2$.
Then:
and:
It remains to investigate $x^2 y$:
Thus we have $6$ elements:
:$e, x, x^2$ which form a cyclic group of order $3$
:$y, x y, x^2 y$ all of which are self-inverse.
Thus in this case $G$ is the symmetric group on $3$ letters $S_3$.
The possibilities are exhausted.
Hence the result.
== Sources ==
* : Chapter $\text {IV}$: Rings and Fields: $25$. Cyclic Groups and Lagrange's Theorem
* : Chapter $2$: Subgroups and Cosets: $\S 41 \zeta$
* : Chapter $5$: Cosets and Lagrange's Theorem: Groups with six elements
* : Chapter $11$: The Sylow Theorems: Example $11.13$
Category:Groups of Order 6
Groups of Order 8

== Theorem ==
Let $G$ be a group of order $8$.
Then $G$ is isomorphic to one of the following:
:$\Z_8$
:$\Z_4 \oplus \Z_2$
:$\Z_2 \oplus \Z_2 \oplus \Z_2$
:$D_4$
:$\Dic 2$
where:
:$\Z_n$ is the cyclic group of order $n$
:$D_4$ is the dihedral group of order $8$
:$\Dic 2$ is the dicyclic group of order $8$.
== Proof ==
The abelian cases are handled by the corollary to Abelian Group Factored by Prime.
Let $G$ be non-abelian.
By Lagrange's theorem the order of non-identity elements in $G$ is either $2$, $4$ or $8$.
 that there exists an order $8$ element.
Then $G$ is generated by this element.
So $G$ is by definition cyclic.
But Cyclic Group is Abelian, contradicting the assumption that $G$ is non-abelian.
So there is no order $8$ element.
By Non-Abelian Order 8 Group has Order 4 Element, there exists at least one order $4$ element in $G$.
Let it be denoted by $a$.
Let $A$ denote the subgroup generated by $a$.
By Lagrange's theorem there are two cosets in $G$: $A$ and $G \setminus A$.
Let $b \in G \setminus A$.
Then $\set {a, b}$ is a generator of $G$.
Now we consider how $a$ and $b$ interact with each other.
Consider the element $x = b a b^{-1}$.
By Subgroup of Index 2 is Normal, $b A b^{-1} = A$.
So $x \in A$.
By Order of Conjugate Element equals Order of Element, the only possible choices are $x = a$ or $x = a^3$.
If $x = a$, then $a$ and $b$ commute.
Since $\set {a, b}$ generates $G$, this makes $G$ an abelian group, which is a contradiction.
So $b a b^{-1} = a^3$.
It suffices to consider the order of $b$:
:If $\order b = 2$, then $G \cong D_4$.
:If $\order b = 4$, then $G \cong \Dic 2$.
== Also see ==
* Classification of Groups of Order up to 15
* Groups of Order 12
== Sources ==
* : Chapter $2$: Group Homomorphism and Isomorphism: $\S 62 \gamma$
* : Chapter $5$: Cosets and Lagrange's Theorem: Groups with eight elements
Category:Order of Groups
Category:Groups of Order 8
Guarini's Problem

== Classic Problem ==
$2$ white knights and $2$ black knights are placed on the corners of a $3 \times 3$ chessboard as follows:
All pieces may move according to the rules of chess.
How can the white knights and black knights change places?
== Solution ==
A knight cannot access the square $\mathrm b 2$, so let us ignore that square throughout the following analysis.
From every square, it is possible to access exactly $2$ other squares by means of a knight's move.
Let us construct a graph whose vertices correspond to the squares and whose edges are the possible moves between those squares.
It is seen to be the cycle graph of order $8$.
:
The solution becomes apparent.
Each knight needs to move along the vertices of the graph from where it starts to the position directly opposite in the graph.
Hence the moves are:
:$\mathrm a 1 \to \mathrm c 2 \to \mathrm a 3 \to \mathrm b 1 \to \mathrm c 3$
:$\mathrm a 3 \to \mathrm b 1 \to \mathrm c 3 \to \mathrm a 2 \to \mathrm c 1$
:$\mathrm c 3 \to \mathrm a 2 \to \mathrm c 1 \to \mathrm b 3 \to \mathrm a 1$
:$\mathrm c 1 \to \mathrm b 3 \to \mathrm a 1 \to \mathrm c 2 \to \mathrm a 3$
Or they could go in the opposite direction:
:$\mathrm a 1 \to \mathrm b 3 \to \mathrm c 1 \to \mathrm a 2 \to \mathrm c 3$
:$\mathrm c 1 \to \mathrm a 2 \to \mathrm c 3 \to \mathrm b 1 \to \mathrm a 3$
:$\mathrm c 3 \to \mathrm b 1 \to \mathrm a 3 \to \mathrm c 2 \to \mathrm a 1$
:$\mathrm a 3 \to \mathrm c 2 \to \mathrm a 1 \to \mathrm b 3 \to \mathrm c 1$
Move each knight in turn along its required path, so as to vacate the square it was on ready for the knight behind it to occupy.
== Also see ==
* Definition:Knight's Tour
== Historical Note ==
== Sources ==
* : Dynamical Chess Puzzles: $341$. The Four Frogs
* : Exchanging the Knights: $101$
Category:Classic Problems
Category:Recreational Chess
Category:Knight's Tours
Guarini's Problem/Historical Note

== Historical Note on Guarini's Problem ==
Guarini's Problem is one of the first recreational chess problems to be posed.
 published it in $1512$.
The solution given here is that by from $1917$.
He couched the problem in the context of frogs hopping between lily pads, presumably for variety or disguise.
He credits appropriately in his solution.
== Sources ==
* : Exchanging the Knights: $101$
Category:Historical Notes
Gyromagnetic Ratio of Proton

== Theorem ==
The gyromagnetic ratio of the proton is given by:
== Proof ==
== Sources ==
* : $2$. Physical Constants and Conversion Factors: Table $2.3$ Adjusted Values of Constants
::which gives the mantissa as $2 \cdotp 675 \, 196 \, 5$, with an uncertainty of $\pm 82$ corresponding to the $2$ least significant figures
Category:Gyromagnetic Ratio
Category:Protons
Gyromagnetic Ratio of Proton Uncorrected for Diamagnetism

== Theorem ==
The gyromagnetic ratio of the proton, adjusted for diamagnetism, $\mathrm {H_2 O}$, is given by:
== Proof ==
== Sources ==
* : $2$. Physical Constants and Conversion Factors: Table $2.3$ Adjusted Values of Constants
Category:Gyromagnetic Ratio
Category:Protons
Gödel's Beta Function Lemma

== Theorem ==
Let $\beta: \N^3 \to \N$ be Gödel's $\beta$ function.
Let $\sequence {x_0, x_1, \dotsc, x_n }$ be a finite sequence of natural numbers.
Then there exist some $a, b \in \N$ such that, for every $i \le n$:
:$\map \beta {a, b, i} = x_i$
== Proof ==
Let $M = \map \max {n, x_0, x_1, \dotsc, x_n}$.
Define $b = M!$, the factorial of $M$.
Let $\sequence {y_1, y_2, \dotsc, y_{n + 1} }$ be the finite sequence defined by:
:$y_i = 1 + \paren {i \times b}$
As $M \ge n$, by definition:
:$M \ge \paren {n + 1} - 1$
Therefore, by Multiples of Factorial Plus One are Coprime, the $\sequence {y_1, \dotsc, y_{n + 1} }$ are pairwise coprime.
Thus, by the Chinese Remainder Theorem, there exists some $a \in \N$ such that:
But then, for each $0 \le i \le n$:
Therefore, by the congruence:
:$\map \rem {a, y_{i + 1} } = x_i$
But:
:$y_{i + 1} = 1 + \paren {\paren {i + 1} \times b}$
Thus:
:$\map \rem {a, 1 + \paren {\paren {i + 1} \times b} } = x_i$
But that is the definition of the $\beta$ function.
Hence the result.
Category:Gödel's Beta Function
Gödel's Beta Function is Arithmetically Definable

== Theorem ==
Let Gödel's $\beta$ function $\beta: \N^3 \to \N$ be defined as:
:$\map \beta {x, y, z} = \map \rem {x, 1 + \paren {z + 1} \times y}$
Then there exists a $\Sigma_1$ WFF of $4$ free variables:
:$\map \phi {r, x, y, z}$
such that:
:$r = \map \beta {x, y, z} \iff \N \models \map \phi {\sqbrk r, \sqbrk x, \sqbrk y, \sqbrk z}$
where $\sqbrk a$ denotes the unary representation of $a \in \N$.
== Proof ==
Follows from:
:Basic Primitive Recursive Functions are Arithmetically Definable
:Addition is Arithmetically Definable
:Multiplication is Arithmetically Definable
:Remainder is Arithmetically Definable
:Substitution of Arithmetically Definable Functions is Arithmetically Definable
Gödel's Incompleteness Theorems

== Theorem ==
=== Gödel's First Incompleteness Theorem ===
=== Gödel's Second Incompleteness Theorem ===
== Also see ==
* Gödel's Completeness Theorem
== Historical Note ==
== Sources ==
* 
* 
* 
Category:Gödel's Incompleteness Theorems
Category:Mathematical Logic
Gödel's Incompleteness Theorems/First

== Theorem ==
Let $T$ be the set of theorems of some recursive set of sentences in the language of arithmetic such that $T$ contains minimal arithmetic.
$T$ cannot be both consistent and complete.
=== Corollary ===
== Proof ==
 that such a $T$ is consistent and complete.
By the Undecidability Theorem, since $T$ is consistent and contains $Q$, it is not recursive.
But, by Complete Recursively Axiomatized Theories are Recursive, since $T$ is complete and is the set of theorems of a recursive set, it is recursive.
The result follows by Proof by Contradiction.
== Also see ==
* Gödel's Second Incompleteness Theorem
== Sources ==
* 
* : $\S 15$: Theorem $6$
* 
* 
Category:Gödel's Incompleteness Theorems
Gödel's Incompleteness Theorems/First/Corollary

== Theorem ==
Let $T$ be the set of theorems of some recursive set of sentences in the language of arithmetic such that $T$ contains minimal arithmetic.
If $T$ is both consistent and complete, it does not contain minimal arithmetic.
== Proof ==
This is simply the contrapositive of Gödel's First Incompleteness Theorem.
== Sources ==
* : $\S 15$: Theorem $6$: Corollary
Category:Gödel's Incompleteness Theorems
Gödel's Incompleteness Theorems/Historical Note

== Historical Note on Gödel's Incompleteness Theorems ==
Gödel's Incompleteness Theorems answered the second of Hilbert's $23$ (then) unsolved problems of mathematics.
Hence it ended attempts, like those of and , to develop the whole of mathematics from a finite set of logical axioms.
It also damages the idea of finding a finite set of basic axioms of physics to define all natural phenomena.
== Sources ==
* 
* 
Category:Historical Notes
Gödel's Incompleteness Theorems/Second

== Theorem ==
Let $T$ be the set of theorems of some recursive set of sentences in the language of arithmetic such that $T$ contains minimal arithmetic.
Let $\map {\mathrm {Cons} } T$ be the propositional function which states that $T$ is consistent.
Then it is not possible to prove $\map {\mathrm {Cons} } T$ by means of formal statements within $T$ itself.
== Proof ==
== Also see ==
* Gödel's First Incompleteness Theorem
== Sources ==
* 
* 
* 
Category:Gödel's Incompleteness Theorems
H-Cobordism Theorem

== Theorem ==
Let $X^n, Y^n$ be two simply connected manifolds.
Let $n \in \N: n \ge 5$ and $\exists W$ such that $W$ is an h-cobordism between $X$ and $Y$.
Then $\exists \psi: W \to X \times \closedint 0 1$ such that $\psi$ is a diffeomorphism.
In particular, $X$ and $Y$ are diffeomorphic.
== Proof ==
Category:Manifolds
Hadamard Conjecture

== Open Question ==
Let $n \in \Z$ be a (strictly) positive integer.
Then there exists a Hadamard matrix for order $4 n$.
== Progress ==
As of March $2024$, There remain the following multiples of $4$ less than $2000$ for which it is not known whether a Hadamard matrix exists:
:$668$, $716$, $892$, $1132$, $1244$, $1388$, $1436$, $1676$, $1772$, $1916$, $1948$, $1964$
== Sources ==
* 
* 
* 
Category:Hadamard Matrices
Category:Open Questions
Hadamard Factorization Theorem

== Theorem ==
Let $f: \C \to \C$ be an entire function of finite order $\omega$.
Let $0$ be a zero of $f$ of multiplicity $m \ge 0$.
Let $\sequence {a_n}$ be the sequence of non-zero zeroes of $f$, repeated according to multiplicity.
Then:
:$f$ has finite rank $p \le \omega$
and:
:there exists a polynomial $g$ of degree at most $\omega$ such that:
::$\ds \map f z = z^m e^{\map g z} \prod_{n \mathop = 1}^\infty E_p \paren {\frac z {a_n} }$
where $E_p$ denotes the $p$th Weierstrass elementary factor.
== Proof ==
By Convergence Exponent is Less Than Order, $f$ has finite exponent of convergence $\tau \le \omega$.
By Relation Between Rank and Exponent of Convergence, $f$ has finite rank $p \leq \omega$.
== Also see ==
* Weierstrass Factorization Theorem
* Order is Maximum of Exponent of Convergence and Degree
* Definition:Hadamard's Canonical Factorization
== Sources ==
* : Chapter III: Further Theory of $\map \zeta s$. Applications: $\S7$: Integral Functions: Theorem $F3$
Category:Entire Functions
Category:Infinite Products
Category:Complex Analysis
Hadamard Matrix/Examples/Arbitrary 2 x 2

== Example of Hadamard Matrix ==
An arbitrary example of a $2 \times 2$ Hadamard matrix is:
:$\begin {pmatrix} 1 & 1 \\ -1 & 1 \end {pmatrix}$
== Proof ==
We multiply $\begin {pmatrix} 1 & 1 \\ -1 & 1 \end {pmatrix}$ by its transpose, and find:
:$\begin {pmatrix} 1 & 1 \\ -1 & 1 \end {pmatrix} \begin {pmatrix} 1 & -1 \\ 1 & 1 \end {pmatrix} = \begin {pmatrix} 2 & 0 \\ 0 & 2 \end {pmatrix} = 2 \begin {pmatrix} 1 & 0 \\ 0 & 1 \end {pmatrix}$
Hence the result by definition of Hadamard matrix.
== Sources ==
* 
* 
Category:Examples of Hadamard Matrices
Hadamard Product over Group forms Group

== Theorem ==
Let $\struct {G, \cdot}$ be a group whose identity is $e$.
Let $\map {\MM_G} {m, n}$ be a $m \times n$ matrix space over $\struct {G, \cdot}$.
Then $\struct {\map {\MM_G} {m, n}, \circ}$, where $\circ$ is Hadamard product, is also a group.
== Proof ==
As $\struct {G, \cdot}$, being a group, is a monoid, it follows from Matrix Space Semigroup under Hadamard Product that $\struct {\map {\MM_G} {m, n}, \circ}$ is also a monoid.
As $\struct {G, \cdot}$ is a group, it follows from Negative Matrix is Inverse for Hadamard Product that all elements of $\struct {\map {\MM_G} {m, n}, \circ}$ have an inverse.
The result follows.
== Also see ==
* Matrix Entrywise Addition forms Abelian Group
Category:Hadamard Product
Category:Examples of Groups
Hahn-Banach Separation Theorem/Hausdorff Locally Convex Space/Complex Case

== Theorem ==
Let $\struct {X, \PP}$ be a Hausdorff locally convex space over $\C$ equipped with its standard topology. 
Let $X^\ast$ be the topological dual space of $\struct {X, \PP}$.
=== Open Convex Set and Convex Set ===
=== Compact Convex Set and Closed Convex Set ===
Category:Hahn-Banach Separation Theorem
Hahn-Banach Separation Theorem/Hausdorff Locally Convex Space/Complex Case/Compact Convex Set and Closed Convex Set

== Theorem ==
Let $\struct {X, \PP}$ be a Hausdorff locally convex space over $\C$ equipped with its standard topology. 
Let $X^\ast$ be the topological dual space of $\struct {X, \PP}$. 
Let $A \subseteq X$ be an compact convex set.
Let $B \subseteq X$ be a closed convex set disjoint from $A$. 
Then there exists $f \in X^\ast$ such that: 
:$\ds \sup_{x \mathop \in A} \map \Re {\map f x} < \inf_{x \mathop \in B} \map \Re {\map f x}$
== Proof ==
Let $X_\R$ be the realification of $X$. 
Applying Hahn-Banach Separation Theorem: Hausdorff Locally Convex Space: Real Case: Compact Convex Set and Closed Convex Set, there exists a continuous $\R$-linear functional $g : X \to \R$ and $c \in \R$ such that: 
:$\ds \sup_{x \mathop \in A} \map g x < \inf_{x \mathop \in B} \map g x$
From Continuous Real Linear Functional on Complex Topological Vector Space is Real Part of Continuous Complex Linear Functional, there exists $f \in X^\ast$ such that: 
:$\map g x = \map \Re {\map f x}$ for each $x \in X$. 
Then, we have: 
:$\ds \sup_{x \mathop \in A} \map \Re {\map f x} < \inf_{x \mathop \in B} \map \Re {\map f x}$
Category:Hahn-Banach Separation Theorem
Hahn-Banach Separation Theorem/Hausdorff Locally Convex Space/Complex Case/Open Convex Set and Convex Set

== Theorem ==
Let $\struct {X, \PP}$ be a Hausdorff locally convex space over $\C$ equipped with its standard topology. 
Let $X^\ast$ be the topological dual space of $\struct {X, \PP}$. 
Let $A \subseteq X$ be an open convex set.
Let $B \subseteq X$ be a convex set disjoint from $A$. 
Then there exists $f \in X^\ast$ and $c \in \R$ such that: 
:$A \subseteq \set {x \in X : \map \Re {\map f x} < c}$
and:
:$B \subseteq \set {x \in X : \map \Re {\map f x} \ge c}$
That is: 
:there exists $f \in X^\ast$ and $c \in \R$ such that $\map \Re {\map f a} < c \le \map \Re {\map f b}$ for each $a \in A$ and $b \in B$.
== Proof ==
Let $X_\R$ be the realification of $X$. 
Applying Hahn-Banach Separation Theorem: Hausdorff Locally Convex Space: Real Case: Open Convex Set and Convex Set, there exists a continuous $\R$-linear functional $g : X \to \R$ and $c \in \R$ such that: 
:$\map g a < c \le \map g b$ for each $a \in A$ and $b \in B$.
From Continuous Real Linear Functional on Complex Topological Vector Space is Real Part of Continuous Complex Linear Functional, there exists $f \in X^\ast$ such that: 
:$\map g x = \map \Re {\map f x}$ for each $x \in X$. 
Then, we have: 
:$\map \Re {\map f a} < c \le \map \Re {\map f b}$ for each $a \in A$ and $b \in B$.
Category:Hahn-Banach Separation Theorem
Hahn-Banach Separation Theorem/Hausdorff Locally Convex Space/Real Case

== Theorem ==
Let $\struct {X, \PP}$ be a Hausdorff locally convex space over $\R$ equipped with its standard topology. 
Let $X^\ast$ be the topological dual space of $\struct {X, \PP}$.
=== Open Convex Set and Convex Set ===
=== Compact Convex Set and Closed Convex Set ===
Category:Hahn-Banach Separation Theorem
Hahn-Banach Separation Theorem/Hausdorff Locally Convex Space/Real Case/Open Convex Set and Convex Set

== Theorem ==
Let $\struct {X, \PP}$ be a Hausdorff locally convex space over $\R$ equipped with its standard topology. 
Let $X^\ast$ be the topological dual space of $\struct {X, \PP}$. 
Let $A \subseteq X$ be an open convex set.
Let $B \subseteq X$ be a convex set disjoint from $A$. 
Then there exists $f \in X^\ast$ and $c \in \R$ such that: 
:$A \subseteq \set {x \in X : \map f x < c}$
and:
:$B \subseteq \set {x \in X : \map f x \ge c}$
That is: 
:there exists $f \in X^\ast$ and $c \in \R$ such that $\map f a < c \le \map f b$ for each $a \in A$ and $b \in B$.
== Proof ==
Fix $a_0 \in A$ and $b_0 \in B$. 
From Sum of Set and Open Set in Topological Vector Space is Open, we have: 
:$A - B$ is open. 
From Translation of Open Set in Topological Vector Space is Open, we have: 
:$A - B + \paren {b_0 - a_0}$ is open.
Note that since $A \cap B = \O$, we have $0 \notin A - B$. 
Hence $b_0 - a_0 \notin A - B + \paren {b_0 - a_0}$.
From , we have that $A - B$ is convex.
From Translation of Convex Set in Vector Space is Convex, we have that $A - B + \paren {b_0 - a_0}$ is convex.
Applying Open Convex Set in Hausdorff Locally Convex Space is Separated from Points outside Set by Continuous Linear Functional, there exists $f \in X^\ast$ such that: 
:$\map f v < \map f {b_0 - a_0}$
for all $v \in A - B + \paren {b_0 - a_0}$.
That is: 
:$\map f {a - b + b_0 - a_0} < \map f {b_0 - a_0}$
for each $a \in A$ and $b \in B$. 
Since $f$ is linear, we have: 
:$\map f a < \map f b$
for each $a \in A$ and $b \in B$.
Then we have: 
:$\map f {a_0} < \map f b$
for all $b \in B$. 
Hence: 
:$\ds \inf_{b \mathop \in B} \map f b > -\infty$
Let:
:$\ds c = \inf_{b \mathop \in B} \map f b$
Then, we have: 
:$\map f a \le c \le \map f b$ 
It remains to show that $\map f a < c$ for each $a \in A$.
Let $a \in A$.
Since $A$ is open, there exists an open neighborhood $V_x$ of ${\mathbf 0}_X$ such that $x + V_x$ is an open neighborhood of $x$ in $A$.
Now, since we have: 
:$\map f {a_0} < \map f b$
for each $b \in B$, we in particular have that $f$ is not constant ${\mathbf 0}_X$. 
Hence there exists $v \in X$ such that: 
:$\map f v \ne 0$
Replacing $v$ by $-v$ if necessary, suppose that: 
:$\map f v > 0$
From Multiple of Vector in Topological Vector Space Converges, we have: 
:$\dfrac v n \to {\mathbf 0}_X$ as $n \to \infty$ 
So for some $N \in \N$, we have: 
:$\dfrac v N \in V_x$
by the definition of a convergent sequence.
Set:
:$v' = \dfrac v N$
Then we have $\map f {v'} > 0$ and $x + v' \in A$.
We therefore have: 
In particular, $\map f a < c$ for each $a \in A$.
Hence we have found $f \in X^\ast$ such that: 
:$\map f a < c \le \map f b$ for each $a \in A$ and $b \in B$. 
Category:Hahn-Banach Separation Theorem
Hahn-Banach Separation Theorem/Normed Vector Space/Complex Case

== Theorem ==
Let $\struct {X, \norm \cdot}$ be a normed vector space over $\C$.
Let $\struct {X^\ast, \norm \cdot_{X^\ast} }$ be the normed dual space of $\struct {X, \norm \cdot}$.
=== Open Convex Set and Convex Set ===
=== Compact Convex Set and Closed Convex Set ===
Category:Hahn-Banach Separation Theorem
Hahn-Banach Separation Theorem/Normed Vector Space/Complex Case/Compact Convex Set and Closed Convex Set

== Theorem ==
Let $\struct {X, \norm \cdot}$ be a normed vector space over $\C$.
Let $\struct {X^\ast, \norm \cdot_{X^\ast} }$ be the normed dual space of $\struct {X, \norm \cdot}$.
Let $A$ be a compact convex set.
Let $B$ be a closed convex set disjoint from $A$.
Then there exists $f \in X^\ast$, $c \in \R$ and $\epsilon > 0$ such that: 
:$A \subseteq \set {x \in X : \map \Re {\map f x} \le c - \epsilon}$
and:
:$B \subseteq \set {x \in X : \map \Re {\map f x} \ge c + \epsilon}$
That is: 
:there exists $f \in X^\ast$, $c \in \R$ and $\epsilon > 0$ such that $\map \Re {\map f a} \le c - \epsilon < c + \epsilon \le \map \Re {\map f b}$ for $a \in A$ and $b \in B$.
== Proof ==
Let $\struct {X_\R, \norm {\, \cdot \,}_\R}$ be the realification of $X$ equipped with the restricted norm. 
Applying Hahn-Banach Separation Theorem: Real Case: Open Convex Set and Convex Set to $\struct {X_\R, \norm {\, \cdot \,}_\R}$, there exists a bounded linear functional $g : X_\R \to \R$, $c \in \R$ and $\epsilon > 0$ such that: 
:$A \subseteq \set {x \in X : \map g x < c - \epsilon}$
and:
:$B \subseteq \set {x \in X : \map g x \ge c + \epsilon}$
From Bounded Real-Valued Linear Functional is Real Part of Unique Bounded Complex-Valued Linear Functional, there exists $f \in X^\ast$ such that: 
:$\map g x = \map \Re {\map f x}$ 
for each $x \in X$. 
Then $f \in X^\ast$ satisfies: 
:$A \subseteq \set {x \in X : \map \Re {\map f x} < c - \epsilon}$
and:
:$B \subseteq \set {x \in X : \map \Re {\map f x} \ge c + \epsilon}$
as desired.
Category:Hahn-Banach Separation Theorem