nyuuzyou/wikis huggingface dataset:
Linearly Independent Set is Contained in some Basis/Infinite Dimensional Case

== Theorem ==
Let $K$ be a field. 
Let $E$ be a vector space over $K$. 
Let $H$ be a linearly independent subset of $E$.
There exists a basis $B$ for $E$ such that $H \subseteq B$.
 
== Proof ==
Let: 
:$\SS = \set {L \supseteq H : L \subseteq E \text { is linearly independent} }$
We have $H \in \SS$, so certainly $\SS \ne \O$.
With view to apply Zorn's Lemma, we show that every non-empty $\subseteq$-chain in $\SS$ has an upper bound.
Let $\CC$ be a non-empty $\subseteq$-chain in $\SS$.
Let: 
:$\ds C = \bigcup \CC$
Let $x_1, x_2, \ldots, x_n \in C$ and $\alpha_1, \alpha_2, \ldots, \alpha_n$ such that: 
:$\ds \sum_{k \mathop = 1}^n \alpha_k x_k = 0$ 
For each $1 \le i \le n$, let $C_i \in \CC$ be such that $x_i \in C_i$. 
Since $\CC$ is a $\subseteq$-chain, there exists $1 \le j \le n$ such that: 
:$C_i \subseteq C_j$ for all $1 \le i \le n$. 
Then $x_i \in C_j$ for each $1 \le i \le n$. 
Since $C_j$ is linearly independent, we have: 
:$\alpha_1 = \alpha_2 = \ldots = \alpha_n = 0$
So $C$ is linearly independent, with $H \subseteq \CC$. 
So $C \in \SS$, while $F \subseteq C$ for each $F \in \CC$.
So $C$ is an upper bound for $\CC$.
So every non-empty $\subseteq$-chain in $\SS$ has an upper bound.
So by Zorn's Lemma, $\SS$ has a $\subseteq$-maximal element $\BB$. 
We show that $\BB$ is a basis for $E$.
From definition 2 of a basis, it suffices to argue that $\BB$ is $\subseteq$-maximal among all linearly independent subsets of $G$. 
Suppose that $\BB'$ is a linearly independent subset of $G$ such that $\BB \subseteq \BB'$. 
Then since $H \subseteq \BB$, we have $H \subseteq \BB'$.
So $\BB' \in \SS$. 
So by the maximality of $\BB$ in $\SS$, we have $\BB = \BB'$.
So $\BB$ is $\subseteq$-maximal among all linearly independent subsets of $G$, and hence is a basis.
Category:Linearly Independent Set is Contained in some Basis
Linearly Independent Solutions of y'' - y = 0

== Theorem ==
The second order ODE:
:$(1): \quad y' ' - y = 0$
has solutions:
:$y_1 = e^x$
:$y_2 = e^{-x}$
which are linearly independent.
== Proof ==
We have that:
Hence it can be seen by inspection that:
are solutions to $(1)$.
Calculating the Wronskian of $y_1$ and $y_2$:
So the Wronskian of $y_1$ and $y_2$ is never zero.
Thus from Zero Wronskian of Solutions of Homogeneous Linear Second Order ODE iff Linearly Dependent:
:$y_1$ and $y_2$ are linearly independent.
== Sources ==
* : $\S 3.15$: Problem $1$
Category:Linear Second Order ODE/y - y = 0
Linearly Independent Solutions to 1st Order Systems

== Theorem ==
The 1st-order homogeneous linear system of differential equations $x' = A \left({t}\right) x$, expressed with the vectors $x', x: \R \to \R^n$ and the matrix function $A: \R \to M_{n \times n} \left({\R}\right)$, has $n$ linearly independent solutions, and if $ \phi_1 , \phi_2, \dots, \phi_n$ are $n$ linearly independent solutions, then $c_1 \phi_1 + c_2 \phi_2 + \cdots + c_n \phi_n$, where $c_i$ are constants, is a general solution.
== Proof ==
Let $v_1, v_2, \dots, v_n$ be linearly independent vectors in $\R^n$, and let $\phi_i$ be solutions to the IVPs $x' = A \left({t}\right) x, \, x \left({t_0}\right) = v_i$ for $i = 1, 2, \dots, n$.
Suppose the solutions are not independent, i.e. $c_1 \phi_1 + c_2 \phi_2 + \cdots + c_n \phi_n = 0$ for some constants $c_i$ not all zero.
Then:
: $c_1 \phi_1 \left({t_0}\right) + c_2 \phi_2 \left({t_0}\right) + \cdots c_n \phi_n \left({t_0}\right) = c_1 v_1 + c_2 v_2 + \cdots + c_n v_n = 0$
meaning the vectors $v_i$ are linearly dependent, a contradiction, so the solutions $\phi_i$ must be linearly independent.
By linearity of the system, every vector function of the form $ x = c_1 \phi_1 + \cdots + c_n \phi_n$ is a solution.
Let $z$ be an arbitrary solution of the system.
Since $\phi_i \left({t_0}\right)$ are linearly independent and count $n$ in number, they form a basis for $\R^n$, hence $z \left({t_0}\right)$ must be a linear combination of those solutions, and then by uniqueness of solutions $z$ is a linear combination of the vector functions $\phi_i$.
This proves this is a general solution.
Category:Ordinary Differential Equations
Linearly Independent Subset also Independent in Generated Subspace

== Theorem ==
Let $G$ be a finitely generated $K$-vector space.
Let $S$ be a linearly independent subset of $G$.
Let $M$ be the subspace of $G$ generated by $S$.
If $M \ne G$, then $\forall b \in G: b \notin M$, the set $S \cup \set b$ is linearly independent.
== Proof ==
Suppose that:
:$\ds \sum_{k \mathop = 1}^n \lambda_k x_k + \lambda b = 0$
where $\sequence {x_n}$ is a sequence of distinct vectors of $S$.
If $\lambda \ne 0$, then $\ds b = -\lambda^{-1} \paren {\sum_{k \mathop = 1}^n \lambda_k x_k} \in M$ which contradicts the definition of $b$.
Hence $\lambda = 0$, and so:
:$\ds \sum_{k \mathop = 1}^n \lambda_k x_k = 0$
Therefore $\lambda_1 = \lambda_2 = \cdots = \lambda_n = \lambda = 0$ as $S$ is linearly independent.
== Sources ==
* : Chapter $\text {V}$: Vector Spaces: $\S 27$. Subspaces and Bases: Theorem $27.11$
* 
Category:Linear Independence
Linearly Independent over the Rational Numbers iff Linearly Independent over the Integers

== Theorem ==
Let $z_1, z_2, \ldots, z_n$ be complex numbers.
Then:
:$z_1, z_2, \ldots, z_n$ are linearly independent over the rational numbers $\Q$
:
:$z_1, z_2, \ldots, z_n$ are linearly independent over the integers $\Z$.
== Proof ==
=== Forward implication ===
Let $z_1, z_2, \ldots, z_n$ be linearly independent over the rational numbers $\Q$.
That is, if $q_1, q_2, \ldots, q_n$ are rational numbers such that:
:$q_1 z_1 + q_2 z_2 + \cdots + q_n z_n = 0$
then:
:$q_1 = q_2 = \cdots = q_n = 0$
Let $a_1, a_2, \ldots, a_n$ be integers such that:
:$a_1 z_1 + a_2 z_2 + \cdots + a_n z_n = 0$
Then, by Integers form Subdomain of Rationals and the assumption, it follows that:
:$a_1 = a_2 = \cdots = a_n = 0$
Hence it follows by definition that $z_1, z_2, \ldots, z_n$ are linearly independent over the integers $\Z$.
=== Backward implication ===
Let $z_1, z_2, \ldots, z_n$ be linearly independent over the integers $\Z$.
That is, if $a_1, a_2, \ldots, a_n$ are integers such that:
:$a_1 z_1 + \cdots + a_n z_n = 0$
then:
:$a_1 = \cdots = a_n = 0$
Let $q_1, q_2, \ldots, q_n$ be rational numbers such that:
:$q_1 z_1 + q_2 z_2 + \cdots + q_n z_n = 0$
Let the lowest common multiple of the denominators of $q_1, q_2, \ldots, q_n$ be $m$.
Then, by the definition of lowest common multiple:
:$m q_1, m q_2, \ldots, m q_n \in \Z$
Thus:
:$m q_1 z_1 + m q_2 z_2 + \cdots + m q_n z_n = 0$
where the coefficients of $z_1, z_2, \ldots, z_n$ are all integers.
Hence, by the assumption:
:$m q_1 = m q_2 = \cdots = m q_n = 0$
Since $m$ is non-zero, it follows that:
:$q_1 = q_2 = \cdots = q_n = 0$
It follows by definition that $z_1, z_2, \ldots, z_n$ are linearly independent over the rational numbers $\Q$.
Category:Complex Analysis
Linearly Ordered Space is Compact iff Complete

== Theorem ==
Let $T = \struct {S, \preceq, \tau}$ be a linearly ordered space.
Then $T$ is a compact space it is complete.
== Proof ==
=== Necessary Condition ===
Let $T$ be a compact space.
Let $A \subseteq S$.
 $A$ has no supremum.
Consider the sets:
:$P_\alpha = \set {x \in S: x \prec \alpha}$ for $\alpha \in A$
:$B_\beta = \set {x \in S: \beta \prec x}$ for $\beta$ an upper bound of $A$.
We have that $P_\alpha$ and $B_\beta$ cover $S$ but contain no finite subcover.
Thus $T$ is not a compact space.
Similarly if $A$ has no infimum.
It follows that $A$ has both a supremum and infimum.
As $A$ is arbitrary, it follows that $S$ is a complete ordered set.
=== Sufficient Condition ===
Let $T$ be a complete ordered set.
Let $\UU$ be an open cover of $S$.
Let $a$ be the infimum of $S$.
Let $B$ be the set of the elements $y \in S$ for which $\hointr a y$ have a finite cover from elements of $\UU$.
Let $\alpha$ be the supremum of $B$.
If $\alpha \in U \in \UU$, then $U \subseteq B$.
So unless $\alpha$ is the supremum of $S$ itself, there exists an open interval $\openint x y \subseteq U$ such that $\alpha \in \openint x y$.
But we have that $\alpha$ be the supremum of $B$.
Hence $\openint x y = \O$.
But that means $y \in B$ which is impossible.
Thus $B = S$.
That is, the set of elements of $S$ which have a finite subcover of $\UU$ is $S$ itself.
Hence $T$ is a compact space by definition.
== Sources ==
* : Part $\text {II}$: Counterexamples: $39$. Order Topology: $7$
Category:Linearly Ordered Spaces
Category:Compact Spaces
Category:Complete Lattices
Linearly Ordered Space is Completely Normal

== Theorem ==
Let $T = \struct {S, \preceq, \tau}$ be a linearly ordered space.
Then $T$ is a completely normal space.
== Proof ==
By Linearly Ordered Space is $T_1$, $T$ is a $T_1$ (Fréchet) space.
By Linearly Ordered Space is $T_5$, $T$ is a $T_5$ space.
Hence the result, by definition of completely normal space.
== Sources ==
* : Part $\text {II}$: Counterexamples: $39$. Order Topology: $6$
Category:Linearly Ordered Spaces
Category:Completely Normal Spaces
Linearly Ordered Space is T1

== Theorem ==
Let $T = \struct {S, \preceq, \tau}$ be a linearly ordered space.
Then $T$ is a $T_1$ (Fréchet) space.
== Proof ==
Let $p \in S$.
By definition of linearly ordered space, the rays:
:$R_1 := \set {x \in S: x \prec p}$
:$R_2 := \set {x \in S: p \prec x}$
are open in $T$.
Thus their union:
:$R_1 \cup R_2 = \set {x \in S: x \prec p \lor p \prec x}$
is also open in $T$.
Consider the complement in $S$ of $R_1 \cup R_2$:
:$\relcomp S {R_1 \cup R_2} = \set {y \in S: y \notin R_1 \cup R_2}$
By definition of closed set, $\relcomp S {R_1 \cup R_2}$ is closed in $T$
But $\relcomp S {R_1 \cup R_2} = \set p$.
Thus $p$ is a closed point of $T$.
Hence the result, by definition of $T_1$ (Fréchet) space.
== Sources ==
* : Part $\text {II}$: Counterexamples: $39$. Order Topology: $6$
Category:Linearly Ordered Spaces
Category:T1 Spaces
Linearly Ordered Space is T5

== Theorem ==
Let $T = \struct {S, \preceq, \tau}$ be a linearly ordered space.
Then $T$ is a $T_5$ space.
== Proof ==
Let $A$ and $B$ be separated sets of $T$.
Let $A^*$ and $B^*$ be defined as:
:$A^* := \ds \bigcup \set {\closedint a b: a, b \in A, \closedint a b \cap B^- = \O}$
:$B^* := \ds \bigcup \set {\closedint a b: a, b \in B, \closedint a b \cap A^- = \O}$
where $A^-$ and $B^-$ denote the closure of $A$ and $B$ in $T$.
Let $A^*$, $B^*$ and $\complement_S \left({A^* \cup B^*}\right)$ be expressed as the union of convex components of $S$:
:$\ds A^* = \bigcup A_\alpha, \quad B^* = \bigcup B_\beta, \quad \relcomp S {A^* \cup B^*} = \bigcup C_\gamma$
where $\relcomp S X$ denotes the complement of $X$ with respect to $S$.
Let $M$ be the linearly ordered set:
:$M = \set {A_\alpha, B_\beta, C_\gamma}$
as defined in Partition of Linearly Ordered Space by Convex Components is Linearly Ordered Set.
Let:
:$S_\alpha$ be the set of strict upper bounds for $A_\alpha$.
:$S_\beta$ be the set of strict upper bounds for $B_\beta$.
From Successor Sets of Linearly Ordered Set Induced by Convex Component Partition:
:each of the sets $A_\alpha \in M$ has an immediate successor in $M$ if $A_\alpha$ intersects the closure of $S_\alpha$
:each of the sets $B_\beta \in M$ has an immediate successor in $M$ if $B_\beta$ intersects the closure of $S_\beta$
:that immediate successor ${C_\alpha}^+$ to $A_\alpha$ is an element in $\set {C_\gamma}$.
For each $\gamma$, let $k_\gamma \in C_\gamma$ be selected and fixed.
Let $A_\alpha \cap {S_\alpha}^- \ne \O$.
Then there exists a unique ${k_\alpha}^+ \in {C_\alpha}^+$.
In such cases, let $I_\alpha = \hointr p { {k_\alpha}^+}$ where $p \in A \cap {S_\alpha}^-$.
The other case is when $A_\alpha \cap {S_\alpha}^- = \O$.
In this case, let $I_\alpha = \O$.
Let $J_\alpha$ be defined similarly for the strict lower bounds of $A_\alpha$, using the same set of points $k_\gamma \in C_\gamma$.
Then:
:For each $\alpha$, let $U_\alpha = J_\alpha \cup A_\alpha \cup I_\alpha$.
:For each $\beta$, let $V_\beta = J_\beta \cup A_\beta \cup I_\beta$.
Each $U_\alpha$ and $V_\beta$ is a convex open set in $S$ containing $A_\alpha$ and $B_\beta$ respectively.
Thus $\ds U = \bigcup U_\alpha$ and $\ds V = \bigcup V_\beta$ are open sets in $T$ containing $A^*$ and $B^*$ respectively.
We have that:
:no $A_\alpha$ intersects no $B_\beta$
:the same $K_\gamma$ being used throughout implies that no $J_\beta$ or $I_\beta$ can intersect any $J_\alpha$ or $I_\alpha$.
Thus no $U_\alpha$ can intersect any $V_\beta$.
Thus $U \cap V = \O$.
So we have two disjoint open sets $U, V$ in $S$ containing the two separated sets $A$ and $B$ respectively.
As $A$ and $B$ are arbitrary, it follows that such $U$ and $V$ can be found whatever $A$ and $B$ are, provided they are separated.
Hence the result by definition of $T_5$ space.
== Sources ==
* : Part $\text {II}$: Counterexamples: $39$. Order Topology: $6$
Category:Linearly Ordered Spaces
Category:T5 Spaces
Linearly Ordered Space is T5/Mistake

== Source Work ==
:
:Part $\text {II}$: Counterexamples
::Section $39$: Order Topology
:::Item $6$
== Mistake ==
:For each $\gamma$, select and fix some point $k_\gamma \in C_\gamma$. Then whenever $A_\alpha \cap \overline S_\alpha \ne \O$, there exists a unique ${k_\alpha}^+ \in {C_\alpha}^+$, the immediate successor of $A_\alpha$ ... otherwise, if $A_\alpha \cap {\overline S_\alpha}^\alpha = \O$, let $I_\alpha = \O$. ...
== Correction ==
There is a minor typo here: there is a superfluous $\alpha$ superscript.
That last clause should say:
:otherwise, if $A_\alpha \cap \overline S_\alpha = \O$, let $I_\alpha = \O$.
== Also see ==
* Complement of Set of Rational Pairs in Real Euclidean Plane is Arc-Connected
== Sources ==
* : Part $\text {II}$: Counterexamples: $39$. Order Topology: $6$
Category:Mistakes/Counterexamples in Topology
Lines Joining Equal and Parallel Straight Lines are Parallel

== Theorem ==
The straight lines joining equal and parallel straight lines at their endpoints, in the same direction, are themselves equal and parallel.
== Proof ==
:
Let $AB, CD$ be equal and parallel.
Let $AC, BD$ join their endpoints in the same direction.
Draw $BC$.
From Parallelism implies Equal Alternate Angles:
:$\angle ABC = \angle BCD$
We have that $AB, BC$ are equal to $DC, CB$ and $\angle ABC = \angle BCD$.
It follows from Triangle Side-Angle-Side Congruence that $AC = BD$.
Also, $\triangle ABC = \triangle DCB$, and thus $\angle ACB = \angle CBD$.
We have that $BC$ falling on the two straight lines $AC, BD$ makes the alternate angles equal.
Therefore from Equal Alternate Angles implies Parallel Lines:
:$AC \parallel BD$
== Sources ==
* : Book $\text{I}$. Propositions
Category:Parallel Lines
Lines Parallel to Same Line not in Same Plane are Parallel to each other

== Theorem ==
== Proof ==
:
Let $AB$ and $CD$ be straight lines which are both parallel to another straight line $EF$ which is in a different plane.
It is to be demonstrated that $AB$ is parallel to $CD$.
Let $G$ be an arbitrary point on $EF$.
Let $GH$ be drawn at right angles to $EF$, in the plane holding both $AB$ and $EF$.
Let $GK$ be drawn at right angles to $EF$, in the plane holding both $CD$ and $EF$.
We have that $EF$ is at right angles to each of the straight lines $GH$ and $GK$.
Therefore from :
:$EF$ is at right angles to the plane through $GH$ and $GK$.
We have that $EF$ is parallel to $AB$.
Therefore from :
:$AB$ is at right angles to the plane through $GH$ and $GK$.
For the same reason:
:$CD$ is at right angles to the plane through $GH$ and $GK$.
Therefore each of the straight lines $AB$ and $CD$ is at right angles to the plane through $GH$ and $GK$.
From :
:$AB$ is parallel to $CD$.
== Sources ==
* : Book $\text{XI}$. Propositions
Category:Parallel Lines
Category:Planes
Lines Through Endpoints of One Side of Triangle to Point Inside Triangle is Less than Sum of Other Sides

== Theorem ==
Given a triangle and a point inside it, the sum of the lengths of the line segments from the endpoints of one side of the triangle to the point is less than the sum of the other two sides of the triangle.
=== Corollary ===
== Proof ==
:
Given a triangle $ABC$ and a point $D$ inside it.
We can construct lines connecting $A$ and $B$ to $D$, and then extend the line $AD$ to a point $E$ on $BC$.
In $\triangle ABE$, $AB + AE>BE$.
Then, $AB + AC = AB + AE + EC > BE + EC$ by Euclid's second common notion.
Similarly, $CE + ED > CD$, so $CE + EB = CE + ED + DB > CD + DB$.
Thus, $AB + AC > BE + EC > CD + DB$.
== Sources ==
* : Book $\text{I}$. Propositions
Category:Triangles
Lines Through Endpoints of One Side of Triangle to Point Inside Triangle is Less than Sum of Other Sides/Corollary

== Theorem ==
The angle between the two line segments from the endpoints of one side to a point inside the triangle is greater than the angle between the other two sides of the triangle.
== Proof ==
:
From External Angle of Triangle is Greater than Internal Opposite:
:$\angle BDC > \angle CED$
Similarly:
: $\angle CEB > \angle BEC$
Since $\angle CED$ is the same angle as $\angle CEB$:
: $\angle BDC > \angle CEB > \angle BEC$
== Sources ==
* : Book $\text{I}$. Propositions
Category:Triangles
Lines are Subspaces of Plane

== Theorem ==
The one-dimensional subspaces of $\R^2$ are precisely the homogeneous lines of plane analytic geometry.
== Proof ==
Follows directly from Equivalent Statements for Vector Subspace Dimension One Less.
== Sources ==
* : Chapter $\text {V}$: Vector Spaces: $\S 28$. Linear Transformations
Category:Linear Algebra
Category:Plane Analytic Geometry
Lines of Electric Force do not Cross

== Theorem ==
Let $\mathbf E$ be an electric field.
Let $L_1$ and $L_2$ be lines of force within $\mathbf E$.
Then $L_1$ and $L_2$ do not cross over each other.
== Proof ==
Follows directly from Vector Lines do not Intersect.
== Sources ==
* : Chapter $1$: Force and energy in electrostatics: $1.2$ The Electric Field
Category:Lines of Electric Force
Lines through Center Square of Order 3 Magic Square are in Arithmetic Sequence

== Theorem ==
Consider the order 3 magic square:
Each of the lines through the center cell contain $3$ integers in arithmetic sequence.
== Proof ==
By observation:
:$\tuple {1, 5, 9}$: common difference $4$
:$\tuple {2, 5, 8}$: common difference $3$
:$\tuple {3, 5, 7}$: common difference $2$
:$\tuple {4, 5, 6}$: common difference $1$
== Sources ==
* : $9$
* : $9$
Category:Magic Squares
Category:Arithmetic Sequences
Liouville's Constant is Transcendental

== Theorem ==
Liouville's constant:
is transcendental.
=== Corollary ===
== Proof ==
Let $q = 10^{n!}$ and write:
:$\ds L = \frac p q + \sum_{k \mathop = n + 1}^\infty \frac 1 {10^{k!} }$
for some suitable $p \in \Z$.
Then:
Thus, by definition, $L$ is a Liouville number.
Therefore, by Liouville's Theorem, $L$ is transcendental.
== Historical Note ==
== Sources ==
* : $0 \cdotp 11000 10000 00000 00000 00010 00000 00000 00000 0 \ldots$
* : Chapter $\text {A}.29$: Liouville ()
* : Chapter $\text {B}.18$: Algebraic and Transcendental Numbers. $e$ is Transcendental
* : $0 \cdotp 11000 \, 10000 \, 00000 \, 00000 \, 00010 \, 00000 \, 00000 \, 00000 \, 0 \ldots$
Category:Transcendental Number Theory
Category:Transcendental Numbers
Liouville's Constant is Transcendental/Corollary

== Corollary to Liouville's Constant is Transcendental ==
All real numbers of the form:
where
:$a_1, a_2, a_3, \ldots \in \set {1, 2, \ldots, 9}$
are transcendental.
== Proof ==
Let $n \in \N$.
For $n = 1$, let $p = a_1$ and $q = 10$. Then:
:$\ds \size {L - \dfrac p q} = \sum_{k \mathop = 2}^\infty \dfrac {a_k} {10^{k!} } < \dfrac 1 {10} = \dfrac 1 q$
For $n > 1$, let $q = 10^{n!}$ and write:
:$\ds L = \frac p q + \sum_{k \mathop = n + 1}^\infty \frac {a_k} {10^{k!} }$
for some suitable $p \in \Z$.
Then:
Thus, by definition, $L$ is a Liouville number.
Therefore, by Liouville's Theorem, $L$ is transcendental.
== Sources ==
* : Chapter $\text {B}.18$: Algebraic and Transcendental Numbers. $e$ is Transcendental
Category:Transcendental Number Theory
Category:Transcendental Numbers
Liouville's Constant is Transcendental/Historical Note

== Historical Note on Liouville's Constant is Transcendental ==
Liouville's constant was proved to be transcendental by in $1844$ as a demonstration that there exist real numbers which are provably transcendental.
This was the simplest of several such numbers that he constructed.
== Sources ==
* : $0 \cdotp 11000 10000 00000 00000 00010 00000 00000 00000 0 \ldots$
* : Chapter $\text {A}.29$: Liouville ()
* : $0 \cdotp 11000 \, 10000 \, 00000 \, 00000 \, 00010 \, 00000 \, 00000 \, 00000 \, 0 \ldots$
Category:Historical Notes
Liouville's Theorem

* Liouville's Lemma:
::concerning diophantine approximations.
* Liouville's Theorem (Hamiltonian Mechanics):
::the phase-space distribution function is constant along the trajectories of the system.
* Liouville's Theorem (Number Theory):
::concerning the transcendental nature of Liouville numbers.
* Liouville's Theorem (Differential Algebra):
::how to tell whether an elementary function has an elementary integral.
* Liouville's Theorem (Conformal Mappings):
::a conformal mapping is a Möbius transformation.
* Liouville's Theorem (Complex Analysis):
::a bounded entire complex function is constant.
Category:Liouville's Theorem
Liouville's Theorem (Complex Analysis)

== Theorem ==
Let $f: \C \to \C$ be a bounded entire function.
Then $f$ is constant.
=== Corollary 1 ===
=== Corollary 2 ===
=== Banach Space ===
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : Chapter $\text {A}.29$: Liouville ()
* 
* 
* 
* 
Category:Liouville's Theorem (Complex Analysis)
Category:Liouville's Theorem
Category:Complex Analysis
Liouville's Theorem (Complex Analysis)/Banach Space

== Theorem ==
Let $\struct {X, \norm {\, \cdot \,} }$ be a Banach space over $\C$.
Let $f : \C \to X$ be an analytic function that is bounded.
Then $f$ is constant. 
== Proof ==
Take $M \ge 0$ such that: 
:$\norm {\map f x} \le M$
for each $x \in X$. 
Let $\struct {X^\ast, \norm {\, \cdot \,}_{X^\ast} }$ be the normed dual space of $\struct {X, \norm {\, \cdot \,} }$.
From Banach Space Valued Function is Analytic iff Weakly Analytic, $f$ is weakly analytic. 
So for each $\phi \in X^\ast$, $\phi \circ f : \C \to \C$ is analytic.
Further, for each $z \in \C$ we have: 
:$\cmod {\map \phi {\map f x} } \le \norm \phi_{X^\ast} \norm {\map f x} \le M \norm \phi_{X^\ast}$
So $\phi \circ f$ is bounded.
From Liouville's Theorem (Complex Analysis), it follows that $\phi \circ f$ is constant for each $\phi \in X^\ast$.
That is, for each $z, w \in \C$ we have: 
:$\map \phi {\map f z} = \map \phi {\map f w}$ 
for each $\phi \in X^\ast$. 
That is, since $\phi$ is linear:
:$\map \phi {\map f z - \map f w} = 0$ 
From Normed Dual Space Separates Points, we have: 
:$\map f z = \map f w$ 
for each $z, w \in \C$.
So $f$ is constant.
== Sources ==
* : $3.2$: Integration of continuous vector-valued functions
Category:Liouville's Theorem (Complex Analysis)
Category:Banach Spaces
Liouville's Theorem (Complex Analysis)/Corollary 1

== Corollary to Liouville's Theorem (Complex Analysis) ==
Let $f: \C \to \C$ be an entire function such that: 
:$\size {\map f z} \ge M$
for all $z \in \C$ for some real constant $M > 0$. 
Then $f$ is constant.
== Proof ==
Note that since:
:$\size {\map f z} \ge M > 0$
we cannot have: 
:$\map f {z_0} = 0$
for any $z_0 \in \C$.
So, by the Quotient Rule for Complex Derivatives, we have: 
:$\dfrac 1 f$ is entire.
We also have that: 
:$\ds \size {\frac 1 {\map f z} } \le \frac 1 M$
for all $z \in \C$. 
So:
:$\dfrac 1 f$ is a bounded entire function.
So, by Liouville's Theorem (Complex Analysis), we have: 
:$\dfrac 1 f$ is constant.
That is, there exists some $C \in \C \setminus \set 0$ such that: 
:$\dfrac 1 {\map f z} = C$
for all $z \in \C$. 
Then, we have: 
:$\map f z = \dfrac 1 C$
for all $z \in \C$. 
So $f$ is constant.
Category:Complex Analysis
Liouville's Theorem (Complex Analysis)/Corollary 2

== Corollary to Liouville's Theorem (Complex Analysis) ==
Let $f: \C \to \C$ be a non-constant entire function.
The maximum modulus:
:$\ds \map M {r, f} := \max_{\cmod z \mathop = r} \cmod {\map f z}$
grows at least linearly in $r$.
== Proof ==
Category:Liouville's Theorem (Complex Analysis)