nyuuzyou/wikis huggingface dataset:
Finite Complement Topology is not Metrizable

== Theorem ==
Let $T = \struct {S, \tau}$ be a finite complement topology on an infinite set $S$.
Then $T$ is not a metrizable space.
== Proof ==
We have:
:Metrizable Space is Hausdorff
:Finite Complement Space is not $T_2$ (Hausdorff)
== Sources ==
* 
Category:Metrizable Topologies
Category:Finite Complement Topology
Finite Connected Simple Graph is Tree iff Size is One Less than Order

== Theorem ==
Let $T$ be a finite connected simple graph of order $n$.
Then $T$ is a (finite) tree the size of $T$ is $n - 1$.
== Proof ==
By definition:
:the order of a (finite) tree is how many nodes it has
and:
:the size of a (finite) tree is how many edges it has.
=== Necessary Condition ===
=== Sufficient Condition ===
== Beware ==
Category:Finite Connected Simple Graph is Tree iff Size is One Less than Order
Category:Tree Theory
Finite Connected Simple Graph is Tree iff Size is One Less than Order/Beware

== Beware ==
Just because a simple graph $G$ has an order one greater than its size does not necessarily make it a tree.
$G$ still has to be connected.
Take this simple graph, for instance:
:
It has order $5$ and size $4$ but is not a tree.
== Sources ==
* : $\S 4.1$: The Minimal Connector Problem: An Introduction to Trees: Problem $6$
Category:Finite Connected Simple Graph is Tree iff Size is One Less than Order
Finite Connected Simple Graph is Tree iff Size is One Less than Order/Lemma

== Theorem ==
Let $G = \struct {V, E}$ be a non-edgeless connected finite simple graph.
Let $G$ have no cycles.
Then:
:$G$ has at least one leaf node.
== Proof ==
Let $G = \struct {V, E}$ be a non-edgeless connected finite simple graph.
Let us select $v_1 \in V$ and some $v_2 \in V$ which is adjacent to $v_1$.
Such will always exist because $G$ is connected and not edgeless.
For $k \ge 2$, either $v_k$ is adjacent to $v_{k - 1}$ and no other, or it is adjacent to $v_{k - 1}$ and some $v_{k + 1}$ where $v_{k + 1} \ne v_{k - 1}$.
Since there are no cycles in $G$, each of $v_1, v_2, \ldots, v_{k + 1}$ are all distinct.
Hence as $G$ is finite, the process will terminate.
Thus $v_{k + 1}$ will be a leaf node.
== Sources ==
* : $\S 2.3.4.1$: Free Trees: Theorem $\mathrm A$
Category:Finite Connected Simple Graph is Tree iff Size is One Less than Order
Finite Connected Simple Graph is Tree iff Size is One Less than Order/Necessary Condition

== Theorem ==
Let $T$ be a tree of order $n$.
Then the size of $T$ is $n-1$.
== Proof ==
By definition:
:the order of a tree is how many nodes it has
:the size of a tree is how many edges it has.
Suppose $T$ is a tree with $n$ nodes. We need to show that $T$ has $n - 1$ edges.
The proof proceeds by strong induction.
Let $T_n$ be a tree with $n$ nodes.
For all $n \in \N_{>0}$, let $\map P n$ be the proposition that a tree with $n$ nodes has $n - 1$ edges.
=== Basis for the Induction ===
$\map P 1$ says that a tree with $1$ node has no edges.
It is clear that $T_1$ is $N_1$, the edgeless graph, which has $1$ node and no edges.
So $\map P 1$ is (trivially) true.
This is our basis for the induction.
=== Induction Hypothesis ===
Now we need to show that, if $\map P j$ is true for all $1 \le j \le k$, then it logically follows that $\map P {k + 1}$ is true.
So this is our induction hypothesis:
:Every tree with $j$ nodes such that $1 \le j \le k$ has $j - 1$ edges.
Then we need to show:
:Every tree with $k + 1$ nodes has $k$ edges.
=== Induction Step ===
==== Induction Step: Proof 1 ====
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
==== Induction Step: Proof 2 ====
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
== Sources ==
* : $\S 4.1$: The Minimal Connector Problem: An Introduction to Trees: Theorem $4.2$
Category:Finite Connected Simple Graph is Tree iff Size is One Less than Order
Finite Connected Simple Graph is Tree iff Size is One Less than Order/Necessary Condition/Induction Step

== Theorem ==
Let the following hold:
:For all $j \le k$, a tree of order $j$ is of size $j - 1$.
Then this holds:
:A tree of order $k + 1$ is of size $k$.
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : $\S 4.1$: The Minimal Connector Problem: An Introduction to Trees: Theorem $4.2$
Category:Finite Connected Simple Graph is Tree iff Size is One Less than Order
Finite Connected Simple Graph is Tree iff Size is One Less than Order/Necessary Condition/Induction Step/Proof 1

== Theorem ==
== Proof ==
Let $T_{k + 1}$ be an arbitrary tree with $k + 1$ nodes.
Take any node $v$ of $T_{k + 1}$ of degree $1$.
Such a node exists from Finite Tree has Leaf Nodes.
Consider $T_k$, the subgraph of $T_{k + 1}$ created by removing $v$ and the edge connecting it to the rest of the tree.
By Connected Subgraph of Tree is Tree, $T_k$ is itself a tree.
The order of $T_k$ is $k$, and it has one less edge than $T_{k + 1}$ by definition.
We have that $T_k$ has $k - 1$ edges.
So $T_{k + 1}$ must have $k$ edges.
== Sources ==
* : $\S 2.3.4.1$: Free Trees: Theorem $\mathrm A$
Category:Finite Connected Simple Graph is Tree iff Size is One Less than Order
Finite Connected Simple Graph is Tree iff Size is One Less than Order/Necessary Condition/Induction Step/Proof 2

== Theorem ==
== Proof ==
Let $T_{k + 1}$ be an arbitrary tree with $k + 1$ nodes.
Remove any edge $e$ of $T_{k + 1}$.
By definition of tree $T_{k + 1}$ has no circuits.
Therefore from Condition for Edge to be Bridge it follows that $e$ must be a bridge.
So removing $e$ disconnects $T_{k + 1}$ into two trees $T_1$ and $T_2$, with $k_1$ and $k_2$ nodes, where $k_1 + k_2 = k + 1$.
Then, , $T_1$ and $T_2$ have $k_1 - 1$ and $k_2 - 1$ edges.
Putting the edge $e$ back again, it can be seen that $T_{k + 1}$ has $\paren {k_1 - 1} + \paren {k_2 - 1} + 1 = k$ edges.
Therefore a tree of order $k + 1$ is of size $k$.
== Sources ==
* : $\S 4.1$: The Minimal Connector Problem: An Introduction to Trees: Theorem $4.2$
Category:Finite Connected Simple Graph is Tree iff Size is One Less than Order
Finite Connected Simple Graph is Tree iff Size is One Less than Order/Sufficient Condition

== Theorem ==
Let $T$ be a finite connected simple graph of order $n$.
Let the size of $T$ be $n - 1$.
Then $T$ is a (finite) tree.
== Proof ==
Let $T$ is a connected simple graph of order $n$ with size $n - 1$.
From Finite Connected Simple Graph with Size One Less than Order has no Circuits:
:$T$ has no circuits.
Hence $T$ is a tree by definition.
== Sources ==
* : $\S 4.1$: The Minimal Connector Problem: An Introduction to Trees: Problem $9$
* : $\S 2.3.4.1$: Free Trees: Theorem $\mathrm A$
Category:Finite Connected Simple Graph is Tree iff Size is One Less than Order
Finite Connected Simple Graph with Size One Less than Order has no Circuits

== Theorem ==
Let $G$ be a finite connected simple graph of order $n$.
Let the size of $T$ be $n - 1$.
Then $G$ has no circuits.
== Proof ==
Let $G$ be a connected simple graph of order $n$ with $n - 1$ edges.
By definition:
:the order of a finite tree is how many nodes it has
:the size of a finite tree is how many edges it has.
 $G$ contains at least one circuit.
From Condition for Edge to be Bridge:
:there exists at least one edge $e$ in $G$ which is not a bridge.
Let $G' := G \setminus \set e$ be $G$ after the edge deletion of $e$.
As $e$ is not a bridge, $G'$ is a graph which is:
:connected
:has $n$ vertices
:$n - 2$ edges.
Let us try and construct this connected simple graph $G'$ with $n$ vertices and $n - 2$ edges.
We start with the edgeless graph $N_n$, and add edges till $G'$ is connected.
We pick any two vertices of $N_n$, label them $u_1$ and $u_2$ for convenience, and use one edge to connect them, labelling that edge $e_1$.
We pick any other vertex, label it $u_3$, and use one edge to connect it to either $u_1$ or $u_2$, labelling that edge $e_2$.
We pick any other vertex, label it $u_4$, and use one edge to connect it to either $u_1, u_2$ or $u_3$, labelling that edge $e_3$.
We continue in this way, until we pick a vertex, label it $u_{n - 1}$, and use one edge to connect it to either $u_1, u_2, \ldots, u_{n - 2}$, labelling that edge $e_{n - 2}$.
That was the last of our edges, and the last vertex still has not been connected.
Therefore a simple graph with $n$ vertices and $n - 2$ edges cannot be connected.
That is, $G'$ cannot exist.
This contradicts our supposition that $G$ contains at least one circuit.
Hence, by Proof by Contradiction, $G$ has no circuits.
Category:Finite Graphs
Category:Connectedness (Graph Theory)
Category:Simple Graphs
Category:Circuits (Graph Theory)
Finite Cyclic Group has Euler Phi Generators

== Theorem ==
Let $C_n$ be a (finite) cyclic group of order $n$.
Then $C_n$ has $\map \phi n$ generators, where $\map \phi n$ denotes the Euler $\phi$ function.
== Proof ==
From List of Elements in Finite Cyclic Group, the elements of $G$ are:
:$\set {g^k: g \in G, 0 \le k < n}$
From Power of Generator of Cyclic Group is Generator iff Power is Coprime with Order, $g^k$ generates $G$ $k \perp n$.
The result follows by definition of the Euler $\phi$ function.
== Sources ==
* : $0$: Some Conventions and some Basic Facts: Exercise $5 \ \text{(ii)}$
* : $\S 3.4$: Cyclic groups: Exercise $1$
Category:Finite Cyclic Groups
Finite Cyclic Group is Isomorphic to Integers under Modulo Addition

== Theorem ==
Let $\struct {G, \circ}$ be a finite group whose identity element is $e$.
Then $\struct {G, \circ}$ is cyclic of order $n$ $\struct {G, \circ}$ is isomorphic with the additive group of integers modulo $n$ $\struct {\Z_n, +_n}$.
== Proof ==
=== Necessary Condition ===
Let $\struct {G, \circ}$ be a cyclic group of order $n$.
From List of Elements in Finite Cyclic Group:
:$G = \set {a^0, a^1, a^2, \ldots, a^n}$
where $a^0 = e, a^1 = a$.
From the definition of integers modulo $n$, $\Z_n$ can be expressed as:
:$\Z_n = \set {\eqclass 0 n, \eqclass 1 n, \ldots, \eqclass {n - 1} n}$
where $\eqclass x n$ is the residue class of $x$ modulo $n$.
Let $\phi: G \to \Z_n$ be the mapping defined as:
:$\forall k \in \set {0, 1, \ldots, n - 1}: \map \phi {a^k} = \eqclass k n$
By its definition it is clear that $\phi$ is a bijection.
Also:
Thus the morphism property of $\phi$ is demonstrated, and $\phi$ is thus a group homomorphism.
By definition, a group isomorphism is a group homomorphism which is also a bijection.
=== Sufficient Condition ===
Now suppose $G$ is a group such that $\phi: \Z_n \to G$ is a group isomorphism.
Let $a = \map \phi {\eqclass 1 n}$.
Let $g \in G$.
Then $g = \map \phi {\eqclass k n}$ for some $\eqclass k n \in \Z_n$.
Therefore:
So every element of $G$ is a power of $a$.
So by definition $G$ is cyclic.
== Sources ==
* : Chapter $\text{II}$: Groups: A Little Number Theory
* : Chapter $2$: Group Homomorphism and Isomorphism: $\S 63$
Category:Cyclic Groups
Category:Modulo Arithmetic
Finite Difference Table/Examples/Arbitrary Example 1

== Example of Finite Difference Table ==
The following is an example of the structure of a finite difference table:
:$\begin {array} {rr|rrr} x & y & \Delta & \Delta^2 & \Delta^3 \\ \hline
x_0 & y_0 & & & \\
 & & \Delta y_0 & & \\
x_1 & y_1 & & \Delta^2 y_0 & \\
 & & \Delta y_1 & & \Delta^3 y_0 \\
x_2 & y_2 & & \Delta^2 y_1 & \\
 & & \Delta y_2 & & \Delta^3 y_1 \\
x_3 & y_3 & & \Delta^2 y_2 & \\
 & & \Delta y_3 & & \\
x_4 & y_4 & & & \\
\end {array}$
Each entry in the rightmost $3$ columns is obtained from the entries in the column immediately to the left:
:subtracting the entry immediately above and to the left
:from the entry immediately below and to the left.
== Sources ==
* 
* 
Category:Examples of Finite Difference Tables
Finite Difference Table/Examples/Arbitrary Example 2

== Example of Finite Difference Table ==
Let $y = \map f x$ be the function defined as:
:$\forall x \in \R: \map f x = x^2 - 2$
Let $x_0 = -2$ and $h = 1$.
The finite difference table of $y$ can be expressed in the form:
:$\begin {array} {rr|rrr} x & y & \Delta & \Delta^2 & \Delta^3 \\ \hline
-2 & 2 & & & \\
 & & -3 & & \\
-1 & -1 & & 2 & \\
 & & -1 & & 0 \\
 0 & -2 & & 2 & \\
 & & 1 & & 0 \\
 1 & -1 & & 2 & \\
 & & 3 & & \\
 2 & 2 & & & \\
\end {array}$
== Sources ==
* 
* 
Category:Examples of Finite Difference Tables
Finite Dimensional Real Vector Space with Euclidean Norm form Normed Vector Space

== Theorem ==
Let $\R^n$ be an $n$-dimensional real vector space.
Let $\norm {\, \cdot \,}_2$ be the Euclidean norm.
Then $\struct {\R^n, \norm {\, \cdot \,}_2}$ is a normed vector space.
== Proof ==
We have that:
:Real Vector Space is Vector Space
:By Euclidean Space is Normed Vector Space, $\norm {\, \cdot \,}_2$ is a norm on $\R^n$
By definition, $\struct {\R^n, \norm {\, \cdot \,}_2}$ is a normed vector space.
== Sources ==
* : $\S 1.2$: Normed and Banach spaces. Normed Spaces
Category:Examples of Normed Vector Spaces
Finite Dimensional Subspace of Normed Vector Space is Closed

== Theorem ==
Let $V$ be a normed vector space.
Let $W$ be a finite dimensional subspace of $V$. 
And the ground field $K$ is complete.
Then $W$ is closed.
== Proof 1 ==
Suppose that $\dim W = n$.
Let: 
:$\set {e_1, e_2, \ldots, e_n}$
be a basis for $W$.
 that $W$ is not a closed set.
Then there exists a convergent sequence $\sequence {w_k}$ in $W$ such that: 
:$w_k \to w$ in $V$
where $w \in V \setminus W$.
Note that:
:$\set {e_1, e_2, \ldots, e_n}$
is linearly independent in $W$, and hence $V$.
Note that since $w \not \in W$, $w$ cannot be written as a linear combination of elements of $\set {e_1, e_2, \ldots, e_n}$.
So:
:$\set {e_1, e_2, \ldots, e_n, w}$
is linearly independent in $V$.
So consider the subspace:
:$W^* = \span \set {e_1, e_2, \ldots, e_n, w}$.
Using the sequence $\sequence {w_n}$ from before, write: 
:$w_k = \tuple {w_k^{\paren 1}, w_k^{\paren 2}, \ldots, w_k^{\paren n}, 0} \in W^*$
and:
:$w = \tuple {0, 0, \ldots, 0, 1} \in W^*$
Since Norms on Finite-Dimensional Real Vector Space are Equivalent, $w_k\to w$ in direct product norm.
Since sequence of vectors converges iff each component converges, we necessarily have: 
:$w_k^{\paren j} \to 0$
for each $1 \le j \le n$.
However, since the $(n+1)$-th component of $w_k$ is 0 and that of $w$ is $1$, we would also have: 
:$0 \to 1$
Clearly this is impossible, so we have derived a contradiction.
So $W$ is necessarily closed.
== Proof 2 ==
Because $K$ is complete, the finite dimensional normed space $W$ is complete.
By Subspace of Complete Metric Space is Closed iff Complete, $W$ is closed.
== Necessity of completeness of the ground field ==
$K = \Q$ is not complete.
$W = \Q$
$V = \Q \sqbrk {\sqrt 2}$
Since
:$\dim_K W = 1$
$W$ is a finite dimensional subspace of $V$. 
But $W$ is not closed in $V$.
== Also see ==
* Finite-Dimensional Subspace of Hausdorff Topological Vector Space is Closed
== Source ==
* 
Category:Functional Analysis
Category:Finite Dimensional Vector Spaces
Finite Direct Product of Modules is Module

== Theorem ==
Let $\struct {R, +_R, \times_R}$ be a ring.
Let $\struct {G_1, +_1, \circ_1}_R, \struct {G_2, +_2, \circ_2}_R, \ldots, \struct {G_n, +_n, \circ_n}_R$ be $R$-modules.
Let:
:$\ds G = \prod_{k \mathop = 1}^n G_k$
be their direct product.
Then $G$ is a module.
== Proof 1 ==
== Proof 2 ==
== Also see ==
* Finite Direct Product of Unitary Modules is Unitary Module
== Sources ==
* : Chapter $\text {V}$: Vector Spaces: $\S 26$. Vector Spaces and Modules: Example $26.5$
Category:Module Theory
Category:Direct Products
Category:Finite Direct Product of Modules is Module
Finite Direct Product of Modules is Module/Proof 2

== Theorem ==
== Proof ==
=== ===
Let $x = \tuple {x_1, x_2, \ldots, x_n}, y = \tuple {y_1, y_2, \ldots, y_n} \in G$.
So $(1)$ holds.
=== ===
Let $x = \tuple {x_1, x_2, \ldots, x_n} \in G$.
So $(2)$ holds.
=== ===
Let $x = \tuple {x_1, x_2, \ldots, x_n} \in G$.
So $(3)$ holds.
Hence the result.
Category:Finite Direct Product of Modules is Module
Finite Direct Product of Unitary Modules is Unitary Module

== Theorem ==
Let $\struct {R, +_R, \times_R}$ be a ring.
Let $\struct {G_1, +_1, \circ_1}_R, \struct {G_2, +_2, \circ_2}_R, \ldots, \struct {G_n, +_n, \circ_n}_R$ be unitary $R$-modules.
Let:
:$\ds G = \prod_{k \mathop = 1}^n G_k$
be their direct product.
Then $G$ is a unitary module.
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : Chapter $\text {V}$: Vector Spaces: $\S 26$. Vector Spaces and Modules: Example $26.5$
Category:Unitary Modules
Category:Direct Products
Category:Finite Direct Product of Unitary Modules is Unitary Module
Finite Direct Product of Unitary Modules is Unitary Module/Proof 2

== Theorem ==
== Proof ==
From Finite Direct Product of Modules is Module we have that $G$ is a module.
It remains to be shown that:
:$\forall x \in G: 1_R \circ x = x$
Let $x = \tuple {x_1, x_2, \ldots, x_n} \in G$.
Then:
Hence the result.
Category:Finite Direct Product of Unitary Modules is Unitary Module
Finite Direct Sum of Noetherian Module is Noetherian

== Theorem ==
Let $A$ be a commutative ring with unity.
Let $n \in \N_{>0}$.
Let $M_1, \ldots, M_n$ be $A$-modules.
Then the direct sum:
:$\ds \bigoplus_{i \mathop = 1}^n M_i$
is an $A$-Noetherian module.
== Proof ==
By Direct Sum of Modules is Module, it is an $A$-module.
Thus we only need to show that it is Noetherian.
We prove it by induction.
For $n = 1$, there is nothing to prove.
Let $n \ge 2$.
Suppose that the claim is true for $n-1$, i.e.:
:$\ds \bigoplus_{i \mathop = 1}^{n-1} M_i$
is Noetherian.
Then, consider the short exact sequence:
:$\ds 0 \longrightarrow \bigoplus_{i \mathop = 1}^{n-1} M_i \stackrel {\alpha} {\longrightarrow} \bigoplus_{i \mathop = 1}^n M_i \stackrel {\beta} {\longrightarrow} M_n \longrightarrow 0$
where:
:$\map \alpha {\tuple {x_1, \ldots, x_{n-1} } } := \tuple {x_1, \ldots, x_{n-1}, 0}$
:$\map \beta { \tuple {x_1, \ldots, x_n} } := x_n$
Recall that $M_n$ is also Noetherian .
Thus by Short Exact Sequence Condition of Noetherian Modules:
:$\ds \bigoplus_{i \mathop = 1}^n M_i$
is Noetherian.
Category:Noetherian Modules
Finite Discontinuity/Examples/Example 1

== Example of Finite Discontinuity ==
Let $f: \R \to \R$ be the real function defined as:
:$\forall x \in \R: \map f x = \map \cos {\dfrac 1 x}$
Then $f$ has a finite discontinuity at $x = 0$ which is non-removable.
:
== Sources ==
* 
* 
Category:Examples of Finite Discontinuities
Finite Discrete Space satisfies all Compactness Properties

== Theorem ==
Let $T = \struct {S, \tau}$ be a finite discrete topological space.
Then $T$ satisfies the following compactness properties:
:$T$ is compact.
:$T$ is Sequentially Compact.
:$T$ is Countably Compact.
:$T$ is Weakly Countably Compact.
:$T$ is a Lindelöf Space
:$T$ is Pseudocompact.
:$T$ is $\sigma$-Compact.
:$T$ is Locally Compact.
:$T$ is Weakly Locally Compact.
:$T$ is Strongly Locally Compact.
:$T$ is $\sigma$-Locally Compact.
:$T$ is Weakly $\sigma$-Locally Compact.
:$T$ is Fully Normal.
:$T$ is Fully $T_4$.
:$T$ is Paracompact.
:$T$ is Countably Paracompact.
:$T$ is Metacompact.
:$T$ is Countably Metacompact.
== Proof ==
A finite discrete space is by definition a topology on a finite set.
A Discrete Space is Fully Normal.
A fully normal space is fully $T_4$ by definition.
The rest of the results follow directly from Finite Space Satisfies All Compactness Properties.
== Sources ==
* : Part $\text {II}$: Counterexamples: $1$. Finite Discrete Topology: $8$
Category:Finite Topological Spaces
Category:Compact Spaces
Category:Discrete Topology
Finite Dissection of Polyhedra

== Theorem ==
Let $A$ and $B$ be polyhedra with the same volume.
Then it is not necessarily the case that there exists a dissection of $A$ into finitely many components that may be reassembled to form $B$.
== Proof ==
== Historical Note ==
== Sources ==
* 
* 
* 
Category:Dissections
Category:Polyhedra
Category:Hilbert 23
Finite Dissection of Polyhedra/Historical Note

== Historical Note on Finite Dissection of Polyhedra ==
The question of the Finite Dissection of Polyhedra was the $3$rd of the Hilbert 23.
It was settled in $1900$ by , who provided a counterexample: two polyhedra of the same volume for which there exists no dissection of one into the other.
This was the first of the Hilbert 23 to be solved.
== Sources ==
* 
* 
* 
Category:Historical Notes
Finite Dissection of Polyhedra/Historical Note/Mistake

== Source Work ==
:
:dissection proof
== Mistake ==
:... the German mathematician showed in $1901$ that there are two polyhedra of the same volume for which it is impossible to reconstruct one from the other by dissection.
== Correction ==
In fact it was $1900$ in which he performed this feat.
This mistake is repeated in entry equidecomposable.
The date of this result is not reported in the second edition.
== Sources ==
* 
* 
Category:Mistakes/The Penguin Dictionary of Mathematics
Finite Field Extension has Finite Galois Group

== Theorem ==
Let $E / F$ be a finite field extension.
Then its Galois group is finite.
== Proof ==
Because $E / F$ is finite, it is finitely generated.
Let $\alpha_1, \ldots, \alpha_n \in E$ with $E = \map F {\alpha_1, \ldots, \alpha_n}$.
By Finite Field Extension is Algebraic, $\alpha_1, \ldots, \alpha_n$ are algebraic over $F$.
Let $f_1, \ldots, f_n$ be their minimal polynomials.
Let $f = f_1\dots f_n$.
By Galois Group Acts Faithfully on Generating Set, $\Gal {E / F}$ acts faithfully on the roots of $f$.
Thus $\Gal {E / F}$ is finite.
Category:Field Extensions
Category:Galois Groups of Field Extensions
Finite Field Extension is Algebraic

== Theorem ==
Let $L / K$ be a finite field extension.
Then $L / K$ is algebraic.
== Proof ==
Let $x \in L$ be arbitrary.
Let $n = \index L K$ be the degree of $L$ over $K$.
From Size of Linearly Independent Subset is at Most Size of Finite Generator, there is a $K$-linear combination of $\set {1, \ldots, x^n}$ equal to $0$.
Say $a_n x^n + \cdots + a_1 x + a_0 = 0$, $a_i \in K$, $i = 0, \ldots, n$.
Therefore $x$ satisfies a polynomial with coefficients in $K$.
That is, $x$ is algebraic.
Since $x \in L$ was chosen arbitrarily, $L / K$ is algebraic.
Category:Field Extensions
Finite Fourier Series

== Theorem ==
Let $\map a n$ be any finite periodic real function on $\Z$ with period $b$.
Let $\xi = e^{2 \pi i/ b}$ be the first $b$th root of unity.
Then:
:$\ds \map a n = \sum_{k \mathop = 0}^{b - 1} \map {a_*} k \xi^{n k}$
where:
:$\ds \map {a_*} n = \frac 1 b \sum_{k \mathop = 0}^{b - 1} \map a k \xi^{-n k}$
== Proof ==
Since $a$ has period $b$, we have:
:$\map a {n + b} = \map a n$
So if we define:
:$\ds \map F z = \sum_{n \mathop \ge 0} \map a n z^n$
we have:
where the last step defines the polynomial $P$.
If we expand $F$ now using partial fractions, we get
Category:Fourier Analysis
Finite Generalized Sum Converges to Summation

== Theorem ==
Let $G$ be a commutative topological semigroup with identity $0_G$.
Let $\set{i_0, i_1, \ldots, i_n}$ be a finite enumeration of a finite set $I$. 
Let $\family{g_i}_{i \in I}$ be an indexed family of elements of $G$.
Then:
:the generalized sum $\ds \sum_{i \mathop \in I} g_i$ 
converges to:
:the summation over finite index $\ds \sum_{i \mathop \in I} g_i$
== Proof ==
Let $\FF$ be the set of finite subsets of $I$.
Let $h = \ds \sum_{i \mathop \in I} g_i$ be the summation over finite index $I$.
Let $U$ be an open subset of $G$ such that $h \in U$.
From Set is Subset of Itself:
:$I \in \FF$
Let: 
:$J \in \FF : I \subseteq J$
Let $h'= \ds \sum_{j \mathop \in J} g_j$ be the summation over finite index $J$.
By definition of $\FF$:
:$J \subseteq I$
By definition of set equality:
:$I = J$
Hence:
:$h'= h \in U$
Since $U$ was arbitrary, by definition of convergance:
:the generalized sum $\ds \sum_{i \mathop \in I} g_i$ 
converges to:
:the summation over finite index $\ds \sum_{i \mathop \in I} g_i$
Category:Generalized Sums