nyuuzyou/wikis huggingface dataset:
Complex Roots of Polynomial with Real Coefficients occur in Conjugate Pairs/Proof 2

== Theorem ==
== Proof ==
Let $\alpha = p + q i$.
Let $p + q i$ be expressed in exponential form as $\alpha = r e^{i \theta}$.
As $\alpha = r e^{i \theta}$ satisfies $\map f \alpha = 0$, it follows that:
:$a_n r^n e^{n i \theta} + a_{n - 1} r^{n - 1} e^{\paren {n - 1} i \theta} + \dotsb + a_1 r e^{i \theta} + a_0 = 0$
Taking the conjugate of both sides:
:$a_n r^n e^{-n i \theta} + a_{n - 1} r^{n - 1} e^{-\paren {n - 1} i \theta} + \dotsb + a_1 r e^{-i \theta} + a_0 = 0$
it follows that $\overline \alpha = p - q i$ is also a root of $f$.
If any of the $a_k$ are complex, then the conjugate of $f$ is not the same polynomial as $f$.
Therefore the result holds only for real coefficients.
== Sources ==
* : $1$: Complex Numbers: Solved Problems: Polynomial Equations: $36$
Category:Complex Roots of Polynomial with Real Coefficients occur in Conjugate Pairs
Complex Roots of Unity/Examples

== Examples of Complex Roots of Unity==
=== Complex Square Roots of Unity ===
=== Complex Cube Roots of Unity ===
=== Complex $4$th Roots of Unity ===
=== Complex $5$th Roots of Unity ===
=== Complex $6$th Roots of Unity ===
=== Complex $7$th Roots of Unity ===
Category:Examples of Complex Roots of Unity
Complex Roots of Unity/Examples/4th Roots

== Example of Complex Roots of Unity ==
The complex $4$th roots of unity are the elements of the set:
:$U_n = \set {z \in \C: z^4 = 1}$
They are:
== Proof ==
By definition, the first complex $4$th root of unity $\alpha$ is given by:
We have that:
:$e^{0 i \pi / 4} = e^0 = 1$
which gives us, as always, the zeroth complex $n$th root of unity for all $n$.
The remaining complex $4$th roots of unity can be expressed as $e^{4 i \pi / 4} = e^{i \pi}$ and $e^{6 i \pi / 4} = e^{3 i \pi / 2}$, but it is simpler to calculate them as follows:
:
== Sources ==
* : $\S 3$. Roots of Unity: Example $1$.
* : $1$: Complex Numbers: Supplementary Problems: The $n$th Roots of Unity: $105 \ \text {(a)}$
* 
Category:Examples of Complex Roots of Unity
Complex Roots of Unity/Examples/5th Roots

== Example of Complex Roots of Unity ==
The complex $5$th roots of unity are the elements of the set:
:$U_n = \set {z \in \C: z^5 = 1}$
They are:
== Proof ==
By definition, the first complex $5$th root of unity $\alpha$ is given by:
We have that:
:$e^{0 i \pi / 5} = e^0 = 1$
which gives us, as always, the zeroth complex $n$th root of unity for all $n$.
The remaining complex $5$th roots of unity follow:
== Sources ==
* : $\S 3$. Roots of Unity: Example $2$.
* : $1$: Complex Numbers: Solved Problems: The $n$th Roots of Unity: $37$
Category:Examples of Complex Roots of Unity
Category:Complex 5th Roots of Unity
Complex Roots of Unity/Examples/6th Roots

== Example of Complex Roots of Unity ==
The complex $6$th roots of unity are the elements of the set:
:$U_n = \set {z \in \C: z^6 = 1}$
They are:
== Proof ==
By definition, the first complex $6$th root of unity $\alpha$ is given by:
We have that:
:$e^{0 i \pi / 6} = e^0 = 1$
which gives us, as always, the zeroth complex $n$th root of unity for all $n$.
The remaining complex $6$th roots of unity follow:
== Illustration ==
== Sources ==
* : $\S 3$. Roots of Unity: Exercise $1$.
Category:Examples of Complex Roots of Unity
Complex Roots of Unity/Examples/6th Roots/Illustration

== Illustration of Complex $6$th Roots of Unity ==
From Complex Roots of Unity are Vertices of Regular Polygon Inscribed in Circle the complete set of complex $6$th roots of unity can be depicted as the vertices of the following regular hexagon:
:
where $\alpha$ is used to denote the first complex $6$th root of unity.
== Sources ==
* : $\S 3$. Roots of Unity: Figure $12$
Category:Examples of Complex Roots of Unity
Complex Roots of Unity/Examples/7th Roots

== Example of Complex Roots of Unity ==
The complex $7$th roots of unity are the elements of the set:
:$U_n = \set {z \in \C: z^7 = 1}$
They are:
== Proof ==
By definition, the first complex $7$th root of unity $\alpha$ is given by:
Hence the result:
:
== Sources ==
* : $1$: Complex Numbers: Supplementary Problems: The $n$th Roots of Unity: $105 \ \text {(b)}$
Category:Examples of Complex Roots of Unity
Complex Roots of Unity/Examples/Cube Roots

== Example of Complex Roots of Unity ==
The complex cube roots of unity are the elements of the set:
:$U_3 = \set {z \in \C: z^3 = 1}$
They are:
The notation $\omega$ for, specifically, the complex cube roots of unity is conventional.
=== Conjugate Form ===
== Proof ==
== Sources ==
* : $\S 3$. Roots of Unity
* : Introduction
* 
* 
* 
* 
* 
Category:Examples of Complex Roots of Unity
Category:Cube Roots of Unity
Complex Roots of Unity/Examples/Cube Roots/Conjugate Form

== Example of Complex Roots of Unity ==
The Cube Roots of Unity can be expressed in the form:
:$U_3 = \set {1, \omega, \overline \omega}$
where:
:$\omega = -\dfrac 1 2 + \dfrac {i \sqrt 3} 2$
:$\overline \omega$ denotes the complex conjugate of $\omega$.
== Proof ==
We have that the Cube Roots of Unity can be expressed as:
:$U_3 = \set {1, \omega, \omega^2}$
Then:
== Sources ==
* : $\S 3$. Roots of Unity
Category:Cube Roots of Unity
Complex Roots of Unity/Examples/Square Roots

== Example of Complex Roots of Unity ==
The complex square roots of unity are the elements of the set:
:$U_2 = \set {z \in \C: z^2 = 1}$
They are:
== Sources ==
* : Chapter $2$: The Symmetric Groups: $\S 81$
Category:Examples of Complex Roots of Unity
Category:Square Roots
Complex Roots of Unity are Vertices of Regular Polygon Inscribed in Circle

== Theorem ==
Let $n \in \Z$ be an integer such that $n \ge 3$.
Let $z \in \C$ be a complex number such that $z^n = 1$.
Let $U_n = \set {e^{2 i k \pi / n}: k \in \N_n}$ be the set of $n$th roots of unity.
Let $U_n$ be plotted on the complex plane.
Then the elements of $U_n$ are located at the vertices of a regular $n$-sided polygon $P$, such that:
:$(1):\quad$ $P$ is circumscribed by a unit circle whose center is at $\tuple {0, 0}$
:$(2):\quad$ one of those vertices is at $\tuple {1, 0}$.
== Proof ==
The above diagram illustrates the $7$th roots of unity.
== Sources ==
* : $\S 3$. Roots of Unity
* : $1$: Complex Numbers: The $n$th Roots of Unity
Category:Geometry of Complex Plane
Category:Complex Roots of Unity
Complex Roots of Unity in Exponential Form

== Theorem ==
Let $n \in \Z$ be an integer such that $n > 0$.
Let $z \in \C$ be a complex number such that $z^n = 1$.
Then:
:$U_n = \set {e^{2 i k \pi / n}: k \in \N_n}$
where $U_n$ is the set of $n$th roots of unity.
That is:
:$z \in \set {1, e^{2 i \pi / n}, e^{4 i \pi / n}, \ldots, e^{2 \paren {n - 1} i \pi / n} }$
Thus for every integer $n$, the number of $n$th roots of unity is $n$.
Setting $\omega := e^{2 i \pi / n}$, $U_n$ can then be written as:
:$U_n = \set {1, \omega, \omega^2, \ldots, \omega^{n - 1} }$
== Proof ==
Let $z \in \set {e^{2 i k \pi / n}: k \in \N_n}$.
Then:
: $z^n \in \set {e^{2 i k \pi}: k \in \N_n}$
Hence $z^n = 1$.
Now suppose $z^n = 1$. We determine the possible values of $z$ using Roots of Complex Number.
Let $z = r e^{i \theta}$.
Then $\cmod {z^n} = 1 \implies \cmod z = 1$.
Similarly, we have $n \theta = 0 \bmod 2 \pi$.
So $\theta = \dfrac {2 k \pi} n$ for $k \in \Z$.
Hence the result.
 
== Also see ==
* Condition for Complex Root of Unity to be Primitive
== Sources ==
* : $\S 3$. Roots of Unity
* : $1$: Complex Numbers: The $n$th Roots of Unity
Category:Complex Roots of Unity
Complex Roots of Unity include 1

== Theorem ==
Let $n \in \Z_{>0}$ be a (strictly) positive integer.
Let $U_n = \set {z \in \C: z^n = 1}$ be the set of complex $n$th roots of unity.
Then $1 \in U_n$.
That is, $1$ is always one of the complex $n$th roots of unity of any $n$.
== Proof ==
By definition of integer power:
:$1^n = 1$
for all $n$.
Hence the result.
== Sources ==
* : $\S 3$. Roots of Unity
Category:Complex Roots of Unity
Complex Roots of Unity occur in Conjugate Pairs

== Theorem ==
Let $n \in \Z_{>0}$ be a (strictly) positive integer.
Let $U_n$ denote the complex $n$th roots of unity:
:$U_n = \set {z \in \C: z^n = 1}$
Let $\alpha \in U_n$ be the first complex $n$th root of unity.
Then:
:$\forall k \in \Z_{>0}, k < \dfrac n 2: \overline {\alpha^k} = \alpha^{n - k}$
That is, each of the complex $n$th roots of unity occur in conjugate pairs:
:$\tuple {\alpha, \alpha^{n - 1} }; \tuple {\alpha^2, \alpha^{n - 2} }; \ldots; \tuple {\alpha^s, \alpha^{n - s} }$
where:
:$s = \dfrac {n - 1} 2$ for odd $n$
:$s = \dfrac {n - 2} 2$ for even $n$.
== Proof ==
Consider the polynomial equation:
:$(1): \quad z^n - 1 = 0$
The complex $n$th roots of unity are:
:$1, \alpha, \alpha^2, \ldots, \alpha^{n - 1}$
From Complex Roots of Polynomial with Real Coefficients occur in Conjugate Pairs, the roots of $(1)$ occur in conjugate pairs.
Let $k \in \Z$ such that $1 \le k \le n$.
Then:
That is, the complex $n$th root of unity which is the other half of the conjugate pair with $\alpha^k$ is $\alpha^{n - k}$.
When $n$ is odd, these pair up as:
:$\tuple {\alpha, \alpha^{n - 1} }; \tuple {\alpha^2, \alpha^{n - 2} }; \ldots; \tuple {\alpha^s, \alpha^{n - s} }$
where $s$ is the largest integer less than $\dfrac n 2$; that last pair can be expressed:
:$\tuple {\alpha^s, \alpha^{s + 1} }$
When $n$ is even:
:$\alpha^s = \alpha^{n - s}$
when $s = \dfrac n 2$.
Indeed:
and so is wholly real.
Thus from Complex Number equals Conjugate iff Wholly Real:
:$\alpha^{n / 2} = \alpha^{n - n / 2}$
Hence the complex $n$th roots of unity pair up as:
:$\tuple {\alpha, \alpha^{n - 1} }; \tuple {\alpha^2, \alpha^{n - 2} }; \ldots; \tuple {\alpha^s, \alpha^{n - s} }$
where $s$ is the largest integer less than $\dfrac n 2$; that last pair can be expressed:
:$\tuple {\alpha^s, \alpha^{s + 2} }$
The result follows.
== Sources ==
* : $\S 3$. Roots of Unity: $(3.12)$.
Category:Complex Roots of Unity
Complex Sequence is Null iff Modulus of Sequence is Null

== Theorem ==
Let $\sequence {z_n}_{n \mathop \in \N}$ be a complex sequence.
Then: 
:$z_n \to 0$
:
:$\cmod {z_n} \to 0$
That is: 
:$\sequence {z_n}_{n \mathop \in \N}$ is a null sequence $\sequence {\cmod {z_n} }_{n \mathop \in \N}$ is a null sequence.
== Proof 1 ==
== Proof 2 ==
Category:Complex Sequence is Null iff Modulus of Sequence is Null
Category:Convergent Complex Sequences
Category:Complex Modulus
Complex Sequence is Null iff Modulus of Sequence is Null/Proof 1

== Theorem ==
== Proof ==
By the definition of convergent sequence, we have: 
:$z_n \to 0$ for each $\epsilon > 0$ there exists $N \in \N$ such that $\cmod {z_n} < \epsilon$ for each $n \ge N$.
Similarly: 
:$\cmod {z_n} \to 0$ for each $\epsilon > 0$ there exists $N \in \N$ such that $\cmod {\paren {\cmod {z_n} } } < \epsilon$ for each $n \ge N$.
From Complex Modulus of Real Number equals Absolute Value, we have: 
:$\cmod {\paren {\cmod {z_n} } } = \cmod {z_n}$
So:
:$\cmod {z_n} \to 0$ for each $\epsilon > 0$ there exists $N \in \N$ such that $\cmod {z_n} < \epsilon$ for each $n \ge N$.
We can therefore see that: 
:$z_n \to 0$
:
:$\cmod {z_n} \to 0$
Category:Complex Sequence is Null iff Modulus of Sequence is Null
Complex Sequence is Null iff Positive Integer Powers of Sequence are Null

== Theorem ==
Let $\sequence {z_n}_{n \mathop \in \N}$ be a complex sequence.
Let $k \in \N$. 
Then: 
:$z_n \to 0$
:
:$z_n^k \to 0$
== Proof ==
=== Necessary Condition ===
Suppose that: 
:$z_n \to 0$
Let $\epsilon > 0$. 
Then from the definition of convergent sequence we can find $N \in \N$ such that for $n \ge N$ we have: 
:$\cmod {z_n} < \epsilon^{1/k}$
then: 
:$\cmod {z_n}^k < \epsilon$ 
From Power of Complex Modulus equals Complex Modulus of Power, this gives: 
:$\cmod {z_n^k} < \epsilon$
for all $n \ge N$. 
So:
:$z_n^k \to 0$ 
=== Sufficient Condition ===
Suppose that: 
:$z_n^k \to 0$ for some $k$.
Let $\epsilon > 0$. 
Then we can find $N \in \N$ such that: 
:$\cmod {z_n^k} < \epsilon^k$
From Power of Complex Modulus equals Complex Modulus of Power, we have:
:$\cmod {z_n}^k < \epsilon^k$
so:
:$\cmod {z_n} < \epsilon$
Since $\epsilon > 0$ was arbitrary, we have: 
:$z_n \to 0$
from the definition of a convergent complex sequence. 
Category:Convergent Complex Sequences
Complex Sine Function is Entire/Proof 1

== Theorem ==
== Proof ==
By the definition of the complex sine function, $\sin$ admits a power series expansion about $0$: 
:$\ds \sin z = \sum_{n \mathop = 0}^\infty \paren {-1}^n \frac {z^{2 n + 1} } {\paren {2 n + 1}!}$
By Complex Function is Entire iff it has Everywhere Convergent Power Series, to show that $\sin$ is entire it suffices to show that this series is everywhere convergent.
From Radius of Convergence from Limit of Sequence: Complex Case, it is sufficient to show that: 
:$\ds \lim_{n \mathop \to \infty} \size {\frac {\paren {-1}^{n + 1} } {\paren {2 n + 3}!} \times \frac {\paren {2 n + 1}!} {\paren {-1}^n} } = 0$
We have: 
Hence the result.
Category:Complex Sine Function is Entire
Complex Sine Function is Entire/Proof 2

== Theorem ==
== Proof ==
Let: 
for all $z \in \C$. 
By Complex Exponential Function is Entire, we have that $f$ is entire. 
By Polynomial is Entire, we have that $g$ and $h$ are entire. 
Therefore, by Composition of Entire Functions is Entire, we have that both $f \circ g$ and $f \circ h$ are entire.
By Linear Combination of Entire Functions is Entire, we then have that: 
:$\dfrac 1 {2 i} \paren {f \circ g - f \circ h}$
is entire.
Note that: 
Therefore, $\sin$ is an entire function.
Category:Complex Sine Function is Entire
Complex Sine Function is Unbounded

== Theorem ==
Let $\sin: \C \to \C$ be the complex sine function. 
Then $\sin$ is unbounded.
== Proof ==
By Complex Sine Function is Entire, we have that $\sin$ is an entire function. 
 that $\sin$ was a bounded function.
Then, by Liouville's Theorem, we would have that $\sin$ is a constant function.
However we have, for instance, by Sine of Zero is Zero: 
:$\sin 0 = 0$
and by Sine of 90 Degrees: 
:$\sin \dfrac \pi 2 = 1$
Therefore, $\sin$ is clearly not a constant function, a contradiction.
We hence conclude, by Proof by Contradiction, that $\sin$ is unbounded.
== Also see ==
* Complex Cosine Function is Unbounded
Category:Sine Function
Complex Singular Point/Examples/Reciprocal of z-2 Squared

== Example of Complex Singular Point ==
Let $f$ be the complex function defined as:
:$\map f z = \dfrac 1 {\paren {z - 2}^2}$
Then $f$ has a singular point at $z = 2$.
== Sources ==
* 
* 
Category:Examples of Complex Singular Points
Complex Subtraction/Examples

== Examples of Complex Subtraction ==
=== Example: $\paren {8 - 6 i} - \paren {2 i - 7}$ ===
=== Example: $\paren {6 - 2 i} - \paren {2 - 5 i}$ ===
=== Example: $\paren {7 + i} - \paren {4 - 2 i}$ ===
</onlyinclude>
Category:Examples of Complex Subtraction
Category:Complex Arithmetic
Complex Subtraction/Examples/(6 - 2i) - (2 - 5i)/Proof 2

== Example of Complex Subtraction ==
== Proof ==
:
By definition of complex subtraction:
:$\paren {6 - 2 i} - \paren {2 - 5 i} = \paren {6 - 2 i} + \paren {-2 + 5 i}$
Let the complex numbers $6 - 2 i$ and $-2 + 5 i$ be represented by the points $P_1$ and $P_2$ respectively in the complex plane.
Complete the parallelogram with $OP_1$ and $OP_2$ as the adjacent sides.
Using Geometrical Interpretation of Complex Addition, the point $P$ represents the complex number $4 + 3 i$, which is the sum of $6 - 2 i$ and $-2 + 5 i$.
Hence, $4 + 3 i$ is the difference of $6 - 2 i$ and $2 - 5 i$.
== Sources ==
* : $1$: Complex Numbers: Solved Problems: Graphical Representation of Complex Numbers. Vectors: $5 \ \text{(b)}$
Category:Examples of Complex Subtraction
Complex Subtraction/Examples/(7 + i) - (4 - 2i)/Proof 2

== Example of Complex Subtraction ==
== Proof ==
:
By definition of complex subtraction:
:$\paren {7 + i} - \paren {4 - 2 i} = \paren {7 + i} + \paren {-4 + 2 i}$
Let the complex numbers $7 + i$ and $-4 + 2 i$ be represented by the points $P_1$ and $P_2$ respectively in the complex plane.
Complete the parallelogram with $OP_1$ and $OP_2$ as the adjacent sides.
Using Geometrical Interpretation of Complex Addition, the point $P$ represents the complex number $3 + 3 i$, which is the sum of $7 + i$ and $-4 + 2 i$.
Hence, $3 + 3 i$ is the difference of $7 + i$ and $2 - 5 i$.
== Sources ==
* : $1$: Complex Numbers: Supplementary Problems: Graphical Representation of Complex Numbers. Vectors: $61 \ \text {(b)}$
Category:Examples of Complex Subtraction
Complex Subtraction/Examples/(8 - 6i) - (2i - 7)

== Example of Complex Subtraction ==
:$\paren {8 - 6 i} - \paren {2 i - 7} = 15 - 8 i$
== Proof ==
== Sources ==
* : $1$: Complex Numbers: Solved Problems: Fundamental Operations with Complex Numbers: $1 \ \text {(c)}$
Category:Examples of Complex Subtraction
Complex Subtraction is Closed

== Theorem ==
The set of complex numbers is closed under subtraction:
:$\forall a, b \in \C: a - b \in \C$
== Proof ==
From the definition of complex subtraction:
:$a - b := a + \paren {-b}$
where $-b$ is the inverse for complex number addition.
From Complex Numbers under Addition form Group, it follows that:
:$\forall a, b \in \C: a + \paren {-b} \in \C$
Therefore complex number subtraction is closed.
== Sources ==
* : Chapter $1$: Integral Domains: $\S 2$. Operations: Example $1$
Category:Complex Subtraction
Complex Vector Space with Dot Product is Hilbert Space

== Theorem ==
Let $\C^d$ be a complex vector space with $d$ dimensions.
Let $\innerprod \cdot \cdot$ be the dot product on $\C^d$.
Then $\C^d$ endowed with $\innerprod \cdot \cdot$ is a Hilbert space.
== Proof ==
Let us explore the inner product norm of $\innerprod \cdot \cdot$:
:$\ds \norm z = \sqrt{ \sum_{i \mathop = 1}^d x_i^2 + y_i^2 }$
and subsequently the metric induced by $\norm \cdot$:
:$\ds \map d {z, z'} = \norm {z - z'} = \sqrt{ \sum_{i \mathop = 1}^d \paren{ x_i - x'_i }^2 + \paren{ y_i - y'_i }^2 }$
== Sources ==
* : $\text{I}$ Hilbert Spaces: $\S 1.$ Elementary Properties and Examples
Category:Examples of Hilbert Spaces
Component Mappings of Set Coproduct are Injective

== Theorem ==
Let $S_1$ and $S_2$ be sets.
Let $\struct {C, i_1, i_2}$ be a coproduct of $S_1$ and $S_2$.
Then $i_1$ and $i_2$ are injections.
== Proof ==
By definition of coproduct:
:for all sets $X$ and mappings $f_1: S_1 \to X$ and $f_1: S_1 \to X$
::there exists a unique mapping $h: C \to X$ such that:
:::$h \circ i_1 = f_1$
:::$h \circ i_2 = f_2$
Let $X := S_1$ and $f_1: S_1 \to X = I_{S_1}$ where $I_{S_1}$ denotes the identity mapping.
We have:
:$h \circ i_1 = I_{S_1}$
From Identity Mapping is Injection, $I_{S_1}$ is an injection.
From Injection if Composite is Injection it follows that $i_1$ is an injection.
By setting $X := S_2$ and $f_2 := I_{S_2}$ it follows in the same way that $i_2$ is also an injection.
Hence the result.
== Sources ==
* : $\S 5$. Induced mappings; composition; injections; surjections; bijections: Exercise $18$
Category:Set Coproducts
Component is not necessarily Path Component

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Let $C$ be a component of $T$.
Then it is not necessarily the case that $C$ is also an path component of $T$.
== Proof ==
Let $C$ be the closed topologist's sine curve embedded in the real Euclidean plane.
From Closed Topologist's Sine Curve is Connected, $C$ is connected in $T$
Therefore $C$ is a component in the subspace of $T$ induced by $C$.
From Closed Topologist's Sine Curve is not Path-Connected, $C$ is not path-connected.
Therefore $C$ is not a path component in the subspace of $T$ induced by $C$.
Hence the result.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $4$: Connectedness
Category:Components
Category:Path Components
Component of Finite Union in Ultrafilter

== Theorem ==
Let $S$ be a non-empty set.
Let $\UU$ be a ultrafilter on $S$.
Let $\set {Y_1, \dots, Y_n}$ be a pairwise disjoint set of subsets of $S$ such that $S = Y_1 \cup \cdots \cup Y_n$.
Then there is a unique $k \in \set {1, \dots, n}$ such that $Y_k \in \UU$.
== Proof ==
Assume that none of the $Y_k$ are empty.
If not, then any empty ones can simply be removed.
=== Uniqueness ===
 that $Y_j, Y_k \in \UU$ with $j \ne k$.
Then since $Y_1, \dots, Y_n$ are a pairwise disjoint:
:$Y_j \cap Y_k = \O$
But by the definition of an ultrafilter, $\UU$ has the finite intersection property.
It follows from this contradiction that no such $Y_j, Y_k \in \UU$ with $j \ne k$ exists.
=== Existence ===
 that $Y_1, \dots, Y_n \notin \UU$.
Since the $Y_k$ are pairwise disjoint:
:$\ds Y_i^c = \bigcup \set {Y_k: k \ne i}$
Then by the definition of an ultrafilter:
:$\ds \forall i: Y_i^c = \bigcup \set {Y_k: k \ne i} \in \UU$
But:
:$\ds \bigcap_{i \mathop = 1}^n \bigcup \lset {Y_k: k \ne i} = \O$
contradicting the supposition that $\UU$ has the finite intersection property.
Thus $Y_k \in \UU$ for some $k$.
== Sources ==
* 
Category:Filter Theory
Component of Locally Connected Space is Open

== Theorem ==
Let $T = \left({S, \tau}\right)$ be a locally connected topological space.
Let $G$ be a component of $T$.
Then $G$ is open.
== Proof ==
By definition of locally connected space, $T$ has a basis of connected sets in $T$.
Thus $S$ is a union of open connected sets in $T$.
By Components are Open iff Union of Open Connected Sets, the components of $T$ are open.
== Also see ==
* Path Component of Locally Path-Connected Space is Open, an analogous result for path components
* Equivalence of Definitions of Locally Connected
Category:Components
Category:Locally Connected Spaces
Component of Point is not always Intersection of its Clopen Sets

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Let $x \in S$.
Let $\map {\operatorname {Comp}_x} T$ denote the component of $x$ in $T$.
Let $K_x = \ds \bigcap_{x \mathop \in K} K$ clopen in $T$.
Then it is not always the case that $\map {\operatorname {Comp}_x} T = K_x$
== Proof ==
Note that from Clopen Set contains Components of All its Points:
:$\map {\operatorname {Comp}_x} T \subseteq K_x$
It remains to be demonstrated that it is not always the case that $K_x \subseteq \map {\operatorname {Comp}_x} T$.
Let $T$ be the nested rectangle space in the Euclidean plane.
Let $L_1$ and $L_2$ be the boundary lines of $T$.
Let $x \in L_1$.
From Boundary Line in Nested Rectangle Space is Component $L_1$ is a component of $T$.
That is:
:$L_1 = \map {\operatorname {Comp}_x} T$ 
From Union of Boundary Lines in Nested Rectangle Space is Quasicomponent, $L_1 \cup L_2$ is a quasicomponent of $T$.
By Quasicomponent is Intersection of Clopen Sets:
:$L_1 \cup L_2 = \ds \bigcap_{x \mathop \in K} K$ clopen in $T$
Thus, while $\map {\operatorname {Comp}_x} T \subseteq L_1 \cup L_2$, it is not the case that $L_1 \cup L_2 \subseteq \map {\operatorname {Comp}_x} T$.
Hence the result.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $4$: Connectedness
Category:Clopen Sets
Category:Connected Sets
Component of Vector/Examples/Acceleration

== Example of Components of Vector ==
Let $\mathbf a$ be an acceleration of a particle $P$ in space.
Then we have:
:$\mathbf a = \dfrac {\d \mathbf v} {\d t} = \dfrac {\d^2 \mathbf r} {\d t^2}$
where:
:$\mathbf v$ is the velocity of $P$ at time $t$
:$\mathbf r$ is the displacement of $P$ at time $t$.
Thus:
:$\mathbf a = \dfrac {\d^2 x} {\d t^2} \mathbf i + \dfrac {\d^2 y} {\d t^2} \mathbf j + \dfrac {\d^2 z} {\d t^2} \mathbf k$
where:
:$\mathbf r = x \mathbf i + y \mathbf j + z \mathbf k$
== Sources ==
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Category:Examples of Components of Vectors
Category:Acceleration
Component of Vector/Examples/Velocity

== Example of Components of Vector ==
Let $\mathbf v$ be a velocity of a particle $P$ in space.
Then we have:
:$\mathbf v = \dfrac {\d \mathbf r} {\d t}$
where $\mathbf r$ is the displacement of $P$ at time $t$.
Thus:
:$\mathbf v = \dfrac {\d x} {\d t} \mathbf i + \dfrac {\d y} {\d t} \mathbf j + \dfrac {\d z} {\d t} \mathbf k$
where:
:$\mathbf r = x \mathbf i + y \mathbf j + z \mathbf k$
== Sources ==
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* 
Category:Examples of Components of Vectors
Category:Velocity