nyuuzyou/wikis huggingface dataset:
Velocity of Periodic Wave is Wavelength times Frequency

== Theorem ==
Let $\phi$ be a periodic wave expressed as:
:$\forall x, t \in \R: \map \phi {x, t} = \map f {x - c t}$
where $c$ is the velocity of $\phi$.
Then:
:$c = \nu \lambda$
where:
:$\nu$ is the frequency of $\phi$
:$\lambda$ is the wavelength of $\phi$.
== Proof ==
== Sources ==
* : Chapter $\text {I}$: The Equation of Wave Motion: $\S 3$
* 
* 
Category:Periodic Waves
Velocity of Point Moving on Surface of Sphere is Perpendicular to Radius

== Theorem ==
Let $P$ be a point moving on the surface of a sphere.
The velocity of $P$ is perpendicular to its radius at $P$.
== Proof ==
Let $S$ be a sphere whose center is at $O$.
By definition of a sphere, all the points on the surface of $S$ are the same distance from its center.
Let $\map {\mathbf v} t$ denote the position vector of $P$ with respect to $O$ at time $t$.
Then the magnitude $\norm {\mathbf v}$ of $\mathbf v$ is contstant.
Hence from Dot Product of Constant Magnitude Vector-Valued Function with its Derivative is Zero:
:$\map {\mathbf v} t \cdot \dfrac {\d \map {\mathbf v} t} {\d t} = 0$
That is, the dot product of the velocity of $P$ with the radius vector of $P$ is zero.
Hence by Dot Product of Perpendicular Vectors, the velocity of $P$ is perpendicular to its radius at $P$.
== Sources ==
* : Chapter $\text {III}$: The Differentiation of Vectors: $2$. Differentiation of Sums and Products
Category:Mechanics
Velocity of Point in Straight Line

== Theorem ==
Let $\mathbf v$ be the velocity of a particle $P$ in space.
Let $P$ be moving along a straight line $\LL$ whose positive direction has been established.
Then the motion of $P$ can be defined by:
:$\mathbf v = \dfrac {\d s} {\d t} \mathbf i$
where $\mathbf i$ denotes the unit vector in the positive direction of $\LL$.
== Proof ==
We have that $P$ moves along a straight line $\LL$.
Then the rate of change of displacement perpendicular to $\LL$ is zero.
Let the $\LL$ be embedded in a Cartesian space $\CC$.
From the components of velocity vector, we have:
:$\mathbf v = \dfrac {\d \mathbf r} {\d t}$
where $\mathbf r$ is the displacement of $P$ at time $t$ expressed as a position vector from a given origin.
Thus:
:$\mathbf v = \dfrac {\d x} {\d t} \mathbf i + \dfrac {\d x} {\d t} \mathbf j + \dfrac {\d x} {\d t} \mathbf k$
where:
:$\mathbf r = x \mathbf i + y \mathbf j + z \mathbf k$
Let us define $\CC$ such that $\LL$ is parallel to the $x$-axis of $\CC$.
Then we have:
== Sources ==
* 
* 
Category:Velocity
Velocity of Point on Rotating Body

== Theorem ==
Let $B$ be a body undergoing rotational motion around an axis $A$.
Let the angular velocity of $B$ about $A$ be $\bsomega$.
Let $O$ be an arbitrary point on $A$.
Let $P$ be a point in $B$ whose position vector $O$ is $\mathbf r$.
Then the velocity $\mathbf v$ of $P$ is given by:
:$\mathbf v = \bsomega \times \mathbf r$
== Proof ==
== Sources ==
* 
* 
Category:Velocity
Category:Angular Velocity
Category:Kinematics
Velocity of Rocket in Outer Space

== Theorem ==
Let $B$ be a rocket travelling in outer space.
Let the velocity of $B$ at time $t$ be $\mathbf v$.
Let the mass of $B$ at time $t$ be $m$.
Let the exhaust velocity of $B$ be constant at $\mathbf b$.
Then the velocity of $B$ at time $t$ is given by:
:$\map {\mathbf v} t = \map {\mathbf v} 0 + \mathbf b \ln \dfrac {\map m 0} {\map m t}$
where $\map {\mathbf v} 0$ and $\map m 0$ are the velocity and mass of $B$ at time $t = 0$.
== Proof ==
From Motion of Rocket in Outer Space, the equation of motion of $B$ is given by:
:$m \dfrac {\d \mathbf v} {\d t} = -\mathbf b \dfrac {\d m} {\d t}$
Hence:
== Sources ==
* : Chapter $\text {B}.8$: Rocket Propulsion in Outer Space
Category:Rocket Science
Velocity with respect to Relative Velocity

== Theorem ==
Let $A$ and $B$ be bodies in space.
Let $\mathbf v_A$ and $\mathbf v_B$ denote the velocities of $A$ and $B$ such that $\mathbf v_A$ and $\mathbf v_B$ are very much smaller than the speed of light.
Let $\mathbf v_{AB}$ denote the velocity of $A$ relative to $B$.
Then:
:$\mathbf v_A = \mathbf v_{AB} + \mathbf v_B$
== Proof ==
The result follows directly.
== Sources ==
* 
* 
Category:Relative Velocity
Vertex Condition for Isomorphic Graphs

== Theorem ==
Let $G_1$ and $G_2$ be isomorphic graphs.
Then the degrees of the vertices of $G_1$ are exactly the same as the degrees of the vertices of $G_2$.
== Proof ==
Let $\phi: \map V {G_1} \to \map V {G_2}$ be an isomorphism.
Let $u \in \map V {G_1}$ be an arbitrary vertex of $G_1$ such that $\map \phi u = v \in \map V {G_2}$.
Let $\map {\deg_{G_1} } u = n$.
We need to show that $\map {\deg_{G_2} } v = n$.
As $\map {\deg_{G_1} } u = n$, there exist $u_1, u_2, \ldots, u_n \in \map V {G_1}$ which are adjacent to $u$.
Every other vertex of $G_1$ is not adjacent to $u$.
Let $\map \phi {u_i} = v_i$ for $1, 2, \ldots, n$.
Because $\phi$ is an isomorphism, each of the vertices $v_1, v_2, \ldots, v_n \in \map V {G_2}$ are adjacent to $v$.
Similarly, every other vertex of $G_2$ is not adjacent to $v$.
Thus $\map {\deg_{G_2} } v = n$.
This applies to all vertices $u \in \map V {G_1}$.
Hence the result.
== Also see ==
* Same Degrees of Vertices does not imply Graph Isomorphism: It does not necessarily follow that if the degrees of the vertices of $G_1$ are exactly the same as the degrees of the vertices of $G_2$, then $G_1$ and $G_2$ are isomorphic.
== Sources ==
* : Chapter $2$: Elementary Concepts of Graph Theory: $\S 2.2$: Isomorphic Graphs: Theorem $2.4$
Category:Graph Isomorphisms
Category:Degrees of Vertices
Vertex Cut/Examples/Arbitrary Example 1

== Examples of Vertex Cuts ==
:
Removing $\set {B, C, F}$ would also remove all the dotted edges and leave this graph with two components:
:one containing the vertices $A$ and $E$
and
:one containing the vertices $D$, $G$, $H$, and $I$.
Thus $\set {B, C, F}$ is a vertex cut.
Category:Vertex Cuts
Vertical Section of Cartesian Product

== Theorem ==
Let $X$ and $Y$ be sets.
Let $A \subseteq X$ and $B \subseteq Y$, so that $A \times B \subseteq X \times Y$. 
Let $x \in X$. 
Then: 
:$\paren {A \times B}_x = \begin{cases}B & x \in A \\ \O & x \not \in A\end{cases}$
where $\paren {A \times B}_x$ is the $x$-vertical section of $A \times B$.
== Proof ==
Let $x \in A$. 
From the definition of the horizontal section, we have: 
:$y \in \paren {A \times B}_x$
:
:$\tuple {x, y} \in A \times B$ 
Since $x \in A$, this equivalent to: 
:$y \in B$
So:
:$y \in \paren {A \times B}_x$ $y \in B$
giving:
:$\paren {A \times B}_x = B$ if $x \in A$.
Now let $x \in X \setminus A$.
So, by the definition of set difference, we have $x \in X$ and $x \not \in A$.
As before, we have:
:$y \in \paren {A \times B}_x$ 
:
:$\tuple {x, y} \in A \times B$
But this is equivalent to: 
:$x \in A$ and $y \in B$.
Since $x \not \in A$, there exists no $y \in \paren {A \times B}_x$.
So:
:$\paren {A \times B}_x = \O$ if $x \not \in A$.
Category:Vertical Section of Sets
Category:Cartesian Product
Vertical Section of Characteristic Function is Characteristic Function of Vertical Section

== Theorem ==
Let $X$ and $Y$ be sets. 
Let $E \subseteq X \times Y$.
Let $x \in X$. 
Then: 
:$\paren {\chi_E}_x = \chi_{E_x}$
where:
:$E_x$ is the $x$-vertical section of $E$
:$\chi_E$ and $\chi_{E_x}$ are the characteristic functions of $E$ and $E_x$ as subsets of $X \times Y$ respectively
:$\paren {\chi_E}_x$ is the $x$-vertical function of $\chi_E$.
== Proof ==
We show that: 
:$\map {\paren {\chi_E}_x} y = \begin{cases}1 & y \in E_x \\ 0 & y \not \in E_x\end{cases}$
at which point we'll be done from the definition of the characteristic function of $E_x$. 
From the definition of the $x$-horizontal section, we have: 
:$\map {\paren {\chi_E}_x} y = \map {\chi_E} {x, y}$ 
From the definition of a characteristic function, we have: 
:$\map {\chi_E} {x, y} = 1$ $\tuple {x, y} \in E$.
From the definition of the $x$-vertical section, we then have: 
:$\map {\chi_E} {x, y} = 1$ $y \in E_x$. 
So:
:$\map {\paren {\chi_E}_x} y = 1$ $y \in E_x$. 
From the definition of a characteristic function, we also have: 
:$\map {\chi_E} {x, y} = 1$ $\tuple {x, y} \not \in E$.
From the definition of the $x$-vertical section, we then have: 
:$\map {\chi_E} {x, y} = 0$ $y \not \in E_x$.
So:
:$\map {\paren {\chi_E}_x} y = 0$ $y \not \in E_x$
giving:
:$\map {\paren {\chi_E}_x} y = \begin{cases}1 & y \in E_x \\ 0 & y \not \in E_x\end{cases}$
which was the demand.
Category:Characteristic Functions
Category:Vertical Section of Functions
Vertical Section of Continuous Function is Continuous

== Theorem ==
Let $X$, $Y$ and $T$ be topological spaces.
Equip the Cartesian product $X \times Y$ with the product topology.
Let $f : X \times Y \to T$ be a continuous mapping. 
Let $x \in X$.
Then the $x$-vertical section $f_x : Y \to T$ is continuous.
== Proof ==
From the definition of the $x$-vertical section, we have: 
:$\map {f_x} y = \map f {x, y}$
for each $y \in Y$. 
Define the map $p_x : Y \to X \times Y$ by: 
:$\map {p_x} y = \tuple {x, y}$ 
for each $y \in Y$. 
We have that: 
:$f_x = f \circ p_x$
From Composite of Continuous Mappings is Continuous, since $f$ is continuous, it suffices to show that $p_x$ is continuous.
From Box Topology on Finite Product Space is Product Topology, the product topology on $X \times Y$ is also the box topology. 
From Continuity Test using Basis, it then suffices to check that: 
:$p_x^{-1} \sqbrk {U \times V}$ is open whenever $U$ is open in $X$ and $V$ is open in $Y$.
Let $U$ be open in $X$ and $V$ be open in $Y$.
Note that: 
:$y \in p_x^{-1} \sqbrk {U \times V}$
:
:$\tuple {x, y} \in U \times V$
which is equivalent to: 
:$x \in U$ and $y \in V$. 
Clearly then if $x \not \in U$, we have $p_x^{-1} \sqbrk {U \times V} = \O$.
We can also read off from this equivalence that if $x \in U$, we have $p_x^{-1} \sqbrk {U \times V} = V$.
From Empty Set is Element of Topology, $\O$ is open in $Y$.
Since $V$ is open in $Y$ by hypothesis, we therefore have that $p_x$ is continuous from Continuity Test using Basis.
Hence the result.
Category:Vertical Section of Functions
Category:Continuous Mappings
Category:Product Topology
Vertical Section of Empty Set

== Theorem ==
Let $X$ and $Y$ be sets. 
Let $x \in X$. 
Then: 
:$\O_x = \O$
where $\O$ is the empty set and $\O_x$ is the $x$-vertical section of $\O$. 
== Proof ==
 suppose that: 
:$y \in \O_x$
Then from the definition of the $x$-vertical section, we have: 
:$\tuple {x, y} \in \O$
This is impossible from the definition of the empty set.
So:
:there exists no $y \in \O_x$
giving:
:$\O_x = \O$
Category:Vertical Section of Sets
Vertical Section of Linear Combination of Functions is Linear Combination of Vertical Sections

== Theorem ==
Let $X$ and $Y$ be sets.
Let $f_1, f_2, \ldots, f_n : X \times Y \to \overline \R$ be functions. 
Let $\alpha_1, \alpha_2, \ldots, \alpha_n$ be real numbers.
Let $x \in X$. 
Then: 
:$\ds \paren {\sum_{k \mathop = 1}^n \alpha_k f_k}_x = \sum_{k \mathop = 1}^n \alpha_k \paren {f_k}_x$
where $f_x$ denotes the $x$-vertical section of the function $f$. 
== Proof ==
Let $y \in Y$. 
We have:
so:
:$\ds \paren {\sum_{k \mathop = 1}^n \alpha_k f_k}_x = \sum_{k \mathop = 1}^n \alpha_k \paren {f_k}_x$
Category:Vertical Section of Functions
Vertical Section of Measurable Function is Measurable

== Theorem ==
Let $\struct {X, \Sigma_X}$ and $\struct {Y, \Sigma_Y}$ be measurable spaces.
Let $f : X \times Y \to \overline \R$ be a $\Sigma_X \otimes \Sigma_Y$-measurable function where $\Sigma_X \otimes \Sigma_Y$ is the product $\sigma$-algebra of $\Sigma_X$ and $\Sigma_Y$. 
Let $x \in X$.
Then: 
:$f_x$ is $\Sigma_Y$-measurable
where $f_x$ is the $x$-vertical section of $f$.
== Proof ==
By the definition of a $\Sigma_X$-measurable function, we have that: 
:$f^{-1} \sqbrk D \in \Sigma_X \otimes \Sigma_Y$ for each Borel set $D \subseteq \R$.
We aim to show that: 
:$\paren {f_x}^{-1} \sqbrk D \in \Sigma_Y$ for each Borel set $D \subseteq \R$.
Let $D \subseteq \R$ be a Borel set. 
From Preimage of Vertical Section of Function is Vertical Section of Preimage, we have: 
:$\paren {f_x}^{-1} \sqbrk D = \paren {f^{-1} \sqbrk D}_x$
From Horizontal Section of Measurable Set is Measurable, we have: 
:$\paren {f^{-1} \sqbrk D}_x \in \Sigma_Y$
so:
:$\paren {f_x}^{-1} \sqbrk D \in \Sigma_Y$
So:
:$\paren {f_x}^{-1} \sqbrk D \in \Sigma_Y$ for each Borel set $D \subseteq \R$.
so:
:$f_x$ is $\Sigma_Y$-measurable.
Category:Vertical Section of Functions
Category:Measurable Functions
Vertical Section of Measurable Set is Measurable

== Theorem ==
Let $\struct {X, \Sigma_X}$ and $\struct {Y, \Sigma_Y}$ be measurable spaces. 
Let $E \in \Sigma_X \otimes \Sigma_Y$ where $\Sigma_X \otimes \Sigma_Y$ is the product $\sigma$-algebra of $\Sigma_X$ and $\Sigma_Y$.
Let $x \in X$. 
Then: 
:$E_x \in \Sigma_Y$
where $E_x$ is the $x$-vertical section of $E$. 
== Proof ==
Let:
:$\FF = \set {E \subseteq X \times Y : E_x \in \Sigma_Y}$
We will show that $\FF$ contains each $S_1 \times S_2$ with $S_1 \in \Sigma_X$ and $S_2 \in \Sigma_Y$. 
We will then show that $\FF$ is a $\sigma$-algebra, at which point we will have: 
:$\map \sigma {\set {S_1 \times S_2 : S_1 \in \Sigma_X, \, S_2 \in \Sigma_Y} } \subseteq \FF$
from Sigma-Algebra Contains Generated Sigma-Algebra of Subset.
From the definition of the product $\sigma$-algebra, we will then have: 
:$\Sigma_X \otimes \Sigma_Y \subseteq \FF$
We will then have the demand.
Let $S_1 \in \Sigma_X$ and $S_2 \in \Sigma_Y$. 
From Vertical Section of Cartesian Product, we have: 
:$\ds \paren {S_1 \times S_2}_x = \begin{cases}S_2 & x \in S_1 \\ \O & x \not \in S_1\end{cases}$
From the definition of a $\sigma$-algebra, we have $\O \in \Sigma_Y$, so in either case we have: 
:$\paren {S_1 \times S_2}_x \in \Sigma_Y$
That is: 
:$S_1 \times S_2 \in \FF$
It remains to show that $\FF$ is a $\sigma$-algebra.
Since $S_1 \times S_2 \in \FF$ for $S_1 \in \Sigma_X$ and $S_2 \in \Sigma_Y$.
Since $X \in \Sigma_X$ and $Y \in \Sigma_Y$, we obtain: 
:$X \times Y \in \FF$
We show that $\FF$ is closed under countable union.
Let $\sequence {E_n}_{n \mathop \in \N}$ be a sequence in $\FF$.
We have $\paren {E_n}_x \in \Sigma_Y$ for each $n \in \N$. 
So, since $\Sigma_Y$ is a $\sigma$-algebra, we have: 
:$\ds \bigcup_{n \mathop = 1}^\infty \paren {E_n}_x \in \Sigma_Y$
From Union of Horizontal Sections is Horizontal Section of Union, we have: 
:$\ds \bigcup_{n \mathop = 1}^\infty \paren {E_n}_x = \paren {\bigcup_{n \mathop = 1}^\infty E_n}_x$ 
So we have: 
:$\ds \paren {\bigcup_{n \mathop = 1}^\infty E_n}_x \in \Sigma_Y$
That is: 
:$\ds \bigcup_{n \mathop = 1}^\infty E_n \in \FF$
We finally show that $\FF$ is closed under complementation.
Let $E \in \FF$.
We then have $E_x \in \Sigma_Y$.
Since $\Sigma_Y$ is closed under complementation, we have $Y \setminus E_x \in \Sigma_Y$. 
From Complement of Vertical Section of Set is Vertical Section of Complement, we have: 
:$Y \setminus E_x = \paren {\paren {X \times Y} \setminus E}_x$
so that: 
:$\paren {\paren {X \times Y} \setminus E}_x \in \Sigma_Y$
giving: 
:$\paren {X \times Y} \setminus E \in \FF$
So $\FF$ is a $\sigma$-algebra.
As discussed, we therefore obtain: 
:$\Sigma_X \otimes \Sigma_Y \subseteq \FF$ 
In particular, for any $E \in \Sigma_X \otimes \Sigma_Y$, we have $E \in \FF$.
That is: 
:for any $E \in \Sigma_X \otimes \Sigma_Y$ we have $E_x \in \Sigma_X$
as was the demand.
== Sources ==
* : $5.1$: Constructions: Lemma $5.1.2$
Category:Vertical Section of Sets
Vertical Section of Simple Function is Simple Function

== Theorem ==
Let $\struct {X, \Sigma_X}$ and $\struct {Y, \Sigma_Y}$ be measurable spaces.
Let $\struct {X \times Y, \Sigma_X \otimes \Sigma_Y}$ be the product measurable space of $\struct {X, \Sigma_X}$ and $\struct {Y, \Sigma_Y}$.
Let $f : X \times Y \to \R$ be a simple function.
Let $x \in X$. 
Then $f_x : Y \to \R$ is a simple function, where $f_x$ is the $x$-vertical section of $f$. 
== Proof ==
Write the standard representation of $f$ as: 
:$\ds f = \sum_{k \mathop = 1}^n a_k \chi_{E_k}$
with:
:$E_1, E_2, \ldots, E_n$ pairwise disjoint $\Sigma_X \otimes \Sigma_Y$-measurable sets
:$a_1, a_2, \ldots, a_n$ real numbers.
We have: 
From Intersection of Vertical Sections is Vertical Section of Intersection, we have that: 
:$\paren {E_1}_x, \paren {E_2}_x, \ldots, \paren {E_n}_x$ are pairwise disjoint.
From Vertical Section of Measurable Set is Measurable, we have that: 
:$\paren {E_1}_x, \paren {E_2}_x, \ldots, \paren {E_n}_x$ are $\Sigma_X$-measurable.
So, we have: 
:$\ds f_x = \sum_{k \mathop = 1}^n a_k \chi_{\paren {E_k}_x}$
with:
:$\paren {E_1}_x, \paren {E_2}_x, \ldots, \paren {E_n}_x$ pairwise disjoint $\Sigma_X$-measurable sets
:$a_1, a_2, \ldots, a_n$ real numbers.
So $f_x$ is a simple function.
Category:Vertical Section of Functions
Category:Simple Functions
Vertical Section preserves Increasing Sequences of Functions

== Theorem ==
Let $X$ and $Y$ be sets.
Let $\sequence {f_n}_{n \mathop \in \N}$ be an increasing sequence of real-valued functions with $f_i : X \times Y \to \overline \R$ for each $i$. 
Let $x \in X$. 
Then the sequence $\sequence {\paren {f_n}_x}_{n \mathop \in \N}$ is increasing, where $\paren {f_n}_x$ denotes the $x$-vertical section of $f_n$. 
== Proof ==
Since $\sequence {f_n}_{n \mathop \in \N}$ is an increasing sequence of real-valued functions, we have: 
:$\map {f_i} {x, y} \le \map {f_j} {x, y}$ for all $i, j$ with $i \le j$. 
for all $\tuple {x, y} \in X \times Y$.
In particular, for fixed $x \in X$, we have: 
:$\map {f_i} {x, y} \le \map {f_j} {x, y}$ for all $i, j$ with $i \le j$ and all $y \in Y$.
From the definition of the $x$-vertical section, we have: 
:$\map {\paren {f_i}_x} y \le \map {\paren {f_j}_x} y$ for all $i, j$ with $i \le j$ and all $y \in Y$.
So: 
:$\sequence {\paren {f_n}_x}_{n \mathop \in \N}$ is an increasing sequence of real-valued functions.
Category:Vertical Section of Functions
Vertical Section preserves Increasing Sequences of Sets

== Theorem ==
Let $X$ and $Y$ be sets.
Let $\sequence {A_n}_{n \mathop \in \N}$ be an increasing sequence in $X \times Y$. 
Let $x \in X$. 
Then: 
:$\sequence {\paren {A_n}_x}_{n \mathop \in \N}$ is an increasing sequence.
== Proof ==
Since $\sequence {A_n}_{n \mathop \in \N}$ is increasing, we have: 
:$A_n \subseteq A_{n + 1}$
for each $n$. 
From Vertical Section preserves Subsets, we have: 
:$\paren {A_n}_x \subseteq \paren {A_{n + 1} }_x$
for each $n$.
So:
:$\sequence {\paren {A_n}_x}_{n \mathop \in \N}$ is an increasing sequence.
Category:Vertical Section of Sets
Category:Increasing Sequences of Sets
Vertical Section preserves Pointwise Limits of Sequences of Functions

== Theorem ==
Let $X$ and $Y$ be sets. 
Let $f : X \times Y \to \overline \R$ be a function.
Let $\sequence {f_n}_{n \mathop \in \N}$ be a sequence of functions converging pointwise to $f$. 
Let $x \in X$. 
Then: 
:$\paren {f_n}_x \to f_x$
pointwise, where: 
:$\paren {f_n}_x$ denotes the $x$-vertical section of $f_n$
:$f_x$ denotes the $x$-vertical section of $f$. 
== Proof ==
From the definition of pointwise convergence, we have: 
:$\ds \map f {x, y} = \lim_{n \mathop \to \infty} \map {f_n} {x, y}$
for each $x \in X$ and $y \in Y$. 
Fix $x \in X$. 
From the definition of the $x$-vertical section, we have: 
:$\map {f_n} {x, y} = \map {\paren {f_n}_x} y$
and:
:$\map f {x, y} = \map {f_x} y$
So:
:$\ds \map {f_x} y = \lim_{n \mathop \to \infty} \map {\paren {f_n}_x} y$
for each $y \in Y$. 
So:
:$\paren {f_n}_x \to f_x$
pointwise.
Category:Vertical Section of Functions
Vertical Section preserves Subsets

== Theorem ==
Let $X$ and $Y$ be sets.
Let $A \subseteq B$ be subsets of $X \times Y$. 
Let $x \in X$. 
Then: 
:$A_x \subseteq B_x$
where $A_x$ is the $x$-vertical section of $A$ and $B_x$ is the $x$-vertical section of $B$. 
== Proof ==
Note that if: 
:$y \in A_x$
from the definition of $x$-vertical section, we have: 
:$\tuple {x, y} \in A$
so:
:$\tuple {x, y} \in B$
So, from the definition of $x$-vertical section, we have: 
:$y \in B_x$
So:
:if $y \in A_x$ then $y \in B_x$.
That is: 
:$A_x \subseteq B_x$
Category:Vertical Section of Sets
Vertices in Locally Finite Graph

== Theorem ==
Let $G$ be a locally finite graph.
Then if $G$ is infinite, it contains an infinite number of vertices.
== Proof ==
Suppose $G = \struct {V, E}$ has a finite number of vertices.
Let $V = \set {v_1, v_2, \ldots, v_n}$ where $n = \card V$ is the cardinality of $V$.
As $G$ is locally finite, each element of $V$ has a finite number of incident edges.
For each $v_k \in V$, let $r_k$ be the degree of $v_k$.
Then:
:$\ds \card E \le \sum_{i \mathop = 1}^n r_k$
where $\card E$ is the cardinality of $E$, that is, the number of edges in $G$.
As $\ds \sum_{i \mathop = 1}^n r_k$ is the sum of a finite number of finite numbers, it is itself finite.
So $\card E$ is itself finite, and so $G$ has:
:a finite number of vertices
:a finite number of edges
and so is by definition a finite graph.
By transposition, if $G$ is infinite, it must contain an infinite number of vertices.
== Note ==
If $G$ is not locally finite, then in order to have a finite number of vertices and an infinite number of edges, it needs to be a multigraph.
Category:Graph Theory
Vertices of Equilateral Triangle in Complex Plane

== Theorem ==
Let $z_1$, $z_2$ and $z_3$ be complex numbers.
Then:
: $z_1$, $z_2$ and $z_3$ represent on the complex plane the vertices of an equilateral triangle
:
:${z_1}^2 + {z_2}^2 + {z_3}^2 = z_1 z_2 + z_2 z_3 + z_3 z_1$
=== Corollary ===
== Proof ==
:
=== Sufficient Condition ===
=== Necessary Condition ===
== Sources ==
* : $\S 3$. Roots of Unity: Exercise $7$
Category:Equilateral Triangles
Category:Geometry of Complex Plane
Category:Vertices of Equilateral Triangle in Complex Plane
Vertices of Equilateral Triangle in Complex Plane/Corollary

== Corollary to Vertices of Equilateral Triangle in Complex Plane ==
Let $u, v \in \C$ be complex numbers.
Then:
: $0$, $u$ and $v$ represent on the complex plane the vertices of an equilateral triangle.
:
:$u^2 + v^2 = u v$
== Proof ==
From Vertices of Equilateral Triangle in Complex Plane:
: $z_1$, $z_2$ and $z_3$ represent on the complex plane the vertices of an equilateral triangle.
:
:${z_1}^2 + {z_2}^2 + {z_3}^2 = z_1 z_2 + z_2 z_3 + z_3 z_1$
Setting $z_1 \gets u$, $z_2 \gets v$, $z_3 \gets 0$ we have:
:$u^2 + v^2 + 0^2 = u v + v \times 0 + 0 \times u$
and the result follows.
== Sources ==
* : $\S 3$. Roots of Unity: Exercise $6$
Category:Equilateral Triangles
Category:Geometry of Complex Plane
Vertices of Equilateral Triangle in Complex Plane/Necessary Condition

== Theorem ==
Let $z_1$, $z_2$ and $z_3$ be complex numbers.
Let $z_1$, $z_2$ and $z_3$ fulfil the condition:
:${z_1}^2 + {z_2}^2 + {z_3}^2 = z_1 z_2 + z_2 z_3 + z_3 z_1$
Then $z_1$, $z_2$ and $z_3$ represent on the complex plane the vertices of an equilateral triangle.
== Proof ==
:
Let:
:${z_1}^2 + {z_2}^2 + {z_3}^2 = z_1 z_2 + z_2 z_3 + z_3 z_1$
Then:
Thus $z_2 - z_1$ and $z_3 - z_1$ are at the same angle to each other as $z_1 - z_3$ and $z_2 - z_1$.
Similarly:
Thus $z_2 - z_1$ and $z_3 - z_1$ are at the same angle to each other as $z_1 - z_2$ and $z_3 - z_2$.
Thus all three angles:
:$\angle z_2 z_1 z_3$
:$\angle z_1 z_3 z_2$
:$\angle z_3 z_2 z_1$
are equal.
By definition, therefore, $\triangle z_1 z_2 z_3$ is equilateral.
== Sources ==
* : $1$: Complex Numbers: Supplementary Problems: Miscellaneous Problems: $139$
Category:Vertices of Equilateral Triangle in Complex Plane