nyuuzyou/wikis huggingface dataset:
Mapping to Annihilator on Algebraic Dual is Bijection

== Theorem ==
Let $G$ be an $n$-dimensional vector space over a field.
Let $J: G \to G^{**}$ be the evaluation isomorphism.
Let $G^*$ be the algebraic dual of $G$.
Let $M$ be an $m$-dimensional subspace of $G$.
Let $M^\circ$ be the annihilator of $M$.
Let $G_m$ denote the set of all $m$-dimensional subspaces of $G$.
Let ${G^*}_{n - m}$ denote the set of all $n - m$-dimensional subspaces of $G^*$.
Let $\phi: G_m \to {G^*}_{n - m}$ be the mapping from $G_m$ to the power set of ${G^*}_{n - m}$ defined as:
:$\forall M \in \powerset G: \map \phi M = M^\circ$
Then $\phi$ is a bijection.
== Proof ==
From Annihilator of Annihilator on Algebraic Dual of Subspace is Image under Evaluation Isomorphism, we have that:
:$M^{\circ \circ} = J \sqbrk M$
From Evaluation Isomorphism is Isomorphism, $J: G \to G^{**}$ is a bijection.
== Sources ==
* : Chapter $\text {V}$: Vector Spaces: $\S 28$. Linear Transformations: Theorem $28.10$
Category:Annihilators
Category:Vector Subspaces
Mapping to Identity is Unique Constant Homomorphism

== Theorem ==
Let $\struct {G, \circ}$ and $\struct {H, *}$ be groups whose identities are $e_G$ and $e_H$ respectively.
Then there exists a unique constant mapping from $G$ to $H$ which is a homomorphism:
:$\phi_{e_H}: G \to H: \forall g \in G: \map {\phi_{e_H} } g = e_H$
== Proof ==
Let $h \in H$ such that $\phi_h: G \to H$ is a (group) homomorphism, where $\phi_h$ is defined as:
:$\forall g \in G: \map {\phi_h} g = h$
Then from Group Homomorphism Preserves Identity:
:$\map {\phi_h} {e_G} = e_H$
and so $h = e_H$.
Hence the result by definition of constant mapping.
It remains to prove that such a constant mapping is in fact a homomorphism.
Let $x, y \in G$.
Thus the morphism property is demonstrated, and $\phi_{e_H}$ is seen to be a homomorphism.
== Sources ==
* : Chapter $\text{II}$: Groups: Morphisms
Category:Group Homomorphisms
Category:Constant Mappings
Mapping to Indiscrete Space is Continuous

== Theorem ==
Let $T_1 = \struct {S_1, \tau_1}$ be any topological space.
Let $T_2 = \struct {S_2, \tau_2}$ be the indiscrete topological space on $S_2$.
Let $\phi: S_1 \to S_2$ be a mapping.
Then $\phi$ is continuous.
== Proof ==
From the definition of continuous:
:$U \in \tau_2 \implies \phi^{-1} \sqbrk U \in \tau_1$
The only elements of $\tau_2$ are $S_2$ and $\O$, from which:
:$\phi^{-1} \sqbrk {S_2} = S_1 \in \tau_1$
:$\phi^{-1} \sqbrk \O = \O \in \tau_1$
== Sources ==
* : $3$: Continuity generalized: topological spaces: Exercise $3.9: 6$
* : Part $\text {II}$: Counterexamples: $4$. Indiscrete Topology: $8$
Category:Indiscrete Topology
Mapping to Power is Endomorphism iff Abelian

== Theorem ==
Let $\struct {G, \circ}$ be a group.
Let $n \in \Z$ be an integer.
Let $\phi: G \to G$ be defined as:
:$\forall g \in G: \map \phi g = g^n$
Then $\struct {G, \circ}$ is abelian $\phi$ is a (group) endomorphism.
== Proof ==
=== Necessary Condition ===
Let $\struct {G, \circ}$ be an abelian group.
Let $a, b \in G$ be arbitrary.
Then:
As $a$ and $b$ are arbitrary, the above holds for all $a, b \in G$.
Thus $\phi$ is a group homomorphism from $G$ to $G$.
So by definition, $\phi$ is a group endomorphism.
=== Sufficient Condition ===
Let $\phi: G \to G$ as defined above be a group endomorphism.
Then:
From Power of Product of Commutative Elements in Group it follows that $G$ is an abelian group.
== Sources ==
* : Chapter $7$: Homomorphisms: Exercise $2$
* : $\S 3.2$: Groups; the axioms: Exercise $(9)$
* : $\S 3$: Exercise $6$
Category:Abelian Groups
Category:Group Endomorphisms
Mapping to Singleton is Surjection

== Theorem ==
Let $S$ be a non-empty set.
Let $f: S \to T$ be a mapping.
Let $T$ be a singleton.
Then $f$ is a surjection.
== Proof ==
Let $T = \set t$.
For $f$ to be a surjection, all we need to do is show:
:$\forall y \in T: \exists x \in S: \map f x = y$.
As $S \ne \O$, $\exists s \in S$.
As $f: S \to T$ is a mapping, it follows that $\map f s \in T$.
So as $\map f s \in T$ it follows that $t = \map f s$.
As $T = \set t$, it follows that $\forall y \in T: \exists x \in S: y = \map f x$.
Hence the result.
Category:Surjections
Category:Singletons
Mapping to Singleton is Unique

== Theorem ==
Let $S$ be a set.
Let $T$ be a singleton.
Then there exists a unique mapping $S \to T$.
== Proof ==
Let $T = \set t$.
Let $f$ and $g$ both be mappings from $S$ to $T$.
From Mapping is Constant iff Image is Singleton:
:$\forall s \in S: \map f s = t$
and:
:$\forall s \in S: \map g s = t$
The result follows by Equality of Mappings.
== Also see ==
* Singleton is Terminal Object of Category of Sets
Category:Singletons
Mapping to Square is Endomorphism iff Abelian

== Theorem ==
Let $\struct {G, \circ}$ be a group.
Let $\phi: G \to G$ be defined as:
:$\forall g \in G: \map \phi g = g \circ g$
Then $\struct {G, \circ}$ is abelian $\phi$ is a (group) endomorphism.
== Proof ==
=== Necessary Condition ===
Let $\struct {G, \circ}$ be an abelian group.
Let $a, b \in G$ be arbitrary.
Then:
As $a$ and $b$ are arbitrary, the above holds for all $a, b \in G$.
Thus $\phi$ is a group homomorphism from $G$ to $G$.
So by definition, $\phi$ is a group endomorphism.
=== Sufficient Condition ===
Let $\phi: G \to G$ as defined above be a group endomorphism.
Then:
Thus, by definition, $G$ is an abelian group.
== Sources ==
* : Chapter $2$: Group Homomorphism and Isomorphism: $\S 60 \delta$
* : $\S 3.2$: Groups; the axioms: Exercise $(9)$
* : Chapter $3$: Elementary consequences of the definitions: Exercise $6$
Category:Abelian Groups
Category:Group Endomorphisms
Mapping whose Domain is Small Class is Small

== Theorem ==
Let $F$ be a mapping.
Let the domain of $F$ be a small class.
Then, $F$ is a small class.
== Proof ==
Let $A$ denote the domain of $F$.
Let $B$ denote the image of $F$.
Since $F$ is a mapping, $F$ is also a relation. 
Therefore:
:$F \subseteq A \times B$
where $A \times B$ denotes the Cartesian product of $A$ and $B$.
$B$ is the image of $A$ under $F$ and is therefore a small class by Image of Small Class under Mapping is Small.
Since $A$ and $B$ are both small, their Cartesian Product is Small.
By Axiom of Subsets Equivalents, it follows that $F$ is small.
== Sources ==
* : $\S 6.15$
Category:Class Theory
Mapping whose Graph is Closed in Chebyshev Product is not necessarily Continuous

== Theorem ==
Let $M_1 = \struct {A_1, d_1}$ and $M_2 = \struct {A_2, d_2}$ be metric spaces.
Let $\AA = A_1 \times A_2$ be the cartesian product of $A_1$ and $A_2$.
Let $d_\infty: \AA \times \AA \to \R$ be the Chebyshev distance on $\AA$:
:$\ds \map {d_\infty} {x, y} = \max \set {\map {d_1} {x_1, y_1}, \map {d_2} {x_2, y_2} }$
where $x = \tuple {x_1, x_2}, y = \tuple {y_1, y_2} \in \AA$.
Let $\Gamma_f$ be the graph of $f$.
Let $f: M_1 \to M_2$ be a mapping such that the $\Gamma_f$ is a closed set of $\struct {A_1 \times A_2, d}$.
Then it is not necessarily the case that $f$ is a continuous mapping.
== Proof ==
Consider the mapping $f: \R \to \R$ defined as:
:$\map f x = \begin {cases} \dfrac 1 x : & x > 0 \\ 0 : & x \le 0 \end {cases}$
It is seen that $\map f x$ is continuous everywhere except at $x = 0$.
Hence from Graph of Continuous Mapping between Metric Spaces is Closed in Chebyshev Product, $\Gamma_f$ contains all its limit points except perhaps at $x = 0$.
We have that:
:$\tuple {0, 0} \in \Gamma_f$
For a bounded, non-continuous real function, the limit point at one of the discontinuities would not lie in the graph.
However, this is not the case here.
The sequence $\sequence {x_n, \map f {x_n} }$ for which $\ds \lim_{n \mathop \to \infty} = 0$ has only one limit point, that is: $\tuple {0, 0}$.
But $\tuple {0, 0}$ lies in $\Gamma_f$.
 there exists another limit point.
This would imply that:
:$\forall \epsilon > 0: \exists N \in \N: n > N \implies \map d {\map f x, \map f {x_n} } < \epsilon$
But $\map f x$ is not bounded.
So for some $x < x_n$ we have $\map f {\map f x, \map f {x_n} } > \epsilon$ if $x, x_n > 0$.
This contradicts what was deduced earlier.
So any limit point $\tuple {x, \map f x}$ for which $x = 0$ is $\tuple {0, 0}$ which is in $\Gamma_f$.
Hence $\Gamma_f$ contains all its limit points.
Hence the result by definition of closed set.
== Sources ==
* : Chapter $2$: Metric Spaces: $\S 6$: Open Sets and Closed Sets: Exercise $4$
Category:Continuous Mappings
Category:Chebyshev Distance
Mapping whose Image of Natural Number n is Subset of Image of Successor

== Theorem ==
Let $f: \N \to A$ be a mapping from the set of natural numbers $\N$ to a class $A$.
Let $f$ have the property that:
:$\forall n \in \N: \map f n \subseteq \map f {n^+}$
where $n^+$ is the successor of $n$.
Then:
:$\forall n, m \in N: n \le m \implies \map f n \subseteq \map f m$
=== Corollary ===
== Proof ==
Let us establish that:
:$n = m \implies \map f n = \map f m$
and so:
:$n = m \implies \map f n \subseteq \map f m$
Hence it is sufficient to demonstrate that:
:$\forall n < m: \map f n \subseteq \map f m$
The proof will proceed by the Principle of Finite Induction on $\N$.
Let $S$ be the set defined as:
:$S := \set {m \in \N: \forall n < m: \map f n \subseteq \map f m}$
That is, $S$ is to be the set of all $n$ such that:
:$\forall n < m: \map f n \subseteq \map f m$
First we note that:
:$\not \exists n \in \N: n < 0$
Thus $0 \in S$ vacuously.
=== Basis for the Induction ===
Consider the natural number $1$.
There is only one $n \in \N$ such that $n < 1$, and that is:
:$n = 0$
By definition of $f$:
:$\map f 0 \subseteq \map f {0^+}$
But by definition of the set of natural numbers:
:$0^+ = 1$
So $1 \in S$.
This is the basis for the induction.
=== Induction Hypothesis ===
It is to be shown that if $k \in S$ where $k \ge 1$, then it follows that $k + 1 \in S$.
This is the induction hypothesis:
:$\forall n < k: \map f n \subseteq \map f k$
It is to be demonstrated that it follows that:
:$\forall n < k^+: \map f n \subseteq \map f {k^+}$
=== Induction Step ===
This is the induction step:
Let $k \in S$.
From the induction hypothesis:
:$\forall n < k: \map f n \subseteq \map f k$
Then from the definition of $f$:
:$\map f k \subseteq \map f {k^+}$
That is:
:$\forall n \le k: \map f n \subseteq \map f {k^+}$
That is:
:$\forall n < k^+: \map f n \subseteq \map f {k^+}$
So $k \in S \implies k^+ \in S$.
The result follows by the Principle of Finite Induction:
:$\forall n, m \in N: n \le m \implies \map f n \subseteq \map f m$
== Sources ==
* : Chapter $3$: The Natural Numbers: $\S 5$ Applications to natural numbers: Exercise $5.2 \ \text{(a)}$
Category:Ordering on Natural Numbers
Mapping whose Image of Natural Number n is Subset of Image of Successor/Corollary

== Theorem ==
Let $f: \N \to A$ be a mapping from the set of natural numbers $\N$ to a class $A$.
Let $f$ have the property that:
:$\forall n \in \N: \map f n \subsetneqq \map f {n^+}$
where $n^+$ is the successor of $n$.
Then:
:$\forall n, m \in N: n < m \implies \map f n \subsetneqq \map f m$
== Proof ==
The proof will proceed by the Principle of Finite Induction on $\N$.
Let $S$ be the set defined as:
:$S := \set {m \in \N: \forall n < m: \map f n \subsetneqq \map f m}$
That is, $S$ is to be the set of all $n$ such that:
:$\forall n < m: \map f n \subsetneqq \map f m$
First we note that:
:$\not \exists n \in \N: n < 0$
Thus $0 \in S$ vacuously.
=== Basis for the Induction ===
Consider the natural number $1$.
There is only one $n \in \N$ such that $n < 1$, and that is:
:$n = 0$
By definition of $f$:
:$\map f 0 \subseteq \map f {0^+}$
But by definition of the set of natural numbers:
:$0^+ = 1$
So $1 \in S$.
This is the basis for the induction.
=== Induction Hypothesis ===
It is to be shown that if $k \in S$ where $k \ge 1$, then it follows that $k + 1 \in S$.
This is the induction hypothesis:
:$\forall n < k: \map f n \subsetneqq \map f k$
It is to be demonstrated that it follows that:
:$\forall n < k^+: \map f n \subsetneqq \map f {k^+}$
=== Induction Step ===
This is the induction step:
Let $k \in S$.
From the induction hypothesis:
:$\forall n < k: \map f n \subsetneqq \map f k$
Then from the definition of $f$:
:$\map f k \subsetneqq \map f {k^+}$
That is:
:$\forall n \le k: \map f n \subsetneqq \map f {k^+}$
That is:
:$\forall n < k^+: \map f n \subsetneqq \map f {k^+}$
So $k \in S \implies k^+ \in S$.
The result follows by the Principle of Finite Induction:
:$\forall n, m \in N: n \le m \implies \map f n \subsetneqq \map f m$
== Sources ==
* : Chapter $3$: The Natural Numbers: $\S 5$ Applications to natural numbers: Exercise $5.2 \ \text{(b)}$
Category:Ordering on Natural Numbers
Mappings Between Residue Classes

== Theorem ==
Let $\eqclass a m$ be the residue class of $a$ (modulo $m$).
Let $\phi: \Z_m \to \Z_n$ be a mapping given by:
:$\map \phi {\eqclass x m} = \eqclass x n$
Then $\phi$ is well defined $m$ is a divisor of $n$.
== Proof ==
For $\phi$ to be well defined, we require that:
:$\forall x, y \in \Z_m: \eqclass x m = \eqclass y m \implies \map \phi {\eqclass x m} = \map \phi {\eqclass y m}$
Now:
:$\eqclass x m = \eqclass y m \implies x - y \divides m$
For $\map \phi {\eqclass x m} = \map \phi {\eqclass y m}$ we require that:
:$\eqclass x n = \eqclass y n \implies x - y \divides n$
Thus $\phi$ is well defined :
:$x - y \divides m \implies x - y \divides n$
That is, $m \divides n$.
== Sources ==
* : Chapter $1$: Equivalence Relations: $\S 19 \alpha$
Category:Residue Classes
Mappings Partially Ordered by Extension

== Theorem ==
Let $S$ and $T$ be sets.
Let $F$ be the set of all mappings from $S$ to $T$.
Let $\RR \subseteq F \times F$ be the relation defined as:
:$\tuple {f, g} \in \RR \iff \Dom f \subseteq \Dom g \land \forall x \in \Dom f: \map f x = \map g x$
That is, $f \mathrel \RR g$ $g$ is an extension of $f$.
Then $\RR$ is an ordering on $F$.
== Proof ==
Let $x \in \Dom f$ such that $\map f x = y$.
Then by definition $x \in \Dom g$ and $\map g x = y$.
Thus by definition of subset, $f \subseteq g$.
We have that Subset Relation is Ordering.
Hence the result.
== Sources ==
* : $\S 14$: Order
Category:Mapping Theory
Category:Order Theory
Mappings in Product of Sets are Surjections

== Theorem ==
Let $S$ and $T$ be sets.
Let $\struct {P, \phi_1, \phi_2}$ be a product of $S$ and $T$.
Then $\phi_1$ and $\phi_1$ are surjections.
=== Family of Sets ===
== Proof ==
From the definition:
:For all sets $X$ and all mappings $f_1: X \to S$ and $f_2: X \to T$ there exists a unique mapping $h: X \to P$ such that:
::$\phi_1 \circ h = f_1$
::$\phi_2 \circ h = f_2$
Let $X = S$ and let $f_1 = I_S$ where $I_S$ is the identity mapping on $S$.
Then we have:
:$\phi_1 \circ h = I_S$
We have from Identity Mapping is Surjection that $I_S$ is a surjection.
From Surjection if Composite is Surjection it follows that $\phi_1$ is a surjection.
Similarly, let $X = T$ and let $f_2 = I_T$ where $I_T$ is the identity mapping on $T$.
Then we have:
:$\phi_2 \circ h = I_T$
We have from Identity Mapping is Surjection that $I_T$ is a surjection.
From Surjection if Composite is Surjection it follows that $\phi_2$ is a surjection.
== Sources ==
* : $\S 5$. Induced mappings; composition; injections; surjections; bijections: Theorem $5.11$
Category:Set Products
Category:Surjections
Mappings in Product of Sets are Surjections/Family of Sets

== Theorem ==
Let $\family {S_i}_{i \mathop \in I}$ be an indexed family of sets.
Let $\struct {P, \family {\phi_i} _{i \mathop \in I} }$ be a product of $S$ and $T$.
Then for all $i \in I$, $\phi_i$ is a surjection.
== Proof ==
From the definition:
:For all sets $X$ and all indexed families $\family {f_i}_{i \mathop \in I}$ of mappings $f_i: X \to S_i$ there exists a unique mapping $h: X \to P$ such that:
::$\forall i \in I: \phi_i \circ h = f_i$
Let:
:$i \in I$
:$X = S_i$
:$f_i = I_{S_i}$
where $I_{S_i}$ is the identity mapping on $S_i$.
Then we have:
:$\phi_i \circ h = I_{S_i}$
From Identity Mapping is Surjection:
:$I_{S_i}$ is a surjection.
From Surjection if Composite is Surjection:
:$\phi_i$ is a surjection.
$i \in I$ is arbitrary so the argument applies whatever its value.
Hence the result.
== Sources ==
* : $\S 6$. Indexed families; partitions; equivalence relations
Category:Set Products
Category:Surjections
Mappings to Algebraic Structure form Similar Algebraic Structure

== Theorem ==
Let $X$ be a nonempty set.
Let $G$ be a magma with respect to the binary operations $\circ_1, \ldots, \circ_n$ on $G$.
Let $G^X$ be the set of all mappings from $X$ to $G$.
Denote also by $\circ_1, \ldots, \circ_n$ the binary operations defined on $G^X$ by pointwise addition.
== Also see ==
* Mappings to R-Algebraic Structure form Similar R-Algebraic Structure
== Sources ==
* : $\text{I}$: Groups: $\S 1$: Semigroups, Monoids and Groups: Exercise $2$
Category:Abstract Algebra
Mappings to R-Algebraic Structure form Similar R-Algebraic Structure

== Theorem ==
Let $X$ be a nonempty set.
Let $R$ be a ring.
Let $\struct {G, \circ}_R$ be an $R$-algebraic structure.
Let $G^X$ be the set of all mappings from $X$ to $G$.
Denote also by $\circ$ the binary operation defined on $G^X$ by pointwise ($R$)-scalar multiplication.
== Also see ==
* Mappings to Algebraic Structure form Similar Algebraic Structure
Category:Abstract Algebra
Mappings to Vector Space form Vector Space

== Theorem ==
Let $X$ be a non-empty set.
Let $V$ be a vector space over a field (or division ring) $K$.
Let $V^X$ denote the set of all mappings from $X$ to $V$.
Let $+$ denote pointwise addition on $V^X$.
Let $\circ$ denote pointwise ($K$)-scalar multiplication on $V^X$.
Then $\struct {V^X, +, \circ}_K$ is a vector space over $K$.
== Proof ==
Category:Vector Spaces
Marginal Total/Examples/Arbitrary Example 1

== Examples of Marginal Totals ==
Consider the contingency table below:
$\quad \begin {array} {r|cc|c}
& \text {Column 1} & \text {Column 2} & \text {Row totals} \\
\hline
\text {Row 1} & 10 & 2 & 12 \\
\text {Row 2} & 5 & 8 & 13 \\
\hline
\text {Column totals} & 15 & 10
\end {array}$
The marginal totals for the rows are in the column labelled $\text {Row totals}$
The marginal totals for the columns are in the row labelled $\text {Column totals}$.
Category:Examples of Marginal Totals
Markov's Inequality

== Theorem ==
Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $A \in \Sigma$.
Let $f : A \to \overline \R$ be an $A$-measurable function.
Then:
:$\ds \map \mu {\set {x \in A: \size {\map f x} \ge t} } \le \frac 1 t \int_A \size f \rd \mu$
for any positive $t \in \R$.
=== Corollary ===
== Proof ==
Let $t > 0$ and define:
:$B = \set {x \in A: \size {\map f x} \ge t}$
Let $\chi_B$ denote the indicator function of $B$ on $A$. 
For any $x \in A$, either $x \in B$ or $x \notin B$.
In the first case:
:$t \map {\chi_B} x = t \cdot 1 = t \le \size {\map f x}$
In the second case:
:$t \map {\chi_B} x = t \cdot 0 = 0 \le \size {\map f x}$
Hence:
:$\forall x \in A: t \map {\chi_B} x \le \size {\map f x}$
By Integral of Integrable Function is Monotone:
:$\ds \int_A t \chi_B \rd \mu \le \int_A \size f \rd \mu$
But by Integral of Integrable Function is Homogeneous:
:$\ds \int_A t \chi_B \rd \mu = t \int_A \chi_B \rd \mu = t \map \mu B$
Dividing through by $t > 0$:
:$\ds \map \mu B \le \frac 1 t \int_A \size f \rd \mu$
== Sources ==
* : $10.12$
* : $\S 10$: Problem $10.8$
Category:Measure Theory
Category:Probability Theory
Category:Markov's Inequality
Markov's Inequality/Corollary

== Corollary ==
Let $\struct {\Omega, \Sigma, \Pr}$ be a probability space.
Let $X$ be a real-valued random variable on $\struct {\Omega, \Sigma, \Pr}$.
Then: 
:$\map \Pr {\size X \ge t} \le \dfrac {\expect {\size X} } t$
for each real $t > 0$.
== Proof ==
From Markov's Inequality, we have: 
:$\ds \map \Pr {\set {\omega \in \Omega : \size {\map X \omega} \ge t} } \le \frac 1 t \int \size X \rd \Pr$
From the definition of expectation, we then have: 
:$\map \Pr {\set {\omega \in \Omega : \size {\map X \omega} \ge t} } \le \dfrac {\expect {\size X} } t$
So:
:$\map \Pr {\size X \ge t} \le \dfrac {\expect {\size X} } t$
Category:Markov's Inequality
Markov Chain/Examples/Simplest

== Example of Markov Chain ==
The simplest Markov chain $\sequence {X_B}$ is the case where the state space $S$ is a Boolean domain, say $S = \set {0, 1}$.
Its transition probabilities are hence $p_{00}$, $p_{01}$, $p_{10}$ and $p_{11}$.
Hence for example:
:$p_{01}$ is the probability that $\sequence {X_B}$ changes from state $0$ to state $1$
:$p_{11}$ is the probability that $\sequence {X_B}$ remains in state $1$ if it is already there.
Its transition matrix is of the form:
:$\begin {pmatrix} p_{00} & p_{01} \\ p_{10} & p_{11} \end {pmatrix}$
where:
Note that:
:if $p_{10} = p_{01} = 1$, then $\sequence {X_B}$ oscillates repeatedly from $0$ to $1$ at each transition time
:if $p_{00} = p_{11} = 1$, then $\sequence {X_B}$ stays in the same state, remaining in the state it started at all transition times.
== Sources ==
* 
* 
Category:Examples of Markov Chains
Martingale/Examples/Simple Random Walk

== Example of Martingale ==
Let $\sequence {X_n}_{n \mathop \ge 0}$ be a simple random walk such that $\map \Pr {X_n \to X_{n + 1} } = \map \Pr {X_n \to X_{n 1 1} } = \dfrac 1 2$.
Then the sequence of $x_i$ at time $t_i$ form a martingale.
== Proof ==
Let $\sequence {X_n}$ be at $x_n$ at time $t_n$.
Then at time $t_{n + 1}$ we have:
:$x_{n + 1} = x_n \pm 1$
each with probability $\dfrac 1 2$.
Hence:
:$\expect {x_{n + 1} } = \dfrac {\paren {x_n + 1} + \paren {x_n - 1} } 2 = x_n$
Hence the result by definition of martingale.
== Sources ==
* 
* 
Category:Examples of Martingales
Martingale Composed with Convex Function is Submartingale

== Theorem ==
Let $\struct {\Omega, \Sigma, \sequence {\FF_t}_{t \ge 0}, \Pr}$ be a continuous-time filtered probability space.
Let $\sequence {X_t}_{t \ge 0}$ be a $\sequence {\FF_t}_{t \ge 0}$-martingale.
Let $f : \R \to \R$ be a convex function such that $\map f {X_t}$ is integrable for each $t \in \hointr 0 \infty$. 
Then $\sequence {\map f {X_t} }_{t \ge 0}$ is a $\sequence {\FF_t}_{t \ge 0}$-submartingale. 
== Proof ==
Since $\sequence {X_t}_{t \ge 0}$ is a martingale, we have: 
:$X_t$ is $\FF_t$-measurable
for each $t \in \hointr 0 \infty$. 
From Convex Real Function is Measurable and Composition of Measurable Mappings is Measurable:
:$\map f {X_t}$ is $\FF_t$-measurable
for each $t \in \hointr 0 \infty$.
So $\sequence {\map f {X_t} }_{t \ge 0}$ is $\sequence {\FF_t}_{t \ge 0}$-adapted. 
Now let $s, t \in \hointr 0 \infty$ be such that $0 \le s < t$. 
Let $\expect {\map f {X_t} \mid \FF_s}$ be a version of the conditional expectation of $\map f {X_t}$ given $\FF_s$.
We then have: 
So $\sequence {\map f {X_t} }_{t \ge 0}$ is a $\sequence {\FF_t}_{t \ge 0}$-submartingale.
== Sources ==
* : Proposition $3.12$
Category:Submartingales
Category:Martingales
Category:Convex Real Functions
Mass of Mole of Isotope of Element

== Theorem ==
Let $S$ be a substance made up entirely of a particular isotope $Q$ of a particular element.
Let one atom of $Q$ contain $n$ neutrons and $p$ protons.
Then one mole of $S$ has a mass of approximately $n + p$ grams.
== Proof ==
Let $m_Q$ grams be the mass of one atom of $Q$.
Let $m_C$ grams be the mass of one atom of carbon-12.
Let $M_Q$ grams be the mass of one mole of $Q$.
Let $M_C$ grams be the mass of one mole of carbon-12.
The mass of a proton and the mass of a neutron are approximately equal.
Together, the protons and neutrons make up the vast majority of the mass of the atom they compose.
Hence the mass of one atom is proportional to the total number of its protons and neutrons.
There are $6$ protons and $6$ neutrons in one atom of carbon-12.
Thus we have:
Before $2018$, the mole was defined as the number of atoms in $12$ grams of carbon-12.
While the mole is no longer so defined, the mass of $1$ mole of carbon-12 is still $12$ grams, for all practical purposes.
Thus we have:
Category:Substance
Mass of Mole of Substance

== Theorem ==
Let $S$ be a substance with molecular weight $W_S$.
Then one mole of $S$ has a mass of $W_S$ grams.
== Proof ==
Let $m_S$ grams be the mean mass of one molecule of $S$.
Let $m_C$ grams be the mass of one atom of carbon-12.
Let $M_S$ grams be the mass of one mole of $S$.
Let $M_C$ grams be the mass of one mole of carbon-12.
Then:
Before $2018$, the mole was defined as the number of atoms in $12$ grams of carbon-12.
While the mole is no longer so defined, the mass of $1$ mole of carbon-12 is still $12$ grams, for all practical purposes.
Thus we have:
Category:Substance