nyuuzyou/wikis huggingface dataset:
Equivalence of Definitions of Stirling Numbers of the Second Kind

== Theorem ==
=== Definition 1 ===
=== Definition 2 ===
in the sense that the coefficients of the falling factorial powers in the summand are uniquely defined by the given recurrence relation.
== Proof ==
The proof proceeds by induction.
For all $n \in \N_{> 0}$, let $\map P n$ be the proposition:
:the coefficients of the falling factorial powers in the expression $\ds x^n = \sum_k {n \brace k} x^{\underline k}$ are uniquely defined by $\ds {n \brace k} = {n - 1 \brace k - 1} + k {n - 1 \brace k}$
where $\ds {n \brace k} = \delta_{n k}$ where $k = 0$ or $n = 0$.
First the case where $n = 0$ is attended to.
We have:
Thus, in the expression:
:$\ds x^0 = \sum_k {0 \brace k} x^{\underline k}$
we have:
:$\ds {0 \brace 0} = 1$
and for all $k \in \Z_{>0}$:
:$\ds {0 \brace k} = 0$
That is:
:$\ds {0 \brace k} = \delta_{0 k}$
Hence the result holds for $n = 0$.
=== Basis for the Induction ===
$\map P 1$ is the case:
We have:
Then:
Thus, in the expression:
:$\ds x^1 = \sum_k {1 \brace k} x^{\underline k}$
we have:
:$\ds {1 \brace 1} = 1$
and for all $k \in \Z$ where $k \ne 1$:
:$\ds {1 \brace k} = 0$
That is:
:$\ds {1 \brace k} = \delta_{1 k}$
Thus $\map P 1$ is seen to hold.
This is the basis for the induction.
=== Induction Hypothesis ===
Now it needs to be shown that, if $\map P r$ is true, where $r \ge 1$, then it logically follows that $\map P {r + 1}$ is true.
So this is the induction hypothesis:
:The coefficients in the expression $\ds x^r = \sum_k {r \brace k} x^{\underline k}$ are uniquely defined by $\ds {r \brace k} = {r - 1 \brace k - 1} + k {r - 1 \brace k}$
from which it is to be shown that:
:The coefficients in the expression $\ds x^{r + 1} = \sum_k {r + 1 \brace k} x^{\underline k}$ are uniquely defined by $\ds {r + 1 \brace k} = {r \brace k - 1} + k {r \brace k}$
=== Induction Step ===
This is the induction step:
Anticipating the expected result, we use $\ds {r + 1 \brace k}$ to denote the coefficients of the $k$th falling factorial power in the expansion of $x^{r + 1}$.
Thus:
Thus the coefficients of the falling factorial powers are defined by the recurrence relation:
:$\ds {r + 1 \brace k} = {r \brace k - 1} + k {r \brace k}$
as required.
So $\map P r \implies \map P {r + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:for all $n \in \Z_{\ge 0}$, the coefficients of the falling factorial powers in the expression $\ds x^n = \sum_k {n \brace k} x^{\underline k}$ are uniquely defined by:
::$\ds {n \brace k} = {n - 1 \brace k - 1} + k {n - 1 \brace k}$
:where $\ds {n \brace k} = \delta_{n k}$ where $k = 0$ or $n = 0$.
== Also see ==
* Equivalence of Definitions of Unsigned Stirling Numbers of the First Kind
* Equivalence of Definitions of Signed Stirling Numbers of the First Kind
== Sources ==
* : $\S 1.2.6$: Binomial Coefficients: Exercise $35$
Category:Stirling Numbers
Equivalence of Definitions of Stopping Time in Discrete Time

== Theorem ==
Let $\struct {\Omega, \Sigma, \sequence {\FF_n}_{n \ge 0}, \Pr}$ be a filtered probability space.
Let $T : \Omega \to \Z_{\ge 0} \cup \set {\infty}$ be a function.
=== Definition 1 ===
 
=== Definition 2 ===
 
== Proof ==
=== Definition 1 implies Definition 2 ===
Suppose that:
:$\set {\omega \in \Omega : \map T \omega \le t} \in \FF_t$
for all $t \in \Z_{\ge 0}$.
Setting $t = 0$ this certainly implies:
:$\set {\omega \in \Omega : \map T \omega = t} \in \FF_t$
Now take $t \ge 1$ a positive integer.
We have: 
:$\set {\omega \in \Omega : \map T \omega = t} = \set {\omega \in \Omega : \map T \omega \le t} \setminus \set {\omega \in \Omega : \map T \omega \le t - 1}$
with:
:$\set {\omega \in \Omega : \map T \omega \le t} \in \FF_t$
and:
:$\set {\omega \in \Omega : \map T \omega \le t - 1} \in \FF_{t - 1}$
Since $\sequence {\FF_n}_{n \ge 0}$ is a filtration, we have:
:$\FF_{t - 1} \subseteq \FF_t$
and so:
:$\set {\omega \in \Omega : \map T \omega \le t - 1} \in \FF_t$
So from Sigma-Algebra Closed under Set Difference, we have:
:$\set {\omega \in \Omega : \map T \omega = t} \in \FF_t$ 
=== Definition 2 implies Definition 1 ===
Suppose that:
:$\set {\omega \in \Omega : \map T \omega = t} \in \FF_t$
for all $t \in \Z_{\ge 0}$.
We can write: 
:$\ds \set {\omega \in \Omega : \map T \omega \le t} = \bigcup_{0 \le s \le t, \, t \in \Z} \set {\omega \in \Omega : \map T \omega = s}$
We have from hypothesis:
:$\set {\omega \in \Omega : \map T \omega = s} \in \FF_s$
for $0 \le s \le t$. 
Since $\sequence {\FF_n}_{n \ge 0}$ is a filtration, we have:
:$\set {\omega \in \Omega : \map T \omega = s} \in \FF_t$
for $0 \le s \le t$. 
Since $\sigma$-algebras are closed under finite union, we have: 
:$\set {\omega \in \Omega : \map T \omega \le t} \in \FF_t$ 
Category:Stopping Times
Equivalence of Definitions of Strict Ordering

== Theorem ==
Let $S$ be a set.
Let $\RR$ be a relation on $S$.
=== Definition 1 ===
=== Definition 2 ===
== Proof ==
Let $\RR$ be transitive.
Then by Transitive Relation is Antireflexive iff Asymmetric it follows directly that:
:$(1): \quad$ If $\RR$ is antireflexive then it is asymmetric
:$(2): \quad$ If $\RR$ is asymmetric then it is antireflexive.
Category:Strict Orderings
Equivalence of Definitions of Strict Ordering on Integers

== Theorem ==
=== Definition 1 ===
=== Definition 2 ===
== Proof ==
Let $x, y \in \Z$ such that $x < y$.
Let $x$ and $y$ be defined as from the formal definition of integers:
:$x = \eqclass {x_1, x_2} {}$ and $y = \eqclass {y_1, y_2} {}$ where $x_1, x_2, y_1, y_2 \in \N$.
=== $(1)$ implies $(2)$ ===
Let $<$ be a strict ordering on integers by definition $1$.
Then by definition:
:$y - x$ is non-negative 
That is:
:$\map \PP {y - x}$
where $\PP$ is the strict positivity property.
Thus:
:$\eqclass {y_1, y_2} {} - \eqclass {x_1, x_2} {}$ is strictly positive
By definition of integer subtraction:
:$\eqclass {y_1, y_2} {} + \eqclass {x_2, x_1} {}$ is strictly positive
and by the formal definition of integers:
:$\eqclass {y_1 + x_2, y_2 + x_1} {}$ is strictly positive
We have that $y_1 + x_2$ and $y_2 + x_1$ are natural numbers.
Thus by definition of natural number ordering:
:$y_1 + x_2 > y_2 + x_1$
Thus $<$ is a strict ordering on integers by definition $2$.
=== $(2)$ implies $(1)$ ===
Let $<$ be a strict ordering on integers by definition $2$.
Then by definition:
:$x_1 + y_2 < x_2 + y_1$
That is:
:$x_2 + y_1 > x_1 + y_2$
Hence by the formal definition of integers:
:$\eqclass {y_1 + x_2, y_2 + x_1} {}$ is strictly positive
By definition of integer addition:
:$\eqclass {y_1, y_2} {} + \eqclass {x_2, x_1} {}$ is strictly positive
By definition of integer subtraction:
:$\eqclass {y_1, y_2} {} - \eqclass {x_1, x_2} {}$ is strictly positive
That is:
:$\map \PP {y - x}$
where $\PP$ is the strict positivity property.
Thus $\le$ is a strict ordering on integers by definition $1$.
Category:Strict Orderings
Category:Orderings on Integers
Equivalence of Definitions of Strict Well-Ordering

== Theorem ==
=== Definition 1 ===
=== Definition 2 ===
== Proof ==
Let $A$ be a class.
Let $B$ be a non-empty subset of $A$.
Let $\prec$ be a relation on $A$.
=== Definition 1 implies Definition 2 ===
Suppose that $\prec$ is a strictly well-founded strict total ordering of $A$.
By the definition of strict total ordering, $\prec$ connects $A$.
=== Definition 2 implies Definition 1 ===
Suppose that $\prec$ connects $A$ and that $\prec$ is a strictly well-founded relation.
That is:
:for any $x, y \in A$, either $x = y$, $x \prec y$, or $y \prec x$
:whenever $T$ is a non-empty subset of $A$, $T$ has a strictly minimal element under $\prec$ .
$\prec$ is transitive:
Let $x, y, z \in A$.
Let $x \prec y$ and $y \prec z$.
Since $\prec$ connects $A$, either $x \prec z$ or $z \prec x$.
By Strictly Well-Founded Relation has no Relational Loops it is not the case that $x \prec y$, $y \prec z$ and $z \prec x$.
Thus we conclude that $x \prec z$.
As this holds for all such $x, y, z$, $\prec$ is transitive.
$\prec$ is antireflexive by Strictly Well-Founded Relation is Antireflexive.
Thus $\prec$ is a strict total ordering.
Category:Well-Orderings
Equivalence of Definitions of Strictly Concave Real Function

== Theorem ==
Let $f$ be a real function which is defined on a real interval $I$.
=== Definition 1 ===
=== Definition 2 ===
=== Definition 3 ===
== Proof ==
Let $f$ be strictly concave real function on $I$ according to definition 1.
That is:
:$\forall x, y \in I: \forall \alpha, \beta \in \R_{>0}, \alpha + \beta = 1: \map f {\alpha x + \beta y} > \alpha \map f x + \beta \map f y$
Make the substitutions $x_1 = x, x_2 = \alpha x + \beta y, x_3 = y$.
As $\alpha + \beta = 1$, we have $x_2 = \alpha x_1 + \paren {1 - \alpha} x_3$.
Thus:
:$\alpha = \dfrac {x_3 - x_2} {x_3 - x_1}, \beta = \dfrac {x_2 - x_1} {x_3 - x_1}$
So:
Again from $(1)$:
Thus:
:$\dfrac {\map f {x_2} - \map f {x_1} } {x_2 - x_1} > \dfrac {\map f {x_3} - \map f {x_1} } {x_3 - x_1} > \dfrac {\map f {x_3} - \map f {x_2} } {x_3 - x_2}$
So:
:$\dfrac {\map f {x_2} - \map f {x_1} } {x_2 - x_1} > \dfrac {\map f {x_3} - \map f {x_1} } {x_3 - x_1}$
demonstrating that $f$ is strictly concave on $I$ according to definition 3, and:
:$\dfrac {\map f {x_2} - \map f {x_1} } {x_2 - x_1} > \dfrac {\map f {x_3} - \map f {x_2} } {x_3 - x_2}$
demonstrating that $f$ is strictly concave on $I$ according to definition 2.
As each step is an equivalence, the argument reverses throughout.
Category:Concave Real Functions
Equivalence of Definitions of Strictly Convex Real Function

== Theorem ==
Let $f$ be a real function which is defined on a real interval $I$.
=== Definition 1 ===
=== Definition 2 ===
=== Definition 3 ===
== Proof ==
Let $f$ be strictly convex real function on $I$ according to definition 1.
That is:
:$\forall x, y \in I: \forall \alpha, \beta \in \R_{>0}, \alpha + \beta = 1: \map f {\alpha x + \beta y} < \alpha \, \map f x + \beta \, \map f y$
Make the substitutions $x_1 = x, x_2 = \alpha x + \beta y, x_3 = y$.
As $\alpha + \beta = 1$, we have:
:$x_2 = \alpha x_1 + \paren {1 - \alpha} x_3$
Thus:
:$\alpha = \dfrac {x_3 - x_2} {x_3 - x_1}$
:$\beta = \dfrac {x_2 - x_1} {x_3 - x_1}$
So:
Again from $(1)$:
Thus:
:$\dfrac {\map f {x_2} - \map f {x_1} } {x_2 - x_1} < \dfrac {\map f {x_3} - \map f {x_1} } {x_3 - x_1} < \dfrac {\map f {x_3} - \map f {x_2} } {x_3 - x_2}$
So:
:$\dfrac {\map f {x_2} - \map f {x_1} } {x_2 - x_1} < \dfrac {\map f {x_3} - \map f {x_1} } {x_3 - x_1}$
demonstrating that $f$ is strictly convex on $I$ according to definition 3, and:
:$\dfrac {\map f {x_2} - \map f {x_1} } {x_2 - x_1} < \dfrac {\map f {x_3} - \map f {x_2} } {x_3 - x_2}$
demonstrating that $f$ is strictly convex on $I$ according to definition 2.
As each step is an equivalence, the argument reverses throughout.
Category:Convex Real Functions
Equivalence of Definitions of Strictly Inductive Semigroup

== Theorem ==
=== Definition 1 ===
=== Definition 2 ===
=== Definition 3 ===
== Proof ==
=== $(1)$ implies $(2)$ ===
Let $\struct {S, \circ}$ be a strictly inductive semigroup by definition $1$.
Then by definition $1$:
Thus $\struct {S, \circ}$ is a strictly inductive semigroup by definition $2$.
=== $(2)$ implies $(1)$ ===
Let $\struct {S, \circ}$ be a strictly inductive semigroup by definition $2$.
Then by definition $2$:
Thus $\struct {S, \circ}$ is a strictly inductive semigroup by definition $1$.
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 16$: The Natural Numbers: Exercise $16.8$
Category:Strictly Inductive Semigroups
Equivalence of Definitions of Strictly Positive Real Number

== Theorem ==
=== Definition $1$ ===
=== Definition $2$ ===
== Proof ==
=== Definition $(1)$ implies Definition $(2)$ ===
Let $x$ be a strictly positive real number by definition $1$.
Thus $x$ is a strictly positive real number by definition $2$.
=== Definition $(2)$ implies Definition $(1)$ ===
Let $x$ be a strictly positive real number by definition $2$.
Thus $x$ is a strictly positive real number by definition $1$.
Category:Strictly Positive Real Numbers
Equivalence of Definitions of Strictly Progressing Mappings

== Theorem ==
Let $g$ be a mapping.
=== Definition 1 ===
=== Definition 2 ===
== Proof ==
=== $(1)$ implies $(2)$ ===
Let $g$ be a strictly progressing mapping by definition $1$.
Then by definition:
:$\forall x \in \Dom g: x \subsetneqq \map g x$
Hence :
:$\forall x \in \Dom g: x \subseteq \map g x$
and so $g$ is a progressing mapping.
Also :
:$\forall x \in \Dom g: x \ne \map g x$
and so $g$ has no fixed points.
Thus $g$ is a strictly progressing mapping by definition $2$.
=== $(2)$ implies $(1)$ ===
Let $g$ be a strictly progressing mapping by definition $2$.
Then by definition:
:$\forall x \in \Dom g: x \subseteq \map g x$
and also:
:$\forall x \in \Dom g: x \ne \map g x$
That is:
:$\forall x \in \Dom g: x \subsetneqq \map g x$
Thus $g$ is a strictly progressing mapping by definition $1$.
Category:Strictly Progressing Mappings
Equivalence of Definitions of Strictly Well-Founded Relation

== Theorem ==
=== Definition 1 ===
=== Definition 2 ===
=== Definition 3 ===
== Proof ==
=== $(1)$ $(2)$ ===
By definition, $y \in T$ is a strictly minimal element (under $\RR$) of $T$ :
:$\forall z \in T: z \not \mathrel \RR y$
Thus it is seen that definition $1$ means exactly the same as definition $2$.
=== $(1)$ iff $(3)$ ===
Let $\RR$ be a Strictly Well-Founded Relation by definition $1$.
From Strictly Well-Founded Relation is Well-Founded, $\RR$ is a well-founded relation on $S$.
From Strictly Well-Founded Relation is Antireflexive, $\RR$ is antireflexive.
Thus $\RR$ is a Strictly Well-Founded Relation by definition $3$.
=== $(3)$ implies $(2)$ ===
Let $\RR$ be a Strictly Well-Founded Relation by definition $3$.
Then by definition:
:$\RR$ is a well-founded relation on $S$
:$\RR$ is antireflexive.
Thus we have by definition of well-founded relation on $S$:
:$\forall T \subseteq S: T \ne \O: \exists z \in T: \forall y \in T \setminus \set z: \tuple {y, z} \notin \RR$
But because $\RR$ is antireflexive:
:$\tuple {z, z} \notin \RR$
Hence it follows that:
:$\forall T \subseteq S: T \ne \O: \exists z \in T: \forall y \in \paren {T \setminus \set z} \cup \set z: \tuple {y, z} \notin \RR$
That is, from Set Difference Union Second Set is Union:
:$\forall T \subseteq S: T \ne \O: \exists z \in T: \forall y \in \paren {T \cup \set z}: \tuple {y, z} \notin \RR$
But as $\set z \subseteq T$, we have from Union with Superset is Superset that $T \cup \set z = T$.
Hence:
:$\forall T \subseteq S: T \ne \O: \exists z \in T: \forall y \in T: \tuple {y, z} \notin \RR$
Thus $\RR$ is a Strictly Well-Founded Relation by definition $2$.
Category:Well-Founded Relations
Equivalence of Definitions of Strictly Well-Ordered Set

== Theorem ==
=== Definition 1 ===
=== Definition 2 ===
== Proof ==
By definition, a strictly totally ordered set is a relational structure such that $\RR$ is a strict total ordering.
By definition, a strict well-ordering $\RR$ is a strict total ordering such that $\RR$ is a strictly well-founded relation.
So $\struct {S, \RR}$ is a relational structure such that $\RR$ is a strict well-ordering on $S$.
Hence both definitions specify:
:a relational structure $\struct {S, \RR}$
such that:
:$\RR$ is a strict total ordering
:$\RR$ is a strictly well-founded relation.
Category:Well-Orderings
Category:Strict Orderings
Equivalence of Definitions of Strongly Locally Compact Space

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
=== Definition 1 ===
=== Definition 2 ===
== Proof ==
=== 1 implies 2 ===
Follows immediately from Topological Closure is Closed.
=== 2 implies 1 ===
Let $x \in S$
Let $U$ be an open neighborhood of $x$.
Let $K \subseteq S$ be a closed compact subspace with $U \subseteq K$.
By Topological Closure is Closed $\overline U$ is closed in $T$.
By Set Closure as Intersection of Closed Sets, the closure $\overline U$ of $U$ in $T$ satisfies $\overline U \subseteq K$.
By Closed Set in Closed Subspace, $\overline U$ is closed in $K$.
By Closed Subspace of Compact Space is Compact, $\overline U$ is compact.
Category:Locally Compact Spaces
Equivalence of Definitions of Submartingale in Discrete Time

== Theorem ==
Let $\struct {\Omega, \Sigma, \sequence {\FF_n}_{n \ge 0}, \Pr}$ be a filtered probability space.
Let $\sequence {X_n}_{n \ge 0}$ be an adapted stochastic process.
=== Definition 1 ===
 
=== Definition 2 ===
 
== Proof ==
=== Definition 1 implies Definition 2 ===
Suppose that:
:$(1): \quad$ $X_n$ is integrable for each $n \in \Z_{\ge 0}$
:$(2): \quad \forall n \in \Z_{\ge 0}: \expect {X_{n + 1} \mid \FF_n} \ge X_n$.
We prove that:
:$\forall n \in \Z_{\ge 0}, \, \forall m \ge n: \expect {X_m \mid \FF_n} \ge X_n$.
Fix $n \in \Z_{\ge 0}$.
We induct on $m$. 
For all $m \in \Z_{\ge 0}$ with $m \ge n$, let $\map P m$ be the proposition:
:$\expect {X_m \mid \FF_n} \ge X_n$ almost surely.
==== Basis for Induction ====
The case $m = n$ is immediate from Conditional Expectation of Measurable Random Variable.
So $\map P n$ is seen to hold.
This is our basis for the induction.
==== Induction Hypothesis ====
Now we need to show that, if $\map P m$ is true, where $m \ge n$, then it logically follows that $\map P {m + 1}$ is true.
So this is our induction hypothesis:
:$\expect {X_m \mid \FF_n} \ge X_n$ almost surely.
Then we need to show:
:$\expect {X_{m + 1} \mid \FF_n} \ge X_n$ almost surely.
==== Induction Step ====
This is our induction step.
From $(2)$, we have: 
:$X_m \le \expect {X_{m + 1} \mid \FF_m}$ almost surely.
So we have by Conditional Expectation is Monotone:
:$\expect {X_m \mid \FF_n} \le \expect {\expect {X_{m + 1} \mid \FF_m} \mid \FF_n}$ almost surely.
Since $n \le m$, we have $\FF_n \subseteq \FF_m$ since $\sequence {\FF_n}_{n \ge 0}$ is a filtration.
Then, by the Tower Property of Conditional Expectation, we have:
:$\expect {\expect {X_{m + 1} \mid \FF_m} \mid \FF_n} = \expect {X_{m + 1} \mid \FF_n}$ almost surely.
So:
:$\expect {X_{m + 1} \mid \FF_n} \ge \expect {X_m \mid \FF_n} \ge X_n$ almost surely.
So $\map P m \implies \map P {m + 1}$ and the result follows by the Principle of Mathematical Induction.
=== Definition 2 implies Definition 1 ===
Suppose that: 
:$(1): \quad$ $X_n$ is integrable for each $n \in \Z_{\ge 0}$
:$(2): \quad \forall n \in \Z_{\ge 0}, \, \forall m \ge n: \expect {X_m \mid \FF_n} \ge X_n$.
We need to prove that:
:$\forall n \in \Z_{\ge 0}: \expect {X_{n + 1} \mid \FF_n} \ge X_n$
This is immediate from setting $m = n + 1$ in $(2)$. 
Category:Submartingales
Equivalence of Definitions of Supermartingale in Discrete Time

== Theorem ==
Let $\struct {\Omega, \Sigma, \sequence {\FF_n}_{n \ge 0}, \Pr}$ be a filtered probability space.
Let $\sequence {X_n}_{n \ge 0}$ be an adapted stochastic process.
=== Definition 1 ===
 
=== Definition 2 ===
 
== Proof ==
=== Definition 1 implies Definition 2 ===
Suppose that:
:$(1): \quad$ $X_n$ is integrable for each $n \in \Z_{\ge 0}$
:$(2): \quad \forall n \in \Z_{\ge 0}: \expect {X_{n + 1} \mid \FF_n} \le X_n$.
We prove that:
:$\forall n \in \Z_{\ge 0}, \, \forall m \ge n: \expect {X_m \mid \FF_n} \le X_n$.
Fix $n \in \Z_{\ge 0}$.
We induct on $m$. 
For all $m \in \Z_{\ge 0}$ with $m \ge n$, let $\map P m$ be the proposition:
:$\expect {X_m \mid \FF_n} \ge X_n$ almost surely.
==== Basis for Induction ====
The case $m = n$ is immediate from Conditional Expectation of Measurable Random Variable.
So $\map P n$ is seen to hold.
This is our basis for the induction.
==== Induction Hypothesis ====
Now we need to show that, if $\map P m$ is true, where $m \ge n$, then it logically follows that $\map P {m + 1}$ is true.
So this is our induction hypothesis:
:$\expect {X_m \mid \FF_n} \le X_n$ almost surely.
Then we need to show:
:$\expect {X_{m + 1} \mid \FF_n} \le X_n$ almost surely.
==== Induction Step ====
This is our induction step.
From $(2)$, we have: 
:$X_m \ge \expect {X_{m + 1} \mid \FF_m}$ almost surely.
So we have by Conditional Expectation is Monotone:
:$\expect {X_m \mid \FF_n} \ge \expect {\expect {X_{m + 1} \mid \FF_m} \mid \FF_n}$ almost surely.
Since $n \le m$, we have $\FF_n \subseteq \FF_m$ since $\sequence {\FF_n}_{n \ge 0}$ is a filtration.
Then, by the Tower Property of Conditional Expectation, we have:
:$\expect {\expect {X_{m + 1} \mid \FF_m} \mid \FF_n} = \expect {X_{m + 1} \mid \FF_n}$ almost surely.
So:
:$X_n \ge \expect {X_m \mid \FF_n} \ge \expect {X_{m + 1} \mid \FF_n}$ almost surely.
So $\map P m \implies \map P {m + 1}$ and the result follows by the Principle of Mathematical Induction.
=== Definition 2 implies Definition 1 ===
Suppose that: 
:$(1): \quad$ $X_n$ is integrable for each $n \in \Z_{\ge 0}$
:$(2): \quad \forall n \in \Z_{\ge 0}, \, \forall m \ge n: \expect {X_m \mid \FF_n} \le X_n$.
We need to prove that:
:$\forall n \in \Z_{\ge 0}: \expect {X_{n + 1} \mid \FF_n} \le X_n$
This is immediate from setting $m = n + 1$ in $(2)$. 
Category:Supermartingales
Equivalence of Definitions of Supremum of Real-Valued Function

== Theorem ==
Let $S \subseteq \R$ be a subset of the real numbers.
Let $f: S \to \R$ be a real function on $S$.
=== Definition 1 ===
=== Definition 2 ===
== Proof ==
=== Definition 1 implies Definition 2 ===
Let $K \in \R$ be a supremum of $f$ by definition 1.
Then from the definition:
:$\text{(a)}: \quad K$ is an upper bound of $\map f x$ in $\R$.
:$\text{(b)}: \quad K \le M$ for all upper bounds $M$ of $f \sqbrk S$ in $\R$.
As $K$ is an upper bound it follows that:
:$(1): \quad \forall x \in S: \map f x \le K$
Now let $\epsilon \in \R_{>0}$.
 $(2)$ were false:
:$\forall x \in S: \map f x \le K - \epsilon$
Then by definition, $K - \epsilon$ is an upper bound of $f$.
But by definition that means $K \le K + \epsilon$.
So by Real Plus Epsilon:
:$K < K$
From this contradiction we conclude that:
:$(2): \quad \exists x \in S: \forall \epsilon \in \R_{>0}: K - \epsilon < \map f x$
Thus $K$ is a supremum of $f$ by definition 2.
=== Definition 2 implies Definition 1 ===
Let $K$ be a supremum of $f$ by definition 2:
:$(1) \quad \forall x \in S: \map f x \le K$
:$(2) \quad \exists x \in S: \forall \epsilon \in \R_{>0}: K - \epsilon < \map f x$
From $(1)$ we have that $K$ is an upper bound of $f$.
 that $K$ is not a supremum of $f$ by definition 1.
Then:
:$\exists M \in \R, M < K: \forall x \in S: \map f x \le M$
Then:
:$\exists \epsilon \in \R_{>0}: M = K - \epsilon$
Hence:
:$\exists \epsilon \in \R_{>0}: \forall x \in S: \map f x \le K - \epsilon$
This contradicts $(2)$.
Thus $K$ is a supremum of $f$ by definition 1.
Category:Real Analysis
Equivalence of Definitions of Surjection

== Theorem ==
=== Definition 1 ===
=== Definition 2 ===
== Proof ==
=== Definition 1 implies Definition 2 ===
Let $f$ be a mapping which fulfills the condition:
:$\forall y \in T: \exists x \in \Dom f: \map f x = y$
From Image is Subset of Codomain:
:$\Img f \subseteq T$
It remains to be proved that:
:$T \subseteq \Img f$
Thus:
Thus by definition of set equality:
:$\Img f = T$
and by definition of image of mapping:
:$f \sqbrk S = T$
Hence $f$ is a surjection by definition 2.
=== Definition 2 implies Definition 1 ===
Let $f$ be a mapping which fulfills the condition:
:$f \sqbrk S = T$
that is:
:$\Img f = T$
Then by definition of set equality:
:$T \subseteq \Img f$
Hence:
So by the definition of the image of $f$:
:$\exists x \in \Dom f: \map f x = y$
Hence $f$ is a surjection by definition 1.
== Sources ==
* : $\S 8$: Functions
* : Chapter $\text I$: Algebraic Structures: $\S 5$: Composites and Inverses of Functions
* : $\S 1.10$: Functions: Remark $10.8 \ \text{(f)}$
* : $\S 21$: The image of a subset of the domain; surjections
* : Appendix $\text{A}.5$: Identity, One-one, and Onto Functions: Proposition $\text{A}.5.1: 2 \ \text{(b, c)}$
Category:Surjections