nyuuzyou/wikis huggingface dataset:
Strict Weak Ordering Induces Partition

== Theorem ==
Let $\struct {S, \prec}$ be a relational structure such that $\prec$ is a strict weak ordering on $S$.
Then $S$ can be partitioned into equivalence classes whose equivalence relation is "is non-comparable".
That is, each of the partitions $A$ of $S$ is a relational structure $\struct {\mathbb S, <}$ such that:
:$\mathbb S$ is the set of these partitions of $S$;
:$<$ is the strict total ordering on $\mathbb S$ induced by $\prec$.
== Proof ==
From the definition of strict weak ordering, we define the symbol $\Bumpeq$ as:
:$a \Bumpeq b := \neg a \prec b \land \neg b \prec a$
that is, $a \Bumpeq b$ means "$a$ and $b$ are non-comparable".
Checking in turn each of the criteria for equivalence:
=== Reflexive ===
As $\prec$ is antireflexive, by definition $\forall a \in S: \neg a \prec a$.
Hence by the Rule of Idempotence $\neg a \prec a \land \neg a \prec a$ and so $\forall a \in S: a \Bumpeq a$.
=== Symmetric ===
We have that $a \Bumpeq b$ is defined as being $\neg a \prec b \land \neg b \prec a$.
It follows from the Rule of Commutation that $\neg b \prec a \land \neg a \prec b$, and so $b \Bumpeq a$.
=== Transitive ===
The relation $a \Bumpeq b$ is defined as being transitive in a strict weak ordering.
Category:Order Theory
Category:Equivalence Relations
Strict Well-Ordering Isomorphic to Unique Ordinal under Unique Mapping

== Theorem ==
Let $S$ be a set.
Let $\struct {S, \prec}$ be a strict well-ordering.
Then there exists a unique ordinal $x$ and unique mapping $f$ such that $f: x \to S$ is an order isomorphism.
== Proof ==
The existence of $x$ and $f$ follows from the Counting Theorem.
The uniqueness of $x$ follows from the Counting Theorem.
The uniqueness of $f$ follows from Order Isomorphism between Wosets is Unique.
== Sources ==
* : $\S 7.51$
Category:Ordinals
Category:Well-Orderings
Strictly Decreasing Mapping is Decreasing

== Theorem ==
A mapping that is strictly decreasing is a decreasing mapping.
== Proof ==
Let $\struct {S, \preceq_1}$ and $\struct {T, \preceq_2}$ be ordered sets.
Let $\phi: \struct {S, \preceq_1} \to \struct {T, \preceq_2}$ be strictly decreasing.
From Strictly Precedes is Strict Ordering:
:$x \preceq_1 y \iff x = y \lor x \prec_1 y$
So:
This leaves us with:
== Also see ==
* Strictly Increasing Mapping is Increasing
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 14$: Orderings
Category:Decreasing Mappings
Strictly Increasing Infinite Sequence of Integers is Cofinal in Natural Numbers

== Theorem ==
Let $S = \left\langle{x_n}\right\rangle$ be an infinite sequence of integers which is strictly increasing.
 
Then $S$ is a cofinal subset of $\left({\Z, \le}\right)$ where $\le$ is the usual ordering on the integers.
== Proof ==
Category:Integers
Strictly Increasing Infinite Sequence of Positive Integers is Cofinal in Natural Numbers

== Definition ==
Let $S = \sequence {x_n}$ be an infinite sequence of positive integers which is strictly increasing.
 
Then $S$ is a cofinal subset of $\struct {\N, \le}$ where $\le$ is the usual ordering on the natural numbers.
== Proof ==
By definition, $S = \sequence {x_n}$ is a subset of $\N$.
The result follows from Subset of Natural Numbers is Cofinal iff Infinite.
Category:Integers
Category:Natural Numbers
Strictly Increasing Mapping Between Wosets Implies Order Isomorphism

== Theorem ==
Let $J$ and $E$ be well-ordered sets.
Let there exist a mapping $k: J \to E$ which is strictly increasing.
Then $J$ is order isomorphic to $E$ or an initial segment of $E$.
== Proof ==
If the sets considered are empty or singletons, the theorem holds vacuously or trivially.
Suppose $J, E$ both have at least two elements.
Let $e_0 = \min E$, the smallest element of $E$.
Define the mapping:
:$h: J \to E$:
:$\map h \alpha = \begin {cases} \map \min {E \setminus h \sqbrk {S_\alpha} } & : h \sqbrk {S_\alpha} \ne E \\ e_0 & : h \sqbrk {S_\alpha} = E \end {cases}$
where $S_\alpha$ is the initial segment determined by $\alpha$ and $h \sqbrk {S_\alpha}$ is the image of $S_\alpha$ under $h$.
By the Principle of Recursive Definition for Well-Ordered Sets, this construction is well-defined and uniquely determined.
Observe that:
This equality also holds if $\map h \alpha = e_0$, by Initial Segment Determined by Smallest Element is Empty.
We claim that $\map h \alpha \preceq \map k \alpha$ for all $\alpha \in J$.
 there is some $a \in J$ such that $\map h a \not \preceq \map k a$.
Then $\map k a \prec \map h a$ by the trichotomy law.
Because $\map h a$ has an element preceding it, $\map h a \ne e_0$.
Thus $\map k a \prec \map \min {E \setminus h \sqbrk {S_a} }$ by the construction of $k$.
Then $\map k a \in h \sqbrk {S_a}$, because it precedes the smallest element that isn't in $h \sqbrk {S_a}$.
Recall that $h \sqbrk {S_a} = S_{\map h a}$.
Then $\map k a \in S_{\map h a}$.
This implies that $\map h a \prec \map k a$, contradicting the assumption that $\map h a \not \preceq \map k a$
From this contradiction we can conclude:
:$\map h \alpha \preceq \map k \alpha$ for all $\alpha \in J$.
 there is some $\alpha \in J$ such that $h \sqbrk {S_\alpha} = E$.
Recall that $h \sqbrk {S_\alpha} = S_{\map h \alpha}$.
Thus, were such an $\alpha$ to exist, then it would succeed all elements in $E$.
It particular, it would also succeed $\map k \alpha$.
But we showed above that $\map h \alpha \preceq \map k \alpha$.
From this contradiction we see that there cannot be any $\alpha \in J$ with $h \sqbrk {S_\alpha} = E$.
Thus the definition of $h$ can be simplified:
:$h: J \to E$:
:$\map h \alpha = \map \min {E \setminus h \sqbrk {S_\alpha} }$
Then the hypotheses of Characterization of Strictly Increasing Mapping on Woset are satisfied.
Thus $h$ is a strictly increasing mapping and its image is $E$ or an initial segment of $E$.
From Strictly Monotone Mapping with Totally Ordered Domain is Injective, $h$ is also injective.
From Injection to Image is Bijection, $h$ is also bijective to its image.
We conclude that there is an order isomorphism from $J$ to $E$, or from $J$ to an initial segment of $E$.
== Sources ==
* : Supplementary Exercises $1.3$
Category:Well-Orderings
Strictly Increasing Mapping is Increasing

== Theorem ==
A mapping that is strictly increasing is an increasing mapping.
== Proof ==
Let $\struct {S, \preceq_1}$ and $\struct {T, \preceq_2}$ be ordered sets.
Let $\phi: \struct {S, \preceq_1} \to \struct {T, \preceq_2}$ be strictly increasing.
From Strictly Precedes is Strict Ordering:
:$x \preceq_1 y \implies x = y \lor x \prec_1 y$
So:
This leaves us with:
== Also see ==
* Strictly Decreasing Mapping is Decreasing
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 14$: Orderings
Category:Increasing Mappings
Strictly Increasing Mapping on Well-Ordered Class

== Theorem ==
Let $\struct {S, \prec}$ be a strictly well-ordered class.
Let $\struct {T, <}$ be a strictly ordered class.
Let $f$ be a mapping from $S$ to $T$.
For each $i \in S$ such that $i$ is not maximal in $S$, let:
: $\map f i < \map f {\map \Succ i}$
where $\map \Succ i$ is the immediate successor element of $i$.
Let:
:$\forall i, j \in S: i \preceq j \implies \map f i \le \map f j$
Then for each $i, j \in S$ such that $i \prec j$:
:$\map f i < \map f j$
== Proof ==
By Non-Greatest Element of Well-Ordered Class has Immediate Successor, $\map \Succ i$ is guaranteed to exist.
Let $i \prec j$.
Let $S_i := \set {q \in S: i \prec q}$.
Then $\map \Succ i$ is the minimal element of $S_i$.
By supposition, $j \in S_i$.
Thus:
:$j \nprec \map \Succ i$
Since Well-Ordering is Total Ordering:
:$\map \Succ i \preceq j$
Thus by supposition:
:$\map f {\map \Succ i} \le \map f j$
Since $\map f i < \map f {\map \Succ i}$:
:$\map f i < \map f j$
by transitivity.
Category:Class Mappings
Category:Increasing Mappings
Strictly Increasing Sequence induces Partition

== Theorem ==
Let $\sequence {r_k}_{0 \mathop \le k \mathop \le n}$ be a strictly increasing finite sequence of natural numbers.
Let:
:$\forall k \in \closedint 1 n: A_k := \closedint {r_{k - 1} + 1} {r_k}$
Then:
:$\set {A_k: k \in \closedint 1 n}$ is a partition of $\closedint {r_0 + 1} {r_n}$.
== Proof ==
First we show that the elements of $\set {A_k: k \in \closedint 1 n}$ are disjoint.
Let $j \in \closedint 1 n$.
We have that:
:$\sequence {r_k}_{0 \mathop \le k \mathop \le n}$ is strictly increasing
and:
:$0 \le j - 1 < j \le n$
Thus by Sum with One is Immediate Successor in Naturally Ordered Semigroup:
:$r_0 \le r_{j - 1} < r_j \le r_n \implies \paren {r_0 + 1} \le \paren {r_{j - 1} + 1} \le r_j \le r_n$
So:
:$\O \subset A_j \subseteq \closedint {r_0 + 1} {r_n}$
Because $\sequence {r_k}_{0 \mathop \le k \mathop \le n}$ is strictly increasing:
:$1 \le j < k \le n \implies A_j \cap A_k = \O$
So the elements of $\set {A_k: k \in \closedint 1 n}$ are disjoint, as we were to show.
It remains to be established that:
:if $m \in \closedint {r_0 + 1} {r_n}$
then:
:$m \in A_k$ for some $k \in \closedint 1 n$.
Let $m \in \closedint {r_0 + 1} {r_n}$.
Consider the set:
:$J = \set {j \in \closedint 0 n: m \le r_j}$
Clearly $j \ne \O$ as $n \in J$.
Let $k$ be the smallest element of $J$.
Then because $r_0 < m$:
:$k \ne 0$
Thus:
:$k \in \closedint 1 n$
By its definition:
:$r_{k - 1} < m \le r_k$
Thus, by Sum with One is Immediate Successor in Naturally Ordered Semigroup:
:$r_{k - 1} + 1 \le m \le r_k$
Therefore $m \in A_k$.
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 18$: Induced $N$-ary Operations: Theorem $18.3$
Category:Increasing Sequences
Strictly Increasing Sequence of Natural Numbers

== Theorem ==
Let $\N_{>0}$ be the set of natural numbers without zero:
:$\N_{>0} = \set {1, 2, 3, \ldots}$
Let $\sequence {n_r}$ be a strictly increasing sequence in $\N_{>0}$.
Then:
:$\forall r \in \N_{>0}: n_r \ge r$
== Proof ==
This is to be proved by induction on $r$.
For all $r \in \N_{>0}$, let $\map P r$ be the proposition $n_r \ge r$.
=== Basis for the Induction ===
When $r = 1$, it follows that $n_1 \ge 1$ as $\N_{>0}$ is bounded below by $1$.
Thus $\map P 1$ is true.
This is our basis for the induction.
=== Induction Hypothesis ===
Now we need to show that, if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.
So our induction hypothesis is that $n_k \ge k$.
Then we need to show that $n_{k + 1} \ge k + 1$.
=== Induction Step ===
This is our induction step:
Suppose that $n_k \ge k$.
Then as $\sequence {n_r}$ is strictly increasing:
:$n_{k + 1} > n_k \ge k$
From Sum with One is Immediate Successor in Naturally Ordered Semigroup, we have:
:$n_{k + 1} > k \implies n_{k + 1} \ge k + 1$
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore $\forall r \in \N_{>0}: n_r \ge r$.
Category:Increasing Sequences
Category:Proofs by Induction
Strictly Increasing Sequence on Ordered Set

== Lemma ==
Let $\struct {S, \preceq}$ be a totally ordered set.
Let $\sequence {r_k}_{p \mathop \le k \mathop \le q}$ be a finite sequence of elements of $\struct {S, \preceq}$.
Then $\sequence {r_k}_{p \mathop \le k \mathop \le q}$ is strictly increasing :
:$\forall k \in \closedint {p + 1} q: r_{k - 1} \prec r_k$
== Proof ==
Let $\sequence {r_k}_{p \mathop \le k \mathop \le q}$ be strictly increasing.
Because $\forall k \in \N_{>0}: k - 1 < k$, it follows directly that:
:$\forall k \in \closedint {p + 1} q: r_{k - 1} \prec r_k$
For the other direction, we use a Proof by Contraposition.
To that end, suppose $\sequence {r_k}_{p \mathop \le k \mathop \le q}$ is not strictly increasing.
Let $K$ be the set of all $k \in \closedint p q$ such that:
:$\exists j \in \closedint p q$ such that $j < k$ and $r_k \preceq r_j$
The set $K$ is not empty because $\sequence {r_k}_{p \mathop \le k \mathop \le q}$ is not strictly increasing.
As $K \subset \N$ and the latter is well-ordered, then so is $K$.
Thus $K$ has a minimal element $m$.
Thus there exists $j \in \closedint p q$ such that:
:$j < m$
and:
:$r_m \preceq r_j$
Because $m - 1 < m$:
:$j \le m - 1$
and so:
:$m - 1 \notin K$
So:
:$r_m \preceq r_j\prec r_{m - 1}$
Since orderings are transitive, it follows
:$r_m \preceq r_{m - 1}$
From Rule of Transposition it follows that
:$\forall k \in \closedint {p + 1} q: r_{k - 1} \prec r_k \implies \sequence {r_k}_{p \mathop \le k \mathop \le q}$ is strictly increasing.
The result follows.
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 18$: Induced $N$-ary Operations: Lemma for Theorem $18.4$
Category:Increasing Sequences
Category:Proofs by Contraposition
Strictly Inductive Semigroup is Inductive Semigroup

== Theorem ==
Let $\struct {S, \circ}$ be a strictly inductive semigroup.
Then $\struct {S, \circ}$ is an inductive semigroup.
== Proof ==
In accordance with our assertion, let $\struct {S, \circ}$ be a strictly inductive semigroup.
By definition of strictly inductive semigroup, there exists $\beta \in S$ such that the only subset of $S$ containing both $\beta$ and $x \circ \beta$ whenever it contains $x$ is $S$ itself.
That is:
:$\exists \beta \in S: \forall A \subseteq S: \paren {\beta \in S \land \paren {\forall x \in A: x \circ \beta \in A} } \implies A = S$
Let $\alpha = \beta$.
Then we may say that there exist $\alpha, \beta \in S$ such that the only subset of $S$ containing both $\alpha$ and $x \circ \beta$ whenever it contains $x$ is $S$ itself.
That is:
:$\exists \alpha, \beta \in S: \forall A \subseteq S: \paren {\alpha \in A \land \paren {\forall x \in A: x \circ \beta \in A} } \implies A = S$
Hence, by identifying $\alpha$ with $\beta$, $\struct {S, \circ}$ fulfils the conditions of being an inductive semigroup.
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 16$: The Natural Numbers: Exercise $16.9 \ \text {(b)}$
Category:Inductive Semigroups
Category:Strictly Inductive Semigroups
Strictly Maximal Element is Maximal Element

== Theorem ==
Let $\struct {S, \RR}$ be a relational structure.
Let $T \subseteq S$ be a subset of $S$.
Let $m \in T$ be a strictly maximal element of $T$ under $\RR$.
Then $m$ is a maximal element of $T$ under $\RR$.
== Proof ==
Let $m \in T$ be a strictly maximal element of $T$ under $\RR$.
Then by definition:
:$\forall x \in T: \tuple {m, x} \notin \RR$
 $m$ is not a maximal element of $T$ under $\RR$.
Then:
:$\exists y \in T: \tuple {m, y} \in \RR$
such that $y \ne m$.
But this contradicts the assertion that $\tuple {m, y} \notin \RR$.
Hence it cannot be the case that $m$ is not a maximal element of $T$ under $\RR$.
That is:
:$m$ is a maximal element of $T$ under $\RR$.
Category:Maximal Elements
Strictly Minimal Element is Minimal Element

== Theorem ==
Let $\struct {S, \RR}$ be a relational structure.
Let $T \subseteq S$ be a subset of $S$.
Let $m \in T$ be a strictly minimal element of $T$ under $\RR$.
Then $m$ is a minimal element of $T$ under $\RR$.
== Proof ==
Let $m \in T$ be a strictly minimal element of $T$ under $\RR$.
Then by definition:
:$\forall x \in T: \tuple {x, m} \notin \RR$
 $m$ is not a minimal element of $T$ under $\RR$.
Then:
:$\exists y \in T: \tuple {y, m} \in \RR$
such that $y \ne m$.
But this contradicts the assertion that $\tuple {y, m} \notin \RR$.
Hence it cannot be the case that $m$ is not a minimal element of $T$ under $\RR$.
That is:
:$m$ is a minimal element of $T$ under $\RR$.
Category:Minimal Elements
Strictly Monotone Mapping is Monotone

== Theorem ==
A mapping that is strictly monotone is a monotone mapping.
== Proof ==
If $\phi$ is strictly monotone, then it is either strictly increasing or strictly decreasing.
If $\phi$ is strictly increasing, then by Strictly Increasing Mapping is Increasing, $\phi$ is increasing.
If $\phi$ is strictly decreasing, then by Strictly Decreasing Mapping is Decreasing, $\phi$ is decreasing.
Thus $\phi$ is monotone by definition.
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 14$: Orderings
Category:Strictly Monotone Mappings
Category:Monotone Mappings
Strictly Monotone Mapping with Totally Ordered Domain is Injective

== Theorem ==
Let $\struct {S, \preceq_1}$ be a totally ordered set.
Let $\struct {T, \preceq_2}$ be an ordered set.
Let $\phi: \struct {S, \preceq_1} \to \struct {T, \preceq_2}$ be a strictly monotone mapping.
Then $\phi$ is injective.
== Proof ==
Hence the result.
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 14$: Orderings: Theorem $14.9: \ 1^\circ$
Category:Strictly Monotone Mappings
Category:Total Orderings
Category:Injections
Strictly Monotone Real Function is Bijective

== Theorem ==
Let $f$ be a real function which is defined on $I \subseteq \R$.
Let $f$ be strictly monotone on $I$.
Let the image of $f$ be $J$. 
Then $f: I \to J$ is a bijection.
== Proof ==
From Strictly Monotone Mapping with Totally Ordered Domain is Injective, $f$ is an injection.
From Surjection by Restriction of Codomain, $f: I \to J$ is a surjection.
== Sources ==
* : $\S 12.9$
Category:Strictly Monotone Real Functions
Category:Bijections
Strictly Order-Preserving and Order-Reversing Mapping on Strictly Totally Ordered Set is Injection

== Theorem ==
Let $\struct {S, \prec_1}$ and $\struct {T, \prec_2}$ be strictly totally ordered sets.
Let $\phi: S \to T$ be a mapping.
Let $\pi: S \to T$ be a mapping with the property that:
:$\forall x, y \in S: x \prec_1 y \iff \map \pi x \prec_2 \map \pi y$
Then $\pi$ is an injection.
== Proof ==
 $\pi$ is not an injection.
Hence:
:$\exists x, y \in S: \map \pi x = \map \pi y$
As $S$ is strictly totally ordered:
:$x \prec_1 y$ or $y \prec_1 x$
, let $x \prec_1 y$.
Then we have:
:$\map \pi x = \map \pi y$
But :
:$\map \pi x \prec_2 \map \pi y$
Because $\prec_2$ is a strict ordering, it follows that:
:$\map \pi x \ne \map \pi y$
It follows by Proof by Contradiction that it cannot be the case that $\pi$ is not an injection.
 
Hence the result.
== Sources ==
* : Part $1$: Not Entirely Naive Set Theory: Chapter $2$: Partial Order Relations: Exercise $20 \ \text {(c)}$
Category:Increasing Mappings
Category:Total Orderings
Category:Strict Orderings
Strictly Positive Hausdorff Measure implies Infinite Lower Dimensional Measure

== Theorem ==
Let $n \in \N_{>0}$.
Let $F \subseteq \R^n$ be a subset of the real Euclidean space.
Let $\map {\HH^s} \cdot$ denote the $s$-dimensional Hausdorff measure.
Let $s \in \R_{\ge 0}$.
Then:
:$\map {\HH^s} F > 0 \implies \forall t \in \hointr 0 s : \map {\HH^t} F = +\infty$
== Proof ==
Let:
:$\exists t \in \hointr 0 s : \map {\HH^t} F < +\infty$
Then by Finite Hausdorff Measure Implies Zero Higher Dimensional Measure:
:$\map {\HH^s} F = 0$
Hence the result by Proof by Contraposition.
Category:Measure Theory
Category:Fractals
Strictly Positive Integer Power Function Strictly Succeeds Each Element

== Theorem ==
Let $\struct {R, +, \circ, \le}$ be an ordered ring with unity.
Let $\struct {R, \le}$ be a directed set with no upper bound.
Let $n \in \N_{>0}$.
Let $f: R \to R$ be defined by:
:$\forall x \in R: \map f x = \circ^n x$
Then the image of $f$ has elements strictly succeeding each elements of $R$.
== Proof ==
Let $b \in R$.
By Directed Set has Strict Successors iff Unbounded Above:
:$\exists c \in R: b < c$
:$\exists d \in R: 1 < d$
By the definition of a directed set:
:$\exists e \in R: d \le e, c \le e$
$\struct {R, +, \circ, \le}$ is an ordered ring, so $\le$ is by definition a transitive relation.
Hence by transitivity:
:$b < e$
and:
:$1 < e$
By Strictly Positive Power of Strictly Positive Element Greater than One Succeeds Element:
:$e \le \map f e$
Thus by transitivity:
:$b < \map f e$
Category:Ring Theory
Strictly Positive Integer Power Function is Unbounded Above

== Theorem ==
Let $\R$ be the real numbers with the usual ordering.
Let $n \in \N_{>0}$.
Let $f: \R \to \R$ be defined by:
:$\map f x = x^n$
Then $f$ is unbounded above.
=== General Case ===
== Proof ==
If $n = 1$, then $f$ is the identity function.
By the Axiom of Archimedes, the real numbers are unbounded above.
Thus by definition of the identity function:
$f$ is unbounded above.
Let $n \ge 2$.
 that $f$ is bounded above by $b \in \R$.
 suppose that $b > 0$.
Then by the definition of an upper bound:
:$\forall x \in \R: x^n \le b$
Let $x > b$.
Then:
:$\dfrac {x^n - b} {x - b} \le 0$
By the Mean Value Theorem, there exists a point $p$ between $b$ and $x$ such that:
:$\map {f'} p = \dfrac {x^n - b} {x - b}$
By Derivative of Power:
:$\map {f'} p = n p^{n - 1}$
Therefore $n p^{n - 1} \le 0$.
By:
:Power of Strictly Positive Real Number is Positive
and:
:Strictly Positive Real Numbers are Closed under Multiplication it follows that:
:$p > 0 \implies n p^{n - 1} > 0$
But this is impossible because it was previously established that $p \ge b > 0$.
From this contradiction it follows that there can be no such $b$.
Hence the result.
== Also see ==
* Strictly Positive Integer Power Function Strictly Succeeds Each Element
Category:Real Analysis
Strictly Positive Integer Power Function is Unbounded Above/General Case

== Theorem ==
Let $\struct {R, +, \circ, \le}$ be a totally ordered ring with unity.
Suppose that $R$ has no upper bound.
Let $n \in \N_{>0}$.
Let $f: R \to R$ be defined by:
:$\map f x = \circ^n x$
Then the image of $f$ is unbounded above in $R$.
== Proof ==
Let $1_R$ be the unity of $R$.
Let $b \in R$.
We will show that $b$ is not an upper bound of the image of $f$.
Since $R$ is totally ordered and unbounded above, there is an element $c \in R$ such that $b < c$ and $1_R < c$.
By Strictly Positive Power of Strictly Positive Element Greater than One Succeeds Element, $c \le \map f c$.
Thus by transitivity of ordering, $b < \map f c$.
Category:Totally Ordered Rings
Category:Rings with Unity
Strictly Positive Integers have same Cardinality as Natural Numbers

== Theorem ==
Let $\Z_{>0} := \set {1, 2, 3, \ldots}$ denote the set of strictly positive integers.
Let $\N := \set {0, 1, 2, \ldots}$ denote the set of natural numbers.
Then $\Z_{>0}$ has the same cardinality as $\N$.
== Proof ==
Consider the mapping $f: \N \to \Z_{>0}$ defined as:
:$\forall x \in \N: \map f x = x + 1$
Then $f$ is trivially seen to be a bijection.
The result follows by definition of cardinality.
== Sources ==
* : Chapter $1$: General Background: $\S 5$ The continuum problem
Category:Cardinality
Category:Natural Numbers
Category:Integers
Strictly Positive Power of Strictly Positive Element Greater than One Succeeds Element

== Theorem ==
Let $\struct {R, +, \circ, \le}$ be an ordered ring with unity.
Let $x \in R$ with $x > 1$ and $x > 0$.
Let $n \in \N_{>0}$.
Then:
:$\circ^n x \ge x$
== Proof ==
The proof proceeds by induction:
If $n = 1$, then $\circ^n x = x$.
So:
:$\circ^n x \ge x$
Suppose that $\circ^n x \ge x$.
Then since $x > 1$:
:$\circ^n x > 1$
By Product of Positive Element and Element Greater than One:
:$x \circ \paren {\circ^n x} > x$
Hence:
:$\circ^{n + 1} x \ge x$
Category:Ordered Rings
Strictly Positive Rational Numbers are Closed under Addition

== Theorem ==
Let $\Q_{>0}$ denote the set of strictly positive rational numbers:
:$\Q_{>0} := \set {x \in \Q: x > 0}$
where $\Q$ denotes the set of rational numbers.
The algebraic structure $\struct {\Q_{>0}, +}$ is closed in the sense that:
:$\forall a, b \in \Q_{>0}: a + b \in \Q_{>0}$
where $+$ denotes rational addition.
== Proof ==
Let $a$ and $b$ be expressed in canonical form:
:$a = \dfrac {p_1} {q_1}, b = \dfrac {p_2} {q_2}$
where $p_1, p_2 \in \Z$ and $q_1, q_2 \in \Z_{>0}$.
As $\forall a, b \in \Q_{>0}$ it follows that $p_1, p_2 \in \Z_{>0}$.
By definition of rational addition:
:$\dfrac {p_1} {q_1} + \dfrac {p_2} {q_2} = \dfrac {p_1 q_2 + p_2 q_1} {q_1 q_2}$
From Integers form Ordered Integral Domain, it follows that:
== Sources ==
* : Chapter $1$: The Real and Complex Number Systems: Introduction
Category:Rational Addition
Strictly Positive Rational Numbers are Closed under Multiplication

== Theorem ==
:$\forall a, b \in \Q_{>0}: a b \in \Q_{>0}$
== Proof ==
Let $a$ and $b$ be expressed in canonical form:
:$a = \dfrac {p_1} {q_1}, b = \dfrac {p_2} {q_2}$
where $p_1, p_2 \in \Z$ and $q_1, q_2 \in \Z_{>0}$.
As $\forall a, b \in \Q_{>0}$ it follows that $p_1, p_2 \in \Z_{>0}$.
By definition of rational multiplication:
:$\dfrac {p_1} {q_1} \times \dfrac {p_2} {q_2} = \dfrac {p_1 \times p_2} {q_1 \times q_2}$
From Integers form Ordered Integral Domain, it follows that:
== Sources ==
* : Chapter $1$: The Real and Complex Number Systems: Introduction
Category:Rational Multiplication
Category:Algebraic Closure