nyuuzyou/wikis huggingface dataset:
Non-Finite Cardinal is equal to Cardinal Product

== Theorem ==
Let $\omega$ denote the minimally inductive set.
Let $x$ be an ordinal such that $x \ge \omega$.
Then:
:$\card x = \card {x \times x}$
where $\times$ denotes the Cartesian product.
=== Corollary ===
== Proof ==
The proof shall proceed by Transfinite Induction on $x$.
Let:
:$\forall y \in x: y < \omega \lor \card y = \card {y \times y}$
There are two cases:
=== Case 1: $\card y = \card x$ for some $y \in x$ ===
If this is so, then:
=== Case 2: $\card y < \card x$ for all $y \in x$ ===
We have that either:
:$y < \omega$
or:
:$\card y = \card {y + 1}$
In either case, we have that:
:$\card {y + 1} < \card x$
and therefore:
:$y + 1 \in x$
Therefore, $x$ is a limit ordinal.
Let $R_0$ denote the canonical order of $\On^2$.
Let $J_0$ be defined as the unique order isomorphism between $\On^2$ and $\On$ as defined in canonical order.
It follows that:
:$\ds \map {J_0} {x \times x} = \bigcup_{y \mathop \in x} \bigcup_{z \mathop \in x} \map {J_0} {y, z}$
But moreover:
Since $J_0$ is a bijection:
Take $\max \set {y, z}$.
It follows that:
Therefore:
:$\card {\map {R_0^{-1} } {y, z} } < \card x$
Thus by Cardinal Inequality implies Ordinal Inequality:
:$\forall y, z \in x: \map {J_0} {y, z} < \card x$
It follows by Supremum Inequality for Ordinals that:
:$\map {J_0} {x \times x} \subseteq \card x$
Hence:
But also by Set Less than Cardinal Product:
:$\card x \le \card {x \times x}$
Thus:
:$\card x = \card {x \times x}$
== Sources ==
* : $\S 10.33$
Category:Ordinals
Category:Cardinals
Non-Finite Cardinal is equal to Cardinal Product/Corollary

== Corollary to Non-Finite Cardinal is equal to Cardinal Product ==
Let $S$ be a set that is equinumerous to its cardinal number.
Let $\card S$ denote the cardinal number of $S$.
Let:
:$\card S \ge \omega$
where $\omega$ denotes the minimally inductive set.
Then:
:$\card {S \times S} = \card S$
== Proof ==
By hypothesis:
:$S \sim \card S$
By Cartesian Product Preserves Cardinality:
:$S \times S \sim \card S \times \card S$
Therefore:
== Sources ==
* : $\S 10.34$
Category:Cardinals
Non-Forking Types have Non-Forking Completions

== Theorem ==
Let $T$ be a complete $\LL$-theory.
Let $\mathfrak C$ be a monster model for $T$.
Let $A\subseteq B$ be subsets of the universe of $\mathfrak C$.
Let $\map \pi {\bar x}$ be an $n$-type over $B$.
If $\pi$ does not fork over $A$, then there is a complete $n$-type $p$ over $B$ such that $\pi \subseteq p$ and $p$ does not fork over $A$.
== Proof ==
Suppose $\pi$ does not fork over $A$.
We will use Zorn's Lemma to find a candidate for the needed complete type.
Consider the collection $\Pi$ of all non-forking sets $\pi'$ of $\LL$-formulas with parameters from $B$ such that $\pi'$ contains $\pi$.
Order $\Pi$ by subset inclusion.
Since a set forks iff a finite subset forks, the union over any chain is still a non-forking set, and hence is an upper bound for the chain.
Thus, by Zorn's Lemma, there is a maximal (with respect to subset inclusion) $p$ in $\Pi$.
 $p$ is not complete.
:By definition, for some $\map \phi {\bar x, \bar b}$, $p$ contains neither $\map \phi {\bar x, \bar b}$ nor $\neg \map \phi {\bar x, \bar b}$.
:Since $p$ is non-forking, by Formula and its Negation Cannot Both Cause Forking, at least one of $p \cup \map \phi {\bar x, \bar b}$ or $p \cup \neg \map \phi {\bar x, \bar b}$ is non-forking as well.
:Hence $p$ is not maximal in $\Pi$, contradicting the choice of $p$.
Thus $p$ is complete.
Category:Model Theory for Predicate Logic
Non-Greatest Element of Well-Ordered Class has Immediate Successor

== Theorem ==
Let $C$ be a well-ordered class under an ordering $\le$.
Let $x \in C$.
Suppose that $x$ is not the greatest element in $C$.
Then $x$ has an immediate successor element in $C$.
== Proof ==
Let $x$ be an element of $C$ which is not the greatest element of $C$.
Let $S$ be the class of successor elements of $x$ in $C$.
We have that $S$ is a subclass of $C$.
Also, $S$ is non-empty because $x$ is not the greatest element.
Thus $S$ is a non-empty subclass of $C$.
We have that $\le$ is a well-ordering on $C$
Hence $S$ has a smallest element, $y$.
By the definition of $S$:
:$x < y$ 
 that for some $z \in C$, $x < z < y$.
Then by the definition of $S$, $z \in S$.
This contradicts the fact that $y$ is the smallest element of $S$.
Thus $y$ is the immediate successor element of $x$.
== Sources ==
* : Chapter $4$: Superinduction, Well Ordering and Choice: Part $\text I$ -- Superinduction and Well Ordering: $\S 1$ Introduction to well ordering: Proposition $1.4$
Category:Well-Orderings
Category:Successor Elements
Non-Homeomorphic Sets may be Homeomorphic to Subsets of Each Other

== Theorem ==
Let $T_1 = \struct {S_1, \tau_1}$ and $T_2 = \struct {S_2, \tau_2}$ be topological spaces.
Let $H_1 \subseteq S_1$ and $H_2 \subseteq S_2$.
Then it is possible for:
:$(1): \quad T_1$ to be homeomorphic to $H_2$
:$(2): \quad T_2$ to be homeomorphic to $H_1$
but:
:$(3): \quad T_1$ and $T_2$ to not be homeomorphic.
== Proof ==
Let $\struct {\R, \tau_d}$ be the real number line under the usual (Euclidean) topology.
Let $S_1 := \closedint 0 1$ be the closed unit interval.
Let $S_2 := \openint 0 1$ be the open unit interval.
Let $H_1 := \openint 0 1$ and $H_2 := \closedint {\dfrac 1 4} {\dfrac 3 4}$.
Then:
:$(1): \quad \struct {S_1, \tau_d}$ is homeomorphic to $\struct {H_2, \tau_d}$ by the mapping $f: S_1 \to H_2$ defined as $\map f x = \dfrac x 2 + \dfrac 1 4$
:$(2): \quad \struct {S_2, \tau_d}$ is trivially homeomorphic to $\struct {H_1, \tau_d}$
From Continuous Image of Compact Space is Compact, if $S_1$ and $S_2$ are homeomorphic then both must be compact.
From Closed Real Interval is Compact in Metric Space, $S_1$ is compact.
But from Open Real Interval is not Compact, $S_2$ is not compact.
Hence:
:$(3): \quad \struct {S_1, \tau_d}$ and $\struct {S_2, \tau_d}$ are not homeomorphic.
== Sources ==
* : $3$: Continuity generalized: topological spaces: Exercise $3.9: 38$
* : Part $\text {II}$: Counterexamples: $32$. Special Subsets of the Real Line: $7$
Category:Homeomorphisms (Topological Spaces)
Non-Injective Mapping may be Strictly Order-Preserving and Order-Reversing

== Theorem ==
Let $\struct {S, \prec_1}$ and $\struct {T, \prec_2}$ be strictly 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 it is not necessarily the case that $\pi$ is an injection.
== Proof ==
Proof by Counterexample:
Let $S = \set {\O, \set a, \set b, \set {a, b} }$ and $T = \set {1, 2, 3}$.
Let $\prec_1$ be the proper subset relation:
:$\forall x, y \in S: x \prec_1 y \iff x \subsetneq y$
Let $\prec_2$ be the usual strict ordering on the integers $1, 2, 3$:
:$\forall x, y \in T: x \prec_2 y \iff x < y$
Let $\pi = \set {\tuple {\O, 1}, \tuple {\set a, 2}, \tuple {\set b, 2}, \tuple {\set {a, b}, 3} }$.
By inspection it is seen that $\pi: S \to T$ is a mapping which is specifically not an injection, as $\map \pi {\set a} = \map \pi {\set b}$.
Equally by inspection it can be seen that:
:$\forall x, y \in S: x \prec_1 y \iff \map \pi x \prec_2 \map \pi y$
== Sources ==
* : Part $1$: Not Entirely Naive Set Theory: Chapter $2$: Partial Order Relations: Exercise $20 \ \text {(b)}$
Category:Increasing Mappings
Non-Integral Value of 3-Göbel Sequence

== Theorem ==
Consider the $3$-Göbel sequence defined recursively as:
:$x_n = \begin {cases}
1 & : n = 0 \\
\ds \paren {1 + \sum_{k \mathop = 0}^{n - 1} {x_k}^3} / n & : n > 0 \end {cases}$
The smallest $n$ such that $x_n$ is not an integer is $88$.
== Proof ==
== Sources ==
* 
* : $43$
* 
Category:Göbel's Sequence
Category:88
Non-Integral Value of Göbel's Sequence

== Theorem ==
Consider Göbel's sequence defined recursively as:
:$x_n = \begin{cases}
1 & : n = 0 \\
\ds \paren {1 + \sum_{k \mathop = 0}^{n - 1} {x_k}^2} / n & : n > 0 \end{cases}$
The smallest $n$ such that $x_n$ is not an integer is $43$.
== Proof ==
== Sources ==
* 
* : $43$
* 
Category:Göbel's Sequence
Category:43
Non-Meager Linear Subspace of Topological Vector Space is Everywhere Dense

== Theorem ==
Let $\GF \in \set {\R, \C}$.
Let $X$ be a topological vector space over $\GF$.
Let $L$ be a proper non-meager linear subspace of $X$. 
Then $L$ is everywhere dense.
== Proof ==
 that $L$ is not everywhere dense.
Then $L^- \ne X$.
From Closure of Linear Subspace of Topological Vector Space is Linear Subspace, $L^-$ is then a proper closed linear subspace of $X$.
From Proper Closed Linear Subspace of Topological Vector Space is Meager, it follows that $L^-$ is meager. 
From Subset of Meager Set is Meager Set, $L$ is meager.
This is contrary to our assumption that $L$ is non-meager.
So $L$ is everywhere dense.
Category:Meager Spaces
Category:Topological Vector Spaces
Non-Negative Additive Function is Monotone

== Theorem ==
Let $\SS$ be an algebra of sets.
Let $f: \SS \to \overline \R$ be an additive function, that is:
:$\forall A, B \in \SS: A \cap B = \O \implies \map f {A \cup B} = \map f A + \map f B$
If $\forall A \in \SS: \map f A \ge 0$, then $f$ is monotone, that is:
:$A \subseteq B \implies \map f A \le \map f B$
== Proof ==
Let $A \subseteq B$.
Then:
Category:Set Systems
Non-Negative Multiple of Positive Element of C*-Algebra is Positive

== Theorem ==
Let $\struct {A, \ast, \norm {\, \cdot \,} }$ be a $\text C^\ast$-algebra.
Let $x \in A$ be positive. 
Let $\alpha \ge 0$.
Then $\alpha x$ is positive.
=== Corollary ===
== Proof ==
Let $\sigma_A$ denote the spectrum in $A$.
Since $x$ is positive, we have:
:$\map {\sigma_A} x \subseteq \hointr 0 \infty$
From Spectral Mapping Theorem for Polynomials, we have:
:$\map {\sigma_A} {\alpha x} = \set {\alpha t : t \in \map {\sigma_A} x}$
Since $t \in \map {\sigma_A} x$ has $t \ge 0$, we have $\alpha t \ge 0$ for each $t \in \map {\sigma_A} x$.
Hence we hae:
:$\map {\sigma_A} {\alpha x} \subseteq \hointr 0 \infty$
Further from $(\text C^\ast 4)$ in the definition of an involution, we have:
:$\paren {\alpha x}^\ast = \overline \alpha x^\ast = \alpha x$ 
since $x$ is Hermitian and $\alpha$ is real.
Hence $\alpha x$ is Hermitian with $\map {\sigma_A} {\alpha x} \subseteq \hointr 0 \infty$.
Hence $\alpha x$ is positive.
Category:Positive Elements of C*-Algebras
Category:Non-Negative Multiple of Positive Element of C*-Algebra is Positive
Non-Negative Multiple of Positive Element of C*-Algebra is Positive/Corollary

== Theorem ==
Let $\struct {A, \ast, \norm {\, \cdot \,} }$ be a $\text C^\ast$-algebra.
Let $\le$ be the canonical ordering of $A$.
Let $x \in A$ be positive. 
Let $\alpha, \beta \in \R$ have $\alpha \le \beta$.
Then we have:
:$\alpha x \le \beta x$
== Proof ==
Since $\alpha \le \beta$, we have that $\beta - \alpha \ge 0$ and hence that $\paren {\beta - \alpha} x$ is positive.
Hence $\beta x - \alpha x$ is positive.
Hence we have $\alpha x \le \beta x$ by the definition of $\le$. 
Category:Non-Negative Multiple of Positive Element of C*-Algebra is Positive
Non-Negative Scalar Multiple of Seminorm on Vector Space is Seminorm

== Theorem ==
Let $\GF \in \set {\R, \C}$. 
Let $X$ be a vector space over $\GF$.
Let $p$ be a seminorm on $X$.
Let $\alpha \in \R_{\ge 0}$.
Let $q = \alpha p$. 
Then $q$ is a seminorm on $X$.
== Proof ==
=== ===
Let $x \in X$ and $k \in \GF$. 
We have:
=== ===
Let $x, y \in X$. 
Then we have: 
Category:Seminorms
Non-Negative Signed Measure is Measure

== Theorem ==
Let $\struct {X, \Sigma}$ be a measurable space.
Let $\mu$ be a signed measure on $\struct {X, \Sigma}$ such that: 
:$\map \mu A \ge 0$
for each $A \in \Sigma$. 
Then $\mu$ is a measure on $\struct {X, \Sigma}$. 
== Proof ==
We verify each of the conditions given in the definition of a measure.
From the definition of a signed measure, $\mu$ is a function $\Sigma \to \overline \R$.
=== Proof of $(1)$ ===
By hypothesis we have: 
:$\map \mu A \ge 0$ 
for each $A \in \Sigma$, so condition $(1)$ is satisfied.
=== Proof of $(2)$ ===
From the definition of a signed measure, we also have that $\mu$ is countably additive, so condition $(2)$ is satisfied.
=== Proof of $(3')$ ===
Finally, from the definition of a signed measure, we have: 
:$\map \mu \O = 0$
So $\mu$ is a measure.
Category:Signed Measures
Non-Null Cauchy Sequence in Non-Archimedean Norm is Eventually Stationary

== Theorem ==
Let $\struct {R, \norm{\,\cdot\,} }$ be a non-Archimedean normed division ring with zero $0_R$
Let $\sequence {x_n}$ be a Cauchy sequence such that $\sequence {x_n}$ does not converge to $0_R$.
Then:
:$\exists N \in \N: \forall n, m \ge N: \norm {x_n} = \norm {x_m}$
=== Corollary ===
== Proof ==
By Cauchy Sequence Is Eventually Bounded Away From Non-Limit then:
:$\exists N_1 \in \N$ and $C \in \R_{\gt 0}: \forall n \ge N_1: \norm {x_n} \gt C$
Since $\sequence {x_n}$ is a Cauchy sequence then:
:$\exists N_2 \in \N: \forall n, m \ge N_2: \norm {x_n - x_m} < C$
Let $N = \max \set {N_1, N_2}$.
Let $n, m \ge N$.
Then:
:$\norm {x_n - x_m} < C < \norm {x_n}$
By Corollary to Three Points in Ultrametric Space have Two Equal Distances then:
:$\norm {x_n} = \norm {x_m}$
The result follows.
== Sources ==
* : $\S 3.2$ Completions: Lemma $3.2.10$
Category:Normed Division Rings
Category:Non-Null Cauchy Sequence in Non-Archimedean Norm is Eventually Stationary
Non-Null Cauchy Sequence in Non-Archimedean Norm is Eventually Stationary/P-adic Norm

== Theorem ==
Let $p$ be a prime number.
Let $\struct {\Q_p, \norm {\,\cdot\,}_p}$ be the $p$-adic numbers.
Let $\sequence {x_n}$ be a Cauchy sequence such that $\sequence {x_n}$ does not converge to $0$.
Then:
:$\exists N \in \N: \forall n, m \ge N: \norm {x_n}_p = \norm {x_m}_p$
== Proof ==
From:
:P-adic Numbers form Non-Archimedean Valued Field
:Non-Null Cauchy Sequence in Non-Archimedean Norm is Eventually Stationary 
it follows that:
:$\exists N \in \N: \forall n, m \ge N: \norm {x_n}_p = \norm {x_m}_p$
== Sources ==
* : $\S 1.4$ The field of $p$-adic numbers $\Q_p$
Category:Non-Null Cauchy Sequence in Non-Archimedean Norm is Eventually Stationary
Non-Palindromes in Base 2 by Reverse-and-Add Process

== Theorem ==
Let the number $22$ be expressed in binary: $10110_2$.
When the reverse-and-add process is performed on it repeatedly, it never becomes a palindromic number.
== Proof ==
It remains to be shown that a binary number of this form does not become a palindromic number.
Let $d_n$ denote $n$ repetitions of a binary digit $d$ in a number.
Thus:
:$10111010000$
can be expressed as:
:$101_3010_4$
Beware that the subscript, from here on in, does not denote the number base.
It is to be shown that the reverse-and-add process applied to:
:$101_n010_{n + 1}$
leads after $4$ iterations to:
:$101_{n + 1}010_{n + 2}$
Thus:
As neither $101_n010_{n + 1}$ nor $101_{n + 1}010_{n + 2}$ are palindromic numbers, nor are any of the intermediate results, the result follows.
== Historical Note ==
== Sources ==
* : $196$
* 
* : $196$
Category:Recreational Mathematics
Category:Palindromic Numbers
Category:Reverse-and-Add
Non-Palindromes in Base 2 by Reverse-and-Add Process/Historical Note

== Historical Note on Non-Palindromes in Base 2 by Reverse-and-Add Process ==
This result, according to in his $1986$ book , was given by , but the source for this has not been identified.
Independently of 's work, and , in the context of a children's mathematics club in , noted the reverse-and-add behaviour of $1010100_2$.
Their subsequent report in Volume $78$, issue $483$ of was used as the basis of the analysis given here in , but it needs to be pointed out that there are a number of mistakes in that source.
== Sources ==
* : $196$
* : $196$
Category:Historical Notes
Non-Palindromes in Base 2 by Reverse-and-Add Process/Mistake

== Source Work ==
:
:The Dictionary
::$43$
== Mistake ==
:In base $2$, $43 = 101011$. This base $2$ number never becomes a palindrome by the reverse-and-add process.
This is false:
The mistake does not originate with . It comes from the original article in from which he lifted the result without checking it:
:Finally, remember that we started with the base $2$ number $101011$, which is $43$ in base $10$. In base $19$, $43$ palindromises in just one step: $43 + 34 = 77$. Palindromising is a property of the expansion, not the number.
For a start, the article in question actually begins with the number $1010100_2$, which is $84_{10}$. However, its reverse-and-add sequence results in:
:$84 \to 132 \to 363$
so the argument continues to hold water.
Secondly, what is that "base $19$" doing there? Surely a misprint for base $10$, which works perfectly well.
== Sources ==
* 
* : $43$
Category:Mistakes/Curious and Interesting Numbers
Non-Prime Euclid Number

== Sequence ==
The sequence of Euclid numbers which are not prime begins:
:$30 \, 031, 510 \, 511, 9 \, 699 \, 691, 223 \, 092 \, 871, 6 \, 469 \, 693 \, 231, 7 \, 420 \, 738 \, 134 \, 811, \ldots$
== Sources ==
* : $30,031$
Category:Euclid Numbers
Non-Reflexive Relation/Examples/Arbitrary Non-Reflexive Relation 1

== Example of Non-Reflexive Relation ==
Let $V_1 = \set {y, z}$.
Let $S$ be the relation on $V_1$ defined as:
:$S = \set {\tuple {y, y}, \tuple {y, z} }$
Then $S$ is neither:
:a reflexive relation, as $\tuple {z, z} \notin S$
nor:
:an antireflexive relation, as $\tuple {y, y} \in S$
Thus $S$ is a non-reflexive relation.
== Sources ==
* : Appendix $\text{A}.2$: Cartesian Products and Relations
Category:Examples of Non-Reflexive Relations
Non-Reflexive Relation/Examples/Arbitrary Non-Reflexive Relation 2

== Example of Non-Reflexive Relation ==
Let $S = \set {a, b}$.
Let $\RR$ be the relation on $S$ defined as:
:$\RR = \set {\paren {a, a} }$
Then $\RR$ is neither:
:a reflexive relation, as $\tuple {b, b} \notin \RR$
nor:
:an antireflexive relation, as $\tuple {a, a} \in \RR$
Thus $\RR$ is a non-reflexive relation.
Category:Examples of Non-Reflexive Relations
Non-Square Positive Integers not Sum of Square and Prime

== Conjecture ==
The sequence of (strictly) positive integers which are not square and not the sum of a square and a prime is believed to be complete:
:$10, 34, 58, 85, 91, 130, 214, 226, 370, 526, 706, 730, 771, 1255, 1351, 1414, 1906, 2986, 3676, 9634, 21679$
== Progress ==
From Square of n such that 2n-1 is Composite is not Sum of Square and Prime, $n^2$ is the sum of a square and a prime $2 n - 1$ composite.
Hence the question is specifically about non-squares.
No prime number is in this sequence, as trivially:
:$p = p + 0^2$
and so is the sum of a prime (itself), and $0^2$, which is square.
Each non-square composite $n \in \Z$ can be tested by subtracting successive squares less than $n$ and investigating whether a prime can result.
In the following, the smallest $m$ such that $n - m^2 = p$ is shown where such a $p$ exists.
Otherwise the nonexistence of such a $p$ is demonstrated explicitly.
As follows:
$10$ cannot be expressed as $10 = m^2 + p$.
Thus $10$ is seen to be in this sequence.
$34$ cannot be expressed as $34 = m^2 + p$.
Thus $34$ is seen to be in this sequence.
Similarly:
$58$ cannot be expressed as $58 = m^2 + p$.
Thus $58$ is seen to be in this sequence.
This establishes the pattern.
The algorithm for determining whether a particular $n$ belongs to this sequence can be defined in pseudocode as follows:
For n := 1, loop indefinitely, incrementing by 1:
 Is n prime? If so, continue to the next n
 Is n square? If so, continue to the next n
 For m := 1, incrementing by 1 until m^2 > n:
 Is n - n^2 prime? If so, continue to the next n
 Next m
 Add n to the sequence
Next n
== Examples ==
== Historical Note ==
Category:Numbers not Sum of Square and Prime
Category:Unproven Hypotheses
Non-Square Positive Integers not Sum of Square and Prime/Examples/10

== Example of Non-Square Positive Integers not Sum of Square and Prime ==
$10$ cannot be expressed as the sum of a square and a prime.
== Proof ==
Testing each $m \in \Z_{>0}$ such that $m^2 < 10$ it is established that there is no solution to $10 - m^2 = p$ where $p$ is prime:
Category:Numbers not Sum of Square and Prime
Category:10
Non-Square Positive Integers not Sum of Square and Prime/Examples/34

== Example of Non-Square Positive Integers not Sum of Square and Prime ==
$34$ cannot be expressed as the sum of a square and a prime.
== Proof ==
Testing each $m \in \Z_{>0}$ such that $m^2 < 34$ it is established that there is no solution to $34 - m^2 = p$ where $p$ is prime:
Category:Numbers not Sum of Square and Prime
Category:34
Non-Square Positive Integers not Sum of Square and Prime/Examples/58

== Example of Non-Square Positive Integers not Sum of Square and Prime ==
$58$ cannot be expressed as the sum of a square and a prime.
== Proof ==
Testing each $m \in \Z_{>0}$ such that $m^2 < 58$ it is established that there is no solution to $58 - m^2 = p$ where $p$ is prime:
Category:Numbers not Sum of Square and Prime
Category:58
Non-Successor Element of Peano Structure is Unique

== Theorem ==
Let $\struct {P, s, 0}$ be a Peano structure.
Then:
:$P \setminus s \sqbrk P$ is a singleton
where:
:$\setminus$ denotes set difference
:$s \sqbrk P$ denotes the image of the mapping $s$.
It follows that the non-successor element $0$ is the only element of $P$ with this property.
== Proof ==
Let $T = P \setminus s \sqbrk P$.
From Axiom $(\text P 4)$ we know that $T \ne \O$.
Now suppose that $t_1 \in T$ and $t_2 \in T$.
 $t_1 \ne t_2$.
Define $A = P \setminus \set {t_2}$.
Thus $t_1 \in A \ne P$.
Moreover, by the nature of $t_2$:
:$x \in A \implies \map s x \in A$
Thus, by the induction axiom $(\text P 5)$, $A = P$.
From this contradiction it follows that $P \setminus s \sqbrk P$ cannot contain two different elements.
== Sources ==
* : Introduction $\S 4$: The natural numbers
Category:Peano's Axioms
Non-Trivial Annihilator Contains Positive Integer

== Theorem ==
Let $\left({R, +, \times}\right)$ be a ring with unity.
Let $A = \operatorname{Ann} \left({R}\right)$ be the annihilator of $R$.
Let $a \in A$ such that $a \ne 0$.
Then $A$ contains at least one strictly positive integer.
== Proof ==
Let the zero of $R$ be $0_R$ and the unity of $R$ be $1_R$.
First we note that:
So:
: $a \in A \implies -a \in A$
But if $a \ne 0$ then either $a > 0$ or $-a > 0$.
That is, either $a$ or $-a$ is positive.
So, if $\operatorname{Ann} \left({F}\right)$ contains at least one non-zero element, it contains at least one strictly positive integer.
Category:Rings with Unity
Non-Trivial Arc-Connected Space is Uncountable

== Definition ==
Let $T$ be a topological space consisting of more than one point.
Let $T$ be arc-connected.
Then $T$ is uncountable.
== Proof ==
From Closed Interval in Reals is Uncountable, the closed unit interval $\closedint 0 1$ consists of an uncountable number of elements.
From the definition of arc-connected, every pair of points in $T$ is either end of the image of an injection from that uncountable set.
From Domain of Injection Not Larger than Codomain it follows that $T$ has a cardinality at least as great as $\closedint 0 1$.
Suppose $T$ were countable.
Then from Subset of Countably Infinite Set is Countable it would follow that the image of an arc in $T$ would also be countable.
But as shown above, it is seen to be uncountable.
So $T$ can not be countable, and is therefore uncountable.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $4$: Connectedness
Category:Arc-Connected Spaces
Category:Uncountable Sets