nyuuzyou/wikis huggingface dataset:
Single Instruction URM Programs/Projection Function

== Theorem ==
The projection functions $\pr_j^k: \N^k \to \N$, defined as:
:$\forall j \in \closedint 1 k: \forall \tuple {n_1, n_2, \ldots, n_k} \in \N^k: \map {\pr_j^k} {n_1, n_2, \ldots, n_k} = n_j$
are each URM computable by a single-instruction URM program.
== Proof ==
The projection functions are computed by the following URM program:
Line		Command
$1$ 		$\map C {j, 1}$
The input $\tuple {n_1, n_2, \ldots, n_j, \ldots, n_k}$ is in $R_1, R_2, \ldots, R_j, \ldots, R_k$ when the program starts.
The program copies $r_j$ to $r_1$ and then stops.
The output $n_j$ is in $R_1$ when the program terminates.
Category:URM Programs
Category:Primitive Recursive Functions
Single Instruction URM Programs/Successor Function

== Theorem ==
The successor function $\Succ: \N \to \N$, defined as:
:$\forall n \in \N: \map \Succ n = n + 1$
is URM computable by a single-instruction URM program.
== Proof ==
The successor function is computed by the following URM program:
Line		Command
$1$ 		$\map S 1$
The input $n$ is in $R_1$ when the program starts.
The program adds $1$ to $r_1$ and then stops.
The output $n + 1$ is in $R_1$ when the program terminates.
Category:URM Programs
Category:Primitive Recursive Functions
Single Instruction URM Programs/Zero Function

== Theorem ==
The zero function $\Zero: \N \to \N$, defined as:
:$\forall n \in \N: \map \Zero n = 0$
is URM computable by a single-instruction URM program.
== Proof ==
The zero function is computed by the following URM program:
Line		Command
$1$ 		$\map Z 1$
This sets the value $0$ into $R_1$ and then stops.
The output $0$ is in $R_1$ when the program terminates.
Category:URM Programs
Category:Primitive Recursive Functions
Single Point Characterization of Simple Closed Contour

== Theorem ==
Let $C$ be a simple closed contour in the complex plane $\C$ with parameterization $\gamma: \closedint a b \to \C$.
Let $t_0 \in \openint a b$ such that $\gamma$ is complex-differentiable at $t_0$.
Let $S \in \set {-1,1}$ and $r \in \R_{>0}$ such that:
:for all $\epsilon \in \openint 0 r$, we have $\map \gamma {t_0} + \epsilon i S \map {\gamma '}{t_0} \in \Int C$
where $\Int C$ denotes the interior of $C$.
If $S = 1$, then $C$ is positively oriented.
If $S = -1$, then $C$ is negatively oriented.
=== Lemma 1 ===
=== Lemma 2 ===
=== Lemma 3 ===
== Proof ==
We show that for all $t_1 \in \openint a b$ where $\gamma$ is complex-differentiable at $t_1$, there exists $r \in \R_{>0}$ such that for all $\epsilon \in \openint 0 r$, it follows that:
:$\map v { t_1, \epsilon} \in \Int C$.
The result then follows by the definitions of positively oriented contour and negatively oriented contour.
By definition of parameterization of contour, there exists $N \in \N$ and a subdivision $\set { c_0 , \ldots , c_N }$ such that $\gamma$ is complex-differentiable at all $t \in \openint {c_k}{ c_{k+1} }$.
Let $k \in \set {1, \ldots, N}$ such that $t_0 \in \openint {c_{k-1} }{ c_{k} }$.
First, suppose $t_1 \in \openint {c_{k-1} }{ c_{k} }$.
From Lemma 1, it follows that there exists $r \in \R_{>0}$ such that for all $\epsilon \in \openint 0 r$:
:$\map v { t_1, \epsilon} \in \Int C$
Set $l := \begin {cases} k+1 & : k < N \\ 1 & : k = N \end {cases}$.
Suppose $t_1 \in \openint {c_{l-1} }{ c_{l} }$.
From Lemma 2, it follows that there exists $r \in \R_{>0}$ such that for all $\epsilon \in \openint 0 r$:
:$\map v { t_1, \epsilon} \in \Int C$
By repeated use of Lemma 2, it follows that for all $n \in \set {k+1, k+2, \ldots, N, 1, 2, \ldots, k-1}$ and for all $t \in \openint {c_{n-1} }{ c_{n} }$, there exists $r \in \R_{>0}$ that fulfills the condition.
Category:Orientation of Complex Contour
Category:Single Point Characterization of Simple Closed Contour
Single Point Characterization of Simple Closed Contour/Lemma 3

== Lemma ==
Let $\tilde \gamma_1 : \closedint a b \to \C$ be an injective piecewise continuously differentiable function such that $\map {\tilde \gamma_1 '}{t} \ne 0$ for all $t \in \openint a b$.
Let $\tilde t_3 , \tilde t_4 \in \openint a b$ with $\tilde t_3 < \tilde t_4$.
Let $\tilde t_5 \in \openint {\tilde t_3}{\tilde t_4}$ such that $\tilde \gamma_1$ is complex-differentiable at all $t \in \openint a b \setminus \set { \tilde t_5 }$.
Let $r_1 \in \R_{>0}$ such that:
:$r_1 = \cmod{ \map {\tilde \gamma_1}{\tilde t_3} - \map {\tilde \gamma_1}{\tilde t_5} } = \cmod { \map {\tilde \gamma_1}{\tilde t_4} - \map {\tilde \gamma_1}{\tilde t_5} }$
Let $\delta \in \R_{>0}$ such that for all $\epsilon \in \openint 0 \delta$:
:$\map {\tilde \gamma_1}{\tilde t_3 + \epsilon} , \map {\tilde \gamma_1}{\tilde t_4 - \epsilon} \in \map { B_{r_1} }{ \map {\tilde \gamma_1}{\tilde t_5} }$
where $\map { B_{r_1} }{ \map {\tilde \gamma_1}{\tilde t_5} }$ denotes an open disk.
Then there exists an injective complex-differentiable function $\tilde \gamma_2 : \closedint a b \to \C$ such that:
:$\map {\tilde \gamma_2} t = \map {\tilde \gamma_1}{t}$ for all $t \in \closedint {a}{\tilde t_3} \cup \closedint {\tilde t_4}{b}$
:$\map {\tilde \gamma_2} t \in \map {B_{r_1} }{ \map {\tilde \gamma_1}{\tilde t_5} }$ for all $t \in \openint{\tilde t_3}{\tilde t_4}$
== Proof ==
Category:Single Point Characterization of Simple Closed Contour
Singleton Class can be Formed from Set

== Theorem ==
Let $V$ be a basic universe.
Let $a \in V$ be a set.
Then the singleton class $\set a$ can be formed, which is a subclass of $V$.
== Proof ==
Using the axiom of specification, let $A$ be the class defined as:
:$A := \set {x: x \in V \land x = a}$
That is:
:$A = \set a$
By the axiom of extension, $\set a$ is the only such class which has $a$ as an element.
== Also see ==
* Definition:Singleton Class
== Sources ==
* : Chapter $2$: Some Basics of Class-Set Theory: $\S 4$ The pairing axiom
Category:Singleton Classes
Singleton Class of Empty Set is Supercomplete

== Theorem ==
Let $\O$ denote the empty set.
Then the singleton $\set \O$ is supercomplete.
== Proof ==
Let $x \in \set \O$ be any element of $\set \O$.
Then it has to be the case that $x = \O$.
Then every element of $\O$ is an element of $\set \O$ vacuously.
That is, $\set \O$ is swelled.
There is one element of $\set \O$, and that is $\O$.
This is a subclass of $\set \O$.
That is, $\set \O$ is transitive.
The result follows by definition of supercomplete class.
== Sources ==
* : Chapter $2$: Some Basics of Class-Set Theory: $\S 4$ The pairing axiom: Note $2$.
Category:Singleton Classes
Category:Empty Set
Category:Supercomplete Classes
Singleton Class of Set is Set

== Theorem ==
Let $x$ be a set.
Then the singleton class $\set x$ is likewise a set.
== Proof ==
Let $x$ and $y$ be sets.
Let $x = y$.
From Doubleton Class of Equal Sets is Singleton Class, the doubleton class $\set {x, y}$ is the singleton class $\set x$.
From the axiom of pairing, the doubleton class $\set {x, y}$ is a set when $x$ and $y$ are sets.
Hence $\set x$ is a set.
== Sources ==
* : Chapter $2$: Some Basics of Class-Set Theory: $\S 4$ The pairing axiom: Corollary $4.1$
Category:Singleton Classes
Category:Singletons
Singleton Classes are Equal iff Sets are Equal

== Theorem ==
Let $a$ and $b$ be sets.
Let $\set a$ and $\set b$ denote the singleton classes of $a$ and $b$.
Then:
:$\set a = \set b \iff a = b$
== Proof ==
Let $a = b$.
Then $\set a$ and $\set b$ contain the same elements.
Hence by the axiom of extension it follows that $\set a = \set b$.
Let $\set a = \set b$.
We have that $a \in \set a$.
As $\set a = \set b$ we also have that $a \in \set b$.
But $b$ is the only element of $\set b$.
Hence it must be the case that $a = b$.
== Sources ==
* : Chapter $2$: Some Basics of Class-Set Theory: $\S 4$ The pairing axiom: Note $3$.
Category:Singleton Classes
Singleton Equality

== Theorems ==
Let $x$ and $y$ be sets.
Then:
:$\set x \subseteq \set y \iff x = y$
:$\set x = \set y \iff x = y$
== Proof ==
Then:
== Sources ==
* : $\S 7.7$
Category:Set Theory
Category:Subsets
Category:Singletons
Singleton Graph is Edgeless

== Theorem ==
The singleton graph $N_1$ is edgeless.
== Proof ==
Recall the definition of $N_1$:
As $N_1$ is simple, the only edges are between distinct vertices.
As $N_1$ has only one vertex, $N_1$ can have no edges.
== Sources ==
* 
Category:Singleton Graph
Category:Edgeless Graphs
Singleton Graph is Unique

== Theorem ==
The singleton graph $N_1$ is unique (up to isomorphism).
== Proof ==
$N_1$ can be expressed as:
:$N_1 := \struct {\set v, \O}$
where:
:$\set v$ is the set of vertices
:$\O$ is the set of edges, which is empty by definition.
Suppose there exists another singleton graph $N_1' = \struct {\set v', \O}$.
Let $\phi: \set v \to \set {v'}$ be the mapping from $N_1$ to $N_1'$ defined as:
:$\map \phi v = v'$
From Mapping from Singleton is Injection, $\phi$ is an injection.
From Mapping to Singleton is Surjection, $\phi$ is an surjection.
Hence $\phi$ is a bijection by definition.
Finally it is noted that:
:for each edge $\set {u, v} \in \map E {N_1}$, there exists an edge $\set {\map \phi u, \map \phi v} \in \map E {N_1'}$
holds vacuously
and:
:for each edge $\set {u, v} \in \map E {N_1'}$, there exists an edge $\set {\map {\phi^{-1} } u, \map {\phi^{-1} } v} \in \map E {N_1}$
also holds vacuously.
Hence $N_1$ is isomorphic to $N_1'$ by definition.
Hence the result.
Category:Singleton Graph
Singleton Partition yields Indiscrete Topology

== Theorem ==
Let $S$ be a set which is not empty.
Let $\PP$ be the (trivial) singleton partition $\set S$ on $S$.
Then the partition topology on $\PP$ is the indiscrete topology.
== Proof ==
By definition, the partition topology on $\PP$ is the set of all unions from $\PP$.
This is (trivially, and from Union of Empty Set) $\set {\O, S}$ which is the indiscrete topology on $S$ by definition.
== Sources ==
* : Part $\text {II}$: Counterexamples: $5$. Partition Topology: $2$
Category:Indiscrete Topology
Category:Partition Topology
Singleton Point is Isolated

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Let $x \in S$.
Then $x$ is an isolated point of the singleton set $\set x$, but not necessarily an isolated point of $T$.
== Proof ==
Let $U \in \tau$ be an open set of $T$ such that $x \in T$.
The fact that such a $U$ exists follows from the fact that:
:from , $S$ is open in $T$
:$x \in S$.
Hence:
:$\set x \subseteq S$
and so from Intersection with Subset is Subset‎:
$\set x \cap U = \set x$
So by definition, $x$ is an isolated point of $\set x$.
From Topological Space is Discrete iff All Points are Isolated, unless $T$ is the discrete space on $S$, not all elements of $T$ are isolated points of $T$.
Category:Isolated Points
Category:Singletons
Singleton Set in Discrete Space is Compact

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space where $\tau$ is the discrete topology on $S$.
Let $x \in S$.
Then $\set x$ is compact.
== Proof 1 ==
== Proof 2 ==
Category:Compact Spaces
Category:Discrete Topology
Category:Singleton Set in Discrete Space is Compact
Singleton Set in Discrete Space is Compact/Proof 1

== Theorem ==
== Proof ==
From Point in Discrete Space is Neighborhood, every point $x \in S$ is contained in an open set $\set x$.
Then from Interior Equals Closure of Subset of Discrete Space we have that $\set x$ equals its closure.
As $\set {\set x}$ is (trivially) an open cover of $\set x$, it follows by definition that $\set x$ is compact.
Category:Singleton Set in Discrete Space is Compact
Singleton Set is Nowhere Dense in Rational Space

== Theorem ==
Let $\struct {\Q, \tau_d}$ be the rational number space under the Euclidean topology $\tau_d$.
Then every singleton subset of $\Q$ is nowhere dense in $\struct {\Q, \tau_d}$.
== Proof ==
Let $x \in \Q$.
By definition of nowhere dense, we need to show that:
:$\paren {\set x^-}^\circ = \O$
where $S^-$ denotes the closure of a set $S$ and $S^\circ$ denotes its interior.
By Real Number is Closed in Real Number Line, $\set x$ is closed in $\struct {\R, \tau_d}$.
$\struct {\Q, \tau_d}$ is a subspace of $\struct {\R, \tau_d}$.
Thus by Closed Set in Topological Subspace, $\set x$ is closed in $\struct {\Q, \tau_d}$.
From Closed Set Equals its Closure, it follows that:
:$\set x^- = \set x$
From Interior of Singleton in Real Number Line is Empty:
:$\set x^\circ = \O$
Hence:
:$\paren {\set x^-}^\circ = \set x^\circ = \O$
Category:Rational Number Space
Category:Singletons
Category:Nowhere Denseness
Singleton Set is not Dense-in-itself

== Theorem ==
Let $T = \left({S, \tau}\right)$ be a topological space.
Let $x \in S$.
Then the singleton set $\left\{{x}\right\}$ is not dense-in-itself.
== Proof ==
From Singleton Point is Isolated, $x$ is isolated in $\left\{{x}\right\}$.
So by definition $\left\{{x}\right\}$ is not dense-in-itself.
Category:Denseness
Category:Singletons
Singleton fulfils Naturally Ordered Semigroup Axioms 1 to 3

== Theorem ==
Let $S$ be a singleton:
:$S = \set s$
for an arbitrary object $s$.
Let $+$ be the operation on $S$ defined as:
:$\forall s \in S: s + s = s$
Let $\le$ be the relation defined on $S$ as:
:$s \le s$
Then the algebraic structure:
:$\struct {S, +, \le}$
is an ordered semigroup which fulfils the axioms:
:
:
:
but:
:does not fulfil 
:$\struct {S, +}$ is not isomorphic to $\struct {\N, +}$.
== Proof ==
Recall the axioms:
=== ===
First it is noted that $\le$ is a reflexive relation.
Hence from Reflexive Relation on Singleton is Well-Ordering:
:$\le$ is a well-ordering.
Hence holds.
=== ===
 holds for $\struct {S, +}$ trivally.
=== ===
We have that:
:$0 \in M$
and:
:$2 \in M$
and trivially holds.
=== ===
We have that $S$ is a singleton.
Hence:
:$\forall a, b \in S: a = b$
and so does not hold.
=== Lack of Isomorphism ===
It remains to demonstrate that $S$ and $\N$ are not isomorphic.
 there exists a (semigroup) isomorphism $\phi$ from $\struct {\N, +}$ to $\struct {S, +}$.
By definition of isomorphism:
:$\phi$ is a homomorphism
:$\phi$ is a bijection.
From Mapping is Constant iff Image is Singleton:
:$\forall n \in \N: \map \phi n = s$
Thus for example:
:$\map \phi 0 = s$
and:
:$\map \phi 1 = s$
and it is immediate that $\phi$ is not an injection.
Hence $\phi$ is not a bijection.
This contradicts our assertion that $\phi$ is an isomorphism.
Hence there can be no such semigroup isomorphism between $\struct {S, +}$ and $\struct {\N, +}$.
Category:Naturally Ordered Semigroup
Singleton in Normed Vector Space is Closed

== Theorem ==
Let $X$ be a vector space over $\R$ or $\C$.
Let $\struct {X, \norm {\, \cdot \,}}$ be a normed vector space.
Let $\set x \subseteq X$ be a singleton.
Then $\set x$ is closed.
== Proof ==
Note that either $\set x = X$ or $\set x \subset X$.
Suppose, $X = \set x$.
Then:
:$X \setminus \set x = \empty$
We have that Empty Set is Open in Normed Vector Space.
$\set x$ is a complement of $\empty$ in $X$.
By definition, $x$ is closed.
Suppose, $X \ne \set x$, i.e. $\set x \subset X$.
Define $U := X \setminus \set x$.
Let $y \in U$.
Let $\epsilon := \norm {x - y}$.
Suppose, $\epsilon > 0$.
By Normed Vector Space is Open in Itself:
:$\forall x \in X : \exists \epsilon \in \R_{>0} : \map {B_\epsilon} x \subseteq X$
Suppose, $z \in \map {B_\epsilon} y$.
Then:
:$\norm {y - z} < \epsilon$.
We have that:
 
Hence, $z \ne x$.
Then $z$ is in the complement of $\set x$ in $X$, i.e. $z \in U$.
In other words:
:$\forall y \in U : \exists \epsilon = \norm {x - y} > 0 : \map {B_\epsilon} x \subseteq U$
So $U$ is open, and $\set x = X \setminus U$ is closed. 
== Sources ==
* : $\S 1.3$: Normed and Banach spaces. Topology of normed spaces
Category:Normed Vector Spaces
Singleton is Chain

== Theorem ==
Let $\struct {S, \preceq}$ be an ordered set.
Let $x \in S$.
Then $\set x$ is a chain of $\struct {S, \preceq}$.
== Proof ==
It suffices to prove that
:$\set x$ is connected
Let $y, z \in \set x$.
By definition of singleton:
:$y = x$ and $z = x$
By definition of reflexivity;
:$y \preceq z$
Thus
:$y \preceq z$ or $z \preceq y$
== Sources ==
* 
Category:Order Theory
Singleton is Connected in Topological Space

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Let $x \in S$.
Then the singleton $\set{x}$ is connected.
== Proof ==
Let $A = \set{x}$.
 
From definition $3$ of a connected set, $A$ is connected in $T$ the subspace $\struct {A, \tau_A}$ is a connected space.
From Topology on Singleton is Indiscrete Topology, $\tau_A$ is the indiscrete topology.
From Indiscrete Space is Connected, $\struct {A, \tau_A}$ is a connected space.
Category:Connected Spaces
Category:Singletons
Singleton is Convex Set

== Theorem ==
Let $V$ be a vector space over $\R$ or $\C$, and let $v \in V$.
Then the singleton $S = \set v$ is a convex set.
== Proof ==
For any $x, y \in S$, we have $x = y = v$.
It follows that:
:$\forall t \in \closedint 0 1: t x + \paren {1 - t} y = v \in S$
Hence $S$ is a convex set.
== Sources ==
* : $\text{I}$ Hilbert Spaces: $\S 2.$ Orthogonality
Category:Vector Spaces
Category:Singletons
Category:Convex Sets (Vector Spaces)
Singleton is Convex Set (Order Theory)

== Theorem ==
Let $\left({S, \preceq}\right)$ be an ordered set.
Let $x \in S$.
Then the singleton $\left\{{x}\right\}$ is convex.
== Proof ==
Let:
:$a, c \in \left\{{x}\right\}$
:$b \in S$
:$a \preceq b \preceq c$
Then $a = c = x$.
Thus $x \preceq b \preceq x$.
Since $\preceq$ is a ordering, it is antisymmetric.
Thus $b = x$, so $b \in \left\{{x}\right\}$.
Since this holds for the only such triple, $\left\{{x}\right\}$ is convex.
Category:Convex Sets (Order Theory)
Category:Singletons
Singleton is Dependent implies Rank is Zero

== Theorem ==
Let $M = \struct {S, \mathscr I}$ be a matroid.
Let $x \in S$.
Let $\set x$ be dependent.
Then:
:$\map \rho {\set x} = 0$
where $\rho$ denotes the rank function of $M$.
=== Corollary ===
== Proof ==
By definition of a dependent subset:
:$\set x \notin \mathscr I$
Then:
Category:Matroid Dependent Subsets
Category:Matroid Rank Functions
Singleton is Dependent implies Rank is Zero/Corollary

== Theorem ==
Let $M = \struct {S, \mathscr I}$ be a matroid.
Let $x \in S$.
Then:
:$x$ is a loop $\map \rho {\set x} = 0$
where $\rho$ denotes the rank function of $M$.
== Proof ==
By definition of a loop:
:$x$ is a loop $\set x \notin \mathscr I$
From Singleton is Dependent implies Rank is Zero:
:if $\set x \notin \mathscr I$ then $\map \rho {\set x} = 0$
From Singleton is Independent implies Rank is One:
:if $\set x \in \mathscr I$ then $\map \rho {\set x} = 1$
It follows that:
:$\set x \notin \mathscr I$ $\map \rho {\set x} = 0$
== Sources ==
* Chapter $1.$ $\S 4.$ Loops and parallel elements
Category:Matroid Dependent Subsets
Category:Matroid Rank Functions
Singleton is Directed and Filtered Subset

== Theorem ==
Let $\struct {S, \precsim}$ be a preordered set.
Let $x$ be an element of $S$.
Then $\set x$ is directed and filtered subset of $S$.
== Proof ==
Let $y, z \in \set x$.
By definition of singleton:
:$ y = x \land z = x$
By definition of reflexivity:
:$y \precsim x \land z \precsim x$
Thus:
:$\exists h \in \set x: y \precsim h \land z \precsim h$
Thus by definition:
:$\set x$ is a directed subset of $S$.
$\set x$ is a filtered subset of $S$ by mutatis mutandis.
== Sources ==
* 
* 
Category:Preorder Theory
Singleton is Finite

== Theorem ==
Let $x$ be arbitrary.
Then $\set x$ is a finite set.
== Proof ==
Define a mapping $f: \set x \to \N_{< 1}$:
:$\map f x = 0$
By definition of singleton:
:$\forall y, z \in \set x: \map f y = \map f z \implies y = z$
By definition:
:$f$ is an injection.
By definition of initial segment of natural numbers:
:$\N_{< 1} = \set 0$
By definition of $f$:
:$\forall n \in N_{< 1}: \exists z \in \set x: \map f z = n$
By definition:
:$f$ is a surjection.
By definition:
:$f$ is a bijection.
By definition of set equivalence:
:$\set x \sim \N_{< 1}$
Thus by definition:
:$\set x$ is a finite set.
== Sources ==
* 
Category:Singletons
Singleton is Independent implies Rank is One

== Theorem ==
Let $M = \struct {S, \mathscr I}$ be a matroid.
Let $x \in S$.
Let $\set x$ be independent.
Then:
:$\map \rho {\set x} = 1$
where $\rho$ denotes the rank function of $M$.
=== Corollary ===
== Proof ==
From Rank of Independent Subset Equals Cardinality:
:$\map \rho {\set x} = \size {\set x}$
From Cardinality of Singleton:
:$\size {\set x} = 1$
The result follows.
Category:Matroid Independent Subsets
Category:Matroid Rank Functions
Singleton is Independent implies Rank is One/Corollary

== Theorem ==
Let $M = \struct{S, \mathscr I}$ be a matroid.
Let $x \in S$.
Then:
:$\set x$ is an independent subset $\map \rho {\set x} = 1$
where $\rho$ denotes the rank function of $M$.
== Proof ==
By definition of an independent subset:
:$x$ is an independent subset $\set x \notin \mathscr I$
From Singleton is Independent implies Rank is One:
:if $\set x \in \mathscr I$ then $\map \rho {\set x} = 1$
From Singleton is Dependent implies Rank is Zero:
:if $\set x \notin \mathscr I$ then $\map \rho {\set x} = 0$
It follows that:
:$\set x \in \mathscr I$ $\map \rho {\set x} = 1$
Category:Matroid Independent Subsets
Category:Matroid Rank Functions
Singleton is Linearly Independent

== Theorem ==
Let $K$ be a division ring.
Let $\struct {G, +_G}$ be a group whose identity is $e$.
Let $\struct {G, +_G, \circ}_K$ be a $K$-vector space whose zero is $0_K$.
Let $x \in G: x \ne e$.
Then $\set x$ is a linearly independent subset of $G$.
== Proof ==
The only sequence of distinct terms in $\set x$ is the one that goes: $x$.
Suppose $\exists \lambda \in K: \lambda \circ x = e$.
From Zero Vector Space Product iff Factor is Zero it follows that $\lambda = 0$.
Hence the result from definition of linearly independent set.
== Sources ==
* : Chapter $\text {I}$: Elementary Definitions: $\S 4$. Vector Spaces
* : Chapter $\text {V}$: Vector Spaces: $\S 27$. Subspaces and Bases
Category:Linear Independence
Singleton is Terminal Object of Category of Sets

== Theorem ==
Let $\mathbf {Set}$ be the category of sets.
Let $S = \set x$ be any singleton set.
Then $S$ is a terminal object of $\mathbf {Set}$.
== Proof ==
Let $T$ be a set, and let $f: T \to S$ be a mapping.
Then since for all $t \in T$, we have $\map f t \in S$, it follows that:
:$\forall t \in T: \map f t = x$
By Equality of Mappings, there is precisely one such mapping $f: T \to S$.
Hence the result, by definition of terminal object.
== Sources ==
* : $\S 2.2$: Example $2.11$: $1$
Category:Category Theory
Category:Singletons
Singleton of Bottom is Ideal

== Theorem ==
Let $\struct {S, \preceq}$ be a bounded below ordered set.
Then
:$\set \bot$ is an ideal in $\struct {S, \preceq}$
where $\bot$ denotes the smallest element in $S$.
== Proof ==
=== Non-empty ===
By definition of singleton:
:$\bot \in \set \bot$
By definition:
:$\set \bot$ is a non-empty set.
=== Directed ===
Thus by Singleton is Directed and Filtered Subset:
:$\set \bot$ is directed.
=== Lower ===
Let $x \in \set \bot, y \in S$ such that
:$y \preceq x$
By definition of singleton:
:$x = \bot$
By definition of smallest element:
:$\bot \preceq y$
By definition of antisymmetry:
:$y = \bot$
Thus by definition of singleton:
:$y \in \set \bot$
Thus by definition:
:$\set \bot$ is a lower section.
Thus by definition:
:$\set \bot$ is an ideal in $\struct {S, \preceq}$
== Sources ==
* 
Category:Order Theory
Singleton of Element is Subset/Proof 2

== Theorem ==
== Proof ==
=== Necessary Condition ===
Let $x \in S$.
We have:
:$\set x = \set {y \in S: y = x}$
From Subset of Set with Propositional Function:
:$\set {x \in S: \map P x} \subseteq S$
Hence:
:$\set x \subseteq S$
=== Sufficient Condition ===
Let $\set x \subseteq S$.
From the definition of a subset:
:$x \in \set x \implies x \in S$
Category:Singleton of Element is Subset
Singleton of Empty Class is Transitive

== Example of Transitive Class ==
Let $\O$ denote the empty class.
Then the singleton $\set \O$ is transitive.
== Proof ==
There is one element of $\set \O$, and that is $\O$.
This is a subclass of $\set \O$.
That is, $\set \O$ is transitive.
== Sources ==
* : Chapter $2$: Some Basics of Class-Set Theory: $\S 10$ Some useful facts about transitivity
Category:Examples of Transitive Classes
Category:Singletons
Category:Empty Class