nyuuzyou/wikis huggingface dataset:
Syndrome Decoding/Examples/(6, 3) code in Z2

== Example of Syndrome Decoding ==
Let $C$ be the linear code:
:$C = \set {000000, 100110, 010101, 110011, 001011, 101101, 011110, 111000}$
Then the Syndrome Decoding table $T$ for $C$ is:
:$\begin{array} {cc}
000000 & 000 \\
100000 & 110 \\
010000 & 101 \\
001000 & 011 \\
000100 & 100 \\
000010 & 010 \\
000001 & 001 \\
100001 & 111 \\
\end{array}$
To find the $8$th row, it is not necessary to hunt directly an element $v$ of $\map V {6, 2}$ of weight $2$ which does not exist anywhere in the other $7$ rows.
What you do is identify the remaining syndrome of $3$ digits that has not been used yet (that is: $111$).
Then you work out what combinations of coset leaders and their own syndromes which when added together make that last syndrome.
The fact that in this case that combination is not unique (here we get $100001$, $010010$, $001100$ and $000111$) means that an element of $\map V {6, 2}$ which is more than $2$ distant from a codeword cannot be uniquely decoded.
== Example ==
== Sources ==
* : Chapter $6$: Error-correcting codes: Example $6.22$
Category:Examples of Linear Codes
Syndrome Decoding/Examples/(6, 3) code in Z2/Example

== Example of Syndrome Decoding on Linear $\tuple {6, 3}$-code in $\Z_2$ ==
Let $C$ be the linear code:
:$C = \set {000000, 100110, 010101, 110011, 001011, 101101, 011110, 111000}$
Then the Syndrome Decoding of $100111$ yields $100110$.
== Proof ==
The standard parity check table for $C$ is:
:$P := \begin{pmatrix}
1 & 1 & 0 & 1 & 0 & 0 \\ 
1 & 0 & 1 & 0 & 1 & 0 \\
0 & 1 & 1 & 0 & 0 & 1 \end{pmatrix}$
The syndrome decoding table $T$ for $C$ is:
:$\begin{array} {cc}
000000 & 000 \\
100000 & 110 \\
010000 & 101 \\
001000 & 011 \\
000100 & 100 \\
000010 & 010 \\
000001 & 001 \\
100001 & 111 \\
\end{array}$
By multiplying $P$ by $\paren {100111}^\intercal$, we get that the syndrome for $100111$ is $001$.
We find $000001$ in column $1$ and subtract it from $100111$.
We obtain $100110$.
== Sources ==
* : Chapter $6$: Error-correcting codes: Example $6.22$
Category:Examples of Linear Codes
Syndrome is Zero iff Vector is Codeword

== Theorem ==
Let $C$ be a linear $\tuple {n, k}$-code whose master code is $\map V {n, p}$
Let $G$ be a (standard) generator matrix for $C$.
Let $P$ be a standard parity check matrix for $C$.
Let $w \in \map V {n, p}$.
Then the syndrome of $w$ is zero $w$ is a codeword of $C$.
== Proof ==
Let $G = \paren {\begin{array} {c|c} \mathbf I & \mathbf A \end{array} }$.
Let $c \in \map V {n, p}$.
Then, by definition of $G$, $c$ is a codeword of $C$ $c$ is of the form $u G$, where $u \in \map V {k, p}$.
Thus $c \in C$ :
where:
:$v = u \mathbf A$
:$\paren {\begin{array} {c|c} u & v \end{array} }$ denotes the $1 \times n$ matrix formed from the $k$ elements of $u$ and the $n - k$ elements of $v$.
Let $w \in \map V {n, p}$.
$w$ can be expressed in the form:
:$w = \paren {\begin{array} {c|c} u_1 & v_1 \end{array} }$
where $u_1 \in \map V {k, p}$.
The syndrome of $v$ is then calculated as:
It follows that the syndrome of $w$ is zero $w$ is the concatenation of $u_1$ and $v_1$, where:
:$v_1^\intercal = \mathbf A^\intercal u_1^\intercal = \paren {u_1 \mathbf A}^\intercal$
Thus the syndrome of $w$ is zero $w$ is a codeword of $C$.
== Sources ==
* : Chapter $6$: Error-correcting codes: Proposition $6.20$
Category:Linear Codes
Syndrome is Zero iff Vector is Codeword/Mistake

== Source Work ==
:
Chapter $6$: Error-correcting codes
::Proposition $6.20$: Proof
== Mistake ==
:Hence, the element $c$ is a codeword $c = u G = u \paren {\begin{array} {c|c} I & A \end{array} } = \paren {\begin{array} {c|c} u & v \end{array} }$, where $v = u A$, and $\paren {\begin{array} {c|c} U & v \end{array} }$ denotes the concatenation of $u$ with $v$, ...''
== Correction ==
Where it says:
:where ... $\paren {\begin{array} {c|c} U & v \end{array} }$ denotes the concatenation of $u$ with $v$
it should have:
: $\paren {\begin{array} {c|c} u & v \end{array} }$
== Sources ==
* : Chapter $6$: Error-correcting codes: Proposition $6.20$
Category:Mistakes/A Course in Group Theory
Synthetic Basis and Analytic Basis are Compatible

== Theorem ==
Let $\struct {S, \tau}$ be a topological space.
Then $\BB$ is an analytic basis for $\tau$ $\tau$ is the topology on $S$ generated by the synthetic basis $\BB$.
== Proof ==
=== Necessary Condition ===
Suppose that $\BB$ is an analytic basis for $\tau$.
We proceed to check the axioms for $\BB$ to be a synthetic basis on $S$.
 states that $S \in \tau$.
Therefore, by the definition of an analytic basis:
:$\ds \exists \SS \subseteq \BB: S = \bigcup \SS$
By Equivalent Conditions for Cover by Collection of Subsets, $\BB$ is a cover for $S$.
That is, axiom $(\text B 1)$ for a synthetic basis is satisfied by $\BB$.
Suppose that $A, B \in \BB$.
By the definition of an analytic basis, $\BB \subseteq \tau$.
Therefore $A, B \in \tau$.
 states that $A \cap B \in \tau$.
Therefore, by the definition of an analytic basis:
:$\ds \exists \AA \subseteq \BB: A \cap B = \bigcup \AA$
That is, axiom $(\text B 2)$ for a synthetic basis is satisfied by $\BB$.
Therefore, $\BB$ is a synthetic basis on $S$.
Let $\tau'$ be the topology on $S$ generated by the synthetic basis $\BB$.
It follows from the definition of an analytic basis that $\tau \subseteq \tau'$.
Since the subset relation is transitive, we can apply for a topology to conclude that:
:$\ds \forall U \in \tau': \exists \AA \subseteq \BB \subseteq \tau: U = \bigcup \AA \in \tau$
That is, $\tau' \subseteq \tau$.
By definition of set equality:
:$\tau = \tau'$
=== Sufficient Condition ===
Suppose that $\BB$ is a synthetic basis on $S$, and that $\tau$ is the topology on $S$ generated by $\BB$.
Then:
:$\ds \BB = \set {\bigcup \set B: \set B \subseteq \BB} \subseteq \tau$
By definition, $\BB$ is an analytic basis for $\tau$.
== Sources ==
* : $3$: Continuity generalized: topological spaces: $3.2$: Bases: Proposition $3.2.4$
Category:Topological Bases
Synthetic Basis formed from Synthetic Sub-Basis

== Theorem ==
Let $X$ be a set.
Let $\SS$ be a synthetic sub-basis on $X$.
Define:
:$\ds \BB = \set {\bigcap \FF: \FF \subseteq \SS, \text{$\FF$ is finite} }$
Then $\BB$ is a synthetic basis on $X$.
== Proof ==
We consider $X$ as the universal set.
Thus, in accordance with Intersection of Empty Set, we take the convention that:
:$\ds \bigcap \O = X \in \BB$
By Set is Subset of Union: General Result, it follows that:
:$\ds X \subseteq \bigcup \BB$
That is, axiom $(\text B 1)$ for a synthetic basis is satisfied.
We have that $\BB \subseteq \powerset X$.
Let $B_1, B_2 \in \BB$.
Then there exist finite $\FF_1, \FF_2 \subseteq \SS$ such that:
:$\ds B_1 = \bigcap \FF_1$
:$\ds B_2 = \bigcap \FF_2$
It follows that:
:$\ds B_1 \cap B_2 = \bigcap \paren {\FF_1 \cup \FF_2}$
By Union is Smallest Superset, $\FF_1 \cup \FF_2 \subseteq \SS$.
We have that $\FF_1 \cup \FF_2$ is finite.
Hence $B_1 \cap B_2 \in \BB$, so it follows by definition that axiom $(\text B 2)$ for a synthetic basis is satisfied.
== Note ==
Note that by this construction, any collection of subsets of $X$ can form a synthetic basis and thus generate a topology on $X$.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $1$: General Introduction
Category:Topological Bases
Category:Sub-Bases
Synthetic Division/Examples/Arbitrary Example 1

== Examples of Synthetic Division ==
Let $\map P x$ be the cubic:
:$\map P x = 4 x^3 - 2 x^2 + 3 x - 1$
Applying Horner's rule:
:$\map P x = \paren {\paren {4 x - 2} x + 3} x - 1$
Let $x = 2$.
Then:
Noting that the coefficient of $x^3$ is $3$, we obtain:
:$4 x^3 - 2 x^2 + 3 x - 1 = \paren {x - 2} \paren {4 x^2 + 6 x + 15} + 29$
== Sources ==
* 
* 
Category:Examples of Synthetic Division
Synthetic Sub-Basis and Analytic Sub-Basis are Compatible

== Theorem ==
Let $\struct {X, \tau}$ be a topological space.
Let $\SS \subseteq \powerset X$, where $\powerset X$ denotes the power set of $X$.
Then $\SS$ is an analytic sub-basis for $\tau$ $\tau$ is the topology on $X$ generated by the synthetic sub-basis $\SS$.
== Proof ==
=== Necessary Condition ===
Follows directly from the definitions of the generated topology and an analytic sub-basis.
=== Sufficient Condition ===
Suppose that $\SS$ is an analytic sub-basis for $\tau$.
Let $\tau'$ be the topology on $X$ generated by the synthetic sub-basis $\SS$.
By definition $1$ of the generated topology and the definition of an analytic sub-basis, we have $\tau \subseteq \tau'$.
By definition $2$ of the generated topology, it follows that $\tau' \subseteq \tau$.
By definition of set equality:
:$\tau = \tau'$
Category:Topological Bases
System of Linear Equations as Continuous Linear Transformation

== Theorem ==
Let $x_1, x_2 \in \R$ be real numbers.
Consider the following system of simultaneous linear equations $\paren S$:
Let $\norm {\, \cdot \,}_2$ be the Euclidean norm.
Let $\norm {\, \cdot \,}$ be the supremum operator norm.
Let $I$ be the $2 \times 2$ identity matrix.
Then:
:$S$ is expressible as $\paren {I - K} x = y$ where $x = \begin{pmatrix} x_1 \\ x_2 \end{pmatrix}$, $y = \begin{pmatrix} 1 \\ 2 \end{pmatrix}$, $K = \begin{pmatrix} \frac 1 2 & \frac 1 3 \\ \frac 1 3 & \frac 1 4 \end{pmatrix}$;
:$\norm K < 1$;
:$S$ admits a unique solution.
== Proof ==
=== $\paren S$ is expressible as $\paren {I - K} x = y$ ===
We have that:
Furthermore:
Hence, $\paren S$ can be rewritten as:
Moreover:
By a simple inspection it is evident that $\paren S$ is expressible as $\paren {I - K} x = y$.
=== $\norm K < 1$ ===
We have that:
Hence:
=== $S$ admits a unique solution ===
By Neumann series theorem, $\paren {I - K}^{-1}$ exists in $\map {CL} {\R^2}$.
Hence:
By Inverse of Nonsingular 2 x 2 Real Square Matrix:
Therefore:
== Sources ==
* : Chapter $\S 2.4$: Composition of continuous linear transformations
Category:Simultaneous Equations
Category:Neumann Series
System of Open Neighborhoods are Equal Iff Singleton Closures are Equal

== Theorem ==
Let $T = \struct{S, \tau}$ be a topological space.
Let $x, y \in S$ such that $x \ne y$.
Let $\map \UU x$ and $\map \UU y$ denote the system of open neighborhoods of $x$ and $y$ respectively.
Let $\set x^-$ and $\set y^-$ denote the topological closures of $\set x$ and $\set y$ respectively.
Then:
:$\map \UU x = \map \UU y$ $\set x^- = \set y^-$
== Proof ==
We have
It remains to show, if:
:$S \setminus \set x^- = S \setminus \set y^-$ 
then:
:$\tau \setminus \map \UU x = \tau \setminus \map \UU y$ 
Let:
:$S \setminus \set x^- = S \setminus \set y^-$
We have
By definition of set equality:
:$\tau \setminus \map \UU x = \tau \setminus \map \UU y$ 
The result follows.
Category:Systems of Open Neighborhoods
System of Open Neighborhoods is a Completely Prime Filter

== Theorem ==
Let $\struct{S, \tau}$ be a topological space.
Let $x \in S$.
Let $\map \UU x$ denote the system of open neighborhoods of $x$ in $\struct{S, \tau}$.
Then:
:$\map \UU x$ is a completely prime filter in the complete lattice $\struct{\tau, \subseteq}$
== Proof ==
=== $\map \UU x$ satisfies ===
We have $x \in S$.
By :
:$S \in \tau$
Hence:
:$S \in \map \UU x$
It follows that $\map \UU x$ satisfies .
=== $\map \UU x$ satisfies ===
Let $U, V \in \map \UU x$.
By :
:$U \cap V\in \tau$
By definition of set intersection:
:$x \in U \cap V$
Hence:
:$U \cap V \in \map \UU x$
From Intersection is Subset:
:$U \cap V \subseteq U, U \cap V \subseteq V$
It follows that $\map \UU x$ satisfies .
=== $\map \UU x$ satisfies ===
Let $U \in \map \UU x$.
Let $v \in \tau$ such that $U \subseteq V$.
By definition of subset:
:$x \in V$
Hence:
:$V \in \map \UU x$
It follows that $\map \UU x$ satisfies .
=== $\map \UU x$ is Proper ===
By definition of system of open neighborhoods:
:$\map \UU x \subseteq \tau$
By empty set:
:$x \notin \O$
Hence:
:$\O \notin \map \UU x$
From Empty Set is Element of Topology:
:$\map \UU x \ne \tau$
It follows that $\map \UU x$ is a proper subset of $\tau$.
=== $\map \UU x$ satisfies Complete Primeness ===
Let $\VV \subseteq \tau$:
:$\bigcup \VV \in \map \UU x$
By definition of system of open neighborhoods:
:$x \in \bigcup \VV$
By definition of set union:
:$\exists V \in \VV : x \in V$
Hence:
:$\exists V \in \VV : V \in \map \UU x$
It follows that $\map \UU x$ is a completely prime filter by definition.
== Also see ==
* Topology forms Complete Lattice
== Sources ==
* : Chapter $1$: Spaces and Lattices of Open Sets, $\S 1$ Sober spaces, Definition $1.3$
Category:Systems of Open Neighborhoods
Category:Completely Prime Filters
System of Simultaneous Equations may have Multiple Solutions

== Theorem ==
Let $S$ be a system of simultaneous equations.
Then it is possible that $S$ may have a solution set which is a singleton.
== Proof ==
Consider this system of simultaneous linear equations:
From its evaluation it has the following solutions:
where $t$ is any number.
Hence the are as many solutions as the cardinality of the domain of $t$.
== Sources ==
* : Chapter $1$: Linear Equations and Matrices: $1.1$ Introduction
Category:Simultaneous Equations
System of Simultaneous Equations may have No Solution

== Theorem ==
Let $S$ be a system of simultaneous equations.
Then it is possible that $S$ may have a solution set which is empty.
== Proof ==
Consider this system of simultaneous linear equations:
From its evaluation it is seen to have no solutions.
Hence the result.
== Sources ==
* : Chapter $1$: Linear Equations and Matrices: $1.1$ Introduction
Category:Simultaneous Equations
System of Simultaneous Equations may have Unique Solution

== Theorem ==
Let $S$ be a system of simultaneous equations.
Then it is possible that $S$ may have a solution set which is a singleton.
== Proof ==
Consider this system of simultaneous linear equations:
From its evaluation it has the following unique solution:
Hence the result.
== Sources ==
* : Chapter $1$: Linear Equations and Matrices: $1.1$ Introduction
Category:Simultaneous Equations
T0 Property is Hereditary

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space which is a $T_0$ (Kolmogorov) space.
Let $T_H = \struct {H, \tau_H}$, where $\O \subset H \subseteq S$, be a subspace of $T$.
Then $T_H$ is a $T_0$ (Kolmogorov) space.
== Proof ==
Let $T$ be a $T_0$ (Kolmogorov) space.
That is:
:$\forall x, y \in S$ such that $x \ne y$, either:
::$\exists U \in \tau: x \in U, y \notin U$
:or:
::$\exists U \in \tau: y \in U, x \notin U$
We have that the set $\tau_H$ is defined as:
:$\tau_H := \set {U \cap H: U \in \tau}$
Let $x, y \in H$ such that $x \ne y$.
Then as $x, y \in S$ we have that:
:$\exists U \in \tau: x \in U, y \notin U$
or:
:$\exists U \in \tau: y \in U, x \notin U$
Then either:
:$U \cap H \in \tau_H: x \in U \cap H, y \notin U \cap H$
or:
:$U \cap H \in \tau_H: y \in U \cap H, x \notin U \cap H$
and so the $T_0$ axiom is satisfied.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $2$: Separation Axioms: Functions, Products, and Subspaces
Category:T0 Spaces
Category:Topological Subspaces
T0 Space is Preserved under Closed Bijection

== Theorem ==
Let $T_A = \struct {S_A, \tau_A}$ and $T_B = \struct {S_B, \tau_B}$ be topological spaces.
Let $\phi: T_A \to T_B$ be a closed bijection.
If $T_A$ is a $T_0$ (Kolmogorov) space, then so is $T_B$.
== Proof ==
Let $T_A$ be a $T_0$ (Kolmogorov) space.
By Bijection is Open iff Closed, $\phi$ is an open bijection.
By Bijection is Open iff Inverse is Continuous, it follows that $\phi^{-1}$ is continuous.
By definition: 
:$\forall x, y \in S_A$, either:
::$\exists U \in \tau_A: x \in U, y \notin U$
:or:
::$\exists U \in \tau_A: y \in U, x \notin U$
:
:$\exists x, y \in S_B: \forall V \in \tau_B: x, y \in V \lor x, y \notin V$
That is:
:$\exists x, y \in S_B: \forall V \in \tau_B: \set {x, y} \subseteq V \lor \set {x, y} \cap V = \O$
From Preimage of Subset is Subset of Preimage:
:$\forall V \in \tau_B: \phi^{-1} \sqbrk {\set {x, y} } \subseteq \phi^{-1} \sqbrk V \lor \phi^{-1} \sqbrk {\set {x, y} } \cap \phi^{-1} \sqbrk V = \O$
that is:
:$\forall V \in \tau_B: \map {\phi^{-1} } x, \map {\phi^{-1} } y \in \phi^{-1} \sqbrk V \lor \map {\phi^{-1} } x, \map {\phi^{-1} } y \notin \phi^{-1} \sqbrk V$
But by definition of continuous mapping, $U = \phi^{-1} \sqbrk V$ is open in $T_A$ $V$ is open in $T_B$.
Thus:
:$\forall U \in \tau_A: \map {\phi^{-1} } x, \map {\phi^{-1} } y \in U \lor \map {\phi^{-1} } x, \map {\phi^{-1} } y \notin U$
This contradicts the condition that $T_A$ is a $T_0$ (Kolmogorov) space.
It follows by Proof by Contradiction that $T_B$ must also be a $T_0$ (Kolmogorov) space.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $2$: Separation Axioms: Functions, Products, and Subspaces
Category:T0 Spaces
T0 Space is Preserved under Homeomorphism

== Theorem ==
Let $T_A = \struct {S_A, \tau_A}$ and $T_B = \struct {S_B, \tau_B}$ be topological spaces.
Let $\phi: T_A \to T_B$ be a homeomorphism.
If $T_A$ is a $T_0$ (Kolmogorov) space, then so is $T_B$.
== Proof ==
By definition of homeomorphism, $\phi$ is a closed continuous bijection.
The result follows from $T_0$ (Kolmogorov) Space is Preserved under Closed Bijection.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $2$: Separation Axioms: Functions, Products, and Subspaces
Category:T0 Spaces
Category:Homeomorphisms (Topological Spaces)
T0 Space is not necessarily T1 Space

== Theorem ==
Let $T = \struct {S, \tau}$ be a be a $T_0$ space.
Then it is not necessarily the case that $T$ is a $T_1$ space.
== Proof 1 ==
== Proof 2 ==
Category:T0 Spaces
Category:T1 Spaces
Category:T0 Space is not necessarily T1 Space
T0 Space is not necessarily T1 Space/Proof 1

== Theorem ==
== Proof ==
Proof by Counterexample:
Let $T = \struct {S, \tau_p}$ be a particular point space such that $S$ is not a singleton.
From Particular Point Space is $T_0$, we have that $T$ is a $T_0$ space.
From Non-Trivial Particular Point Topology is not $T_1$, $T$ is not a $T_1$ space.
The result follows.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $2$: Separation Axioms: Completely Regular Spaces
Category:T0 Space is not necessarily T1 Space
T0 Space is not necessarily T1 Space/Proof 2

== Theorem ==
== Proof ==
Proof by Counterexample:
Let $T$ be the overlapping interval space.
From Existence of Topological Space which satisfies no Separation Axioms but $T_0$, we have that:
:$T$ is a $T_0$ space
but:
:$T$ is not a $T_1$ space.
The result follows.
Category:T0 Space is not necessarily T1 Space
T1/2 Space is T0 Space

== Theorem ==
Let $T = \struct {S, \tau}$ be a $T_{\frac 1 2}$ topological space.
Then $T$ is $T_0$ space.
== Proof ==
By Characterization of T0 Space by Closures of Singletons it suffices to prove that
:$\forall x, y \in S: x \ne y \implies x \notin \set y^- \lor y \notin \set x^-$
where $\set y^-$ denotes the closure of $\set y$.
Let $x, y$ be points of $T$ such that:
:$x \ne y$
:
:$x \in \set y^- \land y \in \set x^-$
We will prove that:
:$x \notin \set x'$
where $\set x'$ denotes the derivative of $\set x$.
:
:$x \in \set x'$
As $S$ is open by definition of topological space
by Characterization of Derivative by Open Sets:
:$\exists z \in S: z \in \set x \cap S \land z \ne x$
By definition of intersection:
:$z \in \set x$
This contradicts $z \ne x$ by definition of singleton.
Thus $x \notin \set x'$.
We will prove that:
:$(1): \quad \lnot \forall G \in \tau: y \in G \implies \set x \cap G \ne \O$
:
:$\forall G \in \tau: y \in G \implies \set x \cap G \ne \O$
As sublemma we will show that:
:$\forall U \in \tau: y \in U \implies \exists r \in S: r \in \set x \cap U \land y \ne r$
Let $U \in \tau$ such that:
:$y \in U$
Then by assumption:
:$\set x \cap U \ne \O$
By definition of empty set:
:$\exists z: z \in \set x \cap U$
By definition of intersection:
:$z \in \set x$
Then by definition of singleton:
:$z = x \ne y$
Thus:
:$\exists r \in S: r \in \set x \cap U \land y \ne r$
Then by Characterization of Derivative by Open Sets:
:$y \in \set x'$
By definition of relative complement:
:$y \notin \relcomp S {\set x'} \land x \in \relcomp S {\set x'}$
By definition of $T_{\frac 1 2}$ space:
:$\set x'$ is closed
By definition of closed set:
:$\relcomp S {\set x'}$ is open
By $x \in \set y^-$ and by Condition for Point being in Closure:
:$\set y \cap \relcomp S {\set x'} \ne \O$
Then by definition of empty set:
:$\exists z: z \in \set y \cap \relcomp S {\set x'}$
By definition of intersection:
:$z \in \set y \land z \in \relcomp S {\set x'}$
By definition of singleton:
:$z = y$
Thus $z \in \relcomp S {\set x'}$ contradicts $y \notin \relcomp S {\set x'}$.
This ends the proof of $(1)$.
Then by $(1)$ and Condition for Point being in Closure:
:$y \notin \set x^-$
This contradicts $y \in \set x^-$.
== Sources ==
* 
* 
Category:T0 Spaces
Category:T1/2 Spaces
T1 Property is Hereditary

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space which is a $T_1$ (Fréchet) space.
Let $T_H = \struct {H, \tau_H}$, where $\O \subset H \subseteq S$, be a subspace of $T$.
Then $T_H$ is a $T_1$ (Fréchet) space.
== Proof ==
Let $T$ be a $T_1$ (Fréchet) space.
That is:
:$\forall x, y \in S$ such that $x \ne y$, both:
::$\exists U \in \tau: x \in U, y \notin U$
:and:
::$\exists U \in \tau: y \in U, x \notin U$
We have that the set $\tau_H$ is defined as:
:$\tau_H := \set {U \cap H: U \in \tau}$
Let $x, y \in H$ such that $x \ne y$.
Then as $x, y \in S$:
:$\exists U \in \tau: x \in U, y \notin U$
and:
:$\exists U \in \tau: y \in U, x \notin U$
Then both:
:$U \cap H \in \tau_H: x \in U \cap H, y \notin U \cap H$
and:
:$U \cap H \in \tau_H: y \in U \cap H, x \notin U \cap H$
and so the $T_1$ axiom is satisfied.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $2$: Separation Axioms: Functions, Products, and Subspaces
Category:T1 Spaces
Category:Topological Subspaces
T1 Space is Preserved under Closed Bijection

== Theorem ==
Let $T_A = \struct {S_A, \tau_A}$ and $T_B = \struct {S_B, \tau_B}$ be topological spaces.
Let $\phi: T_A \to T_B$ be a closed bijection.
If $T_A$ is a $T_1$ (Fréchet) space, then so is $T_B$.
== Proof ==
Let $T_A$ be a $T_1$ (Fréchet) space.
By definition, all points in $T_A$ are closed.
Let $a \in S_A$.
Then $\set a$ is a closed set.
As $\phi$ is a closed mapping it follows directly that $\phi \sqbrk {\set a}$ is closed.
As $\phi$ is a bijection it follows that every point in $S_B$ is the image under $\phi$ of a single point in $S_A$.
Hence every point in $S_B$ is closed.
That is, $T_B$ is a $T_1$ (Fréchet) space.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $2$: Separation Axioms: Functions, Products, and Subspaces
Category:T1 Spaces
T1 Space is Preserved under Homeomorphism

== Theorem ==
Let $T_A = \struct {S_A, \tau_A}$ and $T_B = \struct {S_B, \tau_B}$ be topological spaces.
Let $\phi: T_A \to T_B$ be a homeomorphism.
If $T_A$ is a $T_1$ (Fréchet) space, then so is $T_B$.
== Proof ==
By definition of homeomorphism, $\phi$ is a closed continuous bijection.
The result follows from $T_1$ (Fréchet) Space is Preserved under Closed Bijection.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $2$: Separation Axioms: Functions, Products, and Subspaces
Category:T1 Spaces
Category:Homeomorphisms (Topological Spaces)
T1 Space is T0 Space

== Theorem ==
Let $\struct {S, \tau}$ be a Fréchet ($T_1$) space.
Then $\struct {S, \tau}$ is also a Kolmogorov ($T_0$) space.
== Proof ==
Let $\struct {S, \tau}$ be a $T_1$ space.
Let $x, y \in S: x \ne y$.
From the definition of $T_1$ space:
:Both
::$\exists U \in \tau: x \in U, y \notin U$
:and:
::$\exists V \in \tau: y \in V, x \notin V$
From the Rule of Simplification:
: $\exists U \in \tau: x \in U, y \notin U$
From the Rule of Addition:
:Either
::$\exists U \in \tau: x \in U, y \notin U$
:or:
::$\exists V \in \tau: y \in V, x \notin V$
which is precisely the definition of a Kolmogorov ($T_0$) space.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $2$: Separation Axioms
Category:T0 Spaces
Category:T1 Spaces
T1 Space is T1/2 Space

== Theorem ==
Let $T$ be a $T_1$ topological space.
Then $T$ is $T_{\frac 1 2}$ space.
== Proof ==
By Closure of Derivative is Derivative in T1 Space:
:$\forall A \subseteq T: \paren {A'}^- = A'$
where
:$A'$ denotes the derivative of $A$
:$\paren {A'}^-$ denotes the closure of $A'$
Then by Topological Closure is Closed:
:$\forall A \subseteq T: A'$ is closed
Thus by definition:
:$T$ is $T_{\frac 1 2}$ space
== Sources ==
* 
* 
Category:T1 Spaces
Category:T1/2 Spaces