nyuuzyou/wikis huggingface dataset:
Standard Discrete Metric induces Discrete Topology

== Theorem ==
Let $M = \struct {A, d}$ be the (standard) discrete metric space on $A$.
Then $d$ induces the discrete topology on $A$.
Thus the discrete topology is metrizable.
== Proof ==
Let $a \in A$.
From Subset of Standard Discrete Metric Space is Open, a set $U \subseteq A$ is open in $M$.
So, in particular, $\set a$ is open in $\struct {A, d}$.
This holds for all $a \in A$.
From Metric Induces Topology it follows that $\set a$ is an open set in $\struct {A, \tau_{A, d} }$.
The result follows from Basis for Discrete Topology.
== Sources ==
* : Chapter $3$: Topological Spaces: $\S 2$: Topological Spaces: Exercise $2$
* : Part $\text {II}$: Counterexamples: $1 \text { - } 3$. Discrete Topology: $6$
Category:Standard Discrete Metric
Category:Discrete Topology
Category:Metrizable Topologies
Standard Discrete Metric is Metric

== Theorem ==
The standard discrete metric is a metric.
== Proof ==
Let $d: S \times S \to \R$ denote the standard discrete metric on the underlying set $S$ of some space $\struct {S, d}$.
By definition:
:$\forall x, y \in S: \map d {x, y} = \begin {cases} 0 & : x = y \\ 1 & : x \ne y \end {cases}$
=== Proof of ===
So holds for $d$.
=== Proof of ===
Let $x = z$.
Let $x \ne z$.
Either $x \ne y$ or $y \ne z$, or both.
So:
So in either case:
:$\map d {x, y} + \map d {y, z} \ge \map d {x, z}$
and holds for $d$.
=== Proof of ===
Let $x \ne y$.
So holds for $d$.
=== Proof of ===
So holds for $d$.
== Sources ==
* : Chapter $\text{III}$: Metric Spaces: The Definition
* : Chapter $2$: Metric Spaces: $\S 2$: Metric Spaces: Exercise $7$
* : $2$: Continuity generalized: metric spaces: $2.2$: Examples: Example $2.2.2$
Category:Standard Discrete Metric
Standard Discrete Metric is not Topologically Equivalent to p-Product Metrics

== Theorem ==
For $n \in \N$, let $\R^n$ be an Euclidean space.
Let $p \in \R_{\ge 1}$.
Let $d_p$ be the $p$-product metric on $\R^n$.
Let $d_0$ be the standard discrete metric on $\R^n$.
Then $d_p$ and $d_0$ are not topologically equivalent.
== Proof ==
From Open Ball in Standard Discrete Metric Space it is seen that singletons are open sets in $\struct {\R^n, d_0}$.
However, this is not the case in the $\struct {\R^n, d_p}$.
== Sources ==
* : $2$: Continuity generalized: metric spaces: $2.4$: Equivalent metrics: Example $2.4.5$
Category:P-Product Metrics
Category:Standard Discrete Metric
Standard Equation of Ellipsoid

== Theorem ==
Let $\EE$ be an ellipsoid whose axes of symmetry coincide with the coordinate axes of a Cartesian $3$-space.
Then $\EE$ can be expressed by the equation:
:$\dfrac {x^2} {a^2} + \dfrac {y^2} {b^2} + \dfrac {z^2} {c^2} = 1$
== Proof ==
== Sources ==
* 
* 
Category:Ellipsoids
Standard Generator Matrix for Linear Code/Examples

== Examples of Standard Generator Matrices for Linear Codes ==
=== Linear $\tuple {3, 2}$-code in $\Z_2$ ===
=== Linear $\tuple {5, 3}$-code in $\Z_2$ ===
=== Linear $\tuple {6, 3}$-code in $\Z_2$ ===
=== Linear $\tuple {4, 2}$-code in $\Z_3$: Example 1 ===
=== Linear $\tuple {4, 2}$-code in $\Z_3$: Example 2 ===
=== Linear $\tuple {5, 3}$-code in $\Z_5$ ===
Category:Examples of Linear Codes
Standard Generator Matrix for Linear Code/Examples/(3, 2) code in Z2

== Example of Standard Generator Matrix for Linear Code ==
Let $G$ be the standard generator matrix:
:$G := \begin{pmatrix} 1 & 0 & 1 \\ 0 & 1 & 1 \end{pmatrix}$
$G$ generates the linear code $C$:
:$C = \set {000, 101, 011, 110}$
The minimum distance of $C$ is $2$, so $C$ detects $1$ transmission error and corrects none.
== Proof ==
Multiplying $G$ by the $4$ vectors $00, 01, 10, 11$ in turn gives:
all arithmetic being modulo $2$.
The rest of the result follows by inspection.
== Sources ==
* : Chapter $6$: Error-correcting codes: Example $6.12$
Category:Examples of Linear Codes
Standard Generator Matrix for Linear Code/Examples/(4, 2) code in Z3/Example 1

== Example of Standard Generator Matrix for Linear Code ==
Let $G$ be the standard generator matrix:
:$G := \begin{pmatrix}
1 & 0 & 2 & 1 \\ 
0 & 1 & 1 & 2 
\end{pmatrix}$
$G$ generates the linear code $C$:
:$C = \set {0000, 0112, 0221, 1021, 1100, 1212, 2012, 2121, 2200}$
The minimum distance of $C$ is $2$, so $C$ detects $1$ transmission error and corrects $0$ transmission errors.
== Proof ==
Multiplying $G$ by the $9$ vectors $00, 01, 02, 10, 11, 12, 20, 21, 22$ in turn gives:
all arithmetic being modulo $3$.
As can be seen, $2$ of the codewords have weight $2$, while the other codewords which are not $0000$ have more.
From Minimum Distance of Linear Code is Smallest Weight of Non-Zero Codeword, the minimum distance of $C$ is $2$.
From Error Detection Capability of Linear Code, $C$ can detect $2 - 1 = 1$ transmission errors.
From Error Correction Capability of Linear Code, $C$ can correct $\floor {\dfrac {2 - 1} 2} = 0$ transmission errors.
== Sources ==
* : Chapter $6$: Error-correcting codes: Example $6.14$
Category:Examples of Linear Codes
Standard Generator Matrix for Linear Code/Examples/(4, 2) code in Z3/Example 2

== Example of Standard Generator Matrix for Linear Code ==
Let $G$ be the standard generator matrix:
:$G := \begin{pmatrix}
1 & 0 & 1 & 1 \\ 
0 & 1 & 1 & 2 
\end{pmatrix}$
$G$ generates the linear code $C$:
:$C = \set {0000, 0112, 0221, 1011, 1120, 1202, 2022, 2101, 2210}$
The minimum distance of $C$ is $3$, so $C$ detects $2$ transmission errors and corrects $1$ transmission error.
== Proof ==
Multiplying $G$ by the $9$ vectors $00, 01, 02, 10, 11, 12, 20, 21, 22$ in turn gives:
all arithmetic being modulo $3$.
As can be seen, all the codewords have weight $3$.
From Minimum Distance of Linear Code is Smallest Weight of Non-Zero Codeword, the minimum distance of $C$ is $3$.
From Error Detection Capability of Linear Code, $C$ can detect $3 - 1 = 2$ transmission errors.
From Error Correction Capability of Linear Code, $C$ can correct $\floor {\dfrac {3 - 1} 2} = 1$ transmission errors.
== Sources ==
* : Chapter $6$: Error-correcting codes: Exercise $2 \ \text{(b)}$
Category:Examples of Linear Codes
Standard Generator Matrix for Linear Code/Examples/(5, 3) code in Z2

== Example of Standard Generator Matrix for Linear Code ==
Let $G$ be the standard generator matrix:
:$G := \begin{pmatrix}
1 & 0 & 0 & 0 & 1 \\ 
0 & 1 & 0 & 1 & 0 \\
0 & 0 & 1 & 1 & 1 \end{pmatrix}$
$G$ generates the linear code $C$:
:$C = \set {00000, 00111, 01010, 01101, 10001, 10110, 11011, 11100}$
The minimum distance of $C$ is $3$, so $C$ detects $1$ transmission error and corrects $0$ transmission errors.
== Proof ==
Multiplying $G$ by the $8$ vectors $000, 001, 010, 011, 100, 101, 110, 111$ in turn gives:
all arithmetic being modulo $2$.
As can be seen, $2$ of the codewords have weight $2$, and $5$ have weight $3$.
From Minimum Distance of Linear Code is Smallest Weight of Non-Zero Codeword, the minimum distance of $C$ is $2$.
From Error Detection Capability of Linear Code, $C$ can detect $3 - 1 = 0$ transmission error.
From Error Correction Capability of Linear Code, $C$ can correct $\floor {\dfrac {2 - 1} 2} = 0$ transmission errors.
== Sources ==
* : Chapter $6$: Error-correcting codes: Exercise $2 \ \text{(a)}$
Category:Examples of Linear Codes
Standard Generator Matrix for Linear Code/Examples/(5, 3) code in Z5

== Example of Standard Generator Matrix for Linear Code ==
Let $G$ be the standard generator matrix over $\Z_5$:
:$G := \begin{pmatrix}
1 & 0 & 0 & 2 & 1 \\ 
0 & 1 & 0 & 1 & 3 \\
0 & 0 & 1 & 4 & 1 \\
\end{pmatrix}$
$G$ generates a linear code which detects $1$ transmission error and corrects $0$ transmission errors.
== Proof ==
Let $C$ denote the linear code generated by $G$.
There are $5^3 = 125$ codewords in $C$, so it is impractical to list them all.
We have that:
so $C$ has a minimum distance of at least $2$.
Looking at the first $3$ columns shows there can be no codeword with weight less than $2$.
From Minimum Distance of Linear Code is Smallest Weight of Non-Zero Codeword, the minimum distance of $C$ is $2$.
From Error Detection Capability of Linear Code, $C$ can detect $2 - 1 = 1$ transmission errors.
From Error Correction Capability of Linear Code, $C$ can correct $\floor {\dfrac {2 - 1} 2} = 0$ transmission errors.
== Sources ==
* : Chapter $6$: Error-correcting codes: Exercise $2 \ \text{(c)}$
Category:Examples of Linear Codes
Standard Generator Matrix for Linear Code/Examples/(6, 3) code in Z2

== Example of Standard Generator Matrix for Linear Code ==
Let $G$ be the standard generator matrix:
:$G := \begin{pmatrix}
1 & 0 & 0 & 1 & 1 & 0 \\ 
0 & 1 & 0 & 1 & 0 & 1 \\
0 & 0 & 1 & 0 & 1 & 1 \end{pmatrix}$
$G$ generates the linear code $C$:
:$C = \set {000000, 100110, 010101, 110011, 001011, 101101, 011110, 111000}$
The minimum distance of $C$ is $3$, so $C$ detects $2$ transmission errors and corrects $1$ transmission error.
== Proof ==
Multiplying $G$ by the $8$ vectors $000, 001, 010, 011, 100, 101, 110, 111$ in turn gives:
all arithmetic being modulo $2$.
As can be seen, $2$ of the codewords have weight $3$, and $3$ have weight $4$.
From Minimum Distance of Linear Code is Smallest Weight of Non-Zero Codeword, the minimum distance of $C$ is $3$.
From Error Detection Capability of Linear Code, $C$ can detect $3 - 1 = 2$ transmission errors.
From Error Correction Capability of Linear Code, $C$ can correct $\dfrac {3 - 1} 2 = 1$ transmission error.
== Example ==
== Sources ==
* : Chapter $6$: Error-correcting codes: Example $6.13$
Category:Examples of Linear Codes
Standard Generator Matrix for Linear Code/Examples/(6, 3) code in Z2/Example

== Example of Error Detection in Linear $\tuple {6, 3}$ code in $\Z_2$ ==
Let $C$ be the linear code:
$C = \set {000000, 100110, 010101, 110011, 001011, 101101, 011110, 111000}$
The received word $100111$ has $1$ transmission error and so can be corrected.
The received word $100001$ has $2$ transmission errors and so cannot be corrected.
== Proof ==
$100111$ is at a distance $1$ from the codeword $100110$.
As the minimum distance of $C$ is $3$, there is no codeword that is closer than that.
Hence the transmitted codeword can be inferred as being $100110$.
$100001$ is at a distance $2$ from the codewords $000000$, $110011$ and $101101$.
Hence, while it is detected as having $2$ transmission errors, it is not possible to uniquely determine what the transmitted codeword could have been.
== Sources ==
* : Chapter $6$: Error-correcting codes: Example $6.13$
Category:Examples of Linear Codes
Standard Machinery

== Theorem ==
Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $\map {\LL^1_{\overline \R} } \mu$ be the space of $\mu$-integrable functions.
Let $\map P {f_1, \ldots, f_n}$ be a proposition, where the variables $f_i$ denote $\mu$-measurable functions $f_i: X \to \overline \R$.
Let every occurrence of an $f_i$ be of the form:
:$\ds \int \map \Phi {f_i} \rd \mu$
for a suitable index set $I$ and multilinear mapping $\Phi: \map {\LL^1_{\overline \R} } \mu^I \to \map {\LL^1_{\overline \R} } \mu$.
Denote with $\map \chi \Sigma$ the set of characteristic functions of elements of $\Sigma$, that is:
:$\map \chi \Sigma := \set {\chi_E: X \to \R: E \in \Sigma}$
Then the following are equivalent:
:$(A): \quad \forall f_1, \ldots, f_n \in \map \chi \Sigma: \map P {f_1, \ldots, f_n}$
:$(B): \quad \forall f_1, \ldots, f_n \in \map {\LL^1_{\overline \R} } \mu: \map P {f_1, \ldots, f_n}$
that is, it suffices to verify $P$ holds for characteristic functions.
== Proof ==
Category:Proof Techniques
Category:Measure Theory
Standard Normal Random Variable as Transformation of Normal Random Variable

== Theorem ==
Let $\mu$ be a real number. 
Let $\sigma$ be a positive real number. 
Let $X \sim \Gaussian \mu {\sigma^2}$ where $\Gaussian \mu {\sigma^2}$ is the normal distribution with parameters $\mu$ and $\sigma^2$. 
Then: 
:$\dfrac {X - \mu} \sigma \sim \Gaussian 0 1$
where $\Gaussian 0 1$ is the standard normal distribution. 
== Proof ==
Category:Normal Distribution
Standard Ordered Basis Vectors are Orthogonal

== Theorem ==
Let $\struct {\mathbf V, +, \circ}_{\mathbb F}$ be a vector space over a field $\mathbb F$, as defined by the vector space axioms.
Let $\tuple {\mathbf e_1, \mathbf e_2, \ldots, \mathbf e_n}$ be the standard ordered basis of $\mathbf V$.
Let $\mathbf e_i$ and $\mathbf e_j$ be elements of $\tuple {\mathbf e_1, \mathbf e_2, \ldots, \mathbf e_n}$ such that $i \ne j$.
:$\mathbf e_i$ and $\mathbf e_j$ are orthogonal.
== Proof ==
By definition of standard ordered basis:
:$\mathbf e_i$ is a vector whose $i$th component is $1_{\mathbb F}$ with all other components $0_{\mathbb F}$.
Hence:
:$\mathbf e_i \cdot \mathbf e_j = \delta_{i j}$
where $\delta_{i j}$ denotes the Kronecker delta.
That is:
:$i \ne j \implies \mathbf e_i \cdot \mathbf e_j = 0$
Hence the result by definition of orthogonal.
== Sources ==
* : Chapter $1$ Vector Analysis $1.3$ Scalar or Dot Product
Category:Orthogonality (Linear Algebra)
Category:Standard Ordered Bases
Standard Ordered Basis is Basis

== Theorem ==
Let $\struct {R, +, \circ}$ be a ring with unity whose zero is $0_R$ and whose unity is $1_R$.
Let $n$ be a positive integer.
For each $j \in \closedint 1 n$, let $e_j$ be the ordered $n$-tuple of elements of $R$ whose $j$th entry is $1_R$ and all of whose other entries is $0_R$.
Then $\sequence {e_n}$ is an ordered basis of the $R$-module $R^n$.
This ordered basis is called the standard ordered basis of $R^n$.
The corresponding set $\set {e_1, e_2, \ldots, e_n}$ is called the standard basis of $R^n$.
== Proof ==
== Also see ==
* Definition:Engineering Notation
== Sources ==
* : Chapter $\text {V}$: Vector Spaces: $\S 27$. Subspaces and Bases: Example $27.6$
* : Chapter $7$: Vector Spaces: $\S 34$. Dimension: Example $66$
Category:Standard Ordered Bases
Standard Parity Check Matrix/Examples/(6, 3) code in Z2

== Example of Standard Parity Check Matrix ==
Let $C$ be the linear $\tuple {6, 3}$-code in $\Z_2$ whose standard generator matrix $G$ is given by:
:$G := \begin{pmatrix}
1 & 0 & 0 & 1 & 1 & 0 \\ 
0 & 1 & 0 & 1 & 0 & 1 \\
0 & 0 & 1 & 0 & 1 & 1 \end{pmatrix}$
Its standard parity check matrix $P$ is given by:
:$P := \begin{pmatrix}
1 & 1 & 0 & 1 & 0 & 0 \\ 
1 & 0 & 1 & 0 & 1 & 0 \\
0 & 1 & 1 & 0 & 0 & 1 \end{pmatrix}$
== Proof ==
Expressing $G$ in the form:
:$G = \paren {\begin{array} {c|c} \mathbf I_k & \mathbf A \end{array} }$
it is seen that:
:$\mathbf A = \begin{pmatrix}
1 & 1 & 0 \\ 
1 & 0 & 1 \\
0 & 1 & 1 \end{pmatrix}$
It is noted that $\mathbf A^\intercal$ is:
:$\mathbf A^\intercal = \begin{pmatrix}
1 & 1 & 0 \\ 
1 & 0 & 1 \\
0 & 1 & 1 \end{pmatrix}$
as $\mathbf A$ is symmetrical about the main diagonal.
Then each of the elements of $\Z_2$ is self-inverse, so:
:$-\mathbf A^\intercal = \mathbf A^\intercal$
== Sources ==
* : Chapter $6$: Error-correcting codes: Example $6.18$
Category:Examples of Linear Codes
Standard Parity Check Matrix/Examples/(7, 3) code in Z2

== Example of Standard Parity Check Matrix ==
Let $C$ be the linear $\tuple {7, 3}$-code in $\Z_2$ whose standard generator matrix $G$ is given by:
:$G := \begin{pmatrix}
1 & 0 & 0 & 1 & 1 & 0 & 1 \\ 
0 & 1 & 0 & 1 & 1 & 1 & 0 \\
0 & 0 & 1 & 1 & 0 & 1 & 1 \end{pmatrix}$
Its standard parity check matrix $P$ is given by:
:$P := \begin{pmatrix}
1 & 1 & 1 & 1 & 0 & 0 & 0\\ 
1 & 1 & 0 & 0 & 1 & 0 & 0\\
0 & 1 & 1 & 0 & 0 & 1 & 0 \\
1 & 0 & 1 & 0 & 0 & 0 & 1 \end{pmatrix}$
== Proof ==
Expressing $G$ in the form:
:$G = \paren {\begin{array} {c|c} \mathbf I_k & \mathbf A \end{array} }$
it is seen that:
:$\mathbf A = \begin{pmatrix}
1 & 1 & 0 & 1 \\ 
1 & 1 & 1 & 0 \\
1 & 0 & 1 & 1 \end{pmatrix}$
It is noted that $\mathbf A^\intercal$ is:
:$\mathbf A^\intercal = \begin{pmatrix}
1 & 1 & 1 \\ 
1 & 1 & 0 \\
0 & 1 & 1 \\
1 & 0 & 1 \\
\end{pmatrix}$
as $\mathbf A$ is symmetrical about the main diagonal.
Then each of the elements of $\Z_2$ is self-inverse, so:
:$-\mathbf A^\intercal = \mathbf A^\intercal$
Its Syndrome Decoding table is:
:$\begin{array} {cc}
0000000 & 0000 \\
1000000 & 1101 \\
0100000 & 1110 \\
0010000 & 1011 \\
0001000 & 1000 \\
0000100 & 0100 \\
0000010 & 0010 \\
0000001 & 0001 \\
0000011 & 0011 \\
0000101 & 0101 \\
0000110 & 0110 \\
0000111 & 0111 \\
0001001 & 1001 \\
0001010 & 1010 \\
0001100 & 1100 \\
0001111 & 1111 \\
\end{array}$
The syndrome of $1100011$ is $1101 + 1110 + 0011 = 0000$, so $1100011$ is a codeword of $C$.
The syndrome of $1011000$ is $1101 + 1011 + 1000 = 1110$, so $1011000$ corrects to $1011000 - 0100000 = 1111000$.
The syndrome of $0101110$ is $1110 + 1100 + 0010 = 0000$, so $0101110$ is a codeword of $C$.
The syndrome of $0110001$ is $1110 + 1011 + 0001 = 0100$, so $0110001$ corrects to $0110001 - 0000100 = 0110101$.
The syndrome of $1010110$ is $1101 + 1011 + 0110 = 0000$, so $0101110$ is a codeword of $C$.
== Sources ==
* : Chapter $6$: Error-correcting codes: Exercise $5$
Category:Examples of Linear Codes
Standard Topology of Locally Convex Space has Local Basis of Balanced Convex Absorbing Sets

== Theorem ==
Let $\GF \in \set {\R, \C}$. 
Let $\struct {X, \PP}$ be a locally convex space over $\GF$ with standard topology $\tau$. 
Then there exists a local basis $\BB$ for ${\mathbf 0}_X$ in $\struct {X, \tau}$ such that: 
:each $A \in \BB$ is balanced, convex and absorbing.
== Proof ==
For each $\epsilon > 0$ and $S$ a finite subset of $\PP$, set: 
:$U_{\epsilon, S} = \set {x \in X : \map p x < \epsilon \text { for each } p \in S}$
and:
:$\BB = \set {U_{\epsilon, S} : \epsilon > 0 \text { and } S \subseteq \PP \text { finite} }$
From Open Sets in Standard Topology of Locally Convex Space, $\BB'$ is a local basis for ${\mathbf 0}_X$ in $\struct {X, \tau}$.
From Convex Open Neighborhood of Origin in Topological Vector Space contains Balanced Convex Open Neighborhood:
:for each $A \in \BB'$ there exists a balanced convex open neighborhood $V_A$ for ${\mathbf 0}_X$ such that $U_A \subseteq A$.
Then: 
:$\BB = \set {V_A : A \in \BB'}$
forms a local basis for ${\mathbf 0}_X$ in $\struct {X, \tau}$, consisting of balanced convex open neighborhoods of ${\mathbf 0}_X$.
From Locally Convex Space is Topological Vector Space, $\struct {X, \tau}$ is a topological vector space.
From Convex Subset of Topological Vector Space containing Zero Vector in Interior is Absorbing Set, we therefore obtain that $V_A$ is absorbing for each $A \in \BB'$.
Hence $\BB$ is a local basis for ${\mathbf 0}_X$ in $\struct {X, \tau}$, consisting of balanced convex absorbing sets. 
Category:Locally Convex Spaces
Category:Balanced Sets
Category:Absorbing Sets
Category:Convex Sets (Vector Spaces)
Standard Topology on Locally Convex Space makes Seminorms Continuous

== Theorem ==
Let $\struct {X, \PP}$ be a locally convex space.
Let $\tau$ be the standard topology on $\struct {X, \PP}$. 
Let $p \in \PP$.
Then $p$ is continuous as a function $\struct {X, \PP} \to \R$. 
== Proof ==
It is enough to show that for each $x \in X$ and $\epsilon > 0$, there exists $U_x \in \tau$ such that:
:$\size {\map p x - \map p y} < \epsilon$ for each $y \in U_x$. 
From the definition of the standard topology, we have:
:$U_x = \set {y \in X : \map p {y - x} < \epsilon} \in \tau$
Then from Reverse Triangle Inequality: Seminormed Vector Space, we have:
:$\size {\map p y - \map p x} \le \map p {y - x} < \epsilon$ 
Hence $U_x \in \tau$ works.
Category:Locally Convex Spaces
Standardized Random Variable/Examples/Arbitrary Example 1

== Example of Standardized Random Variables ==
Suppose we wish to compare the marks between examined assessments between students in mathematics and French.
Let these assessments be marked out of $100$.
It is not unusual for the unstandardized marks to be on different intervals, say:
:for mathematics, the marks are in the interval $\closedint 0 {95}$
:for French, the marks are in the interval $\closedint {25} {75}$.
Without standardizing the results, it is not possible to compare the assessments meaningfully.
== Sources ==
* 
* 
Category:Examples of Standardized Random Variables
Star Convex Set is Path-Connected

== Theorem ==
Let $A$ be a star convex subset of a topological vector space $V$ over $\R$ or $\C$.
Then $A$ is path-connected.
== Proof ==
Let $x_1, x_2 \in A$.
Let $a \in A$ be a star center of $A$.
By definition of star convex set:
:$\forall t \in \closedint 0 1: x_1 + \paren {1 - t} a, t x_2 + \paren {1 - t} a \in A$
Define two paths $\gamma_1, \gamma_2: t \in \closedint 0 1 \to A$ by:
:$\map {\gamma_1} t = t x_1 + \paren {1 - t} a$
:$\map {\gamma_2} t = t a + \paren {1 - t} x_2$
By definition of topological vector space, it follows that $\gamma_1$ and $\gamma_2$ are continuous.
Because:
:$\map {\gamma_2} t = \paren {1 - t} x_2 + \paren {1 - \paren {1 - t} } a$
and:
:$\paren {1 - t} \in \closedint 0 1$
it follows that:
:$\map {\gamma_2} t \in A$
Note that:
:$\map {\gamma_1} 0 = x_1$, $\map {\gamma_1} 1 = \map {\gamma_2} 0 = a$
and:
:$\map {\gamma_2} 1 = x_2$
Define $\gamma: \closedint 0 1 \to A$ as the concatenation $\gamma_1 * \gamma_2$.
Then $\gamma$ is a path in $A$ joining $x_1$ and $x_2$.
It follows by definition that $A$ is path-connected.
== Sources ==
* : $9$: The Fundamental Group: $\S 52$: The Fundamental Group
* : $\S 3.1$
Category:Vector Spaces
Category:Path-Connected Sets
Star Convex Set is Simply Connected

== Theorem ==
Let $A$ be a star convex subset of a topological vector space $\struct {V, \tau}$ over $\R$ or $\C$.
Let $\tau_A$ be the subspace topology on $A$ induced by $\tau$.
Then $\struct{ A, \tau_A }$ is simply connected.
== Proof ==
Let $a \in A$ be a star center of $A$.
Define $\mathbb I := \closedint 0 1$ as a closed real interval.
Let $\gamma : \mathbb I \to A$ be a loop in $A$ with base point $a$.
Let $\sigma : \mathbb I \to \set a$ be the constant function.
Constant Function is Continuous shows that $\sigma$ is continuous, so $\sigma$ is a loop with base point $a$.
Define $H: \mathbb I \times \mathbb I \to A$ by:
:$\map H { s, t } = t \map {\gamma} s + \paren { 1-t } a$
By definition of star convex set, we have $\map H { s, t } \in A$ for all $\tuple { s, t } \in \mathbb I \times \mathbb I$.
Define $F : \mathbb I \times \mathbb I \to A \times \mathbb I$ by $ \map F { s, t } = \tuple { \map { \gamma }{ s } , t }$.
As a loop is continuous, it follows that $F$ is continuous.
Define $\tilde G : V \times \mathbb I \to V$ by $\map { \tilde G }{ v, t } = t v + \paren { 1-t } a$.
By definition of topological vector space, it follows that $\tilde G$ is continuous.
Let $G = \tilde G {\restriction_{\paren {A \times \mathbb I} \times A} }$ be the restriction of $\tilde G$ to $\paren {A \times \mathbb I} \times A$.
Restriction of Continuous Mapping is Continuous:Topological Spaces shows that $G$ is continuous.
Composite of Continuous Mappings is Continuous shows that $ H = G \circ F$ is continuous.
For all $s, t \in \mathbb I$, we check that $H$ is a path homotopy between $\gamma$ and $\sigma$:
It follows that all loops with base point $a$ are path-homotopic with $\sigma$.
Relative Homotopy is Equivalence Relation shows that all loops with base point $a$ are path-homotopic with each other.
This implies that the fundamental group $\map { \pi_1 } { A, a }$ is trivial.
Star Convex Set is Path-Connected shows that $A$ is path-connected.
Hence the result.
== Sources ==
* : $9$: The Fundamental Group: $\S 52$: The Fundamental Group
Category:Vector Spaces
Category:Simply Connected Spaces
Star of Closed Subset of Banach *-Algebra is Closed

== Theorem ==
Let $\struct {A, \ast, \norm {\, \cdot \,} }$ be a Banach $\ast$-algebra over $\C$.
Let $C \subseteq A$ be closed.
Let:
:$C^\ast = \set {x^\ast : x \in C}$
Then $C^\ast$ is closed.
== Proof ==
Define $\phi : A \to A$ by $\map \phi a = a^\ast$ for each $a \in A$.
We have $C^\ast = \phi \sqbrk C$. 
From the definition of a Banach $\ast$-algebra, we have $\norm {\map \phi a} = \norm a$.
We have $\map \phi {a - b} = \map \phi a - \map \phi b$ for each $a, b \in A$ so:
:$\norm {a - b} = \norm {\map \phi a - \map \phi b}$ for each $a, b \in A$.
By the proof of Image of Closed Set under Linear Isometry from Banach Space to Normed Vector Space is Closed, $C^\ast$ is then closed.
Category:Banach *-Algebras
Star of Convergent Sequence in Banach *-Algebra Converges

== Theorem ==
Let $\struct {A, \ast}$ be a $\ast$-algebra over $\C$.
Let $\sequence {x_n}_{n \in \N}$ be a convergent sequence in $A$ with:
:$x_n \to x$ 
Then:
:$x_n^\ast \to x^\ast$ 
== Proof ==
We have:
so:
:$x_n^\ast \to x^\ast$
Category:Banach *-Algebras