nyuuzyou/wikis huggingface dataset:
Structure of Correlation Matrix

== Definition ==
Let $\sequence a_n$ and $\sequence b_n$ be sequences of $n$ observations.
Let $\mathbf C$ be the correlation matrix $\sequence a_n$ and $\sequence b_n$.
Then $\mathbf C$ is symmetric with entries on the main diagonal all equal to $1$.
== Proof ==
By definition of correlation matrix:
:$\sqbrk c_{i j} = r_{i j}$
where $r_{i j}$ is the sample correlation coefficient between $a_i$ and $b_j$.
Hence by definition of sample correlation coefficient:
:$\sqbrk c_{i j} = \dfrac {s_{a_i b_j} } {\sqrt {s_{a_i a_i} s_{b_j b_j} } }$
We see that:
demonstrating the symmetric nature of $\mathbf C$.
Then we note that:
Hence the result.
== Sources ==
* 
* 
Category:Correlation Matrices
Category:Symmetric Matrices
Structure of Inverse Completion of Commutative Semigroup

== Theorem ==
Let $\struct {S, \circ}$ be a commutative semigroup.
Let $\struct {C, \circ} \subseteq \struct {S, \circ}$ be the subsemigroup of cancellable elements of $\struct {S, \circ}$.
Let $\struct {T, \circ'}$ be an inverse completion of $\struct {S, \circ}$.
Then:
:$T = S \circ' C^{-1}$
where:
:$C^{-1}$ is the inverse of $C$ in $T$
:$S \circ' C^{-1}$ is the subset product of $S$ with $C^{-1}$.
== Proof ==
Let $a \in C$.
Then:
Thus, as:
:$C^{-1} \subseteq S \circ' C^{-1}$
and:
:$S \subseteq S \circ' C^{-1}$
by Union is Smallest Superset it follows that:
:$S \cup C^{-1} \subseteq S \circ' C^{-1}$
From Subset Product defining Inverse Completion of Commutative Semigroup is Commutative Semigroup:
:$S \circ' C^{-1}$ is a commutative semigroup.
So $S \circ' C^{-1}$ is a semigroup which contains $S \cup C^{-1}$.
By definition of generator of semigroup, it follows that:
:$\gen {S \cup C^{-1} } \subseteq S \circ' C^{-1}$
Let $z = x \circ' y^{-1} \in S \circ' C^{-1}$.
Then by definition:
:$x \in S$
and:
:$y^{-1} \in C^{-1}$
and so by definition of generator of semigroup:
:$x \circ' y^{-1} \in \gen {S \cup C^{-1} }$
Thus:
$S \circ' C^{-1} \subseteq \gen {S \cup C^{-1} }$
By definition of set equality:
:$S \circ' C^{-1} = \gen {S \cup C^{-1} }$
and so by definition of the inverse completion:
:$T = S \circ' C^{-1}$
== Sources ==
* : Chapter $\text {IV}$: Rings and Fields: $\S 20$: The Integers: Theorem $20.1: \ 1^\circ$
Category:Inverse Completions
Category:Commutative Semigroups
Structure of Moore-Penrose Pseudoinverse

== Theorem ==
Let $\mathbf A$ be a matrix of order $m \times n$ over the complex numbers $\C$.
Let $\mathbf A^+$ be the Moore-Penrose pseudoinverse of $\mathbf A$.
Let $\mathbf A = \mathbf U \mathbf D \mathbf V^*$ be a singular value decomposition of $\mathbf A$.
Then:
:$\mathbf A^+ = \mathbf V \mathbf D^+ \mathbf U^*$
where $\mathbf D^+$ is obtained by inverting the non-zero diagonal elements of $\mathbf D$.
In the above, recall that $\mathbf V^*$ denotes the Hermitian conjugate of $\mathbf V$.
== Proof ==
== Sources ==
* 
Category:Moore-Penrose Pseudoinverses
Structure of Recurring Decimal

== Theorem ==
Let $\dfrac 1 m$, when expressed as a decimal expansion, recur with a period of $p$ digits with no nonperiodic part.
Let $\dfrac 1 n$, when expressed as a decimal expansion, terminate after $q$ digits.
Then $\dfrac 1 {m n}$ has a nonperiodic part of $q$ digits, and a recurring part of $p$ digits.
== Proof ==
Let $b \in \N_{>1}$ be the base we are working on.
Note that $b^p \times \dfrac 1 m$ is the result of shifting the decimal point of $\dfrac 1 m$ by $p$ digits.
Hence $b^p \times \dfrac 1 m - \dfrac 1 m$ is an integer, and $\paren {b^i - 1} \dfrac 1 m$ is not an integer for integers $0 < i < p$.
Therefore $m \divides b^p - 1$.
Also note that $b^q \times \dfrac 1 n$ is the result of shifting the decimal point of $\dfrac 1 n$ by $q$ digits.
Hence $b^q \times \dfrac 1 n$ is an integer, and $b^j \times \dfrac 1 n$ is not an integer for integers $0 < j < q$.
Therefore $n \divides b^q$ and $n \nmid b^{q - 1}$.
Write $m x = b^p - 1$ and $n y = b^q$.
Then $\dfrac 1 {m n} = \dfrac {x y} {\paren {b^p - 1} b^q}$.
By Division Theorem:
:$\exists! r, s \in \Z: x y = s \paren {b^p - 1} + r, 0 \le r < b^p - 1$
Then we would have:
:$\dfrac 1 {m n} = \dfrac {s \paren {b^p - 1} + r} {\paren {b^p - 1} b^q} = \dfrac s {b^q} + \dfrac r {\paren {b^p - 1} b^q}$
which is a fraction with a nonperiodic part of $q$ digits equal to $s$, followed by a recurring part of $p$ digits equal to $r$.
To show that the nonperiodic part and recurring part of $\dfrac 1 {m n}$ cannot be shorter, we must show:
:$r \not \equiv s \pmod b$: or else the nonperiodic part could be shortened by at least $1$ digit
:$\dfrac r {b^p - 1} \paren {b^i - 1}$ is not an integer for integers $0 < i < p$: or else the recurring part could be shortened to $i$ digits
To show the first condition, suppose the contrary.
Then:
From GCD with Remainder:
:$\gcd \set {b, b^p - 1} = 1$
Since $x \divides b^p - 1$, by Divisor of One of Coprime Numbers is Coprime to Other:
:$\gcd \set {b, x} = 1$
By Euclid's Lemma:
:$b \divides y$
From $n y = b^q$:
:$n \times \dfrac y b = b^{q - 1}$
so $n \divides b^{q - 1}$, which contradicts the properties of $n$.
To show the second condition, suppose the contrary.
Then:
Both fractions are integers, so $\paren {b^i - 1} \dfrac 1 m$ is also an integer, which contradicts the properties of $\dfrac 1 m$.
Hence the result.
== Sources ==
* : $5$
* : $5$
Category:Number Theory
Structure of Simple Algebraic Field Extension

== Theorem ==
Let $F / K$ be a field extension.
Let $\alpha \in F$ be algebraic over $K$.
Let $\mu_\alpha$ be the minimal polynomial of $\alpha$ over $K$.
Let $K \sqbrk X$ be the ring of polynomial functions.
Let $\gen {\mu_\alpha}$ be the ideal of $K \sqbrk X$ generated by $\mu_\alpha$.
Let $K \sqbrk \alpha$ be the subring of $F$ generated by $K \cup \set \alpha$.
Let $\map K \alpha$ be the subfield of $F$ generated by $K \cup \set \alpha$.
Then:
:$K \sqbrk \alpha = \map K \alpha \simeq K \sqbrk X / \gen {\mu_\alpha}$
Moreover:
:$n := \index {\map K \alpha} K = \deg \mu_\alpha$
and:
:$1, \alpha, \dotsc, \alpha^{n - 1}$ is a basis of $\map K \alpha$ over $K$.
== Proof ==
Define $\phi: K \sqbrk X \to K \sqbrk \alpha$ by:
:$\map \phi f = \map f \alpha$
We have
:$\map \phi f = 0 \iff \map f \alpha = 0 \iff \mu_\alpha \divides f$
where the last equivalence follows from $\mu_\alpha$ being the minimal polynomial of $\alpha$.
Thus:
:$\ker \phi = \set {f \in K \sqbrk X: \mu_\alpha \divides f} =: \gen {\mu_\alpha}$
By the corollary to Field Adjoined Set $\phi$ is surjective.
So by the First Ring Isomorphism Theorem:
:$K \sqbrk X / \gen {\mu_\alpha} \simeq K \sqbrk \alpha$
By Principal Ideal of Principal Ideal Domain is of Irreducible Element iff Maximal:
:$\gen {\mu_\alpha}$ is maximal.
So $K \sqbrk \alpha$ is a field by Maximal Ideal iff Quotient Ring is Field.
Also $K \sqbrk \alpha$ is the smallest ring containing $K \cup \set \alpha$.
So because a field is a ring it is also the smallest field containing $K \cup \set \alpha$.
This shows that $K \sqbrk \alpha = \map K \alpha$.
By Field Adjoined Set:
:$K \sqbrk \alpha = \set {\map f \alpha: f \in K \sqbrk X}$
where $K \sqbrk X$ is the ring of polynomial functions over $K$.
From Division Theorem for Polynomial Forms over Field, for each $f \in K \sqbrk X$ there are $q, r \in K \sqbrk X$ such that:
:$f = q \mu_\alpha + r$
and:
:$\deg r < d =: \deg \mu_\alpha$
Therefore, since $\map {\mu_\alpha} \alpha = 0$, we have 
Therefore $1, \dotsc, \alpha^{d - 1}$ span $K \sqbrk \alpha / K$ as a vector space.
Moreover if $s \in K \sqbrk X$, then $\map s \alpha = 0$ and $\deg s < \deg \mu_\alpha$ implies that $s = 0$.
Thus $1, \dotsc, \alpha^{d - 1}$ are linearly independent.
Therefore no non-zero $K$-combination of $1, \dotsc, \alpha^{d - 1}$ is zero.
Therefore $\set {1, \dotsc, \alpha^{d - 1} }$ is a basis of $K \sqbrk \alpha / K$.
The degree of $K \sqbrk \alpha / K$ is by definition the number of elements of a basis for $K \sqbrk \alpha / K$.
Therefore:
:$d = \deg \mu_\alpha = \index {K \sqbrk \alpha} K$
Category:Field Extensions
Structure of Simple Transcendental Field Extension

== Definition ==
Let $F / K$ be a field extension and $\alpha \in F$.
Let $\map K X$ be the Field of Rational Functions in an indeterminate $X$.
If $\alpha$ is transcendental over $K$ then $\map K \alpha \simeq \map K X$.
== Proof ==
Let $\phi: K \sqbrk X \to K \sqbrk \alpha$ be the Evaluation Homomorphism.
We have that:
:$\map \phi f = \map f \alpha$
Therefore by definition of transcendental element:
:$\map \phi f = 0 \implies f = 0$
Moreover $\phi$ is surjective by the corollary to Field Adjoined Set.
Therefore by the First Ring Isomorphism Theorem:
:$K \sqbrk \alpha \simeq K \sqbrk X$
We have that the construction of the field of quotients $\map K X$ uses only the ring axioms.
Thus it follows that:
:$\map Q {K \sqbrk \alpha} = \map Q {K \sqbrk X}$
where $Q$ maps a integral domain to its field of quotients.
That is:
:$\map K \alpha \simeq \map K X$
Category:Field Extensions
Structure under Left Operation is Semigroup

== Theorem ==
Let $\struct {S, \gets}$ be an algebraic structure in which the operation $\gets$ is the left operation.
Then $\struct {S, \gets}$ is a semigroup.
== Proof ==
We need to verify the semigroup axioms:
By the nature of the right operation, $\struct {S, \to}$ is closed:
:$\forall x, y \in S: x \gets y = y \in S$
whatever $S$ may be.
Hence holds.
From Right Operation is Associative, $\gets$ is associative.
Hence holds.
So $\struct {S, \gets}$ is a semigroup.
== Also see ==
* Structure under Right Operation is Semigroup
Category:Left Operation
Category:Examples of Semigroups
Structure under Right Operation is Semigroup

== Theorem ==
Let $\struct {S, \to}$ be an algebraic structure in which the operation $\to$ is the right operation.
Then $\struct {S, \to}$ is a semigroup.
== Proof ==
We need to verify the semigroup axioms:
By the nature of the right operation, $\struct {S, \to}$ is closed:
:$\forall x, y \in S: x \to y = y \in S$
whatever $S$ may be.
Hence holds.
From Right Operation is Associative, $\to$ is associative.
Hence holds.
So $\struct {S, \to}$ is a semigroup.
== Also see ==
* Structure under Left Operation is Semigroup
== Sources ==
* : Chapter $5$: Semigroups: Exercise $3 \ \text{(i)}$
Category:Right Operation
Category:Examples of Semigroups
Structure with Commutative Idempotent Associative Operations satisfying Absorption Laws is Lattice

== Theorem ==
Let $S$ be a set.
Let $\vee$ and $\wedge$ be binary operations which, when applied to $S$, are both:
:closed operations
:commutative operations
:idempotent operations
:associative operations.
Furthermore, let $\vee$ and $\wedge$ satisfy the absorption laws:
:$\forall a, b \in S: a \vee \paren {a \wedge b} = a = a \wedge {a \vee b}$
Then there exists a unique lattice ordering $\preccurlyeq$ on $S$ such that:
:$\forall a, b \in S$:
::$x \vee y = \sup \set {a, b}$
::$x \wedge y = \inf \set {a, b}$
That is:
:$\struct {S, \vee, \wedge, \preccurlyeq}$ is a lattice.
== Proof ==
We have that:
:$\struct {S, \vee}$ is a commutative idempotent semigroup 
:$\struct {S, \wedge}$ is a commutative idempotent semigroup.
That is, $\struct {S, \vee}$ and $\struct {S, \wedge}$ are semilattices.
We are also given that $\vee$ and $\wedge$ satisfy the absorption laws:
:$\forall a, b \in S: a \vee \paren {a \wedge b} = a = a \wedge {a \vee b}$
Finally we note that from Semilattice has Unique Ordering such that Operation is Supremum, there exists a unique ordering $\preccurlyeq$ on $S$ such that:
:$a \vee b = \sup \set {a, b}$
where $\sup \set {a, b}$ is the supremum of $\set {x, y}$ $\preccurlyeq$.
Hence, by definition, the ordered structure:
:$\struct {S, \vee, \wedge, \preccurlyeq}$
is a lattice.
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 14$: Orderings: Exercise $14.23 \ \text {(b)}$
Category:Lattice Theory
Structure with Element both Identity and Zero has One Element

== Theorem ==
Let $\struct {S, \circ}$ be an algebraic structure.
Let $z \in S$ such that $z$ is both an identity element and a zero element.
Then:
:$S = \set z$
== Proof ==
Let $x \in S$.
Then
and so there is no other element of $S$ but $z$.
== Sources ==
* : Chapter $4$. Groups: Exercise $3$
Category:Identity Elements
Category:Zero Elements
Student's t-Distribution with One Degree of Freedom is Standard Cauchy Distribution

== Theorem ==
The Student's $t$-distribution with one degree of freedom is a special case of a standard Cauchy distribution.
== Proof ==
Let $X$ be a continuous random variables with a Student's $t$-distribution with one degree of freedom.
Then $X$ has probability density function: 
:$\map {f_X} x = \dfrac {\map \Gamma {\frac {k + 1} 2} } {\sqrt {\pi k} \map \Gamma {\frac k 2} } \paren {1 + \dfrac {x^2} k}^{-\frac {k + 1} 2}$
where $k = 1$.
Hence:
which is exactly the standard Cauchy distribution.
== Sources ==
* 
* 
Category:Student's t-Distribution
Category:Standard Cauchy Distribution
Sturm-Liouville Problem

== Theorem ==
Let:
:$P \in C^\infty \map P x > 0$
:$Q \in C^0$
:$-\paren {P y'}' + Q y = \lambda y$
:$\map y a = \map y b = 0$
Then the Sturm-Liouville problem has an infinite sequence of eigenvalues $\set {\lambda^{\paren n} }$, and to each $\lambda^{\paren n}$ corresponds an eigenfunction $y^{\paren n}$, unique up to a constant factor.
== Proof ==
:$J \sqbrk y = \ds \int_a^b \paren {P y'^2 + Q y^2} \rd x$
:$\ds \int_a^b y^2 \rd x = 1$
:$\ds \int_a^b \paren {P y'^2 + Q y^2} \rd x > \int_a^b Q y^2 \rd x \ge M \int_a^b y^2 \rd x = M$
:$M = \min \limits_{a \mathop \le x \mathop \le b} \map Q x$
Assume $a = 0$, $b = \pi$.
Choose $\set {\map {\phi_n} x} = \set {\sin n x}$
For $k \ne l$ we have:
:$\ds \int_0^\pi \sin k x \sin l x \rd x = 0$
:$\ds \int_0^\pi \paren {\sum_{k \mathop = 1}^n \alpha_k \sin k x}^2 \rd x = \dfrac \pi 2 \sum_{k \mathop = 1}^n \alpha_k^2 = 1$
:$\ds \map {J_n} {\alpha_1, \ldots, \alpha_n} = \int_o^\pi \paren {\map P {\sum_{k \mathop = 1}^n \alpha_k \sin k x}'^2 + \map Q {\sum_{k \mathop = 1}^n \alpha_k \sin k x}^2} \rd x$
:$\ds \map {y_n^{\paren 1} } x = \sum_{k \mathop = 1}^n \alpha_k^{\paren 1} \sin k x$
:$\set {\lambda_n^{\paren 1} }$
:$\set {y_n^{\paren 1} }$
:$\map {J_n} {\alpha_1, \ldots, \alpha_n} = \map {J_{n + 1} } {\alpha_1, \ldots, \alpha_n, 0}$
:$\lambda_{n + 1}^{\paren 1} \le \lambda_n^{\paren 1}$
:$\lambda^{\paren 1} = \lim_{n \mathop \to \infty} \lambda_n^{\paren 1}$
:$\ds \lambda_n^{\paren 1} = \int_0^\pi \paren {P y_n'^2 + Q y_n^2} \rd x$
:$\ds \int_0^\pi \paren {P y_n'^2 + Q y_n^2} \rd x \le M$
:$\ds \int_0^\pi Py_n'^2 \rd x \le M + \size {\int_0^\pi Q y_n^2 \rd x} \le M + \max \limits_{a \mathop \le x \mathop \le b } \size {\map Q x} = M_1$
:$\ds \int_0^\pi \map {y_n'^2} x \rd x \le \dfrac {M_1} {\min \limits_{a \mathop \le x \mathop \le b} \map P x} = M_2$
:$\map {y_n} 0 = 0$
:$\ds \size {\map {y_n} x}^2 = \size {\int_0^x \map {y_n'} \zeta \rd \zeta}^2 \le \int_0^x \map {y_n'^2} \zeta \rd \zeta \int_0^x \rd \zeta \le M_2 \pi$
:$\ds \size {\map {y_n} {x_2} - \map {y_n} {x_1} }^2 = \size {\int_{x_1}^{x_2} \map {y_n'} x \rd x}^2 \le \int_{x_1}^{x_2} y_n'^2 \rd x \size {\int_{x_1}^{x_2} \rd x}^2 \le M_2 \size {x_2 - x_1}$
:$\map {y^{\paren 1} } x = \lim_{m \mathop \to \infty} \map {y_{n_m} } x$
:$\int_0^\pi \paren {-\paren {P h'}' + Q_1 h} y \rd x = 0$
:$\map h x \in \map {\DD_2} {0, \pi}$
:$\map h 0 = \map h \pi = \map {h'} 0 = \map {h'} \pi = 0$
:$\map y x \in \map {\DD_2} {0, \pi}$
:$-\paren {P y'}' + Q_1 y = 0$
:$\ds \int_0^\pi \paren {-\paren {P y'}' + Q_1 y} y \rd x = -\int_0^\pi P h' ' y \rd x - \int_0^\pi P' h' y \rd x + \int_0^\pi Q_1 h y \rd x = - \int_0^\pi \paren {- P y + \int_0^x P' y \rd \zeta + \int_0^x \paren {\int_0^\zeta Q_1 y \rd t} \rd \zeta} \rd x = 0$
:$-\paren {P y}' + P' y + \ds \int_0^x Q_1 y \rd \zeta = c_1$
:$-P y' + \ds \int_0^x Q_1 y \rd \zeta = c_1$
:$-\paren {P y'}' + Q_1 y = 0$
:$-\paren {P {y^{\paren 1} }'}' + Q y^{\paren 1} = \lambda^{\paren 1} y^{\paren 1}$
:$\dfrac \partial {\partial \alpha_r} \paren {\map {J_n} {\alpha_1, \ldots, \alpha_n} - \lambda_n^{\paren 1} \ds \int_0^\pi \paren {\sum_{k \mathop = 1}^n \alpha_k \sin k x}^2 \rd x} = 0$
Sturm-Liouville Problem/Unit Weight Function/Lemma

== Theorem ==
Let $\map \alpha x: \R \to \R$ such that $\map \alpha x \in C^2 \closedint a b$.
Suppose:
:$\ds \forall \map h x \in C^2 \closedint a b: \map h a = \map h b = \map {h'} a = \map {h'} b = 0: \int_a^b \map \alpha x \, \map {h} x \rd x = 0$
Then:
:$\forall x \in \closedint a b: \map \alpha x = c_0 + c_1 x$
where $ c_0, c_1 \in \R $ are constants.
== Proof ==
Let $ c_0, c_1$ be defined by the conditions:
:$\ds \int_a^b \paren {\map \alpha x - c_0 - c_1 x} \rd x = 0$
:$\ds \int_a^b \rd x \int_a^x \paren {\map \alpha \xi - c_0 - c_1 \xi} \rd \xi = 0$
Suppose:
:$\ds \map h x = \int_a^x \xi \int_a^\xi \paren {\map \alpha t - c_0 - c_1 t} \rd t$
This form satisfies conditions on $h$ in the theorem.
Then:
On the other hand:
Therefore:
:$\map \alpha x - c_0 - c_1 x = 0$
or:
:$\map \alpha x = c_0 + c_1 x$
Category:Calculus of Variations
Sub-Basis for Initial Topology in terms of Sub-Bases of Target Spaces

== Theorem ==
Let $X$ be a set.
Let $I$ be an indexing set.
Let $\family {\struct {Y_i, \tau_i} }_{i \mathop \in I}$ be an indexed family of topological spaces indexed by $I$.
For each $i \in I$, let $S_i$ be a synthetic basis for $\struct {Y_i, \tau_i}$. 
Let $\family {f_i: X \to Y_i}_{i \mathop \in I}$ be an indexed family of mappings indexed by $I$.
Let $\tau$ be the initial topology on $X$ with respect to $\family {f_i}_{i \mathop \in I}$.
Then: 
:$\SS = \set {f_i^{-1} \sqbrk U : i \in I, \, U \in S_i}$
is a synthetic sub-basis for $\struct {X, \tau}$. 
== Proof ==
Note that by the definition of the initial topology, $\tau$ is generated by the synthetic sub-basis: 
:$\SS' = \set {f_i^{-1} \sqbrk U : i \in I, \, U \in \tau_i}$ 
Since $S_i \subseteq \tau_i$ for each $i \in I$, we have: 
:$\SS \subseteq \SS'$
and hence: 
:$\map \tau \SS \subseteq \map \tau {\SS'} = \tau$
where $\tau$ denotes the generated topology.
Now fix $i \in I$ and $U \in \tau_i$. 
Since $S_i$ is a synthetic sub-basis for $\tau_i$, there exists an indexing set $A$ such that: 
:for each $\alpha \in A$ there exists $n_\alpha \in \N$ and $U_{\alpha, 1}, U_{\alpha, 2}, \ldots, U_{\alpha, n_\alpha}$ such that together: 
::$\ds U = \bigcup_{\alpha \mathop \in A} \, \bigcap_{i \mathop = 1}^{n_\alpha} U_{\alpha, n_\alpha}$
Then we have:
where $f_i^{-1} \sqbrk {U_{\alpha, n_\alpha} } \in \SS$ for each $\alpha \in A$. 
So $f_i^{-1} \sqbrk U$ can be expressed as the set union of finite intersections of elements of $\map \tau \SS$.
Since topologies are closed under unions and finite intersections, for each $i \in I$ and $U \in \tau_i$ we have $f_i^{-1} \sqbrk U \in \map \tau \SS$. 
So we have shown that: 
:$\SS' \subseteq \map \tau \SS$
so that: 
:$\tau = \map \tau {\SS'} \subseteq \map \tau {\SS'}$ 
giving:
:$\map \tau \SS = \map \tau {\SS'} = \tau$
So $\SS$ is a synthetic sub-basis for $\struct {X, \tau}$.
Category:Initial Topology
Category:Topological Bases
Sub-Basis for Real Number Line

== Theorem ==
Let the real number line $\R$ be considered as a topology under the usual (Euclidean) topology.
Then:
:$\BB := \set {\openint \gets a, \openint b \to: a, b \in \R}$ is a sub-basis for $\R$.
== Proof ==
Let $\openint c d$ be an open real interval.
Then by definition:
:$\openint c d = \openint \gets d \cap \openint c \to$
and so $\openint c d$ is the intersection of two elements of $\BB$.
From Open Sets in Real Number Line, any open set of $\R$ is the union of countably many open real intervals.
The result follows by definition of sub-basis.
== Sources ==
* : $3$: Continuity generalized: topological spaces: $3.3$: Sub-bases and weak topologies: Example $3.3.1$
Category:Topological Bases
Category:Real Numbers
Sub-Basis for Topological Subspace

== Theorem ==
Let $\struct {X, \tau}$ be a topological space.
Let $K$ be a sub-basis for $\tau$.
Let $\struct {S, \tau'}$ be a subspace of $\struct {X, \tau}$.
Let $K' = \set {U \cap S: U \in K}$.
That is, $K'$ consists of the $\tau'$-open sets in $S$ corresponding to elements of $K$.
Then $K'$ is a sub-basis for $\tau'$.
== Proof ==
Let $B$ be the basis for $\tau$ generated by $K$.
By Basis for Topological Subspace, $B$ generates a basis, $B'$, for $\tau'$.
We will show that $K'$ generates $B'$.
Let $V' \in B'$.
Then for some $V \in B$:
:$V' = S \cap V$
By the definition of a sub-basis, there is a finite subset $K_V$ of $K$ such that:
:$\ds V = \bigcap K_V$
Let $K_V' = \set {S \cap U: U \in K_V}$.
Then:
:$\ds V' = S \cap V = S \cap \bigcap K_V$
so:
:$\ds V' = \bigcap_{P \mathop \in K_V} \paren {S \cap P} = \bigcap K_V'$
Thus $B'$ is generated by $K'$.
Category:Topological Bases
Category:Topological Subspaces
Sub-basis for Uniformity on Real Number Line

== Theorem ==
Let $\struct {\R, \tau_d}$ be the real number line with the usual (Euclidean) topology.
Let $a, b \in \R$ such that $a < b$.
Let $S_{a b}$ be the set of subsets of $\R$ defined as:
:$S_{a b} = \set {\tuple {x, y}: x, y < b \text{ or } x, y > a}$
Then $S_{ab}$ is a basis for a uniformity $U$ which generates the usual topology on $\R$.
Note that $U$ is clearly not the usual metric uniformity.
== Proof ==
== Sources ==
* : Part $\text {II}$: Counterexamples: $28$. Euclidean Topology: $8$
Category:Real Number Line with Euclidean Topology
Subadditivity of Invariant Metric on Vector Space

== Theorem ==
Let $K$ be a field.
Let $X$ be a vector space over $K$.
Let $d$ be an invariant metric on $X$.
Then:
:$\map d {n x, {\mathbf 0}_X} \le n \map d {x, {\mathbf 0}_X}$
for each $n \in \N$ and $x \in X$. 
== Proof ==
We have:
== Sources ==
* : $1.28$: Theorem
Category:Invariant Metrics on Vector Spaces
Subalgebra Generated by Commuting Elements is Commutative

== Theorem ==
Let $K$ be a field. 
Let $A$ be an algebra over $K$.
Let $S \subseteq A$ be a non-empty set such that:
:for all $x, y \in S$ we have $x y = y x$. 
Let $K \sqbrk S$ be the subalgebra generated by $S$.
Then $K \sqbrk S$ is a commutative algebra.
== Proof ==
From Explicit Form for Generated Subalgebra, we have:
:$K \sqbrk S = \span \set {x_1^{k_1} x_2^{k_2} \ldots x_n^{k_n} : x_1, \ldots, x_n \in S, \, k_1, \ldots, k_n \ge 1}$
and $K \sqbrk S$ is a subalgebra of $A$. 
Let $x, y \in K \sqbrk S$. 
Then:
:$\ds x = \sum_{j \mathop = 1}^N \lambda_j x_{j, 1}^{k_{j, 1} } x_{j, 2}^{k_{j, 2} } \ldots x_{j, n}^{k_{j, n} }$
and:
:$\ds y = \sum_{i \mathop = 1}^M \mu_i x_{i, 1}^{s_{i, 1} } x_{i, 2}^{s_{i, 2} } \ldots x_{i, m}^{k_{i, m} }$
where each $x_{j, k}$ and $x_{i, k}$ is in $S$.
Then we have:
:$\ds x y = \sum_{j \mathop = 1}^N \sum_{i \mathop = 1}^M \lambda_j x_{j, 1}^{k_{j, 1} } x_{j, 2}^{k_{j, 2} } \ldots x_{j, n}^{k_{j, n} } \mu_i x_{i, 1}^{s_{i, 1} } x_{i, 2}^{s_{i, 2} } \ldots x_{i, m}^{k_{i, m} }$
Since $K$ is a field and every element in $S$ commutes, we have:
:$\ds \sum_{j \mathop = 1}^N \sum_{i \mathop = 1}^M \paren {\lambda_j x_{j, 1}^{k_{j, 1} } x_{j, 2}^{k_{j, 2} } \ldots x_{j, n}^{k_{j, n} } } \paren {\mu_i x_{i, 1}^{s_{i, 1} } x_{i, 2}^{s_{i, 2} } \ldots x_{i, m}^{k_{i, m} } } = \sum_{i \mathop = 1}^M \sum_{j \mathop = 1}^N \paren {\mu_i x_{i, 1}^{s_{i, 1} } x_{i, 2}^{s_{i, 2} } \ldots x_{i, m}^{k_{i, m} } } \paren {\lambda_j x_{j, 1}^{k_{j, 1} } x_{j, 2}^{k_{j, 2} } \ldots x_{j, n}^{k_{j, n} } }$
Then:
:$\ds \sum_{i \mathop = 1}^M \sum_{j \mathop = 1}^N \paren {\mu_i x_{i, 1}^{s_{i, 1} } x_{i, 2}^{s_{i, 2} } \ldots x_{i, m}^{k_{i, m} } } \paren {\lambda_j x_{j, 1}^{k_{j, 1} } x_{j, 2}^{k_{j, 2} } \ldots x_{j, n}^{k_{j, n} } } = \paren {\sum_{i \mathop = 1}^M \mu_i x_{i, 1}^{s_{i, 1} } x_{i, 2}^{s_{i, 2} } \ldots x_{i, m}^{k_{i, m} } } \paren {\sum_{j \mathop = 1}^N \lambda_j x_{j, 1}^{k_{j, 1} } x_{j, 2}^{k_{j, 2} } \ldots x_{j, n}^{k_{j, n} } } = y x$
Hence $K \sqbrk S$ is a commutative algebra.
Category:Algebras