nyuuzyou/wikis huggingface dataset:
Basis Representation Theorem for Ordinals

== Theorem ==
Let $x$ and $y$ be ordinals.
Let $x > 1$ and $y > 0$.
Then there exist unique finite sequences of ordinals:
:$\sequence {a_i}, \sequence {b_i}$
both of unique length $n$ such that:
:$(1): \quad \sequence {a_i}$ is a strictly decreasing sequence for $1 \le i \le n$
:$(2): \quad 0 < b_i < x$ for all $1 \le i \le n$
:$(3): \quad \ds y = \sum_{i \mathop = 1}^n x^{a_i} b_i$
== Proof ==
The proof shall proceed by Transfinite Induction (Strong Induction) on $y$.
The inductive hypothesis states that for all $v < y$, there exist unique finite sequences of ordinals:
:$\sequence {c_i}, \sequence {d_i}$
both of unique length $n$ such that:
:$(1): \quad \sequence {c_i}$ is strictly decreasing sequence for $1 \le i \le n$
:$(2): \quad 0 < d_i < x$ for all $1 \le i \le n$.
:$(3): \quad \ds v = \sum_{i \mathop = 1}^n x^{c_i} d_i$
Since $x > 1$, it follows by Unique Ordinal Exponentiation Inequality that:
:$x^z \le y < x^{z + 1}$
for some unique $z$.
By the Division Theorem for Ordinals:
:$y = x^z w + v$ and $v < x^z$
for some unique $z$, $w$, and $v$.
Thus:
So $v < y$.
If $v = 0$, then the statement is proven.
If $v \ne 0$, then we may apply the inductive hypothesis.
By the inductive hypothesis:
:$\ds v = \sum_{i \mathop = 1}^n x^{c_i} d_i$
Therefore:
:$\ds y = x^z w + \sum_{i \mathop = 1}^n x^{c_i} d_i$
Set:
:$a_1 = z$
:$a_{i + 1} = c_i$ for $1 \le i \le n$
:$b_1 = w$
:$b_{i + 1} = d_i$ for $1 \le i \le n$
Since $w \ne 0$ and $d_i \ne 0$, it follows that $b_i \ne 0$ for all $1 \le i \le n+1$.
Moreover, since $z > c_1$ and $\sequence {c_i}$ is strictly decreasing, it follows that also $\sequence {a_i}$ is strictly decreasing.
The equation for the first lemma can be rewritten:
:$\ds y = x^{a_1} b_1 + \sum_{i \mathop = 1}^n x^{a_{i + 1} } b_{i + 1}$
By General Associative Law for Ordinal Sum, it follows that:
:$\ds y = \sum_{i \mathop = 1}^{n + 1} x^{a_i} b_i$
Thus, existence is proven.
Furthermore, since $z$ and $\sequence {c_i}$ are unique, and $w$ and $\sequence {a_i}$ are unique, then $\sequence {a_i}$ and $\sequence {b_i}$ are unique.
== Also see ==
* Definition:Cantor Normal Form
== Sources ==
* : $\S 8.44$
Category:Ordinal Arithmetic
Basis Representation is No Longer than Number

== Theorem ==
Let $b > 1$ and $n > 0$ be natural numbers.
Let the base-$b$ representation of $n$ be:
:$\sqbrk {r_m r_{m - 1} \dotsm r_1 r_0}_b$
Then:
:$n > m$
== Proof ==
Category:Number Bases
Basis Representation is Primitive Recursive

== Theorem ==
Let $\operatorname{basis} : \N^3 \to \N$ be defined as follows:
:$\map {\operatorname{basis} } {b, n, i} = \begin{cases}
1 & : b = 1 \land i < n \\
r_i & : b > 1 \land i \le m \\
0 & : \text {otherwise}
\end{cases}$
where $\sqbrk {r_m r_{m - 1} \dots r_1 r_0}_b$ is the base-$b$ representation of $n$.
== Proof ==
Consider the following function:
:$\map f {b, n, i} = \begin{cases}
n & : i = 0 \\
\map {\operatorname{quot} } {\map f {b, n, i - 1}, b} & : i > 0
\end{cases}$
As $f$ is obtained by Primitive Recursion and:
* Constant Function is Primitive Recursive
* Quotient is Primitive Recursive
Thus, $f$ is primitive recursive.
Observe that $\map f {b, n, 0} = n$ which is the number represented by $\sqbrk {r_m r_{m - 1} \dots r_1 r_0}_b$.
Now, suppose that $\map f {b, n, i}$ is the number represented by $\sqbrk {r_m r_{m - 1} \dots r_{i + 1} r_i}_b$.
Then:
Thus, by Principle of Mathematical Induction:
:$\map f {b, n, i} = \sqbrk {r_m r_{m - 1} \dots r_{i + 1} r_i}$
Define $\map g {b, n, i} = \map {\operatorname{rem} } {\map f {b, n, i}, b}$.
Then $g$ is primitive recursive, since Remainder is Primitive Recursive.
For each $i$:
The result follows from:
* Definition by Cases is Primitive Recursive
* Ordering Relations are Primitive Recursive
* Set Operations on Primitive Recursive Relations
* Constant Function is Primitive Recursive
Category:Primitive Recursive Functions
Basis Test for Adherent Point

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Let $\BB$ be a synthetic basis of $T$.
Let $H \subseteq S$.
Then $x \in S$ is an adherent point of $H$ :
:$\forall U \in \BB : x \in U$ satisfies $H \cap U \ne \O$
== Proof ==
=== Necessary Condition ===
Let $x \in S$ be an adherent point of $H$.
By definition of an adherent point of $H$:
:$\forall U \in \tau : x \in U$ satisfies $H \cap U \ne \O$
By definition of a synthetic basis of $T$:
:$\BB \subseteq \tau$
The result follows.
=== Sufficient Condition ===
Let $x$ satisfy:
:$\forall U \in \BB : x \in U$ satisfies $H \cap U \ne \O$
Let $V$ be any open neighborhood of $x$.
By definition of a synthetic basis of $T$:
:$\exists U \in \BB : x \in U \subseteq V$
Then:
:$H \cap U \ne \O$
From the contrapositive statement of Subsets of Disjoint Sets are Disjoint:
:$H \cap V \ne \O$
Thus $x$ is an adherent point of $H$ by definition.
Category:Adherent Points
Basis Test for Isolated Point

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Let $\BB$ be a synthetic basis of $T$.
Let $H \subseteq S$.
Then $x \in H$ is an isolated point of $H$ :
:$\exists U \in \BB : U \cap H = \set x$
== Proof ==
=== Necessary Condition ===
Let $x \in H$ be an isolated point of $H$.
By definition of an isolated point:
:$\exists U \in \tau: U \cap H = \set x$
By definition of a synthetic basis of $T$:
:$\exists V \in \BB: x \in V \subseteq U$
From Set Intersection Preserves Subsets:
:$V \cap H \subseteq U \cap H = \set x$
From Singleton of Element is Subset:
:$\set x \subseteq V \cap H$
From set equality:
:$V \cap H = \set x$
=== Sufficient Condition ===
Let $U \in \BB : U \cap H = \set x$.
By definition of synthetic basis of $T$:
:$U \in \tau$
Then $x$ is an isolated point of $H$ by definition.
Category:Isolated Points
Basis Test for Limit Point

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Let $\BB$ be a synthetic basis of $T$.
Let $H \subseteq S$.
Then $x \in S$ is a limit point of $H$ :
:$\forall U \in \BB : x \in U$ satisfies $H \cap U \setminus \set x \ne \O$
== Proof ==
=== Necessary Condition ===
Let $x \in S$ be a limit point of $H$.
By definition of a limit point of $H$:
:$\forall U \in \tau : x \in U$ satisfies $H \cap U \setminus \set x \ne \O$
By definition of a synthetic basis of $T$:
:$\BB \subseteq \tau$
The result follows.
=== Sufficient Condition ===
Let $x$ satisfy:
:$\forall U \in \BB : x \in U$ satisfies $H \cap U \setminus \set x \ne \O$
Let $V$ be any open neighborhood of $x$.
By definition of a synthetic basis of $T$:
:$\exists U \in \BB : x \in U \subseteq V$
Then:
:$H \cap U \setminus \set x \ne \O$
From the contrapositive statement of Subsets of Disjoint Sets are Disjoint:
:$H \cap V \setminus \set x \ne \O$
Thus $x$ is a limit point of $H$ by definition.
Category:Limit Points
Basis for Box Topology

== Theorem ==
Let $\family {\struct {S_i, \tau_i} }_{i \mathop \in I}$ be an $I$-indexed family of topological spaces.
Let $S$ be the cartesian product of $\family {S_i}_{i \mathop \in I}$.
That is:
:$\ds S := \prod_{i \mathop \in I} S_i$
Define:
:$\ds \BB := \set {\prod_{i \mathop \in I} U_i: \forall i \in I: U_i \in \tau_i}$
Then $\BB$ is a synthetic basis on $S$.
== Proof ==
Let us check the two conditions for $\BB$ to be a synthetic basis in turn.
=== $(\text B 1)$: Covering ===
From , we have $S_i \in \tau_i$ for all $i \in I$.
Thus $S = \ds \prod_{i \mathop \in I} S_i \in \BB$.
Hence by Set is Subset of Union: General Result, we have:
:$S \subseteq \ds \bigcup \BB$
so $\BB$ is a cover for $S$.
=== $(\text B 2)$: Intersections are Unions ===
Let $A = \ds \prod_{i \mathop \in I} U_i$ and $B = \ds \prod_{i \mathop \in I} V_i$ be in $\BB$.
Then by Cartesian Product of Intersections, we have:
:$A \cap B = \ds \prod_{i \mathop \in I} \paren {U_i \cap V_i}$
By , $U_i \cap V_i \in \tau_i$ for all $i \in I$.
Hence $A \cap B \in \BB$, and in particular, by Union of Singleton:
:$A \cap B = \ds \bigcup \set {A \cap B}$
Having verified both conditions, we conclude $\BB$ is a synthetic basis on $S$.
== Also see ==
* Definition:Box Topology
* Synthetic Basis and Analytic Basis are Compatible
Category:Box Topology
Category:Topological Bases
Basis for Discrete Topology

== Theorem ==
Let $S$ be a set.
Let $\tau$ be the discrete topology on $S$.
Let $\BB$ be the set of all singleton subsets of $S$:
:$\BB := \set {\set x: x \in S}$.
Then $\BB$ is a basis for $T$.
== Proof ==
Let $T = \struct {S, \tau}$ be the discrete space on $S$. 
Let $U \in \tau$.
Then:
:$\ds U = \bigcup_{x \mathop \in U} \set x$
Hence:
:$\forall x \in U: \exists \set x \in \BB: \set x \subseteq U$
Thus $U$ is the union of elements of $\BB$.
Hence by definition $\BB$ is a basis for $T$.
Category:Discrete Topology
Category:Topological Bases
Basis for Either-Or Topology

== Theorem ==
Let $T = \struct {S, \tau}$ be the either-or space.
Let $\BB$ be the set:
:$\BB := \set {\set x: x \in S, x \ne 0} \cup \set {\openint {-1} 1}$
that is, the set of all singleton subsets of $S$ less $\set 0$ and including the open real interval $\openint {-1} 1$.
Then $\BB$ is a basis for $T$.
== Proof ==
Let $U \in \tau$ such that $0 \notin U$.
Then:
:$\ds U = \bigcup_{x \mathop \in U} \set x$
where $x \ne 0$.
Hence for all $x \in U$, we have $\set x \in \BB$.
Thus $U$ is the union of elements of $\BB$.
Now suppose $U \in \tau$ such that $0 \in U$.
Then $\openint {-1} 1 \subseteq U$ by definition.
So one of four cases holds:
:$U = \openint {-1} 1$
:$U = \hointr {-1} 1 = \openint {-1} 1 \cup \set {-1}$
:$U = \hointl {-1} 1 = \openint {-1} 1 \cup \set 1$
:$U = \closedint {-1} 1 = \openint {-1} 1 \cup \set {-1} \cup \set 1$
All of these sets are in $\BB$, so that $U$ is the union of elements of $\BB$.
Hence, by definition, $\BB$ is a basis for $T$.
Category:Either-Or Topology
Category:Topological Bases
Basis for Element of Real Number Line

== Theorem ==
Let $M = \struct {\R, d}$ denote the real number line with the usual (Euclidean) metric.
Let $a \in \R$ be a point in $M$.
Then the set of all open intervals containing $a$ is a basis for the neighborhood system of $a$.
== Proof ==
Let $N$ be a neighborhood of $a$ in $M$.
Then by definition:
:$\exists \epsilon \in \R_{>0}: \map {B_\epsilon} a \subseteq N$
where $\map {B_\epsilon} a$ is the open $\epsilon$-ball at $a$.
The result follows from Open Ball in Real Number Line is Open Interval.
== Sources ==
* : Chapter $2$: Metric Spaces: $\S 4$: Open Balls and Neighborhoods
Category:Real Intervals
Category:Real Number Line with Euclidean Metric
Basis for Euclidean Topology on Real Number Line

== Theorem ==
Let $\R$ be the set of real numbers.
Let $\BB$ be the set of subsets of $\R$ defined as:
:$\BB = \set {\openint a b: a, b \in \R}$
That is, $\BB$ is the set of all open real intervals of $\R$:
:$\openint a b := \set {x \in \R: a < x < b}$
Then $\BB$ forms a basis for the Euclidean topology on $\R$.
== Proof ==
From Real Number Line is Metric Space, one can define an open interval on the set of real numbers in terms of an $\epsilon$-neighborhood.
From Open Real Interval is Open Ball, an open interval $\openint a b$ is the open $\epsilon$-ball $\map {B_\epsilon} \alpha$.
Then from Metric Induces Topology we have that:
:$\BB = \set {\openint a b: a, b \in \R}$
forms a topology on $\R$.
== Sources ==
* : Part $\text {II}$: Counterexamples: $28$. Euclidean Topology
Category:Real Number Line with Euclidean Topology
Category:Topological Bases
Basis for Excluded Point Space

== Theorem ==
Let $T = \struct {S, \tau_{\bar p} }$ be an excluded point space.
Consider the set $\BB$ defined as:
:$\BB = \set {\set x: x \in S \setminus \set p} \cup \set S$
Then $B$ is a basis for $S$.
== Proof ==
Let $H \in \tau_{\bar p}$ be open in $T$.
If $H = S$ then trivially $H$ is the union of elements of $\BB$.
So suppose $H \ne S$.
Then by definition $p \notin H$ and so:
:$\forall y \in H: \exists \set y \in \BB$
Thus:
:$\ds H = \bigcup_{y \mathop \in H} \set y$
So $\BB$ is an analytic basis for $T$.
It could equally well be shown that $\BB$ is also a synthetic basis for $T$.
Category:Excluded Point Topology
Category:Topological Bases
Basis for Finite Submodule of Function Space

== Theorem ==
Let $\struct {R, +, \circ}$ be a ring with unity whose zero is $0_R$ and whose unity is $1_R$.
Let $A$ be a set.
For each $a \in A$, let $f_a: A \to R$ be defined as:
:$\forall x \in A: \map {f_a} x = \begin{cases}
1 & : x = a \\
0 & : x \ne a
\end{cases}$
Then $B = \set {f_a: a \in A}$ is a basis of the Finite Submodule of Function Space $R^{\paren A}$.
== Proof ==
Let $\sequence {a_k}_{1 \mathop \le k \mathop \le n}$ be a sequence of distinct terms of $A$.
Let $\sequence {\lambda_k}_{1 \mathop \le k \mathop \le n}$ be a sequence of scalars.
Then: $\ds \sum_{k \mathop = 1}^n \lambda_k f_{a_k}$ is the mapping whose value at $a_k$ is $\lambda_k$ and whose value at any $x$ not in $\set {a_1, a_2, \ldots, a_n}$ is zero.
Hence $B$ is a generator of $R^{\paren A}$ which is linearly independent.
Thus, by definition, $B$ is a basis of $R^{\paren A}$.
If $A = \closedint 1 n$, then $B$ is the standard basis of $R^n$.
== Sources ==
* : Chapter $\text {V}$: Vector Spaces: $\S 27$. Subspaces and Bases: Example $27.8$
Category:Submodules
Category:Function Spaces
Category:Linear Algebra
Category:Bases of Modules
Basis for Open Ordinal Topology

== Theorem ==
Let $\Gamma$ be a limit ordinal.
Let $\hointr 0 \Gamma$ denote the open ordinal space on $\Gamma$.
Consider the set $\BB$ of subsets of $\hointr 0 \Gamma$ of the form:
:$\openint \alpha {\beta + 1} = \hointl \alpha \beta = \set {x \in \hointr 0 \Gamma: \alpha < x < \beta + 1}$
for $\alpha, \beta \in \hointr 0 \Gamma$.
Then $\BB$ forms a basis for $\hointr 0 \Gamma$.
== Proof ==
== Sources ==
* : Part $\text {II}$: Counterexamples: $40 \text { - } 43$. Ordinal Space
Category:Ordinal Spaces
Basis for Particular Point Space

== Theorem ==
Let $T = \struct {S, \tau_p}$ be a particular point space.
Consider the set $\BB$ defined as:
:$\BB = \set {\set {x, p}: x \in S} \cup \set p$
Then $B$ is a basis for $S$.
== Proof ==
Let $H \in \tau_p$ be open in $T$.
Then:
:$\forall y \in H: \exists \set {y, p} \in \BB$
which also holds when $y = p$ as $\set {y, p} = \set p \in \BB$.
Thus:
:$\ds H = \bigcup_{y \mathop \in H} \set {y, p}$
So $\BB$ is an analytic basis for $T$.
It could equally well be shown that $\BB$ is also a synthetic basis for $T$.
Category:Particular Point Topology
Category:Topological Bases
Basis for Partition Topology

== Theorem ==
Let $S$ be a set.
Let $\PP$ be a partition of $S$.
Let $\tau$ be the partition topology on $S$ defined as:
:$a \in \tau \iff a$ is the union of sets of $\PP$
Then $\PP$ forms a basis of $\tau$.
== Proof ==
Checking the criteria for $\PP$ to be a synthetic basis for $\tau$:
We have that $\ds S = \bigcup \PP$ from the definition of a partition.
Therefore, $\ds S \subseteq \bigcup \PP$ and $\PP$ is a cover for $S$.
Next, let $B_1, B_2 \in \PP$.
Then as $\PP$ is a partition of $S$, we have that $B_1 \cap B_2 = \O$.
But from Union of Empty Set we have that $\O$ is the (vacuous) union of sets of $\PP$.
Hence $\PP$ is a synthetic basis for $\tau$.
== Sources ==
* : Part $\text {II}$: Counterexamples: $5$. Partition Topology
Category:Partition Topology
Category:Topological Bases
Basis for Product of Metric Spaces under Chebyshev Distance

== Theorem ==
Let $M_1 = \struct {A_1, d_1}, M_2 = \struct {A_2, d_2}, \ldots, M_n = \struct {A_n, d_n}$ be metric spaces.
Let $\ds \AA = \prod_{i \mathop = 1}^n A_i$ be the cartesian product of $A_1, A_2, \ldots, A_n$.
Let $d_\infty: \AA \times \AA \to \R$ be the Chebyshev distance on $\AA$:
:$\ds \map {d_\infty} {x, y} = \max_{i \mathop = 1}^n \set {\map {d_i} {x_i, y_i} }$
where $x = \tuple {x_1, x_2, \ldots, x_n}, y = \tuple {y_1, y_2, \ldots, y_n} \in \AA$.
For $i \in \set {1, 2, \ldots, n}$, let $U_i$ be open in $M_i$.
Then $\set {\ds \prod_{i \mathop = 1}^n U_i}$ is a basis for the open sets of $M = \struct {\AA, d_\infty}$.
== Proof ==
Let $U$ be an open set of $M$.
Then for all $a \in U$, we have:
:$\map {B_\delta} a \subseteq U$
for some $\delta \in \R_{>0}$.
Then:
:$\ds \bigcup_{i \mathop = 1}^n \map {B_\delta} {a_i} \subseteq U$
and the result follows.
== Sources ==
* : Chapter $2$: Metric Spaces: $\S 6$: Open Sets and Closed Sets: Exercise $1$
Category:Chebyshev Distance
Category:Open Sets (Metric Spaces)
Basis for R-Module R

== Theorem ==
Let $\struct {R, +, \times}$ be a ring with unity whose unity is $1_R$.
Let $\struct {R, +_R, \circ}_R$ denote the $R$-module $R$.
Then $\set {1_R}$ is a basis for $\struct {R, +_R, \circ}_R$.
== Proof ==
From Dimension of $R$-Module $R$ is $1$ we have that $\struct {R, +_R, \circ}_R$ is $1$-dimensional.
From Standard Ordered Basis is Basis it follows directly that $\set {1_R}$ is a basis for $\struct {R, +_R, \circ}_R$.
== Sources ==
* : Chapter $\text {V}$: Vector Spaces: $\S 28$. Linear Transformations
Category:Module on Cartesian Product
Basis for Set of Linear Transformations

== Theorem ==
Let $R$ be a commutative ring with unity whose zero is $0_R$ and whose unity is $1_R$.
Let $\struct {G, +_G, \circ}_R$ be a unitary $R$-module such that $\map \dim G = n$.
Let $\struct {H, +_H, \circ}_R$ be a unitary $R$-module such that $\map \dim H = m$.
Let $\map {\LL_R} {G, H}$ be the set of all linear transformations from $G$ to $H$.
Let $\sequence {a_n}$ be an ordered basis for $G$.
Let $\sequence {b_m}$ be an ordered basis for $H$.
Let $\phi_{i j}: G \to H$ be the unique linear transformation defined for each $i \in \closedint 1 n, j \in \closedint 1 m$ which satisfies:
:$\forall k \in \closedint 1 n: \map {\phi_{i j} } {a_k} = \delta_{i k} b_j$
where $\delta$ is the Kronecker delta.
Then:
:$\set {\phi_{i j}: i \in \closedint 1 n, j \in \closedint 1 m}$
is a basis for $\map {\LL_R} {G, H}$.
== Proof ==
Let $B = \set {\phi_{i j}: i \in \closedint 1 n, j \in \closedint 1 m}$.
Let $\ds \sum_{j \mathop = 1}^m \sum_{i \mathop = 1}^n \lambda_{i j} \phi_{i j} = 0$.
Then:
:$\ds \forall k \in \closedint 1 n: 0 = \sum_{j \mathop = 1}^m \sum_{i \mathop = 1}^n \lambda_{i j} \map {\phi_{i j} } {a_k} = \sum_{j \mathop = 1}^m \lambda_{k j} b_j$
So:
:$\forall j \in \closedint 1 n: \lambda_{k j} = 0$
Hence $B$ is linearly independent.
Now let $\phi \in \map {\LL_R} {G, H}$.
Let $\tuple {\alpha_{i 1}, \alpha_{i 2}, \ldots, \alpha_{i m} }$ be the sequence of scalars that satisfies:
:$\ds \forall i \in \closedint 1 n: \map \phi {a_i} = \sum_{j \mathop = 1}^m \alpha_{i j} b_j$
Then:
:$\ds \forall k \in \closedint 1 n: \map \phi {a_k} = \map {\paren {\sum_{j \mathop = 1}^m \sum_{i \mathop = 1}^n \alpha_{i j} u_{i j} } } {a_k}$
by a calculation similar to the preceding.
So, by Linear Transformation of Generated Module:
:$\ds u = \sum_{j \mathop = 1}^m \sum_{i \mathop = 1}^n \alpha_{i j} u_{i j}$
Thus $B$ is a generator for $\phi \in \map {\LL_R} {G, H}$.
== Sources ==
* : Chapter $\text {V}$: Vector Spaces: $\S 28$. Linear Transformations: Theorem $28.8$
Category:Linear Transformations
Basis for Topological Subspace

== Theorem ==
Let $T = \struct {A, \tau}$ be a topological space.
Let $\O \subseteq H \subseteq A$ and so let $T_H = \struct {H, \tau_H}$ be a subspace of $T$.
Let $\BB$ be a (synthetic) basis for $T$.
Let $\BB_H$ be defined as:
:$\BB_H = \set {U \cap H: U \in \BB}$
Then $\BB_H$ is a (synthetic) basis for $H$.
== Proof ==
$\BB \subseteq \powerset A$ is a synthetic basis for $T$ :
:$(\text B 1): \quad$ $A$ is a union of sets from $\BB$
:$(\text B 2): \quad$ If $B_1, B_2 \in B$, then $B_1 \cap B_2$ is a union of sets from $\BB$.
Let $A = \mathbb S$ be a union of sets from $\BB$.
Then:
:$\ds A = \bigcup_{S \mathop \in \mathbb S} S$
for some $\mathbb S \subseteq \BB$.
Hence:
But if $S \in \BB$, then $S \cap H \in \BB_H$
So $H$ is a union of sets from $\BB_H$.
In the same way we investigate $U_1$ and $U_2$.
Let $U_1, U_2 \in \BB_H$.
Then $U_1 = B_1 \cap H, U_2 = B_2 \cap H$ for some $B_1, B_2 \in \BB$.
Then $U_1 \cap U_2 = B_1 \cap H \cap B_2 \cap H = \paren {B_1 \cap B_2} \cap H$.
As $\BB$ is a (synthetic) basis for $A$, we have that:
:$\ds B_1 \cap B_2 = \bigcup_{S \mathop \in \mathbb S} S$
for some $\mathbb S \subseteq \BB$.
Hence:
But if $S \in \BB$, then $S \cap H \in \BB_H$.
So $U_1 \cap U_2$ is a union of sets from $\BB_H$.
So $\BB_H$ fulfils the conditions to be a synthetic basis for $H$.
== Sources ==
* : $3$: Continuity generalized: topological spaces: Exercise $3.9: 12$
* : Part $\text I$: Basic Definitions: Section $1$: General Introduction
* : $\S 3$: New Spaces From Old: Subspaces
Category:Topological Subspaces
Category:Topological Bases
Basis for Topology on Cartesian Product

== Theorem ==
Let $T_1 = \struct{A_1, \tau_1}$ and $T_2 = \struct{A_2, \tau_2}$ be topological spaces.
Let $A_1 \times A_2$ be the Cartesian product of $A_1$ and $A_2$.
Let $\PP = \set {U_1 \times U_2 : U_1 \in \tau_1, U_2 \in \tau_2}$
Then $\PP$ is a synthetic basis on $A_1 \times A_2$.
== Proof ==
We need to show that conditions $(\text B 1)$ and $(\text B 2)$ in the definition for basis hold for $\PP$.
=== $(\text B 1)$ ===
Because $A_1 \in \tau_1$ and $A_2 \in \tau_2$, $A_1 \times A_2 \in \PP$.
=== $(\text B 2)$ ===
Let $U_1, V_1 \in \tau_1$ and $U_2, V_2 \in \tau_2$.
From Cartesian Product of Intersections:
:$\paren {U_1 \times U_2} \cap \paren {V_1 \times V_2} = \paren {U_1 \cap V_1} \times \paren {U_2 \cap V_2}$
Because $\paren {U_1 \cap V_1} \in \tau_1$ and $\paren {U_2 \cap V_2} \in \tau_2$:
:$\paren {U_1 \cap V_1} \times \paren {U_2 \cap V_2} \in \PP$
So $\paren {U_1 \times U_2} \cap \paren {V_1 \times V_2}$ is the union of (one) set in $\PP$.
Hence the result.
== Also see ==
* Natural Basis of Product Topology of Finite Product where it is shown that the basis $\PP$ is the natural basis of the product topology on a finite Cartesian product
== Sources ==
* : $3$: Continuity generalized: topological spaces: $3.5$: Products: Definition $3.5.1$
Category:Product Topology
Basis induces Local Basis

== Theorem ==
Let $\struct {S, \tau}$ be a topological space.
Let $\BB$ be an analytic basis for $\tau$.
Let $x \in S$, and denote with $\BB_x$ the set $\set {B \in \BB: x \in B}$.
Then $\BB_x$ is a local basis at $x$.
== Proof ==
Let $x \in U$, with $U$ an open set of $\tau$.
Since $\BB$ is an analytic basis for $\tau$, we have:
:$U = \ds\bigcup \AA$
for some $\AA \subseteq \BB$.
Since $x \in \ds \bigcup \AA$, there is a $B \in \AA$ such that $x \in B$, by definition of set union.
Hence, by definition of $\BB_x$, $B \in \BB_x$.
From Set is Subset of Union: General Result, we also have:
:$B \subseteq U$
Hence $\BB_x$ is a local basis at $x$.
== Also see ==
* Union of Local Bases is Basis
Category:Topology
Basis of Free Module is No Greater than Generator

== Theorem ==
Let $R$ be a commutative ring with unity.
Let $M$ be a free $R$-module with basis $B$.
Let $S$ be a generating set for $M$.
Then:
:$\size B \le \size S$.
That is, there exists an injection from $B$ to $S$.
== Outline of Proof ==
Because $S$ is a spanning set, we can construct a surjective homomorphism $R^{\paren S} \to R^{\paren B}$.
Using Krull's Theorem we divide through by a maximal ideal of $R$ to reduce it to the case where $R$ is a division ring, that is, the case of vector spaces.
We then conclude by comparing cardinalities and using Basis of Vector Space Injects into Generator.
== Proof ==
Because $S$ is a generating set, there is a surjective homomorphism
:$\phi : R^{\paren S} \to M$
where $R^{\paren S}$ is the free $R$-module on $S$.
Because $B$ is a basis, there is an isomorphism
:$\psi : R^{\paren B} \to M$
Thus $f = \psi^{-1} \circ \phi: R^{\paren S} \to R^{\paren B}$ is a surjective module homomorphism.
By Krull's Theorem, there exists a maximal ideal $M \subset R$.
By Maximal Ideal iff Quotient Ring is Field, $R / M$ is a field.
Let $k = R / M$.
Let $\pi: R \to k$ denote the quotient mapping.
By some universal properties, there exists a $k$-module homomorphism $\bar f: k^{\paren S} \to k^{\paren B}$ such that:
:$\pi^B \circ f = \bar f \circ \pi^S$
Where $\pi^B$ denotes direct sum of module homomorphisms.
Because $\pi^B \circ f$ is surjective, so is $\bar f$.
Let:
:$C$ be the canonical basis of $k^{\paren S}$
:$D$ be the canonical basis of $k^{\paren S}$
Because $C$ is a generator of $k^{\paren S}$ and $\bar f$ is surjective, $\map {\bar f} C$ is a generator of $k^{\paren B}$.
We have:
== Also see ==
* Bases of Free Module have Equal Cardinality
Category:Free Modules