nyuuzyou/wikis huggingface dataset:
Clopen Set and Complement form Separation

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Let $H \subseteq S$ be a clopen set of $T$.
Let $\relcomp S H$ be the complement of $H$ relative to $S$.
Then $H$ and $\relcomp S H$ form a separation of $T$.
== Proof ==
By Set with Relative Complement forms Partition, $H$ and $\relcomp S H$ form a partition of $S$.
By Complement of Clopen Set is Clopen, $\relcomp S H$ is also a clopen set of $T$.
By definition of clopen set, both $H$ and $\relcomp S H$ are open in $T$.
Thus $H$ and $\relcomp S H$ are a pair of open sets in $T$ forming a partition of $S$.
Hence the result, by definition of separation.
Category:Clopen Sets
Category:Separations
Clopen Set contains Components of All its Points

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Let $H \subseteq S$ be both closed and open in $T$.
Then $H$ contains the components of all of its points.
== Proof ==
Let $H$ be a clopen set in $T$.
By definition, $H$ is open and so $H \in \tau$.
But as $H$ is also closed, by definition $\relcomp S H \in \tau$ where $\complement_S$ denotes complement relative to $S$.
Thus $H$ and $\relcomp S H$ are both open such that:
:$H \cap \relcomp S H = \O$ from Intersection with Relative Complement is Empty
:$H \cup \relcomp S H = S$ from Union with Relative Complement
and so forming a partition of $T$.
As $H$ and $\relcomp S H$ are both closed, it follows from Closed Set Equals its Closure that:
:$H \cap \paren {\relcomp S H}^- = \O = H^- \cap \relcomp S H$
and so by definition $H$ and $\relcomp S H$ are separated.
 that $H$ does not contain the components of all of its points.
That is:
:$\exists x \in H: \map {\operatorname {Comp}_x} T \nsubseteq H$
where $\map {\operatorname {Comp}_x} T$ is the component of $x$ in $T$.
Then:
:$\exists z \in \map {\operatorname {Comp}_x} T: z \in H \cup \relcomp S H$
This means that $H \cup \relcomp S H$ is connected.
This contradicts the fact that $H$ and $\relcomp S H$ are separated.
The result follows by Proof by Contradiction.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $4$: Connectedness
Category:Clopen Sets
Category:Connected Sets
Clopen Sets in Finite Complement Topology

== Theorem ==
Let $T = \struct {S, \tau}$ be a finite complement topology on an infinite set $S$.
Then the only clopen sets of $T$ are $S$ and $\O$.
== Proof ==
Let $U \in \tau$ be open in $T$.
Then by definition of finite complement topology, $S \setminus U$ is finite.
By definition of open set, $S \setminus U$ is closed.
As $S$ is infinite, it follows that $U$ must also be infinite.
Thus unless $U = S$, $S \setminus U$ can not be open.
Hence the result.
== Sources ==
* : Chapter $\text {I}$: Topological Spaces: $1$. Open Sets and Closed Sets
Category:Finite Complement Topology
Clopen Sets in Indiscrete Topology

== Theorem ==
Let $T = \struct {S, \tau}$ be a discrete topological space.
The only subsets of $S$ which are both closed and open in $T$ are $S$ and $\O$. 
== Proof ==
By definition of indiscrete topological space, the only open sets in $\struct {S, \tau}$ are $S$ and $\O$.
From Open and Closed Sets in Topological Space, both $S$ and $\O$ are both closed and open in $\struct {S, \tau}$ are $S$ and $\O$. 
Hence the result.
== Sources ==
* : Chapter $\text {I}$: Topological Spaces: $1$. Open Sets and Closed Sets
Category:Indiscrete Topology
Category:Clopen Sets
Clopen Sets in Modified Fort Space

== Theorem ==
Let $T = \struct {S, \tau_{a, b} }$ be a modified Fort space.
Let $A \subseteq S$ be both closed and open in $T$.
If $a \in A$, then $b \in A$ as well.
That is, any clopen set of $T$ must contain both or neither of $a$ and $b$.
== Proof ==
Let $A$ be both closed and open in $T$.
Let $S = A \mid B$ be a separation of $S$ into the open sets $A$ and $B$.
Then by definition $B$ is also both closed and open in $T$.
Let $a \in A$.
Suppose $b \in B$.
As $A$ is open it must contain all but a finite number of points of $S$.
Thus $A$ must itself be infinite.
But then $B$ can not be open, as $b \in B$ and there is more than a finite number of points in $A$.
So if $a \in A$ then $b \in A$ as well.
== Sources ==
* : Part $\text {II}$: Counterexamples: $27$. Modified Fort Space: $5$
Category:Clopen Sets
Category:Modified Fort Space
Closable Densely-Defined Linear Operator has Smallest Closed Extension

== Theorem ==
Let $\struct {\HH, \innerprod \cdot \cdot}$ be a Hilbert space over $\C$.
Let $\tuple {\map D T, T}$ be a closable densely-defined linear operator.
Then there exists a closed densely-defined linear extension $\struct {\map D S, S}$ of $T$ such that:
:the graph $\map \GG S$ is $\subseteq$-minimal among the graphs of closed densely-defined linear extensions of $T$. 
Further:
:$\map \GG S = \map \cl {\map \GG T}$
== Proof ==
From Characterization of Closable Densely-Defined Linear Operators in terms of Closure of Graph:
:whenever $\tuple { {\mathbf 0}_\HH, y} \in \map \cl {\map \GG T}$, we have $y = {\mathbf 0}_\HH$.
Define:
:$\map D S = \set {x \in \HH : \tuple {x, y} \in \map \cl {\map \GG T} \text { for some } y \in \HH}$
From Linear Transformation defined from Graph, there exists a linear transformation $S : \map D S \to \HH$ such that $\map \GG S = \map \cl {\map \GG T}$.
Now let $S'$ be another closed densely-defined linear extensions of $T$.
We have $\map \GG T \subseteq \map \GG {S'}$.
Since $\map \GG {S'}$ is closed, we obtain:
:$\map \GG S = \map \cl {\map \GG T} \subseteq \map \GG {S'}$
from Set Closure Preserves Set Inclusion.
== Sources ==
* : $\text {VIII}.1$: Domains, graphs, adjoints and spectrum
Category:Closable Densely-Defined Linear Operators
Category:Closed Linear Transformations
Closable Densely-Defined Operator is Closed iff Equal to Closure

== Theorem ==
Let $\struct {\HH, \innerprod \cdot \cdot}$ be a Hilbert space over $\C$.
Let $\tuple {\map D T, T}$ be a closable densely-defined linear operator.
Let $\tuple {\map D {\overline T}, \overline T}$ be the closure of $\tuple {\map D T, T}$.
Then $\tuple {\map D T, T}$ is closed $T = \overline T$. 
== Proof ==
Let $\struct {\HH \times \HH, \norm {\, \cdot \,}_{\HH \times \HH} }$ be the direct product of $\HH$ with itself equipped with the direct product norm. 
Suppose that $\tuple {\map D T, T}$ is closed.
From Set is Closed iff Equals Topological Closure, we have that $\map \cl {\map \GG T} = \map \GG T$.
From Closable Densely-Defined Linear Operator has Smallest Closed Extension, we have $\map \GG {\overline T} = \map \cl {\map \GG T}$ and hence:
:$\map \GG {\overline T} = \map \GG T$
So $T = \overline T$.
Now suppose that $\overline T = \overline T$. 
Then $\map \GG T = \map \GG {\overline T}$. 
Since $\map \GG {\overline T}$ is closed in $\struct {\HH \times \HH, \norm {\, \cdot \,}_{\HH \times \HH} }$, so is $\map \GG T$.
Hence $T$ is closed.
Category:Closed Linear Transformations
Category:Closable Densely-Defined Linear Operators
Closed *-Subalgebra of C*-Algebra is C*-Algebra

== Theorem ==
Let $\struct {A, \ast, \norm {\, \cdot \,} }$ be a $\text C^\ast$-algebra.
Let $B \subseteq A$ be a closed $\ast$-subalgebra of $A$. 
Then $\struct {B, \ast, \norm {\, \cdot \,} }$ is a $\text C^\ast$-algebra.
== Proof ==
Since $\struct {B, \ast}$ is a $\ast$-subalgebra of $A$, it is in particular a $\ast$-algebra.
From Closed Subspace of Banach Space forms Banach Space, $\struct {B, \ast, \norm {\, \cdot \,} }$ is additionally a Banach space.
We have:
:$\norm {x y} \le \norm x \norm y$
and:
:$\norm {x x^\ast} = \norm x^2$
for each $x \in B$, since these properties hold for all $x \in A$.
Hence $\struct {B, \ast, \norm {\, \cdot \,} }$ is a $\text C^\ast$-algebra.
Category:C*-Algebras
Closed Algebraic Structure/Examples/2^m 3^n under Multiplication

== Examples of Closed Algebraic Structures ==
Let $S$ be the set defined as:
:$S := \set {2^m 3^n: m, n \in \Z}$
Then the algebraic structure $\struct {S, \times}$ is closed.
== Proof ==
Let $a, b \in S$ such that:
where $m_1, m_2, n_1, n_2 \in \Z$.
Then:
== Sources ==
* : Chapter $\text I$: Algebraic Structures: $\S 6$: Isomorphisms of Algebraic Structures: Exercise $6.7 \ \text {(c)}$
Category:Closed Algebraic Structures
Closed Ball contains Smaller Closed Ball

== Theorem ==
Let $M = \struct {A, d}$ be a metric space.
Let $a \in A$.
Let $\epsilon, \delta \in \R_{> 0}$ such that $\epsilon \le \delta$.
Let $\map {B^-_\epsilon} a$ be the closed $\epsilon$-ball on $a$.
Let $\map {B^-_\delta} a$ be the closed $\delta$-ball on $a$.
Then:
:$\map {B^-_\epsilon} a \subseteq \map {B^-_\delta} a$
== Proof ==
By definition of subset:
:$\map {B^-_\epsilon} a \subseteq \map {B^-_\delta} a$
Category:Metric Spaces
Closed Ball contains Smaller Open Ball

== Theorem ==
Let $M = \struct {A, d}$ be a metric space.
Let $a \in A$.
Let $\epsilon, \delta \in \R_{> 0}$ such that $\epsilon \le \delta$.
Let $\map {B_\epsilon} a$ be the open $\epsilon$-ball on $a$.
Let $\map {B^-_\delta} a$ be the closed $\delta$-ball on $a$.
Then:
:$\map {B_\epsilon} a \subseteq \map {B^-_\delta} a$
== Proof ==
By definition of subset:
:$\map {B_\epsilon} a \subseteq \map {B^-_\delta} a$
Category:Metric Spaces
Closed Ball in Euclidean Space is Compact

== Theorem ==
Let $x \in \R_n$ be a point in the Euclidean space $\R^n$.
Let $\epsilon \in \R_{>0}$.
Then the closed $\epsilon$-ball $\map {B_\epsilon^-} x$ is compact.
== Proof ==
From Closed Ball is Closed in Metric Space, it follows that $\map {B_\epsilon^-} x$ is closed in $\R^n$.
For all $a \in \map {B_\epsilon^-} x$ we have $\map d {x, a} \le \epsilon$, where $d$ denotes the Euclidean metric.
Then $\map {B_\epsilon^-} x$ is bounded in $\R^n$.
From the Heine-Borel Theorem, it follows that $\map {B_\epsilon^-} x$ is compact.
Category:Euclidean Space
Closed Ball in Metric Space is Closed Neighborhood

== Theorem ==
Let $M = \struct {A, d}$ be a metric space.
Let $x \in A$.
Let $\epsilon \in \R_{>0}$.
Let $\map {B_\epsilon^-} x$ be the closed $\epsilon$-ball of $x$ in $M$.
Then $\map {B_\epsilon^-} x$ is a closed neighborhood of $x$ in $M$.
== Proof ==
From Closed Ball contains Smaller Open Ball:
:$\map {B_\epsilon} x \subseteq \map {B_\epsilon^-} x$
where $\map {B_\epsilon} x$ denotes the open $\epsilon$-ball of $x$ in $M$.
Hence $\map {B_\epsilon^-} x$ is a neighborhood of $x$ by definition.
From Closed Ball is Closed in Metric Space:
:$\map {B_\epsilon^-} x$ is closed in $M$
It follows that $\map {B_\epsilon^-} x$ is a closed neighborhood of $x$ in $M$ by definition.
Category:Closed Sets (Metric Spaces)
Category:Neighborhoods
Closed Ball in Normed Division Ring is Closed Ball in Induced Metric

== Theorem ==
Let $\struct{R, \norm {\,\cdot\,} }$ be a normed division ring.
Let $d$ be the metric induced by the norm $\norm {\,\cdot\,}$.
Let $a \in R$.
Let $\epsilon \in \R_{>0}$ be a strictly positive real number.
Let $\map {{B_\epsilon}^-} {a; \norm {\,\cdot\,} }$ denote the closed ball in the normed division ring $\struct {R, \norm {\,\cdot\,} }$.
Let $\map {{B_\epsilon}^-} {a; d }$ denote the closed ball in the metric space $\struct {R, d}$.
Then:
:$\map {{B_\epsilon}^-} {a; \norm {\,\cdot\,} }$ = $\map {{B_\epsilon}^-} {a; d }$
== Proof ==
The result follows from Equality of Sets.
== Also see ==
* Open Ball in Normed Division Ring is Open Ball in Induced Metric
* Sphere in Normed Division Ring is Sphere in Induced Metric
Category:Definitions/Normed Division Rings
Closed Ball is Closed/Metric Space

== Theorem ==
Let $M = \struct {A, d}$ be a metric space.
Let $x \in A$.
Let $\epsilon \in \R_{>0}$.
Let $\map {B_\epsilon^-} x$ be the closed $\epsilon$-ball of $x$ in $M$.
Then $\map {B_\epsilon^-} x$ is a closed set of $M$.
== Proof ==
We show that the complement $A \setminus B_\epsilon^- \left({x}\right)$ is open in $M$.
Let $a \in A \setminus \map {B_\epsilon^-} x$.
Then by definition of closed ball:
:$\map d {x, a} > \epsilon$
Put:
:$\delta := \map d {x, a} - \epsilon > 0$
Then:
:$\map d {x, a} - \delta = \epsilon$
Let $b \in \map {B_\delta} a$.
Then:
and so:
:$b \notin \map {B_\epsilon^-} x$
Then:
:$\map {B_\delta} a \subseteq A \setminus \map {B_\epsilon^-} x$
so $A \setminus \map {B_\epsilon^-} x$ is open in $M$.
Hence, by definition of closed set:
:$\map {B_\epsilon^-} x$ is closed in $M$.
Category:Closed Sets (Metric Spaces)
Category:Closed Ball is Closed
Closed Ball is Closed/Normed Vector Space

== Theorem ==
Let $M = \struct {X, \norm {\, \cdot \,}}$ be a normed vector space.
Let $x \in X$.
Let $\epsilon \in \R_{> 0}$.
Let $\map {B_\epsilon^-} x$ be the closed $\epsilon$-ball of $x$ in $M$.
Then $\map {B_\epsilon^-} x$ is a closed set of $M$.
== Proof ==
We show that the complement $X \setminus \map {B_\epsilon^-} x$ is open in $M$.
Let $y \in X \setminus \map {B_\epsilon^-} x$.
Then by definition of closed ball:
:$\norm {x - y} > \epsilon$
Put:
:$\delta := \norm {x - y} - \epsilon > 0$
Then:
:$\norm {x - y} - \delta = \epsilon$
Let $z \in \map {B_\delta} y$.
Then:
and so:
:$z \notin \map {B_\epsilon^-} x$
Then:
:$\map {B_\delta} y \subseteq X \setminus \map {B_\epsilon^-} x$
so $X \setminus \map {B_\epsilon^-} x$ is open in $M$.
Hence, by definition of closed set:
:$\map {B_\epsilon^-} x$ is closed in $M$.
== Sources ==
* : $\S 1.3$: Normed and Banach spaces. Topology of normed spaces
Category:Closed Ball is Closed
Category:Closed Sets (Normed Vector Spaces)
Closed Ball is Connected

== Theorem ==
Let $V$ be a normed vector space with norm $\norm {\,\cdot\,}$ over $\R$ or $\C$.
A closed ball in the metric induced by $\norm {\,\cdot\,}$ is connected.
== Proof ==
Follows from:
* Closed Ball is Path-Connected
* Path-Connected Space is Connected
Category:Normed Vector Spaces
Category:Closed Balls
Category:Connected Spaces
Closed Ball is Convex Set

== Theorem ==
Let $\struct {X, \norm {\, \cdot \,} }$ be a normed vector space.
Let $\map { {B_r}^-} x$ be a closed ball in $X$ with radius $r \in \R_{>0}$ and center $x \in X$.
Then $\map { {B_r}^-} x$ is convex.
== Proof ==
Let $y \in \map { {B_1}^-} {\mathbf 0}$.
From , it follows that:
:$\norm {r y} = r \norm y$
It follows that:
:$y \in \map { {B_1}^-} {\mathbf 0}$, $r y \in \map { {B_r}^-} {\mathbf 0}$
As $\norm {r y - \mathbf 0} = \norm {\paren {r y + x} - x}$, it follows that:
:$r y \in \map { {B_r}^-} {\mathbf 0}$
:
:$r y + x \in \map { {B_r}^-} {\mathbf x}$
It follows that:
:$\map { {B_r}^-} {\mathbf x} = r \map { {B_1}^-} {\mathbf 0} + x$
From Closed Unit Ball is Convex Set, it follows that $\map { {B_1}^-} {\mathbf 0}$ is convex.
From Dilation of Convex Set in Vector Space is Convex, it follows that $r \map { {B_1}^-} {\mathbf 0}$ is convex.
From Translation of Convex Set in Vector Space is Convex, it follows that $r \map { {B_1}^-} {\mathbf 0} + x$ is convex.
== Sources ==
* : $\S 5$: Normed Spaces
Category:Closed Balls
Category:Convex Sets (Vector Spaces)
Closed Ball is Disjoint Union of Open Balls in P-adic Numbers

== Theorem ==
Let $p$ be a prime number.
Let $\struct {\Q_p, \norm {\,\cdot\,}_p}$ be the $p$-adic numbers.
Let $a \in \Q_p$.
For all $\epsilon \in \R_{>0}$:
:let $\map { {B_\epsilon}^-} a$ denote the closed $\epsilon$-ball of $a$
and:
:let $\map {B_\epsilon} a$ denote the open $\epsilon$-ball of $a$. 
Then:
:$(1): \quad \forall n \in Z : \map {B^-_{p^{-n} } } a = \ds \bigcup_{i \mathop = 0}^{p - 1} \map {B_{p^{-n} } } {a + i p^n}$
:$(2): \quad \forall n \in Z : \set {\map {B_{p^{-n} } } {a + i p^n} : i = 0, \dotsc, p - 1}$ is a set of pairwise disjoint open balls.
== Proof ==
Let $n \in \Z$.
From Closed Ball is Disjoint Union of Smaller Closed Balls in P-adic Numbers:
:$\map {B^-_{p^{-n} } } a = \ds \bigcup_{i \mathop = 0}^{p - 1} \map {B^-_{p^{-\paren {n + 1} } } } {a + i p}$
where $\set {\map {B^-_{p^{-\paren {n + 1 } } } } {a + i p^n} : i = 0, \dots, p - 1}$ is a set of pairwise disjoint closed balls.
From Open Ball in P-adic Numbers is Closed Ball:
:$\map {B^-_{p^{-n} } } a = \ds \bigcup_{i \mathop = 0}^{p - 1} \map {B_{p^{-n} } } {a + i p}$
where $\set {\map {B_{p^{-n} } } {a + i p^n} : i = 0, \dots, p - 1}$ is a set of pairwise disjoint Open balls.
The result follows.
== Also see ==
* Sphere is Disjoint Union of Open Balls in P-adic Numbers
Category:Topology of P-adic Numbers
Closed Ball is Disjoint Union of Smaller Closed Balls in P-adic Numbers

== Theorem ==
Let $p$ be a prime number.
Let $\struct {\Q_p, \norm {\,\cdot\,}_p}$ be the $p$-adic numbers.
Let $a \in \Q_p$.
For all $\epsilon \in \R_{>0}$, let $\map { {B_\epsilon}^-} a$ denote the closed $\epsilon$-ball of $a$. 
Let $n, m \in Z$, such that $n < m$.
Then:
:$(1) \quad \map {B^-_{p^{-n} } } a = \ds \bigcup_{i \mathop = 0}^{p^{\paren {m - n} } - 1} \map {B^-_{p^{-m} } } {a + i p^n}$
:$(2) \quad \set {\map {B^-_{p^{-m} } } {a + i p^n} : i = 0, \dots, p^{\paren {m - n} } - 1}$ is a set of pairwise disjoint closed balls
== Proof ==
=== Condition $(1)$ ===
=== Condition $(2)$ ===
Category:Topology of P-adic Numbers
Category:Closed Ball is Disjoint Union of Smaller Closed Balls in P-adic Numbers
Closed Ball is Disjoint Union of Smaller Closed Balls in P-adic Numbers/Disjoint Closed Balls

== Theorem ==
Let $p$ be a prime number.
Let $\struct {\Q_p, \norm {\,\cdot\,}_p}$ be the $p$-adic numbers.
Let $a \in \Q_p$.
For all $\epsilon \in \R_{>0}$, let $\map {B_\epsilon} a$ denote the open $\epsilon$-ball of $a$. 
Then:
:$\forall n \in Z : \set{\map {B^-_{p^{-m} } } {a + i p^n} : i = 0, \dotsc, p^{\paren {m - n}} - 1}$ is a set of pairwise disjoint open balls.
== Proof ==
Let $0 \le i, j \le p^{\paren {m - n}} - 1$.
Let $x \in \map {B^-_{p^{-m} } } {a + i p^n} \cap \map {B^-_{p^{-m} } } {a + j p^n}$
From Characterization of Open Ball in P-adic Numbers:
:$\norm {\paren {x -a} - i p^n}_p \le p^{-m}$
and:
:$\norm {\paren {x -a} - j p^n}_p \le p^{-m}$
We have that P-adic Norm satisfies Non-Archimedean Norm Axioms.
Then:
The result follows.
Category:Closed Ball is Disjoint Union of Smaller Closed Balls in P-adic Numbers
Closed Ball is Disjoint Union of Smaller Closed Balls in P-adic Numbers/Lemma 1

== Theorem ==
Let $p$ be a prime number.
Let $\struct {\Q_p, \norm {\,\cdot\,}_p}$ be the $p$-adic numbers.
Let $n, m \in Z$, such that $n < m$.
Then:
:$\forall y \in \Q_p: \norm y_p \le p^{-n}$ there exists $i \in \Z$ such that:
::$(1)\quad 0 \le i \le p^{\paren {m - n}} - 1$
::$(2)\quad \norm {y - i p^n}_p \le p^{-m}$
== Proof ==
=== Necessary Condition ===
Let $y \in \Q_p$.
Let $\norm y_p \le p^{-n}$.
=== Sufficient Condition ===
Let $y\in \Q_p$.
Let there exist $i \in \Z$ such that:
:$(1) \quad 0 \le i \le p^{\paren {m - n}} - 1$
:$(2) \quad \norm {y - i p^n}_p \le p^{-m}$
Category:Closed Ball is Disjoint Union of Smaller Closed Balls in P-adic Numbers
Closed Ball is Disjoint Union of Smaller Closed Balls in P-adic Numbers/Lemma 1/Necessary Condition

== Theorem ==
Let $p$ be a prime number.
Let $\struct {\Q_p, \norm {\,\cdot\,}_p}$ be the $p$-adic numbers.
Let $n, m \in Z$, such that $n < m$.
Let $y \in \Q_p$.
Let $\norm{y}_p \le p^{-n}$.
Then there exists $i \in \Z$ such that:
::$(1) \quad 0 \le i \le p^\paren {m - n} - 1$
::$(2) \quad \norm {y - i p^n}_p \le p^{-m}$
== Proof ==
We have that P-adic Norm satisfies Non-Archimedean Norm Axioms.
Hence:
From Integers are Dense in Unit Ball of P-adic Numbers:
:$\exists \mathop k \in \Z : \norm{p^{-n} y - k}_p \le p^\paren {n - m}$
From Residue Classes form Partition of Integers:
:$\exists \mathop 0 \le i \le p^\paren {m - n} - 1: p^\paren {m - n} \divides k - i$
By definition of the $p$-adic norm:
:$\norm {k - i}_p \le p^\paren {n - m}$
It follows that:
Category:Closed Ball is Disjoint Union of Smaller Closed Balls in P-adic Numbers
Closed Ball is Disjoint Union of Smaller Closed Balls in P-adic Numbers/Lemma 1/Sufficient Condition

== Theorem ==
Let $p$ be a prime number.
Let $\struct {\Q_p, \norm {\,\cdot\,}_p}$ be the $p$-adic numbers.
Let $y \in \Q_p$
Let $n, m \in Z$, such that $n < m$.
Let there exist $i \in \Z$:
:$(1): \quad 0 \le i \le p^\paren {m - n} - 1$
:$(2): \quad \norm {y - i p^n}_p \le p^{-m}$
Then:
:$\norm y_p \le p^{-n}$
== Proof ==
We have that P-adic Norm satisfies Non-Archimedean Norm Axioms:.
Hence:
By assumption:
:$\norm {y - i p^n} \le p^{-m} \le p^{-n}$
and:
Hence:
:$\max \set {\norm {y - i p^n}_p, \norm {i p^n}_p} \le p^{-n}$
So:
:$\norm y_p \le p^{-n}$
Category:Closed Ball is Disjoint Union of Smaller Closed Balls in P-adic Numbers
Closed Ball is Disjoint Union of Smaller Closed Balls in P-adic Numbers/Union of Closed Balls

== Theorem ==
Let $p$ be a prime number.
Let $\struct {\Q_p, \norm {\,\cdot\,}_p}$ be the $p$-adic numbers.
Let $a \in \Q_p$.
For all $\epsilon \in \R_{>0}$, let $\map { {B_\epsilon}^-} a$ denote the closed $\epsilon$-ball of $a$. 
Let $n, m \in Z$, such that $n < m$.
Then:
:$\map {B^-_{p^{-n} } } a = \ds \bigcup_{i \mathop = 0}^{p^{\paren {m - n}} - 1} \map {B^-_{p^{-m} } } {a + i p^{\paren {m - n} } }$
== Proof ==
==== Lemma ====
Let $0 \le i \le p^{\paren{m - n}} - 1$.
Let $x \in \map {B^{\,-}_{p^{-m} } } {a + i p^{-n} }$
By definition of a closed ball:
:$\norm {x - a - i p^{-n} } \le p^{-m}$
From Lemma:
:$\norm {x - a}_p \le p^{-n}$
By definition of a closed ball:
:$x \in \map {B^-_{p^{-n} } } a$
Since $x$ was arbitrary:
:$\map {B^-_{p^{-m} } } {a + i p^{-n} } \subseteq \map {B^-_{p^{-n} } } a$
Since $i$ was arbitrary: 
:$\ds \bigcup_{i \mathop = 0}^{p^{\paren {m - n}} - 1} \map {B^-_{p^{-m} } } {a + i p^{-n} } \subseteq \map {B^-_{p^{-n} } } a$
Let $x \in \map {B^-_{p^{-n} } } a$.
By definition of a closed ball:
:$\norm {x - a}_p \le p^{-n}$
From Lemma:
:$\exists i \in \N : 0 \le i \le p^{\paren {m - n}} - 1: \norm {x - a - i p^{-n} } \le p^{-m}$
By definition of a closed ball:
:$\exists i \in \N : 0 \le i \le p^{\paren {m - n}} - 1 : x \in \map {B^-_{p^{-m} } } {a + i p^{-n} }$
Hence:
:$\map {B^-_{p^{-n} } } a \subseteq \ds \bigcup_{i \mathop = 0}^{p^{\paren {m - n}} - 1} \map {B^-_{p^{-m} } } {a + i p^{-n} }$
It follows that:
:$\map {B^-_{p^{-n} } } a = \ds \bigcup_{i \mathop = 0}^{p^{\paren {m - n}} - 1} \map {B^-_{p^{-m} } } {a + i p^{-n} }$
Category:Closed Ball is Disjoint Union of Smaller Closed Balls in P-adic Numbers
Closed Ball is Path-Connected

== Theorem ==
Let $V$ be a normed vector space with norm $\norm {\,\cdot\,}$ over $\R$ or $\C$.
A closed ball in the metric induced by $\norm {\,\cdot\,}$ is path-connected.
== Proof ==
Follows from:
* Closed Ball is Convex Set
* Normed Vector Space is Hausdorff Topological Vector Space
* Convex Set is Path-Connected
Category:Normed Vector Spaces
Category:Closed Balls
Category:Path-Connected Spaces
Closed Ball is Simply Connected

== Theorem ==
Let $\struct {V, \norm {\,\cdot\,} }$ be a normed vector space over $\R$ or $\C$.
Let $d: V \times V \to \R_{\ge 0}$ be the metric induced by the norm $\norm {\,\cdot\,}$ on $V$.
Let $\tau$ be the the topology on $V$ induced by the metric $d$.
Let $v \in V$ and $\epsilon \in \R_{>0}$.
Let $\map {B_\epsilon^-} v$ be the closed $\epsilon$-ball of $v$.
Let $\tau_B$ be the subspace topology on $\map {B_\epsilon} v$ induced by $\tau$.
Then $\struct{ \map {B_\epsilon^-} v, \tau_B }$ is simply connected.
== Proof ==
Normed Vector Space is Hausdorff Topological Vector Space shows that $\struct {V, \tau}$ is a topological vector space.
The result now follows from Closed Ball is Convex Set and Convex Set is Simply Connected.
Category:Normed Vector Spaces
Category:Closed Balls
Category:Simply Connected Spaces
Closed Ball of P-adic Number

== Theorem ==
Let $p$ be a prime number.
Let $\struct {\Q_p, \norm {\,\cdot\,}_p}$ be the $p$-adic numbers.
Let $\Z_p$ denote the $p$-adic integers.
Let $a \in \Q_p$.
For all $\epsilon \in \R_{>0}$, let $\map { {B_\epsilon}^-} a$ denote the closed ball of $a$ of radius $\epsilon$. 
Then:
:$\forall n \in Z : \map {B^-_{p^{-n} } } a = a + p^n \Z_p$
where $a + p^n \Z_p$ denotes the left coset of the principal ideal $p^n \Z_p$ containing $a$ in the subring $\Z_p$.
That is, the closed ball $\map { {B_\epsilon}^-} a$ is the set:
:$a + p^n \Z_p = \set{a + p^n z : z \in \Z_p}$
== Proof ==
Let $n \in \Z$.
Then:
From set equality:
:$\map {B^-_{p^{-n} } } a = a + p^n \Z_p$
The result follows.
== Also see ==
* Open Balls of P-adic Number
* Local Basis of P-adic Number
Category:Topology of P-adic Numbers