nyuuzyou/wikis huggingface dataset:
Open Ball in Infinite-Dimensional Normed Vector Space is not Weakly Open

== Theorem ==
Let $\GF \in \set {\R, \C}$. 
Let $\struct {X, \norm {\, \cdot \,} }$ be a normed vector space over $\GF$ with weak topology $w$.
Suppose that $\dim X = \infty$. 
Let $\map {B_r} x$ be the open ball in $X$ with radius $r > 0$ and center $x \in X$. 
Then $\map {B_r} x$ is not weakly open. 
 
== Proof ==
From Translation of Open Set in Normed Vector Space is Open and Dilation of Open Set in Normed Vector Space is Open, it suffices to show that $\map {B_1} 0$ is not weakly open. 
 that $\map {B_1} 0$ is weakly open.
From Initial Topology on Vector Space Generated by Linear Functionals is Locally Convex, $\struct {X, w}$ is a locally convex space induced by: 
:$\set {p_f : f \in X^\ast}$
with $p_f : X \to \hointr 0 \infty$ defined by: 
:$\map {p_f} x = \cmod {\map f x}$ 
for each $x \in X$. 
From Open Sets in Standard Topology of Locally Convex Space, there exists $f_1, \ldots, f_n \in X^\ast$ and $\epsilon > 0$ such that: 
:$\set {y \in X : \cmod {\map {f_k} y} < \epsilon \text { for each } 1 \le k \le n} \subseteq \map {B_1} 0$
Define $T : X \to \GF^n$ by: 
:$T x = \tuple {\map {f_1} x, \map {f_2} x, \ldots, \map {f_n} x}$
for each $x \in X$. 
Suppose that: 
:$\ds \bigcap_{i \mathop = 1}^n \ker f_i = \set {\mathbf 0_X}$
Then $\ker T = \set {\mathbf 0_X}$.
So $T$ is injective, and is therefore a bijection onto $T \sqbrk X$.
So $T$ is a vector space isomorphism between $X$ and $T \sqbrk X$. 
So we have: 
:$\dim X = \dim T \sqbrk X$ 
From Dimension of Image of Vector Space under Linear Transformation is Bounded Above by Dimension of Vector Space, we have: 
:$\dim T \sqbrk X \le n$
and so $\dim X \le n < \infty$, contrary to our assumption that $\dim X = \infty$.
So we must have: 
:$\ds \bigcap_{i \mathop = 1}^n \ker f_i \ne \set {\mathbf 0_X}$
Pick $x \ne \mathbf 0_X$ with: 
:$\ds x \in \bigcap_{i \mathop = 1}^n \ker f_i$
Then for each $\lambda \in \GF$, we have $\cmod {\map {f_i} {\lambda x} } = \cmod \lambda \cmod {\map {f_i} x} = 0 < \epsilon$ for each $1 \le i \le n$.
So $\lambda x \in \map {B_1} 0$ for each $\lambda \in \GF$. 
But then, for each $\lambda \in \GF \setminus \set 0$, we have: 
:$\ds \norm x < \frac 1 {\cmod \lambda}$ 
Taking $\lambda \to \infty$ we obtain $\norm x = 0$.
Hence $x = \mathbf 0_X$ from , contrary to our assumption that $x \ne \mathbf 0_X$.
Category:Open Balls
Category:Weakly Open Sets
Open Ball in Normed Division Ring is Open 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 open ball in the normed division ring $\struct {R, \norm {\,\cdot\,} }$.
Let $\map {B_\epsilon} {a; d }$ denote the open 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 ==
* Closed Ball in Normed Division Ring is Closed Ball in Induced Metric
* Sphere in Normed Division Ring is Sphere in Induced Metric
Category:Open Balls 
Category:Normed Division Rings
Open Ball in Normed Vector Space is Connected

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

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

== 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 $a$
:Let $\map {B^-_\epsilon} a$ denote the closed $\epsilon$-ball of $a$. 
Then:
:$\forall n \in \Z : \map {B_{p^{-n} } } a = \map {B^-_{p^{-\paren {n + 1} } } } a$
== Proof ==
Let $n \in \Z$.
Then:
By set equality:
:$\map {B_{p^{-n} } } a = \map {B^-_{p^{-\paren {n + 1} } } } a$
Category:Topology of P-adic Numbers
Open Ball in Real Number Line is Open Interval

== Theorem ==
Let $\struct {\R, d}$ denote the real number line $\R$ with the usual (Euclidean) metric $d$.
Let $x \in \R$ be a point in $\R$.
Let $\map {B_\epsilon} x$ be the open $\epsilon$-ball at $x$.
Then $\map {B_\epsilon} x$ is the open interval $\openint {x - \epsilon} {x + \epsilon}$.
== Proof ==
Let $S = \map {B_\epsilon} x$ be an open $\epsilon$-ball at $x$.
Let $y \in \map {B_\epsilon} x$.
Then:
As the implications go both ways:
:$\map {B_\epsilon} x \subseteq \openint {x - \epsilon} {x + \epsilon}$
and
:$\map {B_\epsilon} x \supseteq \openint {x - \epsilon} {x + \epsilon}$
By definition of set equality:
:$\map {B_\epsilon} x = \openint {x - \epsilon} {x + \epsilon}$
Hence the result.
== Also see ==
* Open Real Interval is Open Ball
== Sources ==
* : $2$: Continuity generalized: metric spaces: $2.3$: Open sets in metric spaces: Example $2.3.2$
Category:Real Intervals
Category:Open Balls
Category:Real Number Line with Euclidean Metric
Open Ball in Real Number Plane under Chebyshev Distance

== Theorem ==
Let $\R^2$ be the real number plane.
Let $d_\infty: \R^2 \times \R^2 \to \R$ be the Chebyshev Distance on $\R^2$:
:$\ds \map {d_\infty} {x, y} := \max \set {\size {x_1 - y_1}, \size {x_2 - y_2} }$
where $x = \tuple {x_1, x_2}, y = \tuple {y_1, y_2} \in \R^2$.
For $a \in \R^2$, let $\map {B_\epsilon} a$ be the open $\epsilon$-ball at $a$.
Then $\map {B_\epsilon} a$ is the interior of the square centered at $a$ and whose sides are of length $2 \epsilon$ parallel to the coordinate axes.
== Proof ==
Let $a = \tuple {a_1, a_2}$.
From Open Ball in Cartesian Product under Chebyshev Distance:
:$\map {B_\epsilon} {a; d_\infty} = \map {B_\epsilon} {a_1; d} \times \map {B_\epsilon} {a_2; d}$
where $d$ is the usual (Euclidean) topology.
From Open Ball in Real Number Line is Open Interval:
:$\map {B_\epsilon} {a_1; d} \times \map {B_\epsilon} {a_2; d} = \openint {a_1 - \epsilon} {a_1 + \epsilon} \times \openint {a_2 - \epsilon} {a_2 + \epsilon}$
That is:
:$x \in \map {B_\epsilon} {a; d_\infty} \iff \paren {a_2 - \epsilon < x_2 < a_2 + \epsilon} \land \paren {a_2 - \epsilon < x_2 < a_2 + \epsilon}$
from which the result follows.
== Sources ==
* : $2$: Continuity generalized: metric spaces: $2.3$: Open sets in metric spaces: Example $2.3.3$
Category:Open Balls
Category:Chebyshev Distance
Open Ball in Standard Discrete Metric Space

== Theorem ==
Let $M = \struct {A, d}$ be a metric space.
Let $d$ be the standard discrete metric on $M$.
Let $a \in A$.
Let $\map {B_\epsilon} {a; d}$ be an open $\epsilon$-ball of $a$ in $M$.
Then:
:$\map {B_\epsilon} {a; d} = \begin {cases} \set a & : \epsilon \le 1 \\ A & : \epsilon > 1 \end {cases}$
 
== Proof ==
Let $\epsilon \in \R_{>0}: \epsilon \le 1$.
Then:
Let $\epsilon \in \R_{>0}: \epsilon > 1$.
Then:
Hence the result.
== Sources ==
* : $2$: Continuity generalized: metric spaces: $2.3$: Open sets in metric spaces: Example $2.3.4$
Category:Open Balls
Category:Standard Discrete Metric
Open Ball is Convex Set

== Theorem ==
Let $V$ be a normed vector space with norm $\norm {\,\cdot\,}$ over $\R$ or $\C$.
An open ball in the metric induced by $\norm {\,\cdot\,}$ is a convex set.
== Proof ==
Let $v \in V$ and $\epsilon \in \R_{>0}$.
Denote the open $\epsilon$-ball of $v$ as $\map {B_\epsilon} v$.
Let $x, y \in \map {B_\epsilon} v$.
Then $x + t \paren {y - x}$ lies on line segment joining $x$ and $y$ for all $t \in \closedint 0 1$.
The distance between $x + t \paren {y - x}$ and $v$ is:
Hence, $x + t \paren {y - x} \in \map {B_\epsilon} v$.
Thus, by definition, $\map {B_\epsilon} v$ is a convex set.
== Sources ==
* : $\text{I}$ Hilbert Spaces: $\S 2.$ Orthogonality
* $3.1$: Norms
Category:Normed Vector Spaces
Category:Open Balls
Category:Convex Sets (Vector Spaces)
Open Ball is Neighborhood of all Points Inside

== Theorem ==
Let $M = \struct {A, d}$ be a metric space.
Let $a \in A$.
Let $\map {B_\epsilon} a$ be an open $\epsilon$-ball of $a$ in $M$.
Let $x \in \map {B_\epsilon} a$.
Then $\map {B_\epsilon} a$ is a neighborhood of $x$ in $M$.
== Proof ==
From Open Ball of Point Inside Open Ball:
:$\exists \delta \in \R: \map {B_\delta} x \subseteq \map {B_\epsilon} a$
Thus by definition $\map {B_\epsilon} a$ is a neighborhood of $x$ in $M$.
== Sources ==
* : Chapter $2$: Metric Spaces: $\S 4$: Open Balls and Neighborhoods: Lemma $4.5$
Category:Open Balls
Category:Neighborhoods
Open Ball is Open Set/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 an open $\epsilon$-ball of $x$ in $M$.
Then $\map {B_\epsilon} x$ is an open set of $M$.
== Proof ==
Let $y \in \map {B_\epsilon} x$.
From Open Ball of Point Inside Open Ball, there exists $\delta \in \R_{>0}$ such that:
:$\map {B_\delta} y \subseteq \map {B_\epsilon} x$
The result follows from the definition of open set.
== Sources ==
* : Chapter $\text{III}$: Metric Spaces: Compactness
* : $2$: Continuity generalized: metric spaces: $2.3$: Open sets in metric spaces: Example $2.3.9$
Category:Open Ball is Open Set
Category:Open Sets (Metric Spaces)
Open Ball is Open Set/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 an open $\epsilon$-ball of $x$ in $M$.
Then $\map {B_\epsilon} x$ is an open set of $M$.
== Proof ==
Let $y \in \map {B_\epsilon} x$.
From Open Ball of Point Inside Open Ball in Normed Vector Space, there exists $\delta \in \R_{>0}$ such that $\map {B_\delta} y \subseteq \map {B_\epsilon} x$
The result follows from the definition of open set.
== Sources ==
* : $\S 1.3$: Normed and Banach spaces. Topology of normed spaces
Category:Open Ball is Open Set
Category:Open Sets (Normed Vector Spaces)
Open Ball is Open Set/Pseudometric Space

== Theorem ==
Let $M = \struct {A, d}$ be a pseudometric space.
Let $x \in A$.
Let $\epsilon \in \R_{>0}$.
Let $\map {B_\epsilon} x$ be an open $\epsilon$-ball of $x$ in $M$.
Then $\map {B_\epsilon} x$ is an open set of $M$.
== Proof ==
Let $y \in \map {B_\epsilon} x$.
From Open Ball of Point Inside Open Ball, there exists $\delta \in \R_{>0}$ such that:
:$\map {B_\delta} y \subseteq \map {B_\epsilon} x$
The result follows from the definition of open set.
Category:Open Ball is Open Set
Category:Open Sets (Pseudometric Spaces)
Open 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 open $\epsilon$-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 Open Ball is Convex Set and Convex Set is Simply Connected.
Category:Normed Vector Spaces
Category:Open Balls
Category:Simply Connected Spaces
Open Ball is Subset of Open Ball

== Theorem ==
Let $M = \struct {A, d}$ be a metric space.
Let $x, y$ be points of $A$.
Then:
:$\epsilon-\delta \ge \map d {x, y} \implies \map {B_\delta} y \subseteq \map {B_\epsilon} x$
where $\map {B_\epsilon} x$ denotes the open $\epsilon$-ball in $M = \struct {A, d}$.
== Proof ==
Let $\epsilon - \delta \ge \map d {x, y}$.
Then $\epsilon \ge \map d {x, y} + \delta$.
If $z \in \map {B_\delta} y$, then $\map d {y, z} < \delta$.
So:
Thus $z \in \map {B_\epsilon} x$.
So $\map {B_\delta} y \subseteq \map {B_\epsilon} x$.
== Sources ==
* 
* 
Category:Open Balls
Open Ball of Metric Space/Examples/Real Number Line Example

== Example of Open Ball of Metric Space ==
Consider the real number line with the usual (Euclidean) metric $\struct {\R, d}$.
Let $H \subseteq \R$ denote the closed real interval $\closedint 0 1$.
Let $d_H$ denote the metric induced on $H$ by $d$.
Let $\map {B_1} {1; d}$ denote the open ball of $\struct {\R, d}$ of radius $1$ and center is $1$.
Let $\map {B_1} {1; d_H}$ denote the open ball of $\struct {H, d_H}$ of radius $1$ and center is $1$.
Then by definition:
:$\map {B_1} {1; d} = \set {x \in \R: 0 < x < 2} = \openint 0 2$
However:
:$\map {B_1} {1; d_H} = \set {x \in \R: 0 < x \le 1} = \hointl 0 1$.
== Sources ==
* : $2$: Continuity generalized: metric spaces: $2.3$: Open sets in metric spaces: Example $2.3.5$
Category:Examples of Open Balls
Open Ball of Point Inside Open Ball/Metric Space

== Theorem ==
Let $M = \struct {A, d}$ be a metric space.
Let $\map {B_\epsilon} x$ be an open $\epsilon$-ball in $M = \struct {A, d}$.
Let $y \in \map {B_\epsilon} x$.
Then:
:$\exists \delta \in \R: \map {B_\delta} y \subseteq \map {B_\epsilon} x$
That is, for every point in an open $\epsilon$-ball in a metric space, there exists an open $\delta$-ball of that point entirely contained within that open $\epsilon$-ball.
== Proof ==
Let $\delta = \epsilon - \map d {x, y}$.
From the definition of open ball, this is strictly positive, since $y \in \map {B_\epsilon} x$.
If $z \in \map {B_\delta} y$, then $\map d {y, z} < \delta$.
So:
:$\map d {x, z} \le \map d {x, y} + \map d {y, z} < \map d {x, y} + \delta = \epsilon$
Thus $z \in \map {B_\epsilon} x$.
So $\map {B_\delta} y \subseteq \map {B_\epsilon} x$.
=== Illustration ===
== Sources ==
* : $2$: Continuity generalized: metric spaces: $2.3$: Open sets in metric spaces: Lemma $2.3.7$
Category:Open Balls
Category:Open Ball of Point Inside Open Ball
Category:Metric Spaces
Open Ball of Point Inside Open Ball/Metric Space/Illustration

== Illustration of Open Ball of Point Inside Open Ball: Metric Space ==
This diagram illustrates the proof of Open Ball of Point Inside Open Ball: Metric Space in $M = \struct {\R^2, d_2}$:
:
In $\R^2$, we can draw a circle of radius $\delta$ whose center is $y$ lying entirely within the larger circle of radius $\epsilon$ whose center is $x$ .
== Sources ==
* : $2$: Continuity generalized: metric spaces: $2.3$: Open sets in metric spaces: Lemma $2.3.7$
Category:Open Ball of Point Inside Open Ball
Open Ball of Point Inside Open Ball/Normed Vector Space

== Theorem ==
Let $M = \struct {X, \norm {\, \cdot \,} }$ be a normed vector space.
Let $\map {B_\epsilon} x$ be an open $\epsilon$-ball in $M = \struct {X, \norm {\, \cdot \,} }$.
Let $y \in \map {B_\epsilon} x$.
Then:
:$\exists \delta \in \R: \map {B_\delta} y \subseteq \map {B_\epsilon} x$
That is, for every point in an open $\epsilon$-ball in a normed vector space, there exists an open $\delta$-ball of that point entirely contained within that open $\epsilon$-ball.
== Proof ==
Let $\delta = \epsilon - \norm {x - y}$.
From the definition of open ball, this is strictly positive, since $y \in \map {B_\epsilon} x$.
Let $z \in \map {B_\delta} y$ be arbitrary.
We have:
Then we have:
That is:
:$z \in \map {B_\delta} y \implies z \in \map {B_\epsilon} x$
Hence $\map {B_\delta} y \subseteq \map {B_\epsilon} x$.
== Sources ==
* : Chapter $1$: Normed and Banach spaces
Category:Open Balls
Open Ball of Point Inside Open Ball/Pseudometric Space

== Theorem ==
Let $M = \struct {A, d}$ be a pseudometric space.
Let $\map {B_\epsilon} x$ be an open $\epsilon$-ball in $M = \struct {A, d}$.
Let $y \in \map {B_\epsilon} x$.
Then:
:$\exists \delta \in \R: \map {B_\delta} y \subseteq \map {B_\epsilon} x$
That is, for every point in an open $\epsilon$-ball in a pseudometric space, there exists an open $\delta$-ball of that point entirely contained within that open $\epsilon$-ball.
== Proof ==
Let $\delta = \epsilon - \map d {x, y}$.
From the definition of open ball, this is strictly positive, since $y \in \map {B_\epsilon} x$.
If $z \in \map {B_\delta} y$, then $\map d {x, y} < \delta$.
So:
:$\map d {x, z} \le \map d {x, y} + \map d {y, z} < \map d {x, y} + \delta = \epsilon$
Thus $z \in \map {B_\epsilon} x$.
So $\map {B_\delta} y \subseteq \map {B_\epsilon} x$.
Category:Open Balls
Category:Open Ball of Point Inside Open Ball
Open Ball with respect to Seminorm is Convex, Balanced and Absorbing

== Theorem ==
Let $\GF \in \set {\R, \C}$. 
Let $X$ be a vector space over $\GF$.
Let $p$ be a seminorm on $X$.
Let $d_p$ be the pseudometric induced by $p$.
Let $B$ be the open unit ball in $\struct {X, d_p}$. 
That is: 
:$B = \set {x \in X : \map p x < 1}$
Then $B$ is convex, balanced and absorbing. 
== Proof ==
=== Proof that $B$ is convex ===
Let $t \in \closedint 0 1$ and $x, y \in B$. 
Then:
so: 
:$t x + \paren {1 - t} y \in B$. 
=== Proof that $B$ is balanced ===
Let $s \in \GF$ such that $\cmod s \le 1$.
Let $x \in B$. 
Then, we have by we have: 
:$\map p {s x} = \cmod s \map p x \le \map p x < 1$ 
so $s x \in B$.
So, we have: 
:$s B \subseteq B$
for all $s \in \GF$ with $\cmod s \le 1$.
So $B$ is balanced. 
=== Proof that $B$ is absorbing ===
From Characterization of Convex Absorbing Set in Vector Space, it is enough to show that:
:$\ds X = \bigcup_{n \mathop = 1}^\infty n B$
By , we have: 
:$\map p {n x} < n$ $\map p x < 1$
for each $n \in \N$.
So, we have: 
:$n B = \set {x \in X : \map p x < n}$
Clearly we have: 
:$\ds \bigcup_{n \mathop = 1}^n n B \subseteq X$
Now let $x \in X$, then we have: 
:$\ds \map p {\frac x {2 \map p x} } = \frac 1 2 < 1$
from , so: 
:$\ds \frac x {2 \map p x} \in B$
Then we have: 
:$x \in \paren {2 \map p x} B$
Taking $N \in \N$ with $N \ge \map p x$, we have $x \in N B$, and so: 
:$\ds x \in \bigcup_{n \mathop = 1}^\infty n B$
So:
:$\ds X = \bigcup_{n \mathop = 1}^\infty n B$ 
== Sources ==
* : $1.34$: Theorem
Category:Convex Sets (Vector Spaces)
Category:Absorbing Sets
Category:Balanced Sets
Category:Seminorms
Open Balls form Basis for Open Sets of Metric Space

== Theorem ==
Let $M = \struct {A, d}$ be a metric space.
Let $\BB$ be the set of all open balls of $M$.
Then $\BB$ is a basis for the open sets of $M$.
== Proof ==
Let $U$ be an open set of $M$.
Then by definition:
:$\forall y \in U: \exists \epsilon \in \R_{>0}: \map {B_\epsilon} y \subseteq U$
Thus:
:$\ds U = \bigcup_{y \mathop \in U} \map {B_\epsilon} y$
== Also see ==
* Open Balls on Rational Centers form Basis for Usual Topology on Plane
== Sources ==
* : Chapter $2$: Metric Spaces: $\S 6$: Open Sets and Closed Sets: Exercise $1$
Category:Open Sets (Metric Spaces)
Category:Open Balls
Category:Topological Bases
Open Balls form Local Basis for Point of Metric Space

== Theorem ==
Let $M = \struct {A, d}$ be a metric space.
Let $x \in A$.
Let $\BB_x$ be the set of all open balls of $M$ centered on $x$.
That is:
:$\BB_x = \set {\map {B_\epsilon} x : \epsilon \in \R_{>0}}$
Then $\BB$ is a local basis of $x$.
== Proof ==
Let $U$ be an open set of $M$ which has $x$ as an element.
Then by definition of an open set:
:$\exists \epsilon \in \R_{>0}: \map {B_\epsilon} x \subseteq U$
From Open Ball of Metric Space is Open Set, $\BB_x$ is a set of open set which have $x$ as an element.
By definition of a local basis, $\BB_x$ is a local basis of $x$.
Category:Open Balls
Open Balls form Local Basis of Metric Space

== Theorem ==
Let $\struct {M, d}$ be a metric space. 
Let $\tau$ be the topology on $M$ induced by $d$. 
Let $x \in X$. 
For each $\epsilon > 0$, let $\map {B_\epsilon} x$ be the open $\epsilon$-ball around $x$.
Let $\BB_x = \set {\map {B_\epsilon} x : \epsilon > 0}$.
Then $\BB_x$ is a local basis at $x$. 
== Proof ==
Let $U$ be an open neighborhood of $x$. 
From the definition of an open set in $\struct {M, d}$:
:there exists $\epsilon > 0$ such that $\map {B_\epsilon} x \subseteq U$. 
Hence:
:there exists $V \in \BB_x$ such that $V \subseteq U$. 
So $\BB_x$ is a local basis at $x$.
Category:Local Bases
Category:Metric Spaces
Open Balls of P-adic Number

== Theorem ==
Let $p$ be a prime number.
Let $\struct {\Q_p, \norm {\,\cdot\,}_p}$ be the $p$-adic numbers.
Let $\Z_p$ be the $p$-adic integers.
Let $a \in \Q_p$.
For all $\epsilon \in \R_{>0}$, let $\map {B_\epsilon} a$ denote the open ball of $a$ of radius $\epsilon$. 
Then:
:$\forall n \in Z : \map {B_{p^{-n} } } a = a + p^{n + 1} \Z_p$
== Proof ==
Let $n \in \Z$.
From Open Ball in P-adic Numbers is Closed Ball:
:$\map {B_{p^{-n} } } a = \map {B^{\,-}_{p^{-\paren {n + 1} } } } a$
From Closed Balls of P-adic Number:
:$\map { B^{\,-}_{p^{-\paren {n + 1} } } } a = a + p^{n + 1} \Z_p$
The result follows.
== Also see ==
* Closed Balls of P-adic Number
* Local Basis of P-adic Number
Category:Topology of P-adic Numbers
Open Balls of Same Radius form Open Cover

== Therorem ==
Let $M = \struct{A, d}$ be a metric space.
Let $\UU_\epsilon = \set{\map {B_\epsilon} x : x \in A}$
That is, $\UU_\epsilon$ is the set of all open balls of radius $\epsilon > 0$.
Then:
:$\UU_\epsilon$ is an open cover of $M$.
== Proof ==
From Open Ball is Open Set:
:$\UU_\epsilon$ is a set of open subsets
From Center is Element of Open Ball:
:$\forall x \in A : x \in \map {B_\epsilon} x$
By definition, $\UU_\epsilon$ is a cover of $A$.
By definition, $\UU_\epsilon$ is an open cover of $A$.
Category:Open Balls
Category:Covers
Open Balls on Rational Centers form Basis for Usual Topology on Plane

== Theorem ==
Let $\R^2$ be the real number plane with the usual (Euclidean) topology.
Let $S$ be the set defined as:
:$S = \set {\tuple {x, y} \in \R^2: x, y \in \Q}$
That is, let $S$ be the set of all points in $\R^2$ whose coordinates are rational numbers.
Let $\BB$ denote the set defined as:
:$\BB = \set {\map {B_q} s: s \in S, q \in \Q}$
That is, let $\BB$ be the set of all open $q$-balls with centers in $S$ and rational radius.
Then $\BB$ forms a basis for $\R^2$.
== Proof ==
Let $d: \R^2 \times \R^2 \to \R$ be the usual (Euclidean) metric on $\R^2$.
Let $U$ be an open set of $\R^2$.
Let $z = \tuple {x, y} \in U$.
Then by definition of open set:
:$\exists \epsilon \in \R_{>0}: \map {B_\epsilon} z \subseteq U$
By Rationals are Everywhere Dense in Reals:
:$\exists \tuple {x', y'}$ such that:
::$\size {x - x'} < \dfrac \epsilon 5$
::$\size {y - y'} < \dfrac \epsilon 5$
Let $s = \tuple {x', y'}$.
Then:
:$\map d {z, s}< \dfrac {e \sqrt 2} 5 < \dfrac \epsilon 3$
Let $q \in \Q$ such that $\dfrac \epsilon 3 < q \le \dfrac {2 \epsilon} 3$.
Then:
:$z \in \map {B_q} s \subseteq \map {B_\epsilon} z \subseteq U$
As $z$ is arbitrary, it follows that $U$ is the union of open balls with centers in $S$ and rational radius.
Hence the result.
== Sources ==
* : $3$: Continuity generalized: topological spaces: $3.2$: Bases: Example $3.2.7$
* : $3$: Continuity generalized: topological spaces: Exercise $3.9: 8$
Category:Topological Bases
Open Balls whose Distance between Centers is Twice Radius are Disjoint

== Theorem ==
Let $M = \struct {A, d}$ be a metric space.
Let $x, y \in A$ such that $\map d {x, y} = 2 r > 0$.
Let $\map {B_r} x$ denote the open $r$-ball of $x$ in $M$.
Then $\map {B_r} x$ and $\map {B_r} y$ are disjoint.
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : $2$: Continuity generalized: metric spaces: Exercise $2.6: 1$
Category:Open Balls whose Distance between Centers is Twice Radius are Disjoint
Category:Open Balls
Open Balls whose Distance between Centers is Twice Radius are Disjoint/Proof 1

== Theorem ==
== Proof ==
 $\map {B_r} x \cap \map {B_r} y \ne \O$.
Then:
But this contradicts our initial assertion that $\map d {x, y} = 2 r$.
The result follows by Proof by Contradiction.
Category:Open Balls whose Distance between Centers is Twice Radius are Disjoint