nyuuzyou/wikis huggingface dataset:
Open Sets in Vector Spaces with Equivalent Norms Coincide

== Theorem ==
Let $M_a = \struct {X, \norm {\, \cdot \, }_a}$ and $M_b = \struct {X, \norm {\, \cdot \,}_b}$ be normed vector spaces.
Let $U \subseteq X$ be an open set in $M_a$.
Suppose, $\norm {\, \cdot \, }_a$ and $\norm {\, \cdot \,}_b$ are equivalent norms, i.e. $\norm {\, \cdot \, }_a \sim \norm {\, \cdot \,}_b$.
Then $U$ is also open in $M_b$. 
== Proof ==
By definition of equivalent norms:
:$\exists m,M \in \R_{> 0} : m \le M : \forall x \in X: m \norm x_b \le \norm x_a \le M \norm x_b$
Since $U$ is open in $M_a$:
:$\forall x \in U : \exists \epsilon_a \in \R_{> 0} : \map {B_{\epsilon_a}} x \subseteq U$
where $\map {B_{\epsilon_a}} x$ stands for an open ball, defined as:
:$\map {B_{\epsilon_a}} x := \set {\forall y \in X : \norm {x - y}_a < \epsilon_a}$
Define an open ball $\map {B_{\epsilon_b}} x$ as:
:$\map {B_{\epsilon_b}} x := \set {\forall y \in X : \norm {x - y}_b < \epsilon_b}$
Then we have that:
:$m \norm {x - y}_b \le \norm {x - y}_a < \epsilon_a$
So far, $\epsilon_b$ was unspecified.
Define $\epsilon_b := \dfrac {\epsilon_a} m$.
So:
:$\forall x \in U : y \in \map {B_{\epsilon_a}} x \implies y \in \map {B_{\epsilon_b}} x$
with $\epsilon_a = m \epsilon_b$.
Hence:
:$\forall x \in U : \exists \epsilon_b \in \R_{> 0} : \map {B_{\epsilon_b}} x \subseteq U$
== Sources ==
* : $\S 1.4$: Normed and Banach spaces. Sequences in a normed space; Banach spaces
Category:Equivalence Relations
Category:Norm Theory
Category:Vector Spaces
Open Sets in Weak-* Topology of Topological Vector Space

== Theorem ==
Let $\GF \in \set {\R, \C}$. 
Let $X$ be a topological vector space over $\GF$. 
Let $X^\ast$ be the topological dual space of $X$. 
Let $w^\ast$ be the weak-$\ast$ topology on $X^\ast$.
Let $U \subseteq X$. 
Then $U$ is open in $\struct {X^\ast, w^\ast}$ for each $f \in X^\ast$ there exists $x_1, x_2, \ldots, x_n \in X$ and $\epsilon > 0$ such that:
:$\set {g \in X^\ast : \cmod {\map f {x_i} - \map g {x_i} } < \epsilon \text { for each } 1 \le i \le n}$ 
== Proof ==
For each $x \in X$, define $p_x : X^\ast \to \hointr 0 \infty$ by: 
:$\map {p_x} f = \cmod {\map f x}$
for each $f \in X^\ast$, and set:
:$\PP = \set {p_x : x \in X}$
From the definition of the weak-$\ast$ topology, $w^\ast$ is generated by $\set {x^\wedge : x \in X}$.
So from from Initial Topology on Vector Space Generated by Linear Functionals is Locally Convex, $w^\ast$ is the standard topology on the locally convex space $\struct {X^\ast, \PP}$.
From Open Sets in Standard Topology of Locally Convex Space, we obtain that $U$ is open in $\struct {X^\ast, w^\ast}$ for each $f \in X^\ast$ there exists $x_1, x_2, \ldots, x_n \in X$ and $\epsilon > 0$ such that:
:$\set {g \in X^\ast : \cmod {\map f {x_i} - \map g {x_i} } < \epsilon \text { for each } 1 \le i \le n}$
Category:Weak-* Topologies
Open Sets in Weak Topology of Topological Vector Space

== Theorem ==
Let $\GF \in \set {\R, \C}$. 
Let $X$ be a topological vector space over $\GF$ with weak topology $w$. 
Let $U \subseteq X$. 
Then $U$ is open in $\struct {X, w}$ for each $x \in U$ there exists $f_1, f_2, \ldots, f_n \in X^\ast$ and $\epsilon > 0$ such that: 
:$\set {y \in X : \cmod {\map {f_i} {y - x} } < \epsilon \text { for each } 1 \le i \le n} \subseteq U$
== Proof ==
Let $X^\ast$ be the topological dual space of $X$. 
By the definition of the weak topology, $w$ is the initial topology on $X$ with respect to $X^\ast$.
For each $f \in X^\ast$, define $p_f : X \to \hointr 0 \infty$ by: 
:$\map {p_f} x = \cmod {\map f x}$
for each $x \in X$, and set: 
:$\PP = \set {p_f : f \in X^\ast}$ 
Then from Initial Topology on Vector Space Generated by Linear Functionals is Locally Convex, $w$ is the standard topology on the locally convex space $\struct {X, \PP}$.
From Open Sets in Standard Topology of Locally Convex Space, we obtain that $U$ is open in $\struct {X, w}$ for each $x \in U$ there exists $f_1, f_2, \ldots, f_n \in X^\ast$ and $\epsilon > 0$ such that: 
:$\set {y \in X : \map {p_{f_i} } {y - x} < \epsilon \text { for each } 1 \le i \le n} = \set {y \in X : \cmod {\map {f_i} {y - x} } < \epsilon \text { for each } 1 \le i \le n} \subseteq U$
Category:Weak Topologies on Topological Vector Spaces
Open Sets of Cartesian 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 $\ds \prod_{i \mathop = 1}^n U_i$ is open in $M = \struct {\AA, d_\infty}$.
== Proof ==
A set $U$ is open it is the neighborhood of each of its points.
That is:
:$\forall a \in U: \exists \delta \in \R_{>0}: \map {B_\delta} a \subseteq U$
where $\map {B_\delta} a$ denotes the open $\delta$-ball of $a$.
Let $I = \set {1, 2, \ldots, n}$.
For all $i \in I$, let $U_i$ be open in $M_i$.
Then:
For all $i \in I$, let $\map d {x_i, a_i} < \delta$.
Then:
Hence $\ds \prod_{i \mathop \in I} U_i$ is open in $M$.
== Sources ==
* : Chapter $2$: Metric Spaces: $\S 6$: Open Sets and Closed Sets: Exercise $1$
Category:Chebyshev Distance
Category:Open Sets (Metric Spaces)
Open Sets of Double Pointed Topology

== Theorem ==
Let $\struct {S, \tau_S}$ be a topological space.
Let $D$ be a doubleton endowed with the indiscrete topology.
Let $\struct {S \times D, \tau}$ be the double pointed topology on $S$.
Then $X \subseteq S \times D$ is open in $\tau$ for some $U \in \tau$:
:$X = U \times D$
=== Corollary ===
== Proof ==
By definition, $\tau$ is the product topology on $X \times D$.
That is, $\tau$ has as a basis sets of the form:
:$U \times V$
with $U \in \tau$ and $V$ open in $D$.
Since $D$ is endowed with the indiscrete topology, either $V = \O$ or $V = D$.
In the former case, by Cartesian Product is Empty iff Factor is Empty, $U \times V = \O$ for all $U \in \tau$.
In particular, then, it follows that $\tau$ has the basis:
:$\BB = \set {U \times D: U \in \tau}$
since $U = \O \in \tau$, and so the basis elements with $V = \O$ are also accounted for.
It is therefore clear that all $U \times D$ with $U \in \tau$ are open in $\tau$.
That open sets of this form constitute all of $\tau$ amounts to showing that:
:$\tau = \BB$
that is, that $\BB$ is already a topology on $S \times D$.
Category:Double Pointed Topologies
Open Sets of Double Pointed Topology/Corollary

== Theorem ==
Let $\struct {S, \tau_S}$ be a topological space.
Let $D$ be a doubleton endowed with the indiscrete topology.
Let $\struct {S \times D, \tau}$ be the double pointed topology on $S$.
A subset $X \subseteq S \times D$ is closed in $\tau$ for some closed set $C$ of $\tau$:
:$X = C \times D$
== Proof ==
By definition, $X$ is closed its complement $\map \complement X$ is open.
By Open Sets of Double Pointed Topology, it follows by that for some $U \in \tau$:
:$\map \complement X = U \times D$
Then by Cartesian Product Distributes over Set Difference and Complement of Complement, we have that:
:$X = \paren {S \times D} \setminus \paren {U \times D} = \paren {S \setminus U} \times D$
Since $S \setminus U$ is a closed set of $\tau$, the result follows.
Category:Double Pointed Topologies
Open Star Convex Set Minus Countable Set is Path-Connected in Real Euclidean Space

== Theorem ==
Let $k \ge 2$ be a natural number.
Let $A \subset \R^k$ be a star convex set, with star center $a \in A$.
Let $G \subset \R^k$ be finite or countable.
Suppose $a \notin G$.
Then $A \setminus G$ is path-connected.
== Proof ==
Let $x \in A \setminus G$ be an arbitrary point.
By definition of dimension, there is some vector $\vec v \in \R^k$ that is linearly independent from $\vec {x a}$.
Now, consider the mapping $f: \R \to \R^k$:
:$\map {\vec f} k = \vec v + k \vec {x a}$
The vector $\vec {x a}$ is linearly independent from every vector in the image of $\vec f$.
:For, $\paren {\vec v + k \vec {x a} } - k \vec {x a} = \vec v$, which is linearly independent from $\vec {x a}$.
Likewise, every pair of vectors in $\Img f$ is linearly independent.
:For, $\dfrac 1 {k_1 - k_2} \paren {\paren {\vec v + k_1 \vec {x a} } - \paren {\vec v + k_2 \vec {x a} } } = \vec {x a}$
:If either $k_1$ or $k_2$ is $0$, then we are done.
:Otherwise, $\dfrac {k_2} {k_2 - k_1} \paren {\paren {\vec v + k_1 \vec {x a} } - \dfrac {k_1} {k_2} \paren {\vec v + k_2 \vec {x a} } } = \vec v$.
Associate with each $g \in G$ the unique $k \in \R$, if it exists, such that:
:$\exists t \in \R: g = x + t \map {\vec f} k$
That is, the real number $k$ such that $g$ lines on the line extending from $x$ in the direction $\map {\vec f} k$.
By Real Numbers are Uncountably Infinite, $\size \R > \size G$.
Therefore, by Surjection iff Cardinal Inequality, the association is not a surjection.
Thus, there is some $k_0 \in \R$ such that no $g \in G$ lies on the line extending from $x$ in the direction $\map {\vec f} {k_0}$.
By definition of open set, there exists some open ball $\map {B_\epsilon} x \subset A$.
Define the mapping $h: \openint 0 \epsilon \to \R^k$:
:$\map h m = x + m \dfrac {\map {\vec f} {k_0} } {\size {\map {\vec f} {k_0} } }$
In words, map each real number in the interval to the point that is distance $m$ in the direction of $\map {\vec f} {k_0}$ from $x$.
As $\vec {x a}$ and $\map {\vec f} {k_0}$ are linearly independent, it follows that the vectors:
:$\overrightarrow {\map h m a}$
are are pairwise linearly independent.
We can again associate the $g \in G$ with the unique $m$ such that $g$ lies on the line $\map h m + t \overrightarrow {\map h m a}$.
Each of those lines lies in $A$, by definition of star convex.
As Real Interval is Uncountably Infinite, it follows that there is some $m_0$ for which no $g$ lies on the corresponding line.
But then, the concatenation of the lines from $x$ to $\map h {m_0}$, and then from $\map h {m_0}$ to $a$ does not intersect any $g \in G$.
Therefore, the same path exists in $A \setminus G$.
Finally, for any $y, z \in A \setminus G$, the concatenation of the path from $y$ to $a$, followed by the reverse of $z$ to $a$, forms a path from $y$ to $z$.
Therefore, $A \setminus G$ is path-connected.
Category:Euclidean Space
Category:Path-Connected Sets
Open Subgroup is Closed

== Theorem ==
Let $G$ be a topological group.
Let $H\leq G$ be an open subgroup.
Then $H$ is closed.
== Proof ==
By Right and Left Regular Representations in Topological Group are Homeomorphisms, the left cosets of $H$ are open.
By Left Coset Space forms Partition, the complement of $H$ is a union of left cosets of $H$.
Because the complement of $H$ is open, $H$ is closed.
Category:Topology
Category:Group Theory
Category:Topological Groups
Open Subscheme is Scheme

== Theorem ==
Let $\struct {X, \OO_X}$ be a scheme.
Let $U \subset X$ be an open subset.
Then the open subscheme $\struct {U, \OO_X {\restriction U}}$ defined by $U$ is a scheme.
== Proof ==
Let $x \in U$.
By Open Neighborhood contains Affine Open Neighborhood, there is an open subset $V \subset U$ with $x \in V$, such that $\struct {V, \OO_X {\restriction V}}$ is an affine scheme.
By Restriction of Restriction of Functor is Restriction $\OO_X {\restriction U} {\restriction V} = \OO_X {\restriction V}$.
By definition $\struct {U, \OO_X {\restriction U}}$ is a scheme.
Category:Algebraic Geometry
Category:Schemes
Open Subset is Lower Section in Lower Topology

== Theorem ==
Let $T = \struct {S, \preceq, \tau}$ be a transitive relational structure with lower topology.
Let $A \subseteq S$ such that
:$A$ is open.
Then $A$ is a lower section of $S$.
== Proof ==
Define $B = \set {\relcomp S {x^\succeq}: x \in S}$
By definition of lower topology:
:$B$ is sub-basis of $T$.
By definitions of sub-basis and basis:
:$\ds \BB = \set {\bigcap \FF: \FF \subseteq B, \FF \text{ is finite} }$ is a basis.
By definition of basis:
:$\ds \tau \subseteq \set {\bigcup X: X \subseteq \BB}$
Let $x \in A$, $y \in S$ such that:
:$y \preceq x$
By definition of open set:
:$A \in \tau$
By definition of subset:
:$\exists Y \subseteq \BB: A = \bigcup Y$
By definition of union:
:$\exists Z \in Y: x \in Z$
By definition of subset:
:$\exists X \subseteq B: Z = \bigcap X \land X$ is finite.
We will prove that:
:$\forall Q \in X: y \in Q$
Let $Q \in X$.
By definition of subset:
:$Q \in B$
By definition of $B$:
:$\exists z \in S: Q = \relcomp S {z^\succeq}$
By definition of intersection:
:$x \in Q$
By Upper Closure is Upper Section:
:$z^\succeq$ is an upper section.
By Complement of Upper Section is Lower Section:
:$Q$ is a lower section.
Thus by definition of lower section:
:$y \in Q$
By definition of intersection:
:$y \in Z$
Thus by definition of union:
:$y \in A$
== Sources ==
* 
Category:Topological Order Theory
Open Subset of Locally Compact Hausdorff Space is Locally Compact Hausdorff Space

== Theorem ==
Let $\struct {X, \tau_X}$ be a locally compact Hausdorff space.
Let $Y \subseteq X$ be open.
Let $\tau_Y$ be the subspace topology on $Y$ inherited from $\struct {X, \tau_X}$.
Then $\struct {Y, \tau_Y}$ is a locally compact Hausdorff space.
== Proof ==
From $T_2$ Property is Hereditary, $\struct {Y, \tau_Y}$ is Hausdorff.
Let $y \in Y$. 
Let $\CC$ be a neighborhood basis of $y$ in $\struct {X, \tau_X}$ consisting of compact sets. 
Let $U$ be a neighborhood of $y$ in $\struct {Y, \tau_Y}$.
Then there exists an open neighborhood $V$ of $y$ in $\struct {Y, \tau_Y}$ such that:
:$V \subseteq U$
From Open Set in Open Subspace:
:$V$ is open in $\struct {X, \tau_X}$.
Hence from the definition of a neighborhood basis, there exists $C \in \CC$ such that $y \in C \subseteq V$.
Again from Open Set in Open Subspace, we have that:
:$C$ is compact in $\struct {Y, \tau_Y}$. 
Hence $C$ is compact in $\struct {Y, \tau_Y}$ with $C \subseteq U$.
Since $C$ is compact and $U$ is arbitrary, we have that $\CC$ is a neighborhood basis for $y$ in $\struct {Y, \tau_Y}$.
Category:Locally Compact Hausdorff Spaces
Open Subset of Locally Connected Space is Locally Connected/Lemma

== Lemma to Open Subset of Locally Connected Space is Locally Connected ==
Let $T = \struct {S, \tau}$ be a topological space.
Let $U$ be a connected set.
Let $A$ be a subset of $T$ such that $U \subseteq A$.
Let $\tau_A$ be the subspace topology on $A$ induced from $T$.
Then $U$ is connected in the topological subspace $\struct {A, \tau_A}$.
== Proof ==
Let $\tau_U$ denote the subspace topology on $U$ induced from $T$.
By the definition of a connected set, the topological space $\struct {U, \tau_U}$ is connected.
Let ${\tau_U}'$ denote the subspace topology on $U$ induced from $\struct {A, \tau_A}$.
By Subspace of Subspace is Subspace, ${\tau_U}' = \tau_U$.
Thus the topological spaces $\struct {U, {\tau_U}'}$ and $\struct {U, \tau_U}$ are identical.
In particular, since $\struct {U, \tau_U}$ is connected, so is $\struct {U, {\tau_U}'}$.
Applying the definition of a connected set once more, it follows that $U$ is connected when considered as a subset of $A$.
Category:Open Subset of Locally Connected Space is Locally Connected
Open Superset is Open Neighborhood

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Let $U \in \tau$ be an open set of $T$.
Let $H \subseteq U$.
Then $U$ is an open neighborhood of $H$.
== Proof ==
By hypothesis, $U$ is open in $T$.
Also , $H \subseteq U$.
The result follows by definition of open neighborhood.
Category:Topology
Open Unit Interval is Proper Subset of Closed Unit Interval

== Theorem ==
The open unit interval:
:$I_o = \openint 0 1$
is a proper subset of the closed unit interval:
:$I_c = \closedint 0 1$
== Proof ==
Let $x \in I_o$.
Then by definition:
:$0 < x < 1$
and so:
:$0 \le x \le 1$
and so:
:$x \in I_c$.
Thus:
:$I_o \subseteq I_c$
Consider:
:$0 \in I_c$
by definition of closed interval.
But it is not the case that $0 < 0$.
So $0 \notin I_o$ and so $I_c \nsubseteq I_o$.
Hence the result by definition of proper subset.
== Sources ==
* : $\S 2.1$: Sets
Category:Real Intervals
Open Unit Interval on Rational Number Space is Bounded but not Compact

== Theorem ==
Let $\struct {\Q, \tau_d}$ be the rational number space under the Euclidean topology $\tau_d$.
Then:
:$\openint 0 1 \cap \Q$ is totally bounded but not compact
where $\openint 0 1$ is the open unit interval.
== Proof ==
We first show that $\openint 0 1 \cap \Q$ is totally bounded.
Let $\epsilon \in \R_{> 0}$.
By the Archimedean Property of $\R$:
:$\exists n \in \N: \dfrac 1 n < \epsilon$
We pick the numbers $\dfrac i n \in \openint 0 1 \cap \Q$, where $i \in \N$ and $0 < i < n$.
Then for all $x \in \openint 0 1 \cap \Q$ and $x \ge \dfrac 1 n$:
For $x \le \dfrac 1 n$:
This shows that $\openint 0 1 \cap \Q$ totally bounded.
From the Heine-Borel Theorem on a metric space, $\openint 0 1 \cap \Q$ is compact it is both totally bounded and complete.
From Rational Number Space is not Complete Metric Space it follows that $\openint 0 1 \cap \Q$ is not compact.
== Sources ==
* : Part $\text {II}$: Counterexamples: $30$. The Irrational Numbers: $10$
Category:Rational Number Space
Category:Compact Spaces
Category:Totally Bounded Metric Spaces
Open and Closed Balls in P-adic Numbers are Clopen in P-adic Metric

== Theorem ==
Let $p$ be a prime number.
Let $\struct {\Q_p, \norm {\,\cdot\,}_p}$ be the $p$-adic numbers.
Let $a \in \Q_p$.
Let $n \in \Z$.
Then the open ball $\map {B_{p^{-n}}} a$ and closed ball $\map {B^-_{p^{-n}}} a$ are clopen in the $p$-adic metric.
== Proof ==
We begin by proving the theorem for the closed ball $\map {B^-_{p^{-n}}} a$.
From Open Ball in P-adic Numbers is Closed Ball then the theorem will be proved.
From P-adic Numbers form Non-Archimedean Valued Field::
:$\norm {\,\cdot\,}_p$ is a non-Archimedean division ring norm.
By definition the $p$-adic closed ball:
:$\map {B^-_{p^{-n} } } a$ is a closed ball in a normed division ring. 
From Closed Ball of Non-Archimedean Division Ring is Clopen:
:$\map {B^-_{p^{-n} } } a$ is clopen in the $p$-adic metric.
The result follows.
== Sources ==
* : $\S 3.3$ Exploring $\Q_p$, Lemma $3.3.8$
Category:Topology of P-adic Numbers
Open and Closed Balls in P-adic Numbers are Compact Subspaces

== Theorem ==
Let $p$ be a prime number.
Let $\struct {\Q_p, \norm {\,\cdot\,}_p}$ be the $p$-adic numbers.
Let $a \in \Q_p$.
Let $n \in \Z$.
Then the open ball $\map {B_{p^{-n} } } a$ and closed ball $\map {B^-_{p^{-n}}} a$ are compact.
=== Corollary ===
== Proof ==
We begin by proving the theorem for the closed ball $\map {B^-_{p^{-n} } } a$.
From Open Ball in P-adic Numbers is Closed Ball then the theorem will be proved.
Let $d$ denote the subspace metric induced on $\map {B^-_{p^{-n}}} a$ by the $p$-adic Metric.
From Open and Closed Balls in P-adic Numbers are Totally Bounded:
:the closed ball $\map {B^-_{p^{-n}}} a$ is totally bounded in $d$.
From P-adic Numbers form Completion of Rational Numbers with P-adic Norm:
:$\struct {\Q_p, \norm {\,\cdot\,}_p}$ is complete.
From Open and Closed Balls in P-adic Numbers are Clopen in P-adic Metric:
:the closed ball $\map {B^-_{p^{-n}}} a$ is closed in the $p$-adic metric.
From Subspace of Complete Metric Space is Closed iff Complete:
:the closed ball $\map {B^-_{p^{-n}}} a$ is complete in $d$.
From Complete and Totally Bounded Metric Space is Sequentially Compact:
:the closed ball $\map {B^-_{p^{-n}}} a$ is sequentially compact in $d$.
From Sequentially Compact Metric Space is Compact:
:the closed ball $\map {B^-_{p^{-n} } } a$ is compact in $d$.
Hence the closed ball $\map {B^-_{p^{-n} } } a$ is a compact subspace in the $p$-adic metric by definition.
The result follows.
Category:Topology of P-adic Numbers
Open and Closed Balls in P-adic Numbers are Compact Subspaces/P-adic Integers

== Theorem ==
Let $p$ be a prime number.
Let $\struct {\Q_p, \norm {\,\cdot\,}_p}$ be the $p$-adic numbers.
Then the set of $p$-adic integers $\Z_p$ is compact.
== Proof ==
By definition the $p$-adic integers $\Z_p$ is the closed ball $\map {B^-_1} 0$.
From Open and Closed Balls in P-adic Numbers are Compact Subspaces, $\map {B^-_1} 0$ is compact.
== Sources ==
* : $\S 3.3$ Exploring $\Q_p$: Lemma $3.3.8$
Category:Topology of P-adic Numbers
Open and Closed Balls in P-adic Numbers are Totally Bounded

== Theorem ==
Let $p$ be a prime number.
Let $\struct {\Q_p, \norm {\,\cdot\,}_p}$ be the $p$-adic numbers.
Let $a \in \Q_p$.
Let $n \in \Z$.
Then the open ball $\map {B_{p^{-n} } } a$ and closed ball $\map {B^-_{p^{-n} } } a$ are totally bounded subspaces.
== Proof ==
We begin by proving the theorem for the closed ball $\map {B^-_{p^{-n} } } a$.
From Open Ball in P-adic Numbers is Closed Ball then the theorem will be proved.
Let $d$ denote the subspace metric induced by the norm $\norm {\,\cdot\,}_p$ on $\map {B^-_{p^{-n} } } a$.
That is, $d: \map {B^-_{p^{-n} } } a \times \map {B^-_{p^{-n} } } a \to \R_{>0}$ is the metric defined by:
::$\forall x, y \in \map {B^-_{p^{-n} } } a: \map d {x, y} = \norm {x - y}_p$
By the definition of totally bounded it needs to be shown that:
:for every $\epsilon \in \R_{>0}$ there exist finitely many points $x_0, \dots, x_k \in \map {B^-_{p^{-n} } } a$ such that:
::$\ds \inf_{0 \mathop \le i \mathop \le k} \norm {x_i - x}_p \le \epsilon$
:for all $x \in \map {B^-_{p^{-n} } } a$. 
From Sequence of Powers of Reciprocals is Null Sequence:
:$\exists N \in \N: \forall k > N: p^{-k} \le \epsilon$.
Let $m = \max \set {n + 1, N}$.
Then $m > n$.
From Closed Ball is Disjoint Union of Smaller Closed Balls in P-adic Numbers:
:$\map {B^-_{p^{-n} } } a = \ds \bigcup_{i \mathop = 0}^{p^{\paren {m - n} } - 1} \map {B^-_{p^{-m} } } {a + i p^n}$
Let $x \in \map {B^-_{p^{-n} } } a$.
Let $k = p^{\paren {m - n} } - 1$.
Then there exists $0 \le j \le k$ such that:
:$x \in \map {B^-_{p^{-m} } } {a + j p^n}$
That is:
:$\norm {x - a + j p^n}_p \le p^{-m} \le \epsilon$
Hence:
:$\ds \inf_{0 \mathop \le i \mathop \le k} \norm {a + i p^n - x}_p \le \norm {x - a + j p^n}_p \le \epsilon$
Since $x \in \map {B^-_{p^{-n} } } a$ was arbitrary, then: 
:$\forall x \in \map {B^-_{p^{-n} } } a : \ds \inf_{0 \mathop \le i \mathop \le k} \norm {a + i p^n - x}_p \le \epsilon$
The result follows.
== Sources ==
* : $\S 3.3$ Exploring $\Q_p$, Lemma $3.3.8$
Category:Topology of P-adic Numbers
Open and Closed Sets in Indiscrete Topology

== Theorem ==
Let $T = \struct {S, \set {\O, S} }$ be an indiscrete topological space.
Let $H \subseteq S$.
=== Open Sets in Indiscrete Topology ===
=== Closed Sets in Indiscrete Topology ===
=== $F_\sigma$ Sets in Indiscrete Topology ===
=== $G_\delta$ Sets in Indiscrete Topology ===
== Sources ==
* : Part $\text {II}$: Counterexamples: $4$. Indiscrete Topology: $2$
Category:Indiscrete Topology