nyuuzyou/wikis huggingface dataset:
Subset of Hilbert Sequence Space with Non-Empty Interior is not Compact

== Theorem ==
Let $A$ be the set of all real sequences $\sequence {x_i}$ such that the series $\ds \sum_{i \mathop \ge 0} x_i^2$ is convergent.
Let $\ell^2 = \struct {A, d_2}$ be the Hilbert sequence space on $\R$.
Let $H$ be a subset of $\ell^2$ whose interior is non-empty.
Then $H$ is not compact in $\ell^2$.
== Proof ==
Let $x \in H^\circ$, where $H^\circ$ denotes the interior of $H$.
By definition, $H^\circ$ is an open set of $\ell^2$ containing $x$.
Again by definition, $H$ is a neighborhood of $x$.
But from Point in Hilbert Sequence Space has no Compact Neighborhood, $x$ has no compact neighborhood in $\ell^2$.
Thus $H$ cannot be compact in $\ell^2$.
== Sources ==
* : Part $\text {II}$: Counterexamples: $36$. Hilbert Space: $4$
Category:Hilbert Sequence Space
Subset of Indiscrete Space is Compact

== Theorem ==
Let $T = \struct {S, \set {\O, S} }$ be an indiscrete topological space.
Let $H \subseteq S$.
$H$ is compact in $T$.
== Proof ==
The subspace $T_H = \struct {H, \set {\O, S \cap H} }$ is trivially also an indiscrete space.
The only open cover of $T_H$ is $\set H$ itself.
The only subcover of $H$ is, trivially, also $\set H$, which is finite.
So $H$ is (trivially) compact in $T$.
== Sources ==
* : Part $\text {II}$: Counterexamples: $4$. Indiscrete Topology: $3$
Category:Indiscrete Topology
Category:Compact Spaces
Subset of Indiscrete Space is Compact and Sequentially Compact

== Theorem ==
Let $T = \struct {S, \set {\O, S} }$ be an indiscrete topological space.
Let $H \subseteq S$.
=== Subset of Indiscrete Space is Compact ===
=== Subset of Indiscrete Space is Sequentially Compact ===
== Sources ==
* : Part $\text {II}$: Counterexamples: $4$. Indiscrete Topology: $3$
Category:Indiscrete Topology
Category:Compact Spaces
Category:Sequentially Compact Spaces
Subset of Indiscrete Space is Dense-in-itself

== Theorem ==
Let $T = \struct {S, \set {\O, S} }$ be an indiscrete topological space.
Let $H \subseteq S$ be a subset of $S$ containing more than one point.
Then $H$ is dense-in-itself.
== Proof ==
Let $x \in H$.
Then as $H$ is not singleton, $\exists y \in H: y \ne x$.
Then every neighborhood of $x$ contains $y$, as the only open set of $T$ is $S$, which also contains both $x$ and $y$.
Hence $x$ is not isolated by definition.
$x$ is general, so all points in $H$ are similarly not isolated.
Hence the subset $H$ is dense-in-itself by definition.
== Sources ==
* : Part $\text {II}$: Counterexamples: $4$. Indiscrete Topology: $5$
Category:Indiscrete Topology
Category:Denseness
Subset of Indiscrete Space is Everywhere Dense

== Theorem ==
Let $T = \struct {S, \set {\O, S} }$ be an indiscrete topological space.
Let $H \subseteq S$ such that $H \ne \O$.
Then $H$ is everywhere dense.
== Proof ==
From Limit Points of Indiscrete Space, every point of $T$ is a limit point of $H$.
Hence $H$ is everywhere dense by definition.
== Sources ==
* : Part $\text {II}$: Counterexamples: $4$. Indiscrete Topology: $7$
Category:Indiscrete Topology
Category:Denseness
Subset of Indiscrete Space is Sequentially Compact

== Theorem ==
Let $T = \struct {S, \set {\O, S} }$ be an indiscrete topological space.
Let $H \subseteq S$.
$H$ is sequentially compact in $T$.
== Proof ==
From Sequence in Indiscrete Space converges to Every Point, every sequence in $T$ converges to every point of $S$.
So every infinite sequence has a subsequence which converges to every point in $S$.
Hence $H$ is (trivially) sequentially compact in $T$.
== Sources ==
* : Part $\text {II}$: Counterexamples: $4$. Indiscrete Topology: $3$
Category:Indiscrete Topology
Category:Sequentially Compact Spaces
Subset of Join Semilattice on Total Ordering is Closed

== Theorem ==
Let $\struct {S, \preccurlyeq}$ be a totally ordered set.
Let $\struct {S, \vee, \preccurlyeq}$ be the join semilattice with respect to $\preccurlyeq$.
Let $T \subseteq S$ be an arbitrary subset of $S$.
Then $\struct {T, \vee, \preccurlyeq}$ is closed under $\vee$.
== Proof ==
Let $x \in T$.
Then from Supremum of Singleton:
:$\sup \set x = x$
That is:
:$x \vee x = x$
and so:
:$x \vee x \in T$
Let $x, y \in T$.
From Supremum of Doubleton in Totally Ordered Set:
:$\sup \set {x, y} \in \set {x, y}$
That is:
:$x \vee y \in \set {x, y}$
But:
:$\set {x, y} \subseteq T$
Hence by definition of subset:
:$x \vee y \in T$
That is:
:$\struct {T, \vee, \preccurlyeq}$ is closed under $\vee$.
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 14$: Orderings: Exercise $14.22 \ \text {(a)}$
Category:Join Semilattices
Category:Total Orderings
Subset of Linear Code with Even Weight Codewords

== Theorem ==
Let $C$ be a linear code.
Let $C^+$ be the subset of $C$ consisting of all the codewords of $C$ which have even weight.
Then $C^+$ is a subgroup of $C$ such that either $C^+ = C$ or such that $\order {C^+} = \dfrac {\order C} 2$.
== Proof ==
Note that the zero codeword is in $C^+$ as it has a weight of $0$ which is even.
Let $c$ and $d$ be of even weight, where $c$ and $d$ agree in $k$ ordinates.
Let $\map w c$ denote the weight of $c$.
Then:
which is even.
Since the negative of a vector $\mathbf v$ in $\Z_2$ equals $\mathbf v$, it follows that the inverse of $c \in C$ is also in $C$.
It follows from the Two-Step Subgroup Test that $C^+$ is a subgroup of $C$.
Let $C \ne C^+$.
Then $C$ contains a codeword $c$ of odd weight.
Let $C^-$ denote the subset of $C$ consisting of all the codewords of $C$ which have odd weight.
Adding $c$ to each codeword of $C^+$ gives distinct codewords of odd weight, so:
:$\order {C^-} \ge \order {C^+}$
Similarly, adding $c$ to each codeword of $C^-$ gives distinct codewords of even weight, so:
:$\order {C^-} \le \order {C^+}$
As $C = C^+ \cup C^-$ it follows that:
:$\order C = 2 \order {C^+}$
Hence the result.
== Sources ==
* : Chapter $6$: Error-correcting codes: Exercise $3$
Category:Linear Codes
Subset of Linearly Independent Set is Linearly Independent

== Theorem ==
A subset of a linearly independent set is also linearly independent.
== Proof ==
Let $G$ be an unitary $R$-module.
Then $\sequence {a_n}$ is a linearly independent sequence $\set {a_1, a_2, \ldots, a_n}$ is a linearly independent set of $G$.
So suppose that $\set {a_1, a_2, \ldots, a_n}$ is a linearly independent set of $G$.
Then clearly $\sequence {a_n}$ is a linearly independent sequence of $G$.
Conversely, let $\sequence {a_n}$ be a linearly independent sequence of $G$.
Let $\sequence {b_m}$ be a sequence of distinct terms of $\set {a_1, a_2, \ldots, a_n}$.
Let $\sequence {\mu_m}$ be a sequence of scalars such that $\ds \sum_{j \mathop = 1}^m \mu_j b_j = 0$.
For each $k \in \closedint 1 n$, let:
:$\lambda_k = \begin{cases}
\mu_j & : j \text { is the unique index such that } a_k = b_j \\
0 & : a_k \notin \set {b_1, b_2, \ldots, b_m}
\end{cases}$
Then:
:$\ds 0 = \sum_{j \mathop = 1}^m \mu_j b_j = \sum_{k \mathop = 1}^n \lambda_k a_k$
Thus:
:$\forall k \in \closedint 1 n: \lambda_k = 0$
As $\set {\mu_1, \ldots, \mu_m} \subseteq \set {\lambda_1, \ldots, \lambda_n}$, it follows that:
:$\forall j \in \closedint 1 m: \mu_j = 0$
and so $\sequence {b_m}$ has been shown to be a linearly independent sequence.
Hence the result.
== Sources ==
* : Chapter $\text {V}$: Vector Spaces: $\S 27$. Subspaces and Bases
* : Chapter $7$: Vector Spaces: $\S 33$. Definition of a Basis
Category:Unitary Modules
Category:Linear Independence
Subset of Linearly Ordered Space which is Order-Complete and Closed but not Compact

== Theorem ==
Let $X = \hointr 0 1 \cup \openint 2 3 \cup \set 4$.
Let $\preceq$ be the ordering on $X$ induced by the usual ordering of the real numbers.
Let $\tau$ be the $\preceq$ order topology on $X$.
Let $Y = \hointr 0 1 \cup \set 4$.
Let $\tau'$ be the $\tau$-relative subspace topology on $Y$.
Then:
:$\struct {Y, \preceq}$ is a complete lattice
:$Y$ is closed in $X$
but:
:$\struct {Y, \tau'}$ is not compact.
== Proof ==
First it is demonstrated that $\struct {Y, \preceq}$ is a complete lattice.
Let $\phi: Y \to \closedint 0 1$ be defined as:
:$\map \phi y = \begin{cases}
y & : y \in \hointr 0 1 \\
1 & : y = 4 \end{cases}$
Then $\phi$ is a order isomorphism.
We have that $\closedint 0 1$ is a complete lattice.
Next is is shown that $\struct {Y, \preceq}$ is closed in $X$.
Let $x \in X \setminus Y$.
Then:
:$x \in \openint 2 3$
Thus:
:$x \in \openint {\dfrac {x + 2} 2} {\dfrac {x + 3} 2} \in \tau'$
Since the complement of $Y$ is open, $Y$ is closed.
Finally it is shown that $\struct {Y, \tau'}$ is not compact.
Let:
:$\AA = \set {x^\preceq: x \in \hointr 0 1} \cup \set {\paren {\dfrac 5 2}^\succeq}$
where:
:$x^\preceq$ denotes the lower closure of $x$
:$x^\succeq$ denotes the upper closure of $x$
Then $\AA$ is an open cover of $Y$ with no finite subcover.
Category:Order Topology
Subset of Locally Finite Set of Subsets is Locally Finite

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Let $\FF \subseteq \powerset S$ be a set of subsets of $S$.
Let $\GG \subseteq \FF$.
If $\FF$ is locally finite then $\GG$ is locally finite.
== Proof ==
We prove the contrapositive statement:
:If $\GG$ is not locally finite then $\FF$ is not locally finite.
Let $\GG$ not be locally finite.
By definition of locally finite:
:$\exists x \in S : \forall N \subseteq S : N$ is a neighborhood of $x : N$ intersects an infinite number of sets in $\GG$.
By definition of subset:
:$\forall X \in \GG : X \in \FF$
Hence:
:$\exists x \in S : \forall N \subseteq S : N$ is a neighborhood of $x : N$ intersects an infinite number of sets in $\FF$.
By definition, $\FF$ is not locally finite.
The result follows from Rule of Transposition.
Category:Locally Finite Sets of Subsets
Subset of Meager Set is Meager Set

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Let $A$ be meager in $T$.
Let $B \subseteq A$.
Then $B$ is meager in $T$.
== Proof ==
Since $A$ is meager in $T$:
:there exists a countable collection of sets $\set {U_n: n \in \N}$ nowhere dense in $T$ such that $\ds A = \bigcup_{n \mathop \in \N} U_n$.
Then, we have: 
From Intersection is Subset:
:$U_n \cap B \subseteq U_n$
From Subset of Nowhere Dense Subset is Nowhere Dense:
:$U_n \cap B$ is nowhere dense in $T$.
Then, we see that:
:$B$ can be written as the union of nowhere dense sets in $T$.
That is:
:$B$ is meager in $T$.
Category:Meager Spaces
Subset of Metric Space contains Limits of Sequences iff Closed

== Theorem ==
Let $M = \struct {A, d}$ be a metric space.
Let $H \subseteq A$.
Then $H$ is closed in $M$ :
:for each sequence $\sequence {a_n}$ of points of $H$ that converges to a point $a \in A$, it follows that $a \in H$.
== Proof ==
=== Necessary Condition ===
Let $H$ be closed in $M$.
Suppose that:
:$\ds \lim_{n \mathop \to \infty} a_n = a$
and:
:$\forall n \in \N_{>0}: a_n \in H$
If the set $\set {a_1, a_2, \ldots}$ is infinite then every neighborhood of $a$ contains infinitely many points of $H$.
Thus $a$ is a limit point of $H$.
So by definition of closed set, $a \in H$.
On the other hand, if $\set {a_1, a_2, \ldots}$ is finite, then for some $N \in \N$:
:$n, m > N \implies a_n = a_m$
Since:
:$\ds \lim_{n \mathop \to \infty} a_n = a$
Then:
:$\forall n > N: \map d {a_n, a} = 0$
Thus:
:$a_n = a$
and so:
:$a \in H$
=== Sufficient Condition ===
Let $H$ be a set such that:
:for each sequence $\sequence {a_n}$ such that $\ds \lim_{n \mathop \to \infty} a_n = a$, it follows that $a \in H$.
Let $b$ be a limit point of $H$.
By , $b$ is the limit of a convergent sequence of points of $H$.
By hypothesis, $b \in H$.
Thus $H$ is a closed set by definition.
 
== Sources ==
* : Chapter $2$: Metric Spaces: $\S 6$: Open Sets and Closed Sets: Theorem $6.8$
Category:Metric Spaces
Category:Closed Sets
Subset of Metric Space is Closed iff contains all Zero Distance Points

== Theorem ==
Let $M = \struct {A, d}$ be a metric space.
Let $H \subseteq A$.
Then $H$ is closed in $M$ :
:$\forall x \in A: \map d {x, H} = 0 \implies x \in H$
where $\map d {x, H}$ denotes the distance between $x$ and $H$.
== Proof ==
=== Necessary Condition ===
Let $H$ be closed in $M$.
Let $x \in A: \map d {x, H} = 0 \implies x \in H$.
By Existence of Sequence in Subset of Metric Space whose Limit is Infimum there exists a sequence $\sequence {a_n}$ of points of $H$ such that:
:$\ds \lim_{n \mathop \to \infty} \map d {x, a_n} = 0$
So every neighborhood of $x$ contains points of $H$.
If some $a_n \in H$ then $x \in H$.
Otherwise each $a_n$ is different from $x$
Then $x$ is the limit of $\sequence {a_n}$ and hence of $H$.
Thus $x \in H$ by definition of closed set.
=== Sufficient Condition ===
Suppose $H$ is such that:
:$\forall x \in A: \map d {x, H} = 0 \implies x \in H$
From Point at Zero Distance from Subset of Metric Space is Limit Point or Element:
:$\map d {x, H} = 0$
 either:
:$x$ is a limit point of $H$
:$x \in H$
Let $x$ be limit point of $H$.
Then by hypothesis $x \in H$.
That is:
:$H$ contains all its limit points
So by definition $H$ is a closed set of $M$.
== Also see ==
* Point at Distance Zero from Closed Set is Element, which presents this result as:
::If $H$ is closed in $M$, then $\map d {x, H} = 0$ $x \in H$.
== Sources ==
* : Chapter $2$: Metric Spaces: $\S 6$: Open Sets and Closed Sets: Theorem $6.9$
Category:Metric Spaces
Category:Closed Sets
Subset of Metric Space is Subset of its Closure

== Theorem ==
Let $M = \struct {A, d}$ be a metric space.
Let $H \subseteq A$ be a subset of $A$.
Then:
:$H \subseteq H^-$
where $H^-$ denotes the closure of $H$.
== Proof ==
By definition of closure:
:$H^- = H' \cup H^i$
where:
:$H'$ denotes the set of limit points of $H$
:$H^i$ denotes the set of isolated points of $H$.
Let $a \in H$.
If $a$ is a limit point of $H$ then $a \in H'$.
Suppose $a \notin H'$.
Then by definition of limit point:
:$\neg \forall \epsilon \in \R_{>0}: \set {x \in A: 0 < \map d {x, a} < \epsilon} \ne \O$
That is:
:$\exists \epsilon \in \R_{>0}: \set {x \in A: 0 < \map d {x, a} < \epsilon} = \O$
and so as $\map d {a, a} = 0$:
:$\exists \epsilon \in \R_{>0}: \set {x \in A: \map d {x, a} < \epsilon} = \set a$
Thus, by definition, $a$ is an isolated point of $H$.
So $a \in H'$ or $a \in H^i$.
Hence by definition of set union:
:$a \in H \subseteq H^-$
By definition of subset:
:$H \subseteq H \subseteq H^-$
and hence the result by definition of closure.
== Sources ==
* : Chapter $2$: Metric Spaces: $\S 6$: Open Sets and Closed Sets: Exercise $6$
Category:Set Closures
Subset of Module Containing Identity is Linearly Dependent

== Theorem ==
Let $G$ be a group whose identity is $e$.
Let $\struct {R, +, \circ}$ be a ring whose zero is $0_R$.
Let $\struct {G, +_G, \circ}_R$ be an $R$-module.
Let $H \subseteq G$ such that $e \in H$.
Then $H$ is a linearly dependent set.
== Proof ==
From Scalar Product with Identity, $\forall \lambda: \lambda \circ e = e$.
Let $H \subseteq G$ such that $e \in H$.
Consider any sequence $\sequence {a_k}_{1 \mathop \le k \mathop \le n}$ in $H$ which includes $e$.
So, let $a_j = e$ for some $j \in \closedint 1 n$.
Let $c \in R \ne 0_R$.
Consider the sequence $\sequence {\lambda_k}_{1 \mathop \le k \mathop \le n}$ of elements of $R$ defined as:
:$\lambda_k = \begin{cases}
c & : k \ne j \\
0_R & : k= j
\end{cases}$
Then:
Thus there exists a sequence $\sequence {\lambda_k}_{1 \mathop \le k \mathop \le n}$ in which not all $\lambda_k = 0_R$ such that:
:$\ds \sum_{k \mathop = 1}^n \lambda_k \circ a_k = e$
Hence the result.
== Sources ==
* : Chapter $\text {V}$: Vector Spaces: $\S 27$. Subspaces and Bases
* : Chapter $7$: Vector Spaces: $\S 33$. Definition of a Basis
Category:Module Theory
Category:Linear Dependence
Subset of Natural Numbers is Cofinal iff Infinite

== Theorem ==
Consider the ordered set $\struct {\N, \le}$, where $\le$ is the usual ordering on the natural numbers.
Let $S \subseteq \N$.
Then $S$ is cofinal it is infinite.
== Proof ==
From Rule of Transposition, we may replace the only if statement by its contrapositive.
Therefore, the following suffices:
=== Implication ===
Suppose $S$ is infinite.
Let $n \in \N$ be an arbitrary natural number.
We have that $\N_n = \set {m \in \N: m \le n}$ is a finite set.
By Subset of Finite Set is Finite, it is not the case that $S \subseteq \N_n$.
Hence, as $\le$ is a total ordering, there exists a $s \in S$ such that $n \le s$.
Hence, as $n$ was arbitrary, $S$ is cofinal.
=== Contrapositive Implication ===
Suppose now that $S$ is finite.
As $\le$ is a total ordering, $S$ has a maximal element.
Let this maximal element of $S$ be $N$.
Then as $N + 1 \not \le N$, from transitivity of $\le$ and maximality of $N$:
:$\forall s \in S: N + 1 \not \le s$
Hence $S$ is not cofinal.
Category:Order Theory
Category:Natural Numbers
Subset of Natural Numbers is either Finite or Denumerable

== Theorem ==
Let $S$ be a subset of the natural numbers $\N$.
Then $S$ is either finite or denumerable.
== Proof ==
Let $\omega$ denote the set of natural numbers as defined by the von Neumann construction.
By the Well-Ordering Principle, $\omega$ is well-ordered by the $\le$ relation.
Thus from the Well-Ordering Principle, $S$ has a smallest element.
Let this smallest element of $S$ be denoted $s_0$.
Also from the Well-Ordering Principle, every subset of $S$ also has a smallest element.
== Sources ==
* : Chapter $3$: The Natural Numbers: $\S 7$ Denumerable classes: Exercise $7.1$
Category:Natural Numbers
Subset of Natural Numbers under Max Operation is Monoid

== Theorem ==
Let $S \subseteq \N$ be a subset of the natural numbers $\N$.
Let $\struct {S, \max}$ denote the algebraic structure formed from $S$ and the max operation.
Then $\struct {S, \max}$ is a monoid.
Its identity element is the smallest element of $S$.
== Proof ==
By the Well-Ordering Principle, $\N$ is a well-ordered set.
By definition, every subset of a well-ordered set is also well-ordered.
Thus $S$ is a well-ordered set.
The result follows from Max Operation on Woset is Monoid.
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 8$: Compositions Induced on Subsets: Example $8.4$
Category:Natural Numbers
Category:Examples of Monoids
Category:Max Operation
Subset of Naturals is Finite iff Bounded

== Theorem ==
Let $X$ be a subset of the natural numbers $\N$.
Then $X$ is finite it is bounded.
== Proof ==
A subset of the natural numbers is also a subset of the real numbers $\R$.
By definition, a bounded subset of $\R$ is bounded below in $\R$ and bounded above in $\R$.
By the Well-Ordering Principle, $X$ is bounded below
Thus $X$ is bounded $X$ is bounded above.
That is, :
:$\exists p \in \N: \forall x \in X: x \le p$
=== Necessary Condition ===
Let $X$ be finite.
Then $X = \set {x_1, x_2, \ldots, x_n}$ for some $n \in \N$.
Let $p = x_1 + x_2 + \ldots + x_n$.
Thus $x \in X \implies x \le p$.
Thus $X$ is bounded above by $p$.
So by definition $X$ is bounded.
=== Sufficient Condition ===
Let $X \subseteq \N$ be bounded.
Then for some $p \in \N$, it is contained in the initial segment $\N_p = \set {0, 1, \ldots, p}$ of $\N$.
It follows from Subset of Finite Set is Finite that $X$ is finite.
== Sources ==
* : Chapter $1, \S 2$
Category:Subsets
Category:Natural Numbers
Subset of Normed Vector Space is Everywhere Dense iff Closure is Normed Vector Space

== Theorem ==
Let $\struct {X, \norm {\, \cdot \,}}$ be a normed vector space.
Let $D \subseteq X$ be a subset of $X$.
Let $D^-$ be the closure of $D$.
Then $D$ is dense $D^- = X$.
== Proof ==
=== Necessary Condition ===
=== Sufficient Condition ===
== Sources ==
* : $\S 1.4$: Normed and Banach spaces. Sequences in a normed space; Banach spaces
Category:Normed Vector Spaces
Category:Denseness
Category:Set Closures
Subset of Normed Vector Space is Everywhere Dense iff Closure is Normed Vector Space/Necessary Condition

== Theorem ==
Let $\struct {X, \norm {\, \cdot \,} }$ is a normed vector space.
Let $D \subseteq X$ be a subset of $X$.
Let $D^-$ be the closure of $D$.
Let $D$ be dense.
Then:
:$D^- = X$
== Proof ==
Let $x \in X \setminus D$.
Suppose $D$ is dense in $X$.
Then:
:$\forall n \in N : \exists d_n \in D : d_n \in \map {B_{\frac 1 n}} x$
where $\ds \map {B_{\frac 1 n}} x$ is an open ball.
Let $\sequence {d_n}_{n \mathop \in \N}$ be a sequence in $D$.
Then:
:$\forall n \in \N : \norm {x - d_n} < \frac 1 n$
Hence, $x$ is a limit point of $D$.
In other words, $x \in D^-$.
We have just shown that:
:$x \in X \setminus D \implies x \in D^-$
Hence:
:$X \setminus D \subseteq D^-$.
By definition of closure:
:$D \subseteq D^-$
Therefore:
Thus:
:$X = D^-$.
== Sources ==
* : Chapter $1$: Normed and Banach spaces
Category:Normed Vector Spaces
Category:Denseness
Category:Set Closures
Subset of Normed Vector Space is Everywhere Dense iff Closure is Normed Vector Space/Sufficient Condition

== Theorem ==
Let $\struct {X, \norm {\, \cdot \,} }$ is a normed vector space.
Let $D \subseteq X$ be a subset of $X$.
Let $D^-$ be the closure of $D$.
Let $D^- = X$.
Then $D$ is dense.
== Proof ==
Let $X = D^-$.
We have to show, that for every $x \in X$ there is an open ball with an element from $D^-$.
We have that $X = D \cup \paren {X \setminus D}$.
Suppose $x \in X \setminus D$.
Then $x \in D^- \setminus D$.
Hence, $x$ is a limit point of $D$.
Therefore, there is a sequence $\sequence {d_n}_{n \mathop \in \N}$ in $D$ which converges to $x$.
Thus:
:$\forall \epsilon \in \R_{> 0} : \exists N \in \N : \norm {x - d_N} < \epsilon$
In other words:
:$\ds d_N \in D \implies d_N \in \map {B_\epsilon} x$
Therefore:
:$d_N \in D \cap \map {B_\epsilon} x$.
Suppose $x \in D$.
Let $\epsilon > 0$.
Then $x \in \map {B_\epsilon} x \cap D$.
From both parts and definition we conclude that $D^-$ is dense in $X$.
== Sources ==
* : Chapter $1$: Normed and Banach spaces
Category:Normed Vector Spaces
Category:Denseness
Category:Set Closures
Subset of Nowhere Dense Subset is Nowhere Dense

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Let $A \subseteq S$ be nowhere dense in $T$.
Let $B \subseteq A$. 
Then $B$ is nowhere dense in $T$.
== Proof ==
 it is not the case that $B$ is nowhere dense in $T$.
Then by definition of nowhere dense:
:$B^-$ contains some open set of $T$ which is non-empty.
From Set Closure Preserves Set Inclusion, we have:
:$B^- \subseteq A^-$
So:
:$A^-$ contains some open set of $T$ which is non-empty.
So $A$ is not nowhere dense in $T$.
This contradicts our assertion that $A$ is nowhere dense in $T$:
The result follows by Proof by Contradiction.
Category:Nowhere Denseness