avewright/proofwiki-math huggingface dataset:
Theorem: Particular Point Space less Particular Point is Discrete

Let $T = \struct {S, \tau_p}$ be a particular point space, whose particular point is $p$.
Let $H = S \setminus \set p$ where $\setminus$ denotes set difference.
Then the topological subspace $T_H = \struct {H, \tau_H}$ induced on $H$ by $\tau_p$ is a discrete space.

Proof:
Let $H = S \setminus \set p$.
Let $V \subseteq H$ be any subset of $H$.
As $p \notin V$, $V$ is a closed set of $T$.
Thus, by definition of closed set $S \setminus V$ is open in $T$.
By definition of subspace topology, $\paren {S \setminus V} \cap H$ is open in $T_H$.
From Intersection with Set Difference is Set Difference with Intersection:
:$\paren {S \setminus V} \cap H = \paren {S \cap H} \setminus V$
and so from Intersection with Subset is Subset:
:$\paren {S \setminus V} \cap H = H \setminus V$
Thus $H \setminus V$ is open in $H$.
But $H \setminus V$ is itself a subset of $H$, and so a closed set of $T$.
Thus, by the same argument, $V$ is open in $H$.
As $V$ is arbitrary, it follows that all subsets of $H$ are open in $H$.
The result follows by definition of discrete space.
{{qed}}
Category:Particular Point Topologies
Category:Discrete Topologies
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Theorem: Zero is Accumulation Point of Sequence in Sierpiński Space

Let $T = \struct {\set {0, 1}, \tau_0}$ be a Sierpiński space.
The sequence in $T$:
:$\sigma = \sequence {0, 1, 0, 1, \ldots}$
has $0$ as an accumulation point.

Proof:
By definition, $\alpha$ is an accumulation point of $\sigma$ {{iff}}:
:$\forall U \in \tau_0: \alpha \in U \implies \set {n \in \N: x_n \in U}$ is infinite.
Both $\set 0$ and $\set {0, 1}$ contain $0$, which occurs an infinite number of times in $\sigma$.
Hence, by definition, $0$ is an accumulation point of $\sigma$.
{{qed}}
Theorem: 1 is Limit Point of Sequence in Sierpiński Space

Let $T = \struct {\set {0, 1}, \tau_0}$ be a Sierpiński space.
The sequence in $T$:
:$\sigma = \sequence {0, 1, 0, 1, \ldots}$
has $1$ as a limit.

Proof:
By definition, $\alpha$ is a limit of $\sigma$ {{iff}}:
:$\forall U \in \tau_0: \alpha \in U \implies \set {n \in \N: x_n \notin U}$ is finite.
The only open set of $T$ containing $1$ is $\set {0, 1}$.
It contains all but a finite number (that is: $0$) elements of $\sigma$.
Hence, by definition, $1$ is a limit of $\sigma$.
{{qed}}
Theorem: Infinite Particular Point Space is not Metacompact

Let $T = \struct {S, \tau_p}$ be an infinite particular point space.
Then $T$ is not metacompact.

Proof:
{{AimForCont}} $T$ is metacompact.
From Metacompact Space is Countably Metacompact it follows that $T$ is countably metacompact.
But we have that Infinite Particular Point Space is not Countably Metacompact.
Hence the result by Proof by Contradiction.
{{qed}}
Category:Particular Point Topologies
Category:Examples of Metacompact Spaces
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Theorem: Infinite Particular Point Space is not Countably Paracompact

Let $T = \struct {S, \tau_p}$ be an infinite particular point space.
Then $T$ is not countably paracompact.

Proof:
{{AimForCont}} $T$ is countably paracompact.
From Countably Paracompact Space is Countably Metacompact it follows that $T$ is countably metacompact.
But we have that Infinite Particular Point Space is not Countably Metacompact.
Hence the result by Proof by Contradiction.
{{qed}}
Category:Particular Point Topologies
Category:Examples of Countably Paracompact Spaces
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Theorem: Closed Extension Topology is not T2

Let $T = \struct {S, \tau}$ be a topological space.
Let $T^*_p = \struct {S^*_p, \tau^*_p}$ be the closed extension space of $T$.
Then $T^*_p$ is not a $T_2$ (Hausdorff) space.

Proof:
{{AimForCont}} $T^*_p$ is not a $T_2$ space.
From $T_2$ Space is $T_1$ Space, $T^*_p$ is a $T_1$ space.
But this contradicts Closed Extension Topology is not $T_1$
Hence by Proof by Contradiction $T^*_p$ can not be a $T_2$ space.
{{qed}}
Theorem: Closed Extension Topology is not T3

Let $T = \struct {S, \tau}$ be a topological space.
Let $T^*_p = \struct {S^*_p, \tau^*_p}$ be the closed extension space of $T$.
Then $T^*_p$ is not a $T_3$ space.

Proof:
By Underlying Set of Topological Space is Closed, $S$ is closed in $T$.
By Closed Sets of Closed Extension Topology, $S$ is closed in $T^*_p$.
{{Defof|Closed Extension Space}} gives:
:$p \notin S$
:Every open set in $T^*_p$ is either $\O$ or it contains $p$.
Thus no open set containing $S$ is disjoint from $\set p$.
This shows that $T^*_p$ is not a $T_3$ space.
{{qed}}
Theorem: Condition for Closed Extension Space to be T4 Space

Let $T = \struct {S, \tau}$ be a topological space.
Let $T^*_p = \struct {S^*_p, \tau^*_p}$ be the closed extension space of $T$.
Then:
:$T^*_p$ is a $T_4$ space {{iff}} $T$ is a $T_4$ space vacuously
and $T^*_p$ in this case is also a $T_4$ space vacuously.

Proof:
=== Sufficient Condition ===
Suppose $T^*_p$ is $T_4$.
Then for any two disjoint closed sets $A, B \subseteq S$ there exist disjoint open sets $U, V \in \tau^*_p$ containing $A$ and $B$ respectively.
However, for any non-empty set $U \in \tau^*_p$, $p \in U$.
Hence no non-empty open sets in $T^*_p$ are disjoint.
Therefore $T^*_p$ is a $T_4$ space vacuously: there do not exist two disjoint closed sets.
{{qed|lemma}}
Theorem: Condition for Closed Extension Space to be T5 Space

Let $T = \struct {S, \tau}$ be a topological space.
Let $T^*_p = \struct {S^*_p, \tau^*_p}$ be the closed extension space of $T$.
Then:
:$T^*_p$ is a $T_5$ space {{iff}} $T$ is a $T_5$ space vacuously
and $T^*_p$ in this case is also a $T_5$ space vacuously.

Proof:
=== Sufficient Condition ===
Let $T^*_p$ be a $T_5$ space.
Then for any two separated sets $A, B \subseteq S$ there exist disjoint open sets $U, V \in \tau^*_p$ containing $A$ and $B$ respectively.
However, for any non-empty set $U \in \tau^*_p$:
:$p \in U$
Hence no non-empty open sets in $T^*_p$ are separated.
Therefore $T^*_p$ is a $T_5$ space vacuously: there do not exist two separated sets.
{{qed|lemma}}
Theorem: Either-Or Topology is T4

Let $T = \struct {S, \tau}$ be the either-or space.
Then $T$ is a $T_4$ space.

Proof:
Follows from:
:Either-Or Topology is $T_5$
:$T_5$ Space is $T_4$.
{{qed}}
Theorem: Limit Points in Fort Space

Let $T = \struct {S, \tau_p}$ be a Fort space.
Let $x \in S$ such that $x \ne p$.
Then $p$ is the only limit point of $x$.

Proof:
From {{Defof|Fort Space}}, we have $\relcomp S {\set x} \in \tau_p$.
For any $y \ne x$, $y \in \relcomp S {\set x}$.
Therefore $\relcomp s {\set x}$ is an open neighborhood of $y$.
From {{Defof|Relative Complement}} we also have $x \notin \relcomp S {\set x}$.
Hence $y$ is not a limit point of $x$.
By {{Defof|Limit Point of Point}}, $x$ cannot be a limit point of $x$.
Therefore $x$ has no limit points in $S$.
{{qed}}
Theorem: Real Number Line is First-Countable

Let $\struct {\R, \tau_d}$ be the real number line with the usual (Euclidean) topology.
Then $\struct {\R, \tau_d}$ is a first-countable space.

Proof:
From Real Number Line is Second-Countable we have that $\struct {\R, \tau_d}$ is a second-countable space.
The result follows from Second-Countable Space is First-Countable.
{{qed}}
Theorem: Alexandroff Extension is Compact

Let $T = \struct {S, \tau}$ be a non-empty topological space.
Let $p$ be a new element not in $S$.
Let $S^* := S \cup \set p$.
Let $T^* = \struct {S^*, \tau^*}$ be the Alexandroff extension on $S$.
Then $T^*$ is a compact topological space.

Proof:
Let $\UU$ be an open cover of $T^*$.
At least one $V \in \UU$ contains $p$.
Because $p \notin S$, $V$ is not an open set of $T$.
Therefore, by definition of the Alexandroff extension, $V$ must be the complement relative to $S^*$ of a closed, compact subset $\relcomp {S^*} V$ of $T$.
Because $\relcomp {S^*} V$ is compact, it can be covered by a finite number of sets in $\UU$.
So:
:$T^*$ can be covered by $V$
and:
:that finite number of sets in $\UU$ are a cover for $\relcomp {S^*} V$.
That is, $\UU$ has a finite subcover.
Hence the result by definition of compact topological space.
{{qed}}
Theorem: Condition for Alexandroff Extension to be T1

Let $T = \struct {S, \tau}$ be a non-empty topological space.
Let $p$ be a new element not in $S$.
Let $S^* := S \cup \set p$.
Let $T^* = \struct {S^*, \tau^*}$ be the Alexandroff extension on $S$.
Then $T^*$ is a $T_1$ space {{iff}} $T$ is a $T_1$ space.

Proof:
=== Necessary Condition ===
Let $T = \struct {S, \tau}$ be a $T_1$ space.
By definition, $T$ is a $T_1$ space {{iff}} all points of $S$ are closed in $T$.
We have that $S$ is open in $T$ by definition of a topology.
Thus by definition of the Alexandroff extension, $S$ is open in $T^*$.
So as $S = S^* \setminus \set p$ is open in $T^*$, it follows that $\set p$ is closed in $T^*$.
That is, $p$ is a closed point of $T^*$.
Now let $x \in S^*$ such that $x \ne p$.
As $T$ is a $T_1$ space, $\set x$ is closed in $T$.
From Finite Topological Space is Compact, $\set x$ is a compact subset of $T$.
Thus by definition of the Alexandroff extension, $S^* \setminus \set x$ is open in $T^*$.
Thus by definition $\set x$ is closed in $T^*$.
That is, $x$ is a closed point of $T^*$.
Thus it has been shown that all points in $T^*$ are closed points of $T^*$.
Thus by definition $T^*$ is a $T_1$ space.
{{qed|lemma}}
Theorem: Condition for Alexandroff Extension to be T2

Let $T = \struct {S, \tau}$ be a non-empty topological space.
Let $p$ be a new element not in $S$.
Let $S^* := S \cup \set p$.
Let $T^* =\struct {S^*, \tau^*}$ be the Alexandroff extension on $S$.
Then $T^*$ is a $T_2$ (Hausdorff) space {{iff}} $T$ is a locally compact Hausdorff space.

Proof:
=== Necessary Condition ===
Let $T = \struct {S, \tau}$ be a locally compact Hausdorff space.
Let $x, y \in S$.
Then as $T$ is a $T_2$ space, there exist two disjoint open sets $U, V \in \tau$ containing $x$ and $y$ respectively.
But by definition of the Alexandroff extension on $S$, $U$ and $V$ are also open sets of $T^*$.
That covers the case where both $x \ne p$ and $y \ne p$.
{{WLOG}}, let $y = p$.
As $T$ is a locally compact Hausdorff space, $x$ has a compact neighborhood $N_x$.
By definition of neighborhood, there exists an open set $U$ of $T$ containing $x$.
By definition of the Alexandroff extension, $U$ is an open set of $T^*$.
By Compact Subspace of Hausdorff Space is Closed, $N_x$ is closed in $T$.
Let $V = S^* \setminus N_x$.
By definition of $V$, $x \notin V$ and $p \in V$.
By definition of the Alexandroff extension, $V$ is an open set of $T^*$.
By Intersection with Complement is Empty iff Subset:
:$U \cup V = \O$
So we have two open sets $U, V$ of $T^*$ such that $x \in U, p \in V$ and $U \cup V = \O$.
Thus by definition $T^*$ is a $T_2$ space.
{{qed|lemma}}
Theorem: Alexandroff Extension which is T2 is also T4

Let $T = \struct {S, \tau}$ be a non-empty topological space.
Let $p$ be a new element not in $S$.
Let $S^* := S \cup \set p$.
Let $T^* = \struct {S^*, \tau^*}$ be the Alexandroff extension on $S$.
Let $T^*$ be a $T_2$ (Hausdorff) space.
Then $T^*$ is a $T_4$ space.

Proof:
We have:
:Alexandroff Extension is Compact
:Compact Hausdorff Space is $T_4$.
{{qed}}
Theorem: Alexandroff Extension of Rational Number Space is not T2

Let $\struct {\Q, \tau_d}$ be the rational number space under the Euclidean topology $\tau_d$.
Let $p$ be a new element not in $\Q$.
Let $\Q^* := \Q \cup \set p$.
Let $T^* = \struct {\Q^*, \tau^*}$ be the Alexandroff extension on $\struct {\Q, \tau_d}$.
Then $T^*$ is not a $T_2$ (Hausdorff) space.

Proof:
From Condition for Alexandroff Extension to be $T_2$, $T^*$ is a $T_2$ space {{iff}} $\struct {\Q, \tau_d}$ is a locally compact Hausdorff Space.
But from Rational Number Space is not Locally Compact Hausdorff Space, $\struct {\Q, \tau_d}$ is not a locally compact Hausdorff Space.
Hence the result.
{{qed}}
Theorem: Alexandroff Extension of Rational Number Space is T1

Let $\struct {\Q, \tau_d}$ be the rational number space under the Euclidean topology $\tau_d$.
Let $p$ be a new element not in $\Q$.
Let $\Q^* := \Q \cup \set p$.
Let $T^* = \struct {\Q^*, \tau^*}$ be the Alexandroff extension on $\struct {\Q, \tau_d}$.
Then $T^*$ is a $T_1$ space.

Proof:
From Condition for Alexandroff Extension to be $T_1$, $T^*$ is a $T_1$ space {{iff}} $\struct {\Q, \tau_d}$ is also a $T_1$ space.
From Rational Numbers form Metric Space, $\struct {\Q, d}$ is a metric space.
From Metric Space is $T_1$, $\struct {\Q, \tau_d}$ is a $T_1$ space.
Hence the result.
{{qed}}
Theorem: Separation Properties of Alexandroff Extension of Rational Number Space

Let $\struct {\Q, \tau_d}$ be the rational number space under the Euclidean topology $\tau_d$.
Let $p$ be a new element not in $\Q$.
Let $\Q^* := \Q \cup \set p$.
Let $T^* = \struct {\Q^*, \tau^*}$ be the Alexandroff extension on $\struct {\Q, \tau_d}$.
Then $T^*$ satisfies no Tychonoff separation axioms higher than a $T_1$ space.

Proof:
From Alexandroff Extension of Rational Number Space is $T_1$, $T^*$ is a $T_1$ space.
From Alexandroff Extension of Rational Number Space is not T2, $T^*$ is not a $T_2$ (Hausdorff) space.
From $T_{2 \frac 1 2}$ Space is $T_2$, $T^*$ is not a $T_{2 \frac 1 2}$ space.
{{Recall|Semiregular Space|semiregular space}}
{{:Definition:Semiregular Space}}
As $T^*$ is not a $T_2$ (Hausdorff) space, $T^*$ is not a semiregular space.
From Regular Space is $T_{2 \frac 1 2}$, $T^*$ is not a regular space.
From Urysohn Space is $T_{2 \frac 1 2}$ Space, $T^*$ is not an Urysohn space.
{{Recall|Regular Space|regular space|index = 1}}
{{:Definition:Regular Space/Definition 1}}
From $T_1$ Space is $T_0$, $T^*$ is a $T_0$ space.
But as $T^*$ is not a regular space, it follows that $T^*$ is not a $T_3$ space.
From $T_{3 \frac 1 2}$ Space is $T_3$, $T^*$ is not a $T_{3 \frac 1 2}$ space.
From Completely Regular Space is Regular, $T^*$ is not a completely regular space.
From Normal Space is Completely Regular, $T^*$ is not a normal space.
{{Recall|Normal Space|normal space|index = 1}}
{{:Definition:Normal Space/Definition 1}}
As noted above, from Alexandroff Extension of Rational Number Space is $T_1$, $T^*$ is a $T_1$ space.
But as $T^*$ is not a normal space, it follows that $T^*$ is not a $T_4$ space.
From Completely Normal Space is Normal, $T^*$ is not a completely normal space.
From $T_5$ Space is $T_4$, $T^*$ is not a $T_5$ space.
From Perfectly Normal Space is Completely Normal, $T^*$ is not a perfectly normal space.
From Perfectly $T_4$ Space is $T_5$, $T^*$ is not a perfectly $T_4$ space.
{{qed}}
Theorem: Particular Point of Alexandroff Extension of Rational Number Space is Dispersion Point

Let $\struct {\Q, \tau_d}$ be the rational number space under the Euclidean topology $\tau_d$.
Let $p$ be a new element not in $\Q$.
Let $\Q^* := \Q \cup \set p$.
Let $T^* = \struct {\Q^*, \tau^*}$ be the Alexandroff extension on $\left({\Q, \tau_d}\right)$.
Then $p$ is a dispersion point of $T^*$.

Proof:
By definition, $p$ is a dispersion point of $T^*$ {{iff}}:
:$\Q^*$ is a connected set in $T^*$
:$\Q^* \setminus \set p$ is totally disconnected in $T^*$.
From Alexandroff Extension of Rational Number Space is Connected, $\Q^*$ is a connected set in $T^*$.
It remains to be shown that $\Q^* \setminus \set p$ is totally disconnected in $T^*$.
By definition, $\Q^* \setminus \set p$ is the rational number space $\struct {\Q, \tau_d}$.
The result follows from Rational Numbers are Totally Disconnected.
{{qed}}
Theorem: Alexandroff Extension of Rational Number Space is Connected

Let $\struct {\Q, \tau_d}$ be the rational number space under the Euclidean topology $\tau_d$.
Let $p$ be a new element not in $\Q$.
Let $\Q^* := \Q \cup \set p$.
Let $T^* = \struct {\Q^*, \tau^*}$ be the Alexandroff extension on $\struct {\Q, \tau_d}$.
Then $T^*$ is a connected space.

Proof:
By definition $T^*$ is a connected space of $T^*$ {{iff}} it admits no separation.
{{AimForCont}} $T^*$ does admit a separation.
That is, there exist open sets $A, B \in \tau^*$ such that $A, B \ne \O$, $A \cup B = \Q^*$ and $A \cap B = \O$.
That is, both $A$ and $B = \relcomp {\Q^*} A$ are open in $T^*$.
{{WLOG}}, Let $p \in A$.
Then $B$ is a closed and compact subset of $\Q$.
From Compact Set of Rational Numbers is Nowhere Dense, $B$ is nowhere dense.
From Set is Closed iff Equals Topological Closure, $B$ equals its closure.
Thus by definition of nowhere dense, the interior of $B$ is empty.
We have that $B$ is open in $T^*$ and $p \notin B$.
Thus it follows by definition of Alexandroff extension that $B$ is open in $\struct {\Q, \tau_d}$.
But from Interior of Open Set, $B$ equals its interior.
This contradicts the deduction above that the interior of $B$ is empty.
Hence by Proof by Contradiction $B$ cannot be open in $T^*$.
That is, $A$ and $B$ cannot form a separation of $T^*$.
Thus $T^*$ is a connected space by definition.
{{qed}}
Theorem: Alexandroff Extension of Rational Number Space is Biconnected

Let $\struct {\Q, \tau_d}$ be the rational number space under the Euclidean topology $\tau_d$.
Let $p$ be a new element not in $\Q$.
Let $\Q^* := \Q \cup \set p$.
Let $T^* = \struct {\Q^*, \tau^*}$ be the Alexandroff extension on $\struct {\Q, \tau_d}$.
Then $T^*$ is a biconnected space.

Proof:
From Particular Point of Alexandroff Extension of Rational Number Space is Dispersion Point, $p$ is a dispersion point of $T^*$.
The result follows from Set with Dispersion Point is Biconnected.
{{qed}}
Theorem: Alexandroff Extension of Rational Number Space is Sequentially Compact

Let $\struct {\Q, \tau_d}$ be the rational number space under the Euclidean topology $\tau_d$.
Let $p$ be a new element not in $\Q$.
Let $\Q^* := \Q \cup \set p$.
Let $T^* = \struct {\Q^*, \tau^*}$ be the Alexandroff extension on $\struct {\Q, \tau_d}$.
Then $T^*$ is a sequentially compact space.

Proof:
The strategy here is to demonstrate that every sequence in $T^*$ is either contained in a compact subspace of $T^*$, or must contain a subsequence which converges to $p$.
{{ProofWanted}}
Theorem: Hilbert Sequence Space is Complete Metric Space

Let $A$ be the set of all real sequences $\left\langle{x_i}\right\rangle$ such that the series $\ds \sum_{i \mathop = 0}^\infty x_i^2$ is convergent.
Let $\ell^2 = \struct {A, d_2}$ be the Hilbert sequence space on $\R$.
Then $\ell^2$ is a complete metric space.

Proof:
We have that Hilbert Sequence Space is Metric Space.
It remains to be shown that it is complete.
Recall that from Real Number Line is Complete Metric Space, $\R$ is a complete metric space.
Let $x^1, x^2, x^3, \ldots$ be a Cauchy sequence $\ell^2$.
Then for each $i \in \N_{>0}$, we have that $\sequence { {x_i}^j}_{j \mathop \in \N_{>0} }$ is a Cauchy sequence in the complete metric space $\R$.
Thus $\sequence { {x_i}^j}_{j \mathop \in \N_{>0} }$ converges to a point in $\R$, say $x_i$.
Then if $x = \sequence {x_i}$, the points $x - x_j$ eventually belong to $H$.
So $x = \paren {x - x_j} + x^j$ must also belong to $H$.
Thus $\map {d_2} {x, x^j} = 0$.
Hence the result.
{{qed}}
Theorem: Hilbert Sequence Space is Separable

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 $\R$.
Then $\ell^2$ is a separable space.

Proof:
Consider the set $H$ of all points of $\ell^2$ which have finitely many rational coordinates and all the rest zero.
$H$ is countable, since
:Rational Numbers are Countably Infinite
:Cartesian Product of Countable Sets is Countable
:Countable Union of Countable Sets is Countable
It remains to show that $H$ is everywhere dense in $\ell^2$.
Indeed, the result then follows by definition of separable space.
To this end, let $\sequence {x_i} \in \ell^2$.
For each $n \in \N$, define $\sequence {q^{\paren n} _i} \in H$ by:
:<nowiki>$q^{\paren n}_i := \begin {cases} \dfrac {\floor {2^{n + i} x} } {2^{n + i} } & : i < n \\ 
0 & : i \ge n \end {cases}$</nowiki>
where $\floor \cdot$ denotes the floor function.
Then we have:
{{begin-eqn}}
{{eqn | l = \map {d_2} {\sequence {x_i}, \sequence {q^{\paren n}_i} }^2
      | r = \sum_{i \mathop \ge 0} \paren {x_i - q^{\paren n}_i}^2
      | c = {{Defof|Hilbert Sequence Space}}
}}
{{eqn | r = \sum_{i \mathop = 0}^{n - 1} \paren {x_i - \frac {\floor {2^{n + i} x} } {2^{n + i} } }^2 + \sum_{i \mathop \ge n} {x_i}^2
}}
{{eqn | r = \sum_{i \mathop = 0}^{n - 1} \paren {\frac {2^{n + i} x - \floor {2^{n + i} x} } {2^{n + i} } }^2 + \sum_{i \mathop \ge n} {x_i}^2
}}
{{eqn | o = \le
      | r = \sum_{i \mathop = 0}^{n - 1} \paren {\frac 1 {2^{n + i} } }^2 + \sum_{i \mathop \ge n} {x_i}^2
      | c = in view of {{Defof|Floor Function}}
}}
{{eqn | o = \le
      | r = \sum_{i \mathop \ge 0} \paren {\frac 1 {2^{n + i} } }^2 + \sum_{i \mathop \ge n} {x_i}^2
}}
{{eqn | r = \frac 1 {4^n} \sum_{i \mathop \ge 0} \frac 1 {4^i} + \sum_{i \mathop \ge n} {x_i}^2
}}
{{eqn | r = \frac 1 {4^n} \cdot \frac 4 3 + \sum_{i \mathop \ge n} {x_i}^2
      | c = Sum of Infinite Geometric Sequence
}}
{{eqn | o = \to
      | r = 0
      | c = as $n \to \infty$, since Tail of Convergent Series tends to Zero
}}
{{end-eqn}}
{{finish|Explain why this proves that $H$ is everywhere dense in $\ell^2$}}
Theorem: Hilbert Sequence Space is Second-Countable

Let $\ell^2$ be the Hilbert sequence space on $\R$.
Then $\ell^2$ is a second-countable space.

Proof:
From Hilbert Sequence Space is Separable, $\ell^2$ is a separable space.
We also have that Hilbert Sequence Space is Metric Space.
The result follows from Separable Metric Space is Second-Countable.
{{qed}}
Theorem: Hilbert Sequence Space is Lindelöf

Let $\ell^2$ be the Hilbert sequence space on $\R$.
Then $\ell^2$ is a Lindelöf space.

Proof:
From Hilbert Sequence Space is Second-Countable, $\ell^2$ is a second-countable space.
The result follows from Second-Countable Space is Lindelöf.
{{qed}}
Theorem: Hilbert Sequence Space is not Locally Compact Hausdorff Space

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 $\R$.
Then $\ell^2$ is not a locally compact Hausdorff space.

Proof:
From Hilbert Sequence Space is Metric Space, $\ell^2$ is a metric space.
From Metric Space is $T_2$, $\ell^2$ is a Hausdorff space.
Let $x = \sequence {x_i} \in A$ be a point of $\ell^2$.
From Point in Hilbert Sequence Space has no Compact Neighborhood, $x$ has no compact neighborhood.
Hence the result by definition of locally compact Hausdorff space.
{{qed}}
Theorem: Compact Subset of Hilbert Sequence Space is Closed

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 compact subset of $\ell^2$.
Then $H$ is closed in $\ell^2$.

Proof:
From Hilbert Sequence Space is Metric Space, $\ell^2$ is a metric space.
From Metric Space is $T_2$, $\ell^2$ is a Hausdorff space.
The result follows from Compact Subspace of Hausdorff Space is Closed.
{{qed}}
Theorem: Subset of Hilbert Sequence Space with Non-Empty Interior is not Compact

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 $\ell^2$.
{{qed}}