nyuuzyou/wikis huggingface dataset:
Hilbert-Waring Theorem/Variant Form/Particular Cases/2

== Particular Case of the Hilbert-Waring Theorem -- Variant Form: $k = 2$ ==
The Hilbert-Waring Theorem -- Variant Form states that:
The case where $k = 2$ is proved by Lagrange's Four Square Theorem‎:
:$G \left({2}\right) = 4$
That is, every sufficiently large positive integer can be expressed as the sum of at most $4$ squares.
== Historical Note ==
Category:Hilbert-Waring Theorem
Hilbert-Waring Theorem/Variant Form/Particular Cases/3

== Particular Case of the Hilbert-Waring Theorem -- Variant Form: $k = 3$ ==
The Hilbert-Waring Theorem -- Variant Form states that:
The case where $k = 3$ is:
Every sufficiently large positive integer can be expressed as the sum of a number of positive cubes.
The exact number is the subject of ongoing research, but at the time of writing ($11$th February $2017$) it is known that it is between $4$ and $7$.
That is:
:$4 \le \map G 3 \le 7$
== Progress ==
From Cube Modulo 9, a cube is congruent to $0$, $1$, or $2$ modulo $9$.
So it takes at least $4$ cubes to make an integer $k \equiv \pm 4 \pmod 9$.
Hence $\map G 3$ is at least $4$.
For numbers less than $1.3 \times 10^9$, it has been demonstrated that $1 \, 290 \, 740$ is the highest number to require $6$ cubes.
It is also known that the number of numbers between $n$ and $2 n$ requiring $5$ cubes falls with increasing $n$. has expressed a belief that this decrease is rapid enough to suggest that $\map G 3 = 4$.
From the work of , , and , $7 \, 373 \, 170 \, 279 \, 850$ is the largest number known that is not the sum of $4$ cubes. They offer an argument that this may indeed be the largest possible.
== Historical Note ==
== Sources ==
* 
Category:Hilbert-Waring Theorem
Category:Unsolved Problems
Hilbert-Waring Theorem/Variant Form/Particular Cases/3/Historical Note

== Particular Case of the Hilbert-Waring Theorem -- Variant Form: $k = 3$: Historical Note ==
It had been established by $1912$ that it takes no more than $9$ positive cubes to represent any positive integer as a sum.
The value of $G \left({3}\right)$ is still the subject of research.
 showed in $1943$ that the upper bound of $G \left({3}\right)$ is $7$.
Recent research, however, suggests that $G \left({3}\right)$ may indeed be $4$.
Category:Historical Notes
Hilbert-Waring Theorem/Variant Form/Particular Cases/4

== Particular Case of the Hilbert-Waring Theorem -- Variant Form: $k = 4$ ==
The Hilbert-Waring Theorem -- Variant Form states that:
The case where $k = 4$ is:
Every sufficiently large positive integer can be expressed as the sum of at most $16$ $4$th powers.
That is:
:$\map G 4 = 16$
== Proof ==
== Also see ==
* Numbers of form $31 \times 16^n$ are sum of $16$ $4$th Powers
== Historical Note ==
== Sources ==
* : $16$
* : $16$
Category:Hilbert-Waring Theorem
Category:16
Hilbert-Waring Theorem/Variant Form/Particular Cases/4/Historical Note

== Particular Case of the Hilbert-Waring Theorem -- Variant Form: $k = 4$: Historical Note ==
 showed in $1939$ that $G \left({4}\right) = 16$.
$13792$ is the largest number to require $17$ fourth powers.
, and showed in $2000$ that every number between $13793$ and $10^{245}$ requires at most $16$.
, and extended 's $1939$ result to show that every number above $10^{220}$ requires no more than $16$.
== Sources ==
* 
Category:Historical Notes
Hilbert-Waring Theorem/Variant Form/Particular Cases/7

== Particular Case of the Hilbert-Waring Theorem -- Variant Form: $k = 7$ ==
The Hilbert-Waring Theorem -- Variant Form states that:
The case where $k = 7$ is:
Every sufficiently large positive integer can be expressed as the sum of a number of positive $7$th powers.
The exact number is the subject of ongoing research, but at the time of writing ($20$th December $2018$) it is known that it is between $8$ and $33$.
That is:
:$8 \le \map G 3 \le 33$
== Historical Note ==
Category:Hilbert-Waring Theorem
Category:Unsolved Problems
Hilbert-Waring Theorem/Variant Form/Particular Cases/7/Historical Note

== Particular Case of the Hilbert-Waring Theorem -- Variant Form: $k = 7$: Historical Note ==
 reports in his of $1986$ that:
:All sufficiently large numbers are the sum of at most $137$ seventh powers.
At time of writing that may have been adequate, but since then the subject has moved on considerably.
The statement has been removed from later editions.
== Sources ==
* : $137$
Category:Historical Notes
Hilbert Cube is Arc-Connected

== Theorem ==
Let $M = \struct {I^\omega, d_2}$ be the Hilbert cube.
Then $M$ is an arc-connected space.
== Proof ==
Let $x = \sequence {x_i}$ and $y = \sequence {y_i}$.
Consider the mapping $f: \closedint 0 1 \to I^\omega$ defined as:
:$\forall t \in \closedint 0 1: \map f t = t x + \paren {1 - t} y = \sequence {t x_i + \paren {1 - t} y_i}$
which is convergent.
Then $f$ is an injective path joining $x$ to $y$.
Hence the result.
== Sources ==
* : Part $\text {II}$: Counterexamples: $38$. Hilbert Cube: $5$
Category:Hilbert Cube
Category:Arc-Connected Spaces
Hilbert Cube is Compact

== Theorem ==
Let $M = \struct {I^\omega, d_2}$ be the Hilbert cube.
Then $M$ is a compact space.
== Proof ==
Let $M'$ be the metric space defined as:
:$M' = \ds \prod_{k \mathop \in \N} \closedint 0 1$
under the product topology.
By definition, $\closedint 0 1$ is the closed unit interval under the usual (Euclidean) topology.
From Hilbert Cube is Homeomorphic to Countable Infinite Product of Real Number Unit Intervals, $M$ is homeomorphic to $M'$.
From Closed Real Interval is Compact in Metric Space, $\closedint 0 1$ is a compact space.
But $M'$ is also a compact space by Tychonoff's Theorem.
From Compactness is Preserved under Continuous Surjection, if $M'$ is compact then so is $M$.
=== Hilbert Cube is Compact in 2-Sequence Space with 2-Norm ===
== Sources ==
* : Part $\text {II}$: Counterexamples: $38$. Hilbert Cube: $4$
Category:Hilbert Cube
Category:Compact Spaces
Category:Hilbert Cube is Compact
Hilbert Cube is Completely Normal

== Theorem ==
Let $M = \struct {I^\omega, d_2}$ be the Hilbert cube.
Then $M$ is a completely normal space.
== Proof ==
We have that Hilbert Cube is Metric Space.
The result follows from Metric Space is Completely Normal.
== Sources ==
* : Part $\text {II}$: Counterexamples: $38$. Hilbert Cube: $2$
Category:Hilbert Cube
Category:Separable Spaces
Hilbert Cube is Homeomorphic to Countable Infinite Product of Real Number Unit Intervals

== Theorem ==
Let $M_1 = \struct {I^\omega, d_2}$ be the Hilbert cube:
:$M_1 = \ds \prod_{k \mathop \in \N} \closedint 0 {\dfrac 1 k}$
under the same metric as that of the Hilbert sequence space:
:$\ds \forall x = \sequence {x_i}, y = \sequence {y_i} \in I^\omega: \map {d_2} {x, y} := \paren {\sum_{k \mathop \ge 0} \paren {x_k - y_k}^2}^{\frac 1 2}$
Let $M_2$ be the metric space defined as:
:$M_2 = \ds \prod_{k \mathop \in \N} \closedint 0 1$
under the product topology.
Then $M_1$ and $M_2$ are homeomorphic.
== Proof ==
Let $x = \sequence {x_i}$ be an arbitrary element of $M_1$.
Let $f: M_1 \to M_2$ be the mapping defined as:
:$\forall x \in M_1: \map f x = \tuple {x_1, 2 x_2, 3 x_3, \ldots}$
Then $f$ is seen to be a bijection.
It remains to be shown that an open set in $M_1$ is mapped to an open set in $M_2$ by $f$.
== Sources ==
* : Part $\text {II}$: Counterexamples: $38$. Hilbert Cube: $1$
Category:Hilbert Cube
Hilbert Cube is Metric Space

== Theorem ==
Let $M = \struct {I^\omega, d_2}$ be the Hilbert cube.
Then $M$ is a metric space.
== Proof ==
As defined, $M$ is a subspace of the Hilbert sequence space $\ell^2$.
We have that Hilbert Sequence Space is Metric Space.
The result follows from Subspace of Metric Space is Metric Space.
== Sources ==
* : Part $\text {II}$: Counterexamples: $38$. Hilbert Cube: $2$
Category:Hilbert Cube
Hilbert Cube is Second-Countable

== Theorem ==
Let $M = \struct {I^\omega, d_2}$ be the Hilbert cube.
Then $M$ is a second-countable space.
== Proof ==
From Hilbert Cube is Separable, $M$ is a separable space.
We also have that Hilbert Cube is Metric Space.
The result follows from Separable Metric Space is Second-Countable.
== Sources ==
* : Part $\text {II}$: Counterexamples: $38$. Hilbert Cube: $3$
Category:Hilbert Cube
Category:Second-Countable Spaces
Hilbert Cube is Separable

== Theorem ==
Let $M = \struct {I^\omega, d_2}$ be the Hilbert cube.
Then $M$ is a separable space.
== Proof ==
Consider the set $H$ of all points of $M$ which have finitely many rational coordinates and all the rest zero.
Then $H$ forms a countable subset of $A$ which is (everywhere) dense.
The result follows by definition of separable space.
== Sources ==
* : Part $\text {II}$: Counterexamples: $38$. Hilbert Cube: $3$
Category:Hilbert Cube
Category:Separable Spaces
Hilbert Matrix/Examples/3x3

== Examples of Hilbert Matrices ==
The order $3$ Hilbert matrix is:
:$H_3 = \begin {bmatrix}
1 & \tfrac 1 2 & \tfrac 1 3 \\
\tfrac 1 2 & \tfrac 1 3 & \tfrac 1 4 \\
\tfrac 1 3 & \tfrac 1 4 & \tfrac 1 5 \\
\end {bmatrix}$
=== Determinant ===
== Sources ==
* 
* 
Category:Examples of Hilbert Matrices
Hilbert Matrix/Examples/3x3/Determinant/Mistake

== Source Work ==
:
:Hilbert matrix
== Mistake ==
:For $n = 3$,
::$H_3 = \begin {pmatrix} 1 & \tfrac 1 2 & \tfrac 1 3 \\ \tfrac 1 2 & \tfrac 1 3 & \tfrac 1 4 \\ \tfrac 1 3 & \tfrac 1 4 & \tfrac 1 5 \\ \end {pmatrix}$
:and $\map \det {H_3} = 1 / 12 \, 160$.
== Correction ==
The actual value is:
:$\map \det {H_3} = 1 / 2160$
See 
== Sources ==
* 
* 
Category:Mistakes/The Penguin Dictionary of Mathematics
Hilbert Matrix/Examples/6x6

== Examples of Hilbert Matrices ==
The order $6$ Hilbert matrix is:
:$\begin {bmatrix}
1 & \tfrac 1 2 & \tfrac 1 3 & \tfrac 1 4 & \tfrac 1 5 & \tfrac 1 6 \\
\tfrac 1 2 & \tfrac 1 3 & \tfrac 1 4 & \tfrac 1 5 & \tfrac 1 6 & \tfrac 1 7 \\
\tfrac 1 3 & \tfrac 1 4 & \tfrac 1 5 & \tfrac 1 6 & \tfrac 1 7 & \tfrac 1 8 \\
\tfrac 1 4 & \tfrac 1 5 & \tfrac 1 6 & \tfrac 1 7 & \tfrac 1 8 & \tfrac 1 9 \\
\tfrac 1 5 & \tfrac 1 6 & \tfrac 1 7 & \tfrac 1 8 & \tfrac 1 9 & \tfrac 1 {10} \\
\tfrac 1 6 & \tfrac 1 7 & \tfrac 1 8 & \tfrac 1 9 & \tfrac 1 {10} & \tfrac 1 {11}
\end {bmatrix}$
Category:Examples of Hilbert Matrices
Hilbert Matrix is Cauchy Matrix

== Theorem ==
A Hilbert matrix is a special case of a Cauchy matrix.
== Proof ==
By definition of Hilbert matrix, the element $a_{i j}$ is:
:$a_{i j} = \dfrac 1 {i + j - 1}$
For all $i, j \in \Z$ such that $1 \le i \le n$ and $1 \le j \le n$, let:
:$x_i = i$
:$y_j = j - 1$
Then:
:$a_{i j} = \dfrac 1 {x_i + y_j}$
The result follows by definition of a Cauchy matrix.
== Sources ==
* : $\S 1.2.3$: Sums and Products: Exercise $45$
Category:Hilbert Matrices
Category:Cauchy Matrices
Hilbert Proof System Instance 2 Independence Results

== Theorem ==
Let $\mathscr H_2$ be Instance 2 of the Hilbert proof systems.
Then the following independence results hold:
=== Independence of $(A1)$ ===
=== Independence of $(A2)$ ===
=== Independence of $(A3)$ ===
=== Independence of $(A4)$ ===
=== $RST \, 4$ is Derivable ===
== Sources ==
* : $\S 4.6$: Independence
Category:Hilbert Proof System Instance 2
Hilbert Proof System Instance 2 Independence Results/Independence of A1

== Theorem ==
Let $\mathscr H_2$ be Instance 2 of the Hilbert proof systems.
Then:
Axiom $(A1)$ is independent from $(A2)$, $(A3)$, $(A4)$.
== Proof ==
Denote with $\mathscr H_2 - (A1)$ the proof system resulting from $\mathscr H_2$ by removing axiom $(A1)$.
Consider $\mathscr C_2$, Instance 2 of constructed semantics.
We will prove that:
* $\mathscr H_2 - (A1)$ is sound for $\mathscr C_2$;
* Axiom $(A1)$ is not a tautology in $\mathscr C_2$
which leads to the conclusion that $(A1)$ is not a theorem of $\mathscr H_2 - (A1)$.
=== Soundness of $\mathscr H_2 - (A1)$ for $\mathscr C_2$ ===
Starting with the axioms:
Next it needs to be shown that the rules of inference of $\mathscr H_2$ preserve $\mathscr C_2$-tautologies.
==== Rule $RST \, 1$: Rule of Uniform Substitution ====
By definition, any WFF is assigned a value $0$, $1$ or $2$.
Thus, in applying Rule $RST \, 1$, we are introducing $0$, $1$ or $2$ in the position of a propositional variable.
But all possibilities of assignments of $0$s, $1$s and $2$s to such propositional variables were shown not to affect the resulting value $0$ of the axioms.
Hence Rule $RST \, 1$ preserves $\mathscr C_2$-tautologies.
==== Rule $RST \, 2$: Rule of Substitution by Definition ====
Because the definition of $\mathscr C_2$ was given in terms of Rule $RST \, 2$, it cannot affect any of its results.
==== Rule $RST \, 3$: Rule of Detachment ====
Suppose $\mathbf A$ and $\mathbf A \implies \mathbf B$ both take value $0$.
Then using Rule $RST \, 2$, definition $(2)$, we get:
:$\neg \mathbf A \lor \mathbf B$
taking value $0$ by assumption.
But $\neg \mathbf A$ takes value $1$ by definition of $\neg$.
So from the definition of $\lor$ it must be that $\mathbf B$ takes value $0$.
Hence Rule $RST \, 3$ also produces only WFFs of value $0$.
==== Rule $RST \, 4$: Rule of Adjunction ====
Suppose $\mathbf A$ and $\mathbf B$ take value $0$.
Then:
proving that Rule $RST \, 4$ also produces only $0$s from $0$s.
Hence $\mathscr H_2 - (A1)$ is sound for $\mathscr C_2$.
=== $(A1)$ is not a $\mathscr C_2$-tautology ===
Recall axiom $(A1)$, the Rule of Idempotence:
:$(p \lor p) \implies p$
Under $\mathscr C_2$, we apply a single definitional abbreviation and have the following:
:$\begin{array}{|cccc|c|c|} \hline
\neg & (p & \lor & p) & \lor & p \\
\hline
1 & 0 & 0 & 0 & 0 & 0 \\
0 & 1 & 1 & 1 & 0 & 1 \\
1 & 2 & 0 & 2 & 2 & 2 \\
\hline
\end{array}$
Hence according to the definition of $\mathscr C_2$, $(A1)$ is not a tautology.
Therefore $(A1)$ is independent from $(A2)$, $(A3)$, $(A4)$.
== Sources ==
* : $\S 4.6$: Independence
Category:Hilbert Proof System Instance 2
Hilbert Proof System Instance 2 Independence Results/RST4 is Derivable

== Theorem ==
Let $\mathscr H_2$ be Instance 2 of the Hilbert proof systems.
Then:
Rule of inference $RST \, 4$ is derivable from $RST \, 1, RST \, 2, RST \, 3$ and the axioms $(A1)$ through $(A4)$.
== Proof ==
Recall the statement of $RST \, 4$:
:If $\mathbf A$ and $\mathbf B$ are theorems of $\mathscr H_2$, then so is $\mathbf A \land \mathbf B$.
Suppose that $\mathbf A$ and $\mathbf B$ are theorems of $\mathscr H_2$.
From Rule of Conjunction/Sequent Form/Formulation 2, we have as a theorem:
:$p \implies \paren{ q \implies \paren{ p \land q } }$
which can be recursively checked to not have used $RST \, 4$ anywhere in a proof.
Applying $RST \, 1$, this becomes:
:$\mathbf A \implies \paren{ \mathbf B \implies \paren{ \mathbf A \land \mathbf B } }$
Using $RST \, 3$ twice, the result follows.
== Sources ==
* : $\S 4.7$: The Derivation of Formulae: $DR \, 2$
Category:Hilbert Proof System Instance 2
Hilbert Proof System Instance 2 is Consistent

== Theorem ==
Instance 2 of the Hilbert proof systems $\mathscr H_2$ is consistent.
== Proof ==
Consider Instance 1 of a constructed semantics, denoted $\mathscr C_1$.
Note that $\neg p$ is not a tautology for $\mathscr C_1$.
We will establish that every $\mathscr H_2$-theorem is a $\mathscr C_1$-tautology.
That is, that $\mathscr H_2$ is sound for $\mathscr C_1$.
Starting with the axioms:
Next it needs to be shown that the rules of inference of $\mathscr H_2$ preserve $\mathscr C_1$-tautologies.
=== Rule $RST \, 1$: Rule of Uniform Substitution ===
By definition, any WFF is assigned a value $1$ or $2$.
Thus, in applying Rule $RST \, 1$, we are introducing $1$ or $2$ in the position of a propositional variable.
But all possibilities of assignments of $1$s and $2$s to such propositional variables were shown not to affect the resulting value $2$ of the axioms.
Hence Rule $RST \, 1$ preserves $\mathscr C_1$-tautologies.
=== Rule $RST \, 2$: Rule of Substitution by Definition ===
Because the definition of $\mathscr C_1$ was given in terms of Rule $RST \, 2$, it cannot affect any of its results.
=== Rule $RST \, 3$: Rule of Detachment ===
Suppose $\mathbf A$ and $\mathbf A \implies \mathbf B$ both take value $2$.
Then using Rule $RST \, 2$, definition $(2)$, we get:
:$\neg \mathbf A \lor \mathbf B$
taking value $2$ by assumption.
But $\neg \mathbf A$ takes value $1$ by definition of $\neg$.
So from the definition of $\lor$ it must be that $\mathbf B$ takes value $2$.
Hence Rule $RST \, 3$ also produces only WFFs of value $2$.
=== Rule $RST \, 4$: Rule of Adjunction ===
Suppose $\mathbf A$ and $\mathbf B$ take value $2$.
Then:
proving that Rule $RST \, 4$ also produces only $2$s from $2$s.
Hence $\mathscr H_2$ is sound for $\mathscr C_1$.
In particular:
:$\not\vdash_{\mathscr H_2} \neg p$
Hence $\mathscr H_2$ is consistent.
== Sources ==
* : $\S 4.5$: Consistency
Category:Hilbert Proof System Instance 2
Hilbert Sequence Space is Arc-Connected

== 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 $\R$.
Then $\ell^2$ is arc-connected.
== Proof ==
Let $x = \sequence {x_i}$ and $y = \sequence {y_i}$.
Consider the mapping $f: \closedint 0 1 \to \ell^2$ defined as:
:$\forall t \in \closedint 0 1: \map f t = t x + \paren {1 - t} y = \sequence {t x_i + \paren {1 - t} y_i}$
which is convergent.
Then $f$ is an injective path joining $x$ to $y$.
Hence the result.
== Sources ==
* : Part $\text {II}$: Counterexamples: $36$. Hilbert Space: $5$
Category:Hilbert Sequence Space
Category:Arc-Connected Spaces
Hilbert Sequence Space is Complete Metric Space

== Theorem ==
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.
== Sources ==
* : Part $\text {II}$: Counterexamples: $36$. Hilbert Space: $1$
Category:Hilbert Sequence Space
Category:Complete Metric Spaces
Hilbert Sequence Space is Homeomorphic to Countable Infinite Product of Real Number Spaces

== 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 $\struct {\R, \tau_d}$ denote the real number line under the Euclidean topology.
Let $\R^\omega = \ds \prod_{i \mathop \in \N} \struct {\R, \tau_d}$ denote the countable-dimensional real Cartesian space under the product topology.
Then $\ell^2$ is homeomorphic to $\R^\omega$.
== Proof ==
== Sources ==
* : Part $\text {II}$: Counterexamples: $36$. Hilbert Space: $6$
Category:Hilbert Sequence Space
Hilbert Sequence Space is Lindelöf

== Theorem ==
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.
== Sources ==
* : Part $\text {II}$: Counterexamples: $36$. Hilbert Space: $2$
Category:Hilbert Sequence Space
Category:Lindelöf Spaces
Hilbert Sequence Space is Metric Space

== 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 $\R$.
Then $\ell^2$ is a metric space.
== Proof 1 ==
== Proof 2 ==
Category:Hilbert Sequence Space
Category:Hilbert Sequence Space is Metric Space
Hilbert Sequence Space is Metric Space/Proof 2

== Theorem ==
== Proof ==
By definition of the Hilbert sequence space on $\R$:
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.
Then $\ell^2 := \struct {A, d_2}$ where $d_2: A \times A: \to \R$ is the real-valued function defined as:
:$\ds \forall x = \sequence {x_i}, y = \sequence {y_i} \in A: \map {d_2} {x, y} := \paren {\sum_{k \mathop \ge 0} \paren {x_k - y_k}^2}^{\frac 1 2}$
From Convergence of Square of Linear Combination of Sequences whose Squares Converge we have that $\ds \sum_{k \mathop \ge 0} \paren {x_k - y_k}^2$ does actually converge.
=== Proof of ===
So holds for $d_2$.
=== Proof of ===
Let $z = \sequence {z_i} \in A$.
So holds for $d_2$.
=== Proof of ===
So holds for $d_2$.
=== Proof of ===
So holds for $d_2$.
== Sources ==
* : $2$: Continuity generalized: metric spaces: $2.2$: Examples: Example $2.2.16$
Category:Hilbert Sequence Space is Metric Space
Hilbert Sequence Space is Second-Countable

== Theorem ==
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.
== Sources ==
* : Part $\text {II}$: Counterexamples: $36$. Hilbert Space: $2$
Category:Hilbert Sequence Space
Category:Second-Countable Spaces
Hilbert Sequence Space is Separable

== 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 $\R$.
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:
:$q^{\paren n} _i :=
\begin {cases} \frac {\floor {2^{n+i} x} }{2^{n+i} } & : i < n
\\ 0 & : i \ge n \end{cases}$
where $\floor \cdot$ denotes the floor function.
Then, we have:
== Sources ==
* : Part $\text {II}$: Counterexamples: $36$. Hilbert Space: $2$
Category:Hilbert Sequence Space
Category:Separable Spaces
Hilbert Sequence Space is not Locally Compact Hausdorff Space

== 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 $\R$.
Then $\ell^2$ is not a locally compact Hausdorff space.
== Proof ==
From Metric Space fulfils all Separation Axioms, $\ell^2$ 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.
== Sources ==
* : Part $\text {II}$: Counterexamples: $36$. Hilbert Space: $3$
Category:Hilbert Sequence Space
Category:Locally Compact Hausdorff Spaces
Hilbert Sequence Space is not Sigma-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 $\R$.
Then $\ell^2$ is not $\sigma$-compact.
== Proof ==
By Compact Subset of Hilbert Sequence Space is Nowhere Dense, a compact subset of $\ell^2$ is nowhere dense in $\ell^2$.
We have that Hilbert Sequence Space is Complete Metric Space.
Hence $\ell^2$ is non-meager.
It follows that $\ell^2$ is not $\sigma$-compact.
== Sources ==
* : Part $\text {II}$: Counterexamples: $36$. Hilbert Space: $4$
Category:Hilbert Sequence Space
Category:Sigma-Compact Spaces