nyuuzyou/wikis huggingface dataset:
Characterization of Euclidean Borel Sigma-Algebra/Closed equals Compact

== Theorem ==
Let $\CC^n$ and $\KK^n$ be the collections of closed and compact subsets of the Euclidean space $\struct {\R^n, \tau}$, respectively.
Then:
:$\map \sigma {\CC^n} = \map \sigma {\KK^n}$
where $\sigma$ denotes generated $\sigma$-algebra.
== Proof ==
By the Heine-Borel Theorem:
:$\KK^n \subseteq \CC^n$.
Thus from Generated Sigma-Algebra Preserves Subset:
:$\map \sigma {\KK^n} \subseteq \map \sigma {\CC^n}$.
Next, let $\map {B_n^-} {\mathbf 0}$ be the closed ball of radius $n$ around $\mathbf 0$ in $\R^n$ for all $n \in \N$.
Observe that:
:$\R^n = \ds \bigcup_{n \mathop \in \N} \map {B_n^-} {\mathbf 0}$.
Now let $U \in \CC^n$ be a closed subset of $\R^n$.
Then from Intersection with Subset is Subset and Intersection Distributes over Union:
:$\ds U = U \cap \R^n = U \cap \bigcup_{n \mathop \in \N} \map {B_n^-} {\mathbf 0} = \bigcup_{n \mathop \in \N} \paren {U \cap \map {B_n^-} {\mathbf 0} }$
From Intersection of Closed Sets is Closed in Topological Space, $U \cap \map {B_n^-} {\mathbf 0}$ is closed for all $n \in \N$.
By definition, $\map {B_n^-} {\mathbf 0}$ is bounded.
Thus, by the Heine-Borel Theorem, $U \cap \map {B_n^-} {\mathbf 0}$ is compact.
Thus, any closed set is the countable union of compact sets.
By the third axiom for a $\sigma$-algebra, this means that:
:$\CC^n \subseteq \map \sigma {\KK^n}$
Now the definition of generated $\sigma$-algebra ensures that:
:$\map \sigma {\CC^n} \subseteq \map \sigma {\KK^n}$
Hence the result, by definition of set equality.
Category:Characterization of Euclidean Borel Sigma-Algebra
Characterization of Euclidean Borel Sigma-Algebra/Open equals Closed

== Theorem ==
Let $\OO^n$ and $\CC^n$ be the collections of open and closed subsets of the Euclidean space $\struct {\R^n, \tau}$, respectively.
Then:
:$\map \sigma {\OO^n} = \map \sigma {\CC^n}$
where $\sigma$ denotes generated $\sigma$-algebra.
== Proof ==
Recall that a closed set is by definition the relative complement of an open set.
Hence Sigma-Algebra Generated by Complements of Generators applies to yield:
:$\map \sigma {\OO^n} = \map \sigma {\CC^n}$
Category:Characterization of Euclidean Borel Sigma-Algebra
Characterization of Euclidean Borel Sigma-Algebra/Rectangle equals Rational Rectangle

== Theorem ==
Let $\JJ_{ho}^n$ be the collection of half-open rectangles in $\R^n$.
Let $\JJ^n_{ho, \text{rat} }$ be the collection of half-open rectangles in $\R^n$ with rational endpoints.
Then:
:$\map \sigma {\JJ_{ho}^n} = \map \sigma {\JJ^n_{ho, \text{rat} } }$
where $\sigma$ denotes generated $\sigma$-algebra.
== Proof ==
From Generated Sigma-Algebra Preserves Subset:
:$\map \sigma {\JJ_{ho, \text{rat} }^n} \subseteq \map \sigma {\JJ_{ho}^n}$
For the converse, it will suffice to show:
:$\JJ_{ho}^n \subseteq \map \sigma {\JJ_{ho, \text{rat} }^n}$
by definition of generated $\sigma$-algebra.
So, let $\horectr {\mathbf a} {\mathbf b}$ be a half-open $n$-rectangle.
Let $\sequence {\mathbf a_m}_{m \mathop \in \N}$ be a sequence in $\Q^n$ with limit $\mathbf a$.
Also, let this sequence be such that $m_1 > m_2 \implies \mathbf a_{m_1} > \mathbf a_{m_2}$, in the component-wise ordering.
Also, choose $\mathbf b' \in \Q^n$ such that $\mathbf b' > \mathbf b$, again in the component-wise ordering.
Then, for any $m \in \N$:
:$\horectr {\mathbf a_m} {\mathbf b'} \in \JJ_{ho, \text{rat} }^n$
By Sigma-Algebra Closed under Countable Intersection, it follows that:
:$\ds \bigcap_{m \mathop \in \N} \horectr {\mathbf a_m} {\mathbf b'} \in \map \sigma {\JJ_{ho, \text{rat} }^n}$
Now observe, for $\mathbf x \in \R^n$:
Next, let $\sequence {\mathbf b_m}_{m \mathop \in \N}$ be an increasing sequence in $\Q^n$ with limit $\mathbf b$.
Also, let $\mathbf a' \in \Q^n$ be such that $\mathbf a' < \mathbf a$.
Again, it follows that $\horectr {\mathbf a'} {\mathbf b_m} \in \JJ_{ho, \text{rat} }^n$.
Thus, by the third axiom for a $\sigma$-algebra:
:$\ds \bigcup_{m \mathop \in \N} \horectr {\mathbf a'} {\mathbf b_m} \in \map \sigma {\JJ^n_{ho} }$
Similar to the above approach, for any $\mathbf x \in \R^n$:
Hence, it follows that:
:$\ds \bigcup_{m \mathop \in \N} \horectr {\mathbf a'} {\mathbf b_m} = \horectr {\mathbf a'} {\mathbf b}$
whence the latter is in $\map \sigma {\JJ^n_{ho, \text{rat} } }$.
Hence by Sigma-Algebra Closed under Intersection:
:$\horectr {\mathbf a} {\mathbf b'} \cap \horectr {\mathbf a'} {\mathbf b} \in \map \sigma {\JJ^n_{ho, \text{rat} } }$
and finally (the proof of) Half-Open Rectangles Closed under Intersection yields:
:$\horectr {\mathbf a} {\mathbf b'} \cap \horectr {\mathbf a'} {\mathbf b} = \horectr {\mathbf a} {\mathbf b}$
since $\mathbf a' < \mathbf a$ and $\mathbf b < \mathbf b'$, thus finishing the proof.
Category:Characterization of Euclidean Borel Sigma-Algebra
Characterization of Euler's Number by Inequality

== Theorem ==
Let $a$ be a (strictly) positive real number.
Then:
:$a = e \iff \forall x \in \R: a^x \ge x + 1$
where $e$ denotes Euler's number.
== Proof ==
=== Forward Implication ===
Proved in Exponential Function Inequality.
=== Reverse Implication ===
Consider $\map {f_a} x = a^x - x - 1$.
Then we need to prove:
:$a = e \impliedby \forall x \in \R: \map {f_a} x \ge 0$
By Linear Combination of Derivatives, Derivative of Power of Constant, Derivative of Identity Function, and Derivative of Constant:
:$\map {f'_a} x = a^x \ln a - 1$
By Linear Combination of Derivatives, Derivative of Power of Constant, Derivative of Constant Multiple, and Derivative of Constant:
:$\map {f_a} x = a^x \paren {\ln a}^2$
Now, we divide the theorem into two cases:
==== Case 1: $0 < a \le 1$ ====
Consider $x = 1$.
Then:
Thus the is not true.
==== Case 2: $a > 1$ ====
We have:
Hence, by Real Function with Strictly Positive Second Derivative is Strictly Convex, we have that $f_a$ is strictly convex.
Trying to find its minimum, we solve:
:$\map {f'_a} x = 0$
Therefore:
At this point:
It is to be noted that when $a = e$, the minimum is:
:$\dfrac {1 + \map \ln {\ln e} } {\ln e} - 1 = 0$
meaning that the only solution to $\map {f_e} x = 0$ is $x = 0$.
Also, from Exponential Function Inequality, we have:
:$\forall x \in \R: e^x \ge x + 1$
Substituting $x = \map \ln {\ln a}$:
:$\ln a \ge \map \ln {\ln a} + 1$
From Logarithm is Strictly Increasing and Logarithm of 1 is 0:
:$\ln a > 0$
Thus:
:$\dfrac {\ln a} {\ln a} \ge \dfrac {\map \ln {\ln a} + 1} {\ln a}$
Therefore:
:$\dfrac {1 + \map \ln {\ln a} } {\ln a} - 1 \le 0$
Then, we solve the case of equality:
From earlier, we have that the only solution is $\map \ln {\ln a} = 0$, when $a = e$.
Therefore, for other values of $a$, the minimum is negative.
Category:Euler's Number
Characterization of Even Cover

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Let $\UU$ be a cover of $S$.
Let $\tau_{S \times S}$ denote the product topology on the cartesian product $S \times S$.
Then $\UU$ is an even cover of $T$ there exists an open neighborhood $V$ of the diagonal $\Delta_S$ of $S \times S$ in the product space $\struct {S \times S, \tau_{S \times S} }$:
:$\set{\map V x : x \in S}$ is a refinement of $\UU$
where:
:$V$ is seen as a relation on $S \times S$
:$\map V x$ denotes the image of $x$ under $V$.
== Proof ==
=== Necessary Condition ===
Let $\UU$ be an even cover of $T$.
By definition of even cover, there exists a neighborhood $N$ of the diagonal $\Delta_S$ of $S \times S$ in the product space $\struct {S \times S, \tau_{S \times S} }$:
:$\set{\map N x : x \in S}$ is a refinement of $\UU$
By definition of a neighborhood, there exists an open set $V$ of $\struct {S \times S, \tau_{S \times S} }$:
:$\Delta_S \subseteq V \subseteq N$
From Set is Open iff Neighborhood of all its Points and Set is Neighborhood of Subset iff Neighborhood of all Points of Subset:
:$V$ is an open neighborhood of the $\Delta_S$
It remains to show that $\set{\map V x : x \in S}$ is a refinement of $\UU$.
Let $x \in S$.
We have:
:$\tuple{x, x} \in \Delta_S \subseteq V$
By definition of image:
:$x \in \map V x$
From Corollary to Image under Subset of Relation is Subset of Image under Relation:
:$\map V x \subseteq \map N x$
By definition of refinement:
:$\exists U \in \UU : \map N x \subseteq U$
From Subset Relation is Transitive:
:$\map V x \subseteq U$
Since $x$ was arbitrary, we have:
:$\forall x \in S : x \in \map V x$
and
:$\forall x \in S : \exists U \in \UU : \map V x \subseteq U$
It follows that $\set{\map V x : x \in S}$ is a cover of $S$ and a refinement of $\UU$ by definition.
=== Sufficient Condition ===
Let $V$ be an open neighborhood of the diagonal $\Delta_S$ of $S \times S$ in the product space $\struct {S \times S, \tau_{S \times S} }$:
:$\set{\map V x : x \in S}$ is a refinement of $\UU$
By definition of open neighborhood:
:$V$ is a neighborhood of $\Delta_S$ in $\struct {S \times S, \tau_{S \times S}}$
It follows that $\UU$ is an even cover of $T$ by definition.
Category:Definitions/Covers
Characterization of Exponential Integral Function/Formulation 1

== Theorem ==
Let $\Ei: \R_{>0} \to \R$ denote the exponential integral function:
:$\map \Ei x = \ds \int_{t \mathop = x}^{t \mathop \to +\infty} \frac {e^{-t} } t \rd t$
Then:
:$\ds \map \Ei x = -\gamma - \ln x + \int_0^x \frac {1 - e^{-u} } u \rd u$
where $\gamma$ denotes the Euler-Mascheroni constant.
== Proof ==
We have, by Derivative of $e^{a x}$: 
:$\map {\dfrac \d {\d u} } {1 - e^{-u} } = e^{-u}$
By Primitive of Reciprocal:
:$\ds \int \frac {\d u} u = \ln u + C$
So: 
We have: 
Also applying Definite Integral to Infinity of $e^{-x} \ln x$ gives:
:$\ds \int_0^x \frac {1 - e^{-u} } u \rd u = \paren {1 - e^{-x} } \ln x + \gamma + \int_x^\infty e^{-u} \ln u \rd u$ 
We have, by Primitive of $e^{a x}$:
:$\ds \int e^{-u} \rd u = -e^{-u} + C$
We have by Derivative of Logarithm Function:
:$\dfrac \d {\d u} \paren {\ln u} = \dfrac 1 u$
We therefore have: 
It remains to evaluate the limit on the . 
We have, for $u > 1$:
:$e^{-u} \ln u > 0$
and by Bounds of Natural Logarithm: 
:$e^{-u} \ln u < u e^{-u} - e^{-u}$
We have, from Limit to Infinity of $x^n e^{-a x}$: 
:$\ds \lim_{u \mathop \to \infty} \paren {u e^{-u} - e^{-u} } = 0$
So, by the Squeeze Theorem: 
:$\ds \lim_{u \mathop \to \infty} \paren {e^{-u} \ln u} = 0$
So: 
:$\ds \int_0^x \frac {1 - e^{-u} } u \rd u = \ln x + \gamma + \map \Ei x$
Rearranging gives: 
:$\ds \map \Ei x = -\gamma - \ln x + \int_0^x \frac {1 - e^{-u} } u \rd u$
== Sources ==
* : $\S 35$: Exponential Integral $\ds \map \Ei x = \int_x^\infty \frac {e^{-u} } u \rd u$: $35.7$
Category:Characterization of Exponential Integral Function
Characterization of Extended Real Sigma-Algebra

== Theorem ==
Let $\map \BB \R$ be the Borel $\sigma$-algebra on $\R$.
Let $\overline \BB$ be the extended real $\sigma$-algebra.
Define $\SS := \powerset {\set {-\infty, +\infty} }$, where $\PP$ denotes power set.
Then:
:$\overline \BB = \set {B \cup S: B \in \map \BB \R, S \in \SS}$
== Proof ==
Let $\overline B \in \overline \BB$.
Then by Extended Real Sigma-Algebra Induces Borel Sigma-Algebra on Reals, we have:
:$\overline B \cap \R \in \map \BB \R$
We also have, by definition of the extended real numbers $\overline \R$, that:
:$\overline \R \setminus \R = \set {-\infty, +\infty}$
and therefore, $\overline B \setminus \R \subseteq \set {-\infty, +\infty}$.
Here, $\setminus$ signifies set difference.
By Set Difference Union Intersection:
:$\overline B = \paren {\overline B \setminus \R} \cup \paren {\overline B \cap \R}$
Therefore, any $\overline B \in \overline \BB$ is of the purported form $B \cup S$ with $B \in \map \BB \R$ and $S \in \SS$.
It remains to show that any such set is in fact an element of $\overline \BB$.
Since any $B \in \map \BB \R$ is naturally also in $\overline \BB$, it suffices to show that:
:$\SS \subseteq \overline \BB$
by applying Sigma-Algebra Closed under Union.
From Closed Set Measurable in Borel Sigma-Algebra, it will now suffice to show that:
:$\O, \set {-\infty}, \set {+\infty}, \set {-\infty, +\infty}$
are all closed sets in $\overline \R$.
That they are follows from Extended Real Number Space is Hausdorff and Finite Subspace of Hausdorff Space is Closed.
== Sources ==
* : $\S 8$, $\S 8$: Problem $2$
Category:Extended Real Numbers
Category:Sigma-Algebras
Characterization of Fineness of Topology in terms of Topological Bases

== Theorem ==
Let $X$ be a set.
Let $\tau$ and $\tau'$ be topologies on $X$. 
Let $\BB \subseteq \tau$ be a basis for $\tau$ and $\BB' \subseteq \tau'$ be a basis for $\tau'$.
:$(1): \quad$ $\tau'$ is finer than $\tau$
:$(2): \quad$ for each $x \in X$ and each $B \in \BB$ containing $x$, there exists $B' \in \BB'$ such that $x \in B' \subseteq B$. 
== Proof ==
=== $(1)$ implies $(2)$ ===
Let $x \in X$ and $B \in \BB$ contain $x$.
Since $\tau'$ is finer than $\tau$, we have that $B \in \tau'$. 
Since $\BB'$ is a basis for $\tau'$, there exists a subset $\set {B'_\alpha : \alpha \in I} \subseteq \BB'$ such that: 
:$\ds B = \bigcup_{\alpha \mathop \in I} B'_\alpha$
from Open Set is Union of Elements of Basis.
Since $x \in B$, we have $x \in B'_\alpha$ for some $\alpha \in I$.
We have $B'_\alpha \subseteq B$ from Set is Subset of Union. 
Setting $B = B'_\alpha$, we have therefore found $B' \in \BB'$ such that $x \in B' \subseteq B$, proving $(2)$. 
=== $(2)$ implies $(1)$ ===
Let $U \in \tau$. 
We aim to prove that $U \in \tau'$. 
Let $x \in U$.
Since $\BB$ is a basis for $\tau$, there exists a subset $\set {B_\alpha : \alpha \in I} \subseteq \BB$ such that: 
:$\ds U = \bigcup_{\alpha \mathop \in I} B_\alpha$
from Open Set is Union of Elements of Basis.
Then for some $\alpha \in I$, we have $x \in B_\alpha \subseteq U$. 
By $(2)$, it follows that there exists $B' \in \BB'$ such that: 
:$x \in B' \subseteq B_\alpha$
in particular: 
:$x \in B' \subseteq U$
We have now shown that for each $x \in U$, there exists $B'_x \in \BB'$ such that: 
:$x \in B'_x \subseteq U$ 
We argue that: 
:$\ds U = \bigcup_{x \in U} B'_x$
Clearly: 
:$\ds U \subseteq \bigcup_{x \in U} B'_x$
since $x \in B'_x$ for each $x \in U$, while: 
:$\ds \bigcup_{x \in U} B'_x \subseteq U$ 
sine $B'_x \subseteq U$ for each $x \in U$.
So we have: 
:$\ds U = \bigcup_{x \in U} B'_x$
So $U$ is the union of elements of $\BB'$. 
In particular, $U$ is the union of elements of $\tau'$. 
So, since topologies are closed under union, we have $U \in \tau'$.
So we have $\tau \subseteq \tau'$ as desired.
Category:Topological Bases
Characterization of Finite Rank Operators

== Theorem ==
Let $\GF \in \set {\R, \C}$. 
Let $\struct {\HH, \innerprod \cdot \cdot_\HH}$ and $\struct {\KK, \innerprod \cdot \cdot_\KK}$ be Hilbert spaces over $\GF$.
Let $T \in \map {B_{00} } {\HH, \KK}$ be a bounded finite rank operator.
Let $n = \map \dim {\operatorname {ran} T}$ be the rank of $T$.
Then there are orthonormal vectors $e_1, \ldots, e_n \in \HH$ and vectors $g_1, \ldots, g_n \in \KK$ such that:
:$\forall h \in \HH: T h = \ds \sum_{i \mathop = 1}^n \innerprod h {e_i}_\HH g_i$
== Proof ==
Let: 
:$\set {e_1, e_2, \ldots, e_n}$
be a basis of $T \sqbrk \HH$.
By Gram-Schmidt Orthogonalization, there exists an orthonormal subset such that: 
:$\set {g_1, g_2, \ldots, g_n}$ 
that is a basis of $T \sqbrk \HH$.
So we have $T x \in \map \span {g_1, g_2, \ldots, g_n}$ for each $x \in \HH$.
So, we can write: 
:$\ds T x = \sum_{i \mathop = 1}^n \map {\alpha_i} x g_i$
for functions $\alpha_i : \HH \to \GF$. 
We show that each $\alpha_i$ is a bounded linear functional. 
For each $\lambda, \mu \in \GF$ and $x, y \in \HH$ such that: 
So:
:$\ds \sum_{i \mathop = 1}^n \paren {\map {\alpha_i} {\lambda x + \mu y} - \paren {\lambda \map {\alpha_i} x + \mu \map {\alpha_i} y} } g_i = \mathbf 0_\KK$
From Orthogonal Set is Linearly Independent Set, $g_1, g_2, \ldots, g_n \in \KK$ are linearly independent. 
So we have: 
:$\map {\alpha_i} {\lambda x + \mu y} = \lambda \map {\alpha_i} x + \mu \map {\alpha_i} y$
for each $i$. 
So each $\alpha_i$ is a linear functional. 
From Pythagoras's Theorem (Inner Product Space), we have: 
:$\ds \norm {T x}^2_\KK = \sum_{i \mathop = 1}^n \cmod {\map {\alpha_i} x}^2$ 
for each $x \in \HH$. 
So we have: 
:$\cmod {\map {\alpha_i} x} \le \norm {T x}_\KK$
for each $x \in \HH$.
From Fundamental Property of Norm on Bounded Linear Transformation, we have: 
:$\cmod {\map {\alpha_i} x} \le \norm T_{\map \BB {\HH, \KK} } \norm x_\HH$
for each $x \in \HH$. 
So each $\alpha_i$ is a bounded linear functional.
From Riesz Representation Theorem (Hilbert Spaces), there exists $e_i$ such that: 
:$\map {\alpha_i} x = \innerprod x {e_i}$
for each $x \in \HH$. 
Then we have: 
:$\ds T x = \sum_{i \mathop = 1}^n \innerprod x {e_i}_\HH g_i$
for each $x \in \HH$. 
== Sources ==
* $\text {II}.4$ Exercise $8$
Category:Linear Transformations on Hilbert Spaces
Category:Finite Rank Operators
Characterization of Generalized Hilbert Sequence Space

== Theorem ==
Let $\alpha$ be an infinite cardinal number.
Let $I$ be an indexed set of cardinality $\alpha$.
Let $H^\alpha = \struct{A, d_2}$ be the generalized Hilbert sequence space of weight $\alpha$ where:
:$A$ denotes the set of all real-valued functions $x : I \to \R$ such that:
::$(1)\quad \set{i \in I: x_i \ne 0}$ is countable
::$(2)\quad$ the generalized sum $\ds \sum_{i \mathop \in I} x_i^2$ is a convergent net.
Let $H = \struct{\ell^2, d_{\ell^2}}$ denote the Hilbert sequence space, where:
:$\ell^2$ denotes the real $2$-sequence space, that is, the set of all real sequences $\sequence {x_i}$ such that the series $\ds \sum_{n \mathop = 0}^\infty x_n^2$ is convergent
Let $x_1, x_2, \ldots, x_m : I \to \R$ be real-valued functions.
Then:
:$x_1, x_2, \ldots, x_m \in A$ 
:there exists an enumeration $\set{j_0, j_1, j_2, \ldots}$ of a countably infinite subset of $I$ such that $\forall k \in \closedint 1 m$:
::$(1)\quad\set{i \in I : \paren{x_k}_i \ne 0} \subseteq \set{j_0, j_1, j_2, \ldots}$
::$(2)\quad\sequence{\paren{x_k}_{j_n}} \in \ell^2$ 
In which case:
:$\forall k \in \closedint 1 m : \ds \sum_{i \mathop \in I} \paren{\paren{x_k}_i}^2 = \sum_{n \mathop = 0}^\infty \paren{\paren{x_k}_{j_n}}^2$
== Proof ==
=== Necessary Condition ===
Let $x_1, x_2, \ldots, x_m \in A$.
=== Sufficient Condition ===
Let $\set{j_0, j_1, j_2, \ldots}$ be a countably infinite subset of $I$ such that $\forall k \in \closedint 1 m$:
::$(1)\quad\set{i \in I : \paren{x_k}_i \ne 0} \subseteq \set{j_0, j_1, j_2, \ldots}$
::$(2)\quad\sequence{\paren{x_k}_{j_n}} \in \ell^2$ 
Category:Generalized Hilbert Sequence Spaces
Category:Characterization of Generalized Hilbert Sequence Space
Characterization of Generalized Hilbert Sequence Space/Necessary Condition

== Theorem ==
Let $\alpha$ be an infinite cardinal number.
Let $I$ be an indexed set of cardinality $\alpha$.
Let $H^\alpha = \struct{A, d_2}$ be the generalized Hilbert sequence space of weight $\alpha$ where:
:$A$ denotes the set of all real-valued functions $x : I \to \R$ such that:
::$(1)\quad \set{i \in I: x_i \ne 0}$ is countable
::$(2)\quad$ the generalized sum $\ds \sum_{i \mathop \in I} x_i^2$ is a convergent net.
Let $H = \struct{\ell^2, d_{\ell^2}}$ denote the Hilbert sequence space, where:
:$\ell^2$ denotes the real $2$-sequence space, that is, the set of all real sequences $\sequence {x_i}$ such that the series $\ds \sum_{n \mathop = 0}^\infty x_n^2$ is convergent
Let $x_1, x_2, \ldots, x_m \in A$.
Then:
there exists an enumeration $\set{j_0, j_1, j_2, \ldots}$ of a countably infinite subset of $I$ such that $\forall k \in \closedint 1 m$:
::$(1)\quad\set{i \in I : \paren{x_k}_i \ne 0} \subseteq \set{j_0, j_1, j_2, \ldots}$
::$(2)\quad\sequence{\paren{x_k}_{j_n}} \in \ell^2$ 
In which case:
:$\forall k \in \closedint 1 m : \ds \sum_{i \mathop \in I} \paren{\paren{x_k}_i}^2 = \sum_{n \mathop = 0}^\infty \paren{\paren{x_k}_{j_n}}^2$
== Proof ==
By definition of $A$:
:$\forall k \in \closedint 1 m : \set{i \in I : \paren{x_k}_i \ne 0}$ is countable
From Infinite Set has Countably Infinite Subset, let:
:$I' \subseteq I$ be countably infinite
Let:
:$J = I' \cup \ds \bigcup_{k \mathop = 1}^m \set{i \in I : \paren{x_k}_i \ne 0}$
From Countable Union of Countable Sets is Countable:
:$J$ is countable
From the contrapositive statement of Subset of Finite Set is Finite:
:$J$ is countably infinite
From Countably Infinite Set has Enumeration, let:
:$\set{j_0, j_1, j_2, \ldots}$ be an enumeration of $J$
From Set is Subset of Union:
:$\forall k \in \closedint 1 m : \set{i \in I : \paren{x_k}_i \ne 0} \subseteq \set{j_0, j_1, j_2, \ldots}$
By definition of $A$:
:$\forall k \in \closedint 1 m : \ds \sum_{i \mathop \in I} \paren{x_k}_i^2$ converges
We have:
Hence:
:$\forall k \in \closedint 1 m : \sequence{\paren{x_k}_{j_n}} \in \ell^2$
Category:Characterization of Generalized Hilbert Sequence Space
Characterization of Generalized Hilbert Sequence Space/Sufficient Condition

== Theorem ==
Let $\alpha$ be an infinite cardinal number.
Let $I$ be an indexed set of cardinality $\alpha$.
Let $H^\alpha = \struct{A, d_2}$ be the generalized Hilbert sequence space of weight $\alpha$ where:
:$A$ denotes the set of all real-valued functions $x : I \to \R$ such that:
::$(1)\quad \set{i \in I: x_i \ne 0}$ is countable
::$(2)\quad$ the generalized sum $\ds \sum_{i \mathop \in I} x_i^2$ is a convergent net.
Let $H = \struct{\ell^2, d_{\ell^2}}$ denote the Hilbert sequence space, where:
:$\ell^2$ denotes the real $2$-sequence space, that is, the set of all real sequences $\sequence {x_i}$ such that the series $\ds \sum_{n \mathop = 0}^\infty x_n^2$ is convergent
Let $x_1, x_2, \ldots, x_m : I \to \R$ be real-valued functions.
Let there exist an enumeration $\set{j_0, j_1, j_2, \ldots}$ of a countably infinite subset of $I$ such that $\forall k \in \closedint 1 m$:
::$(1)\quad\set{i \in I : \paren{x_k}_i \ne 0} \subseteq \set{j_0, j_1, j_2, \ldots}$
::$(2)\quad\sequence{\paren{x_k}_{j_n}} \in \ell^2$ 
Then:
:$x_1, x_2, \ldots, x_m \in A$ 
In which case:
:$\forall k \in \closedint 1 m : \ds \sum_{i \mathop \in I} \paren{\paren{x_k}_i}^2 = \sum_{n \mathop = 0}^\infty \paren{\paren{x_k}_{j_n}}^2$
== Proof ==
By definition of $\ell^2$:
:$\forall k \in \closedint 1 m : \ds \sum_{n \mathop = 0}^\infty \paren{\paren{x_k}_{j_n}}^2 < \infty$
We have:
Category:Characterization of Generalized Hilbert Sequence Space