nyuuzyou/wikis huggingface dataset:
Boundary of Union of Separated Sets equals Union of Boundaries

== Theorem ==
Let $T$ be a topological space.
Let $A, B$ be subsets of $T$.
Let $A$ and $B$ are separated.
Then:
:$\map \partial {A \cup B} = \partial A \cup \partial B$
where:
:$\partial A$ denotes the boundary of $A$
:$A \cup B$ denotes the union of $A$ and $B$.
== Proof ==
By definition of separated sets:
:$(1): \quad A^- \cap B = A \cap B^- = \O$
By Separated Sets are Disjoint:
:$A \cap B = \O$
We will prove that:
:$\partial A \cap \partial B \subseteq \map \partial {A \cup B}$
Let $x \in \partial A \cap \partial B$.
Then by definition of intersection:
:$x \in \partial A \land x \in \partial B$
Hence by Boundary is Intersection of Closure with Closure of Complement:
:$x \in A^- \cap \paren {\relcomp T A}^- \land x \in B^- \cap \paren {\relcomp T B}^-$
where:
:$A^-$ denotes the closure of $A$
:$\relcomp T A = T \setminus A$ denotes the relative complement of $A$ in $T$.
Then by definition of intersection:
:$x \in A^- \land x \in B^-$
Therefore by definition of empty set, intersection, and $(1)$:
:$x \notin B \land x \notin A$
Then by definition of union:
:$x \notin A \cup B$
By definition of relative complement:
:$x \in \relcomp T {A \cup B}$
By definition of closure:
:$\relcomp T {A \cup B} \subseteq \paren {\relcomp T {A \cup B} }^-$
Then by definition of subset:
:$x \in \paren {\relcomp T {A \cup B} }^-$
By Closure of Finite Union equals Union of Closures:
:$A^- \cup B^- = \paren {A \cup B}^-$
By definition of union:
:$x \in \paren {A \cup B}^-$
Then by definition of intersection:
:$x \in \paren {A \cup B}^- \cap \paren {\relcomp T {A \cup B} }^-$
Thus by Boundary is Intersection of Closure with Closure of Complement:
:$x \in \map \partial {A \cup B}$
This ends the proof of inclusion.
Thus by Union with Superset is Superset the result:
:$\partial A \cup \partial B = \map \partial {A \cup B}$
== Sources ==
* 
* 
Category:Separated Sets
Category:Set Boundaries
Bounded Above Subset of Real Numbers/Examples

== Examples of Bounded Above Subsets of Real Numbers ==
=== Example: $\hointl \gets 2$ ===
=== Example: $\openint 0 1$ ===
=== Example: $\set {-1, 0, 2, 5}$ ===
=== Example: $\openint 3 \to$ ===
=== Example: $\closedint 0 1$ ===
=== Example: $\openint \gets \to$ ===
=== Example: $\openint \gets 0$ ===
=== Example: $\hointl 0 1$ ===
Category:Examples of Bounded Above Sets
Bounded Above Subset of Real Numbers/Examples/Closed Interval from 0 to 1

== Example of Bounded Above Subset of Real Numbers ==
Let $I$ be the closed real interval defined as:
:$I := \closedint 0 1$
Then $I$ is bounded above by, for example, $1$, $2$ and $3$, of which $1$ is the supremum.
$I$ is also the greatest element of $I$.
== Sources ==
* : $\S 2$: Continuum Property: Exercise $\S 2.10 \ (3) \ \text{(v)}$
Category:Examples of Bounded Above Sets
Bounded Above Subset of Real Numbers/Examples/Closed Interval from Minus Infinity to 2

== Example of Bounded Above Subset of Real Numbers ==
The subset $I$ of the real numbers $\R$ defined as:
:$I = \hointl \gets 2$
is bounded above by, for example, $2$, $3$ and $4$, of which the supremum is $2$.
$2$ is also the greatest element of $I$.
The set of all upper bounds of $I$ is:
:$\closedint 2 \to$
== Sources ==
* : $\S 10$: The well-ordering principle
* : $\S 2$: Continuum Property: Exercise $\S 2.10 \ (3) \ \text{(ii)}$
Category:Examples of Bounded Above Sets
Bounded Above Subset of Real Numbers/Examples/Finite Set of Reals

== Example of Bounded Above Subset of Real Numbers ==
Let $I$ be the set defined as:
:$I := \set {-1, 0, 2, 5}$
Then $I$ is bounded above by, for example, $5$, $6$ and $7$, of which the supremum is $5$.
$5$ is also the greatest element of $I$.
== Sources ==
* : $\S 2$: Continuum Property: Exercise $\S 2.10 \ (3) \ \text{(iii)}$
Category:Examples of Bounded Above Sets
Bounded Above Subset of Real Numbers/Examples/Left Half Open Interval from 0 to 1

== Example of Bounded Above Subset of Real Numbers ==
Let $I$ be the left half-open real interval defined as:
:$I := \hointl 0 1$
Then $I$ is bounded above.
$1$ is an upper bound of $I$.
== Sources ==
* : $1$: Review of some real analysis: $\S 1.1$: Real Numbers: Example $1.1.1 \ \text{(c)}$
Category:Examples of Bounded Above Sets
Bounded Above Subset of Real Numbers/Examples/Open Interval from 0 to 1

== Example of Bounded Above Subset of Real Numbers ==
Let $I$ be the open real interval defined as:
:$I := \openint 0 1$
Then $I$ is bounded above by, for example, $1$, $2$ and $3$, of which $1$ is the supremum.
However, $I$ does not have a greatest element.
== Proof ==
 $x \in I$ is the greatest element of $I$.
Then $x < \dfrac {x + 1} 2 < 1$ by Mediant is Between.
Thus $\dfrac {x + 1} 2 \in I$ but $x < \dfrac {x + 1} 2$.
So $x$ is not the greatest element of $I$ after all.
So by Proof by Contradiction it follows that $I$ has no greatest element.
== Sources ==
* : $\S 2$: Continuum Property: Exercise $\S 2.10 \ (3) \ \text{(i)}$
Category:Examples of Bounded Above Sets
Bounded Above Subset of Real Numbers/Examples/Open Interval from 3 to Infinity

== Example of Unbounded Above Subset of Real Numbers ==
Let $I$ be the unbounded open real interval defined as:
:$I := \openint 3 \to$
Then $I$ is not bounded above.
Hence $I$ does not admit a supremum, and so does not have a greatest element.
== Sources ==
* : $\S 2$: Continuum Property: Exercise $\S 2.10 \ (3) \ \text{(iv)}$
Category:Examples of Bounded Above Sets
Bounded Above Subset of Real Numbers/Examples/Real Numbers

== Example of Unbounded Above Subset of Real Numbers ==
Let $\R$ denote the set of real numbers.
$\R$ is not bounded above.
== Proof ==
Follows directly from the Axiom of Archimedes.
== Sources ==
* : $1$: Review of some real analysis: $\S 1.1$: Real Numbers: Example $1.1.1 \ \text{(a)}$
Category:Examples of Bounded Above Sets
Bounded Above Subset of Real Numbers/Examples/Strictly Negative Real Numbers

== Example of Bounded Above Subset of Real Numbers ==
Let $I$ be the unbounded open real interval defined as:
:$I := \openint \gets 0$
Then $I$ is bounded above.
$0$ is an upper bound of $I$.
== Sources ==
* : $1$: Review of some real analysis: $\S 1.1$: Real Numbers: Example $1.1.1 \ \text{(b)}$
Category:Examples of Bounded Above Sets
Bounded Below Real-Valued Function/Examples/Tangent on 0 to pi by 2

== Example of Bounded Below Real-Valued Function ==
The real tangent function on the half-open interval $\hointr 0 {\dfrac \pi 2}$:
:$\forall x \in \hointr 0 {\dfrac \pi 2}: \tan x$
is bounded below by $0$, but unbounded above.
== Sources ==
* : Chapter $\text I$: Functions and Limits: $\S 3$: Bounds of a Function
Category:Examples of Bounded Below Real-Valued Functions
Bounded Below Real Sequence/Examples/Strictly Positive Integers

== Example of Bounded Below Real Sequence ==
Let $\sequence {s_n}$ be the real sequence defined as:
:$s_n = n$
That is:
:$\sequence {s_n}$ is the sequence of strictly positive integers.
Then $\sequence {s_n}$ is bounded below.
== Proof ==
All strictly positive integers are by definition greater than zero.
Hence $0$ is a lower bound of $\sequence {s_n}$.
Hence the result by definition of bounded below.
== Sources ==
* 
* 
Category:Examples of Bounded Below Real Sequences
Bounded Below Subset of Real Numbers/Examples

== Examples of Bounded Below Subsets of Real Numbers ==
=== Example: $\openint 1 \to$ ===
=== Example: $\set {x \in \R: x > 0}$ ===
=== Example: $\hointl {-\infty} 2$ ===
=== Example: $\openint 0 1$ ===
=== Example: $\set {-1, 0, 2, 5}$ ===
=== Example: $\openint 3 \infty$ ===
=== Example: $\closedint 0 1$ ===
=== Example: $\openint \gets \to$ ===
=== Example: $\openint \gets 0$ ===
=== Example: $\hointl 0 1$ ===
Category:Examples of Bounded Below Sets of Real Numbers
Bounded Below Subset of Real Numbers/Examples/Closed Interval from 0 to 1

== Example of Bounded Below Subset of Real Numbers ==
Let $I$ be the closed real interval defined as:
:$I := \closedint 0 1$
Then $I$ is bounded below by, for example, $0$, $-1$ and $-2$, of which $0$ is the infimum.
$I$ is also the smallest element of $I$.
== Sources ==
* : $\S 2$: Continuum Property: Exercise $\S 2.10 \ (4) \ \text{(v)}$
Category:Examples of Bounded Below Sets of Real Numbers
Bounded Below Subset of Real Numbers/Examples/Closed Interval from Minus Infinity to 2

== Example of Bounded Below Subset of Real Numbers ==
Let $I$ be the open real interval defined as:
:$I := \openint 0 1$
Then $I$ is not bounded below.
Hence $I$ does not admit an infimum, and so does not have a smallest element.
== Sources ==
* : $\S 2$: Continuum Property: Exercise $\S 2.10 \ (4) \ \text{(ii)}$
Category:Examples of Bounded Below Sets of Real Numbers
Bounded Below Subset of Real Numbers/Examples/Finite Set of Reals

== Example of Bounded Below Subset of Real Numbers ==
Let $I$ be the set defined as:
:$I := \set {-1, 0, 2, 5}$
Then $I$ is bounded below by, for example, $-1$, $-2$ and $3$, of which the infimum is $-1$.
$5$ is also the smallest element of $I$.
== Sources ==
* : $\S 2$: Continuum Property: Exercise $\S 2.10 \ (4) \ \text{(iii)}$
Category:Examples of Bounded Below Sets of Real Numbers
Bounded Below Subset of Real Numbers/Examples/Left Half Open Interval from 0 to 1

== Example of Bounded Below Subset of Real Numbers ==
Let $I$ be the left half-open real interval defined as:
:$I := \hointl 0 1$
Then $I$ is bounded below.
$0$ is an lower bound of $I$.
== Sources ==
* : $1$: Review of some real analysis: $\S 1.1$: Real Numbers: Example $1.1.1 \ \text{(c)}$
Category:Examples of Bounded Below Sets of Real Numbers
Bounded Below Subset of Real Numbers/Examples/Open Interval from 0 to 1

== Example of Bounded Below Subset of Real Numbers ==
Let $I$ be the open real interval defined as:
:$I := \openint 0 1$
Then $I$ is bounded below by, for example, $0$, $-1$ and $-2$, of which $0$ is the infimum.
However, $I$ does not have a smallest element.
== Proof ==
 $x \in I$ is the smallest element of $I$.
Then $0 < \dfrac x 2 < x$ by Mediant is Between.
Thus $\dfrac x 2 \in I$ but $x > \dfrac x 2$.
So $x$ is not the smallest element of $I$ after all.
So by Proof by Contradiction it follows that $I$ has no smallest element.
== Sources ==
* : $\S 2$: Continuum Property: Exercise $\S 2.10 \ (4) \ \text{(i)}$
Category:Examples of Bounded Below Sets of Real Numbers
Bounded Below Subset of Real Numbers/Examples/Open Interval from 0 to Infinity

== Example of Bounded Below Subset of Real Numbers ==
The subset $T$ of the real numbers $\R$ defined as:
:$T = \set {x \in \R: x > 0}$
is bounded below, but unbounded above.
Let $H > 0$ in $T$ be proposed as an upper bound.
Then it is seen that $H + 1 \in T$ and so $H$ is not an upper bound at all.
Examples of lower bounds of $T$ are:
:$-27, 0$
Its infimum is $0$.
== Sources ==
* : Chapter $1$: The Real and Complex Number Systems: Real Numbers: $1.35$. Example $\text{(b)}$
* : $\S 2$: Continuum Property: $\S 2.4$ Examples: $\text {(iii)}$
Category:Examples of Bounded Below Sets of Real Numbers
Bounded Below Subset of Real Numbers/Examples/Open Interval from 1 to Infinity

== Example of Bounded Below Subset of Real Numbers ==
The subset $S$ of the real numbers $\R$ defined as:
:$S = \openint 1 \to$
is bounded below.
Examples of lower bounds of $S$ are:
:$-7, 1, \dfrac 1 2$
The set of all lower bounds of $S$ is:
:$\hointl {-\infty} 1$
== Sources ==
* : $\S 10$: The well-ordering principle
Category:Examples of Bounded Below Sets of Real Numbers
Bounded Below Subset of Real Numbers/Examples/Open Interval from 3 to Infinity

== Example of Bounded Below Subset of Real Numbers ==
Let $I$ be the unbounded open real interval defined as:
:$I := \openint 3 \to$
Then $I$ is bounded below by, for example, $3$, $2$ and $1$, of which the infimum is $3$.
However, $I$ does not have a smallest element.
== Sources ==
* : $\S 2$: Continuum Property: Exercise $\S 2.10 \ (4) \ \text{(iv)}$
Category:Examples of Bounded Below Sets of Real Numbers
Bounded Below Subset of Real Numbers/Examples/Real Numbers

== Example of Unbounded Below Subset of Real Numbers ==
Let $\R$ denote the set of real numbers.
$\R$ is not bounded below.
== Proof ==
Let $x \in \R$.
 $\R$ is bounded below.
Then there exists $x \in \R$ such that $x$ is a lower bound for $\R$.
But then:
:$x - 1 \in \R$ such that $x - 1 < x$
and so $x$ is not a lower bound for $\R$.
Hence by Proof by Contradiction $x$ is not a lower bound for $\R$.
== Sources ==
* : $1$: Review of some real analysis: $\S 1.1$: Real Numbers: Example $1.1.1 \ \text{(a)}$
Category:Examples of Bounded Below Sets of Real Numbers
Bounded Below Subset of Real Numbers/Examples/Strictly Negative Real Numbers

== Example of Unbounded Below Subset of Real Numbers ==
Let $I$ be the unbounded open real interval defined as:
:$I := \openint \gets 0$
Then $I$ is unbounded below.
== Sources ==
* : $1$: Review of some real analysis: $\S 1.1$: Real Numbers: Example $1.1.1 \ \text{(b)}$
Category:Examples of Bounded Below Sets of Real Numbers
Bounded Class is Set

== Theorem ==
Let $B$ be a class.
Let it be assumed that $B$ is a subclass of a basic universe $V$.
Let $B$ be bounded by a set $x$.
Then $B$ is itself a set.
== Proof ==
By definition, every element of $B$ is a subset of $x$.
Then every element of $B$ is an element of the power set $\powerset x$ of $x$.
Thus $B$ is a subclass of $\powerset x$.
By the Axiom of Powers, $\powerset x$ is a set.
That is, $\powerset x$ is an element of $V$.
As $V$ is a swelled class, then by definition, then every subclass of $\powerset x$ is an element of $V$.
That includes $B$.
That is, $B$ is an element of $V$.
Thus, by definition of $V$, $B$ is a set.
== Sources ==
* : Chapter $3$: The Natural Numbers: $\S 4$ A double induction principle and its applications
Category:Bounded Classes
Bounded Complex-Valued Continuous Functions on Topological Space form Unital C*-Algebra

== Theorem ==
Let $X$ be a topological space. 
Let $\map {\CC_b} {X; \C} := \map {\CC_b} X$ be the space of bounded continuous functions on $X$ valued in $\C$.
Let $\norm {\,\cdot\,}_\infty$ be the supremum norm on $\map {\CC_b} X$.
For each $f \in \map {\CC_b} X$, define $\overline f : X \to \C$ by:
:$\map {\overline f} x = \overline {\map f x}$
for each $x \in X$. 
Equip $\map {\CC_b} X$ with pointwise multiplication. 
Then $\struct {\map {\CC_b} X, \norm {\,\cdot\,}_\infty}$ is a unital $\text C^\ast$-algebra.
== Proof ==
From Bounded Continuous Functions on Topological Space form Banach Space, $\struct {\map {\CC_b} X, \norm {\,\cdot\,}_\infty}$ is a Banach space over $\C$.
From Product of Continuous Functions on Topological Ring is Continuous, $f g$ is continuous for each $f, g \in \map {\CC_b} X$.
Further, from Product of Bounded Mappings on Normed Algebra is Bounded, $f g$ is bounded.
For each $f, g \in \map {\CC_b} X$ and $x \in X$ we have:
:$\cmod {\map f x \map g x} = \cmod {\map f x} \cmod {\map g x}$
so that:
:$\norm {f g}_\infty \le \norm f_\infty \norm g_\infty$
Hence $\struct {\map {\CC_b} X, \norm {\,\cdot\,}_\infty}$ is a Banach algebra with pointwise multiplication.
Further, the function ${\mathbf 1}_{\map {\CC_b} X} : X \to \C$ defined by:
:$\map { {\mathbf 1}_{\map {\CC_b} X} } x = 1$
for each $x \in X$, satisfies:
:$\map f x \map { {\mathbf 1}_{\map {\CC_b} X} } x = \map { {\mathbf 1}_{\map {\CC_b} X} } x \map f x = 1$
From Constant Mapping is Continuous, ${\mathbf 1}_{\map {\CC_b} X}$ is continuous.
Further:
:$\cmod { \map { {\mathbf 1}_{\map {\CC_b} X} } x} = 1$
for each $x \in X$.
Hence ${\mathbf 1}_A \in \map {\CC_b} X$ with:
:$\norm { {\mathbf 1}_{\map {\CC_b} X} }_\infty = 1$
So ${\mathbf 1}_{\map {\CC_b} X}$ is an identity element for $\map {\CC_b} X$, making $\map {\CC_b} X$ a unital Banach algebra.
That $\overline {\, \cdot \,}$ is an involution on $\map {\CC_b} X$ follows from Complex Numbers form Unital C*-Algebra.
Finally, from Complex Numbers form Unital C*-Algebra we have:
:$\cmod {\map f x \map {\overline f} x} = \cmod {\map f x}^2$
Taking supremums we have:
:$\norm {f \overline f}_\infty = \norm f^2$ 
for each $f \in \map {\CC_b} X$.
So $\overline {\, \cdot \,}$ satisfies the $\text C^\ast$ identity.
Hence $\struct {\map {\CC_b} X, \norm {\,\cdot\,}_\infty}$ is a unital $\text C^\ast$-algebra.
Category:C*-Algebras
Bounded Function Continuous on Open Interval is Darboux Integrable

== Theorem ==
Let $f$ be a real function defined on an interval $\closedint a b$ such that $a < b$.
Let $f$ be continuous on $\openint a b$.
Let $f$ be bounded on $\closedint a b$.
Then $f$ is Darboux integrable on $\closedint a b$.
== Proof ==
By Condition for Darboux Integrability, it suffices to show that, for a given strictly positive $\epsilon$, a subdivision $S$ of $\closedint a b$ exists such that:
:$\map U S – \map L S < \epsilon$
where $\map U S$ and $\map L S$ are respectively the upper Darboux sum and lower Darboux sum of $f$ on $\closedint a b$ with respect to the subdivision $S$.
Since $f$ is bounded, a strictly positive bound $K$ exists for $f$ on $\closedint a b$.
Let a strictly positive $\epsilon$ be given, and choose a $\delta$ that satisfies:
:$0 < \delta < \min \left({\dfrac \epsilon {6 K}, \dfrac {b - a} 2}\right)$
We have that $f$ is continuous on $\openint a b$.
As $\delta > 0$, $\closedint {a + \delta} {b - \delta}$ is a subset of $\openint a b$.
Thus $f$ is continuous on the interval $\closedint {a + \delta} {b - \delta}$.
By Continuous Real Function is Darboux Integrable, $f$ is Darboux integrable on $\closedint {a + \delta} {b - \delta}$.
Since $f$ is Darboux integrable on $\closedint {a + \delta} {b - \delta}$, there exists a subdivision $S_\delta$ of $\closedint {a + \delta} {b - \delta}$ that satisfies:
:$\map U {S_\delta} – \map L {S_\delta} < \dfrac \epsilon 3$
where $\map U {S_\delta}$ and $\map L {S_\delta}$ are respectively the upper Darboux sum and lower Darboux sum of $f$ on $\closedint {a + \delta} {b - \delta}$ with respect to the subdivision $S_\delta$.
Define the following subdivision of $\closedint a b$:
:$S = S_\delta \cup \set {a, b}$
The upper Darboux sum of $f$ on $\closedint a b$ with respect to $S$ is per definition:
:$\map U S = M_a \delta + \map U {S_\delta} + M_b \delta$
where:
:$M_a$ is the supremum of $f$ on $\closedint a {a + \delta}$
:$M_b$ is the supremum of $f$ on $\closedint {b - \delta} b$.
$M_a$ and $M_b$ exist by the least upper bound property of the real numbers because $f$ is bounded on $\closedint a {a + \delta}$ and $\closedint {b - \delta} b$.
The lower Darboux sum of $f$ on $\closedint a b$ with respect to $S$ is per definition:
:$\map L S = m_a \delta + \map L {S_\delta} + m_b \delta$
where:
:$m_a$ is the infimum of $f$ on $\closedint a {a + \delta}$
:$m_b$ is the infimum of $f$ on $\closedint {b - \delta} b$
$m_a$ and $m_b$ exist by the Continuum Property, because $f$ is bounded on $\closedint a {a + \delta}$ and $\closedint {b - \delta} b$.
Define the sum:
Define the sum:
Therefore, $U'$ and $L'$ satisfy:
:$U' \ge \map U S$
:$L' \le \map L S$
From these two inequalities follows:
Hence:
:$\map U S – \map L S < \epsilon$
Category:Integral Calculus
Bounded Function is of Exponential Order Zero

== Theorem ==
Let $f: \hointr 0 \to \to \mathbb F$ be a function, where $\mathbb F \in \set {\R, \C}$.
Let $f$ be continuous everywhere on its domain, except possibly for some finite number of discontinuities of the first kind in every finite subinterval of $\hointr 0 \to$.
Let $f$ be bounded.
Then $f$ is of exponential order $0$.
== Proof ==
Let $U$ be an upper bound of $f$.
Let $L$ be a lower bound of $f$.
Let $K > \max \set {\size U, \size L}$.
Then: 
The result follows from the definition of exponential order with $M = 1$, $K = K$, and $a = 0$.
== Sources ==
* : Chapter $1$: The Laplace Transform: Functions of Exponential Order
Category:Exponential Order
Bounded Generalized Sum is Absolutely Convergent

== Theorem ==
Let $V$ be a Banach space.
Let $\family {v_i}_{i \mathop \in I}$ be an indexed family of elements of $V$.
Let $\FF$ denote the set of finite subsets of $I$.
Then:
:the generalized sum $\ds \sum_{i \mathop \in I} v_i$ is absolutely net convergent
:there exists $M \in \R_{\mathop \ge 0}$ such that for all $F \in \FF$:
::the summation $\ds \sum_{i \mathop \in F} \norm{v_i} \le M$
== Proof ==
=== Necessary Condition ===
Let the generalized sum $\ds \sum_{i \mathop \in I} v_i$ be absolutely net convergent.
Let: 
:$M = \ds \sum_{i \mathop \in I} \norm{v_i}$
:$\exists F_0 \in \FF : \sum_{i \mathop \in F_0} \norm{v_i} > M$
Let:
:$(1) \quad \epsilon \in \R_{\mathop > 0} : \epsilon < \paren{\ds \sum_{i \mathop \in F_0} \norm{v_i} } - M$
By definition of absolutely net convergence:
:$(2) \quad \exists F \in \FF : \forall E \in \FF : E \supseteq F \implies \ds \size{\sum_{i \mathop \in E} \norm{v_i} - M} < \epsilon$
Let:
:$E = F \cup F_0$
From Set is Subset of Union:
:$E \supseteq F$
We have:
We have a contradiction, so it follows that:
:$\forall F \in \FF : \ds \sum_{i \mathop \in F} \norm{v_i} \le M$
=== Sufficient Condition ===
Let there exist $M \in \R_{\mathop \ge 0}$ such that for all $F \in \FF$: 
:the summation $\ds \sum_{i \mathop \in F} \norm{v_i} \le M$.
Let $S = \set{\ds \sum_{i \mathop \in F} \norm{v_i} : F \in \FF}$
From Least Upper Bound Property, let:
:$c = \sup S$
From Characterizing Property of Supremum of Subset of Real Numbers:
:$(3)\quad \forall F \in \FF : \ds \sum_{i \mathop \in F} \norm{v_i} \le c$
and
:$(4)\quad \forall \epsilon \in \R_{> 0}: \exists F \in \FF : \ds \sum_{i \mathop \in F} \norm{v_i} > c - \epsilon$
Let $\epsilon \in \R_{> 0}$.
From $(4)$:
:$\exists F \in \FF : \ds \sum_{i \mathop \in F} \norm{v_i} > c - \epsilon$
Let $E \in \FF : E \supseteq F$.
We have:
Since $E$ was arbitrary, it follows:
:$\forall E \in \FF : E \supseteq F \leadsto \ds \sum_{i \mathop \in E} \norm{v_i} \in \openint {c - \epsilon} {c + \epsilon}$
Since $\epsilon$ was arbitrary, it follows:
:$\forall \epsilon \in \R_{> 0} : \exists F \in \FF : \forall E \in \FF : E \supseteq F \leadsto \ds \sum_{i \mathop \in E} \norm{v_i} \in \openint {c - \epsilon} {c + \epsilon}$
It follows that the generalized sum $\ds \sum_{i \mathop I} \norm{v_i}$ is convergent to $c$.
The result follows.
Category:Banach Spaces
Category:Generalized Sums
Bounded Lattice has Both Greatest Element and Smallest Element

== Theorem ==
Let $\struct {S, \vee, \wedge, \preceq}$ be a bounded lattice.
Then:
:$(1): \quad \struct {S, \preceq}$ has a smallest element, namely:
::::$\bot := \sup \O$
:$(2): \quad \struct {S, \preceq}$ has a greatest element, namely:
::::$\top := \inf \O$
== Proof ==
By definition of bounded lattice:
:$\bot = \sup \O$ and $\top = \inf \O$ exist
From Supremum of Empty Set is Smallest Element:
:$\bot = \sup \O$ $\bot$ is the smallest element of $\struct {S, \preceq}$
From Infimum of Empty Set is Greatest Element:
:$\top = \inf \O$ $\top$ is the greatest element of $\struct {S, \preceq}$
Category:Bounded Lattices