nyuuzyou/wikis huggingface dataset:
Absolute Value of Cut is Zero iff Cut is Zero

== Definition ==
Let $\alpha$ be a cut.
Let $\size \alpha$ denote the absolute value of $\alpha$.
Then:
:$\size \alpha = 0^*$ $\alpha = 0^*$
where $0^*$ denotes the rational cut associated with the (rational) number $0$.
== Proof ==
Let $\alpha 0^*$.
Then by definition $\size \alpha = \alpha = 0^*$.
Let $\alpha \ne 0^*$.
Then either:
:$\alpha > 0^*$ in which case $\size \alpha = \alpha > 0^*$
or:
:$\alpha < 0^*$ in which case $\size \alpha = -\alpha > 0^*$
In either case $\size \alpha \ne 0^*$.
The result follows.
== Sources ==
* : Chapter $1$: The Real and Complex Number Systems: Dedekind Cuts: $1.24$. Definition
Category:Cuts
Category:Absolute Value Function
Absolute Value of Divergent Infinite Product

== Theorem ==
Let $\struct {\mathbb K, \norm {\,\cdot\,} }$ be a valued field.
The infinite product $\ds \prod_{n \mathop = 1}^\infty a_n$ diverges to $0$ $\ds \prod_{n \mathop = 1}^\infty \norm {a_n}$ diverges to $0$.
== Proof ==
=== 1 implies 2 ===
==== Case 1 ====
Let there be infinitely many $n \in \N$ such that $a_n = 0$.
Then there are infinitely many $n \in \N$ such that $\size {a_n} = 0$.
Thus $\ds \prod_{n \mathop = 1}^\infty \norm {a_n}$ diverges to $0$.
==== Case 2 ====
Let $n_0 \in \N$ such that $a_n \ne 0$ for $n \ge n_0$.
Then $\norm {a_n} \ne 0$ for $n \ge n_0$.
By definition of divergent product:
:$\ds \lim_{N \mathop \to \infty} \prod_{n \mathop = n_0}^N a_n = 0$
By Absolute Value of Limit of Sequence:
:$\ds \lim_{N \mathop \to \infty} \prod_{n \mathop = n_0}^N \norm {a_n} = 0$
Thus $\ds \prod_{n \mathop = 1}^\infty \norm {a_n}$ diverges to $0$.
=== 2 implies 1 ===
==== Case 1 ====
Let there be infinitely many $n \in \N$ such that $\norm {a_n} = 0$.
Then there are infinitely many $n \in \N$ such that $a_n = 0$.
Thus $\ds \prod_{n \mathop = 1}^\infty a_n$ diverges to $0$.
==== Case 2 ====
Let $n_0 \in \N$ such that $\norm {a_n} \ne 0$ for $n \ge n_0$.
Then $a_n \ne 0$ for $n \ge n_0$.
By definition of divergent product:
:$\ds \lim_{N \mathop \to \infty} \prod_{n \mathop = n_0}^N \norm {a_n} = 0$
By Absolute Value of Limit of Sequence:
:$\ds \lim_{N \mathop \to \infty} \prod_{n \mathop = n_0}^N a_n = 0$
Thus $\ds \prod_{n \mathop = 1}^\infty a_n$ diverges to $0$.
== Also see ==
* Absolute Value of Infinite Product, for related results
Category:Infinite Products
Absolute Value of Even Power

== Theorem ==
Let $x \in \R$ be a real number.
Let $n \in \Z$ be an integer.
Then:
:$\size {x^{2 n} } = x^{2 n}$
== Proof ==
From Even Power is Non-Negative:
:$x^{2 n} \ge 0$
The result follows from the definition of absolute value.
Category:Absolute Value Function
Absolute Value of Function is Composite with Absolute Value Function

== Theorem ==
Let $S$ be a set.
Let $\R$ denote the real number line.
Let $f: S \to \R$ be real-valued function.
Let $\size f$ denote the absolute value of $f$, that is, $\size f$ is the mapping defined by:
:$\forall s \in S : \map {\size f} s = \size{\map f s}$
Then:
:$\size f = \size{\,\cdot\,} \circ f$
where:
:$\size{\,\cdot\,} : \R \to \R$ denotes the absoute value function
:$\size{\,\cdot\,} \circ f$ denotes the composite mapping of $\size{\,\cdot\,}$ and $f$
== Proof ==
We have:
By definition of equality of mappings:
:$\size f = \size{\,\cdot\,} \circ f$
Category:Real-Valued Functions
Absolute Value of Infinite Product

== Theorem ==
Let $\struct {\mathbb K, \norm{\,\cdot\,}}$ be a valued field.
Let $\sequence{a_n}$ be a sequence in $\mathbb K$.
=== Convergence ===
=== Absolute Convergence ===
=== Divergence to zero ===
== Also see ==
* Absolute Value of Uniformly Convergent Product
Category:Convergence
Category:Infinite Products
Absolute Value of Integer is Primitive Recursive

== Theorem ==
For every $n \in \N$, let $n$ code the integer $k_n$.
Let $a : \N \to \N$ be defined as:
:$\map a n = \size {k_n}$
Then, $a$ is primitive recursive.
== Proof ==
Let $a : \N \to \N$ be defined as:
:$\map a n = \map {\operatorname{quot}} {n + 1, 2}$
By:
* Constant Function is Primitive Recursive
* Quotient is Primitive Recursive
* Successor Function is Primitive Recursive
it follows that $a$ is primitive recursive.
By definition of code number for integer, either:
:$n = 2 k_n - 1$, with $k_n > 0$
or:
:$n = - 2 k_n$, with $k_n \le 0$
In the first case, with $k_n > 0$, we have:
:$k_n = \size {k_n}$
Then:
In the second case, with $k_n \le 0$, we have:
:$k_n = - \size {k_n}$
Then:
Category:Primitive Recursive Functions
Absolute Value of Integer is not less than Divisors

== Theorem ==
A (non-zero) integer is greater than or equal to its divisors in magnitude:
:$\forall c \in \Z_{\ne 0}: a \divides c \implies a \le \size a \le \size c$
=== Corollary ===
== Proof ==
Suppose $a \divides c$ for some $c \ne 0$.
From Negative of Absolute Value:
: $a \le \size a$
Then:
== Also see ==
* Non-Zero Integer has Finite Number of Divisors: a direct corollary
== Sources ==
* : $\S 11.2$: The division algorithm
* : Chapter $2$: Divisibility Theory in the Integers: $2.2$ The Greatest Common Divisor: Theorem $2 \text{-} 2 \ (6)$
* : Appendix $1$: Some Results from the Elementary Theory of Numbers: Corollary $2$
* : $1$ Divisibility: $1.1$ The Euclidean algorithm and unique factorization: $\text {(iii)}$
Category:Number Theory
Category:Divisors
Absolute Value of Integer is not less than Divisors/Corollary

== Corollary to Absolute Value of Integer is not less than Divisors ==
Let $a, b \in \Z_{>0}$ be (strictly) positive integers.
Let $a \divides b$.
Then:
:$a \le b$
== Proof ==
Follows directly from Absolute Value of Integer is not less than Divisors.
== Sources ==
* : $\S 0.1$. Arithmetic: Theorem $1$
* : Chapter $3$: The Integers: $\S 10$. Divisibility: Theorem $17 \ \text{(iii)}$
Category:Number Theory
Category:Divisors
Absolute Value of Martingale is Submartingale

== Theorem ==
Let $\struct {\Omega, \Sigma, \sequence {\FF_t}_{t \ge 0}, \Pr}$ be a continuous-time filtered probability space.
Let $\sequence {X_t}_{t \ge 0}$ be a $\sequence {\FF_t}_{t \ge 0}$-martingale.
Then $\sequence {\size {X_t} }_{t \ge 0}$ is a $\sequence {\FF_t}_{t \ge 0}$-submartingale.
== Proof ==
From Characterization of Integrable Functions:
:$\size {X_t}$ is integrable for each $t \in \hointr 0 \infty$. 
From Absolute Value Function is Convex, $x \mapsto \size x$ is a convex function.
From Martingale Composed with Convex Function is Submartingale, we have: 
:$\sequence {\size {X_t} }_{t \ge 0}$ is a $\sequence {\FF_t}_{t \ge 0}$-submartingale.
== Sources ==
* : $3.3$: Continuous Time Martingales and Supermartingales
Category:Martingales
Category:Submartingales
Absolute Value of Measurable Function is Measurable

== Theorem ==
Let $\struct {X, \Sigma}$ be a measurable space. 
Let $f : X \to \overline \R$ be a $\Sigma$-measurable function.
Then: 
:$\size f$ is a $\Sigma$-measurable function.
== Proof ==
From Characterization of Measurable Functions, it suffices to show that for each real number $t \in \R$, we have: 
:$\set {x \in X : \size {\map f x} \le t} \in \Sigma$
If $t < 0$, we have: 
:$\set {x \in X : \size {\map f x} \le t} = \O$
So, from Properties of Algebras of Sets, we have: 
:$\set {x \in X : \size {\map f x} \le t} \in \Sigma$
if $t < 0$. 
If $t \ge 0$, we can write: 
Since $f$ is $\Sigma$-measurable, we have that both: 
:$\set {x \in X : -t \le \map f x} \in \Sigma$
and:
:$\set {x \in X : \map f x \le t} \in \Sigma$
from Characterization of Measurable Functions.
From Properties of Algebras of Sets, the intersection of any two sets in $\Sigma$ is contained in $\Sigma$. 
So:
:$\set {x \in X : -t \le \map f x} \cap \set {x \in X : \map f x \le t} \in \Sigma$ 
if $t \ge 0$. 
That is: 
:$\set {x \in X : \size {\map f x} \le t} \in \Sigma$
if $t \ge 0$.
So:
:$\set {x \in X : \size {\map f x} \le t} \in \Sigma$
for all $t \in \R$.
Category:Measurable Functions
Absolute Value of Negative

== Theorem ==
Let $x \in \R$ be a real number.
Then:
:$\size x = \size {-x}$
where $\size x$ denotes the absolute value of $x$.
== Proof ==
Let $x \ge 0$.
By definition of absolute value:
:$\size x = x$
We have that:
:$-x < 0$
and so by definition of absolute value:
:$\size {-x} = -\paren {-x} = x$
Now let $x < 0$.
By definition of absolute value:
:$\size x = -x$
We have that:
:$-x > 0$
and so by definition of absolute value:
:$\size {-x} = -x$
In both cases it is seen that:
:$\size x = \size {-x}$
Category:Absolute Value Function
Absolute Value of Pearson Correlation Coefficient is Less Than or Equal to 1

== Theorem ==
Let $X$ and $Y$ be random variables. 
Let the variances of $X$ and $Y$ exist and be finite. 
Then: 
:$\size {\map \rho {X, Y} } \le 1$
where $\map \rho {X, Y}$ denotes the Pearson correlation coefficient of $X$ and $Y$. 
== Proof ==
So:
:$\size {\map \rho {X, Y} } \le 1$
== Sources == 
* 
* 
* : $4.6$: Covariance and Correlation: Theorem $4.6.3$
Category:Pearson Correlation Coefficient
Absolute Value of Power

== Theorem ==
Let $x$, $y$ be real numbers.
Let $x^y$, $x$ to the power of $y$, be real.
Then:
:$\size {x^y} = \size x^y$
== Proof ==
If $x = 0$, the theorem clearly holds, by the definition of powers of zero.
Suppose $x \ne 0$.
We use the interpretation of real numbers as wholly real complex numbers.
Likewise we interpret the absolute value of $x$ as the modulus of $x$.
Then $x$ can be expressed in polar form:
:$x = r e^{i\theta}$
where $r = \size x$ and $\theta$ is an argument of $x$.
Then:
Category:Absolute Value Function
Absolute Value of Real-Valued Random Variable is Real-Valued Random Variable

== Theorem ==
Let $\struct {\Omega, \Sigma, \Pr}$ be a probability space. 
Let $X$ be a real-valued random variable. 
Then $\size X$ is a real-valued random variable.
 
== Proof ==
Since $X$ is a real-valued random variable, $X$ is $\Sigma$-measurable.
From Absolute Value of Measurable Function is Measurable, $\size X$ is $\Sigma$-measurable.
So $\size X$ is a real-valued random variable.
Category:Random Variables
Absolute Value of Signed Measure Bounded Above by Variation

== Theorem ==
Let $\struct {X, \Sigma}$ be a measurable space. 
Let $\mu$ be a signed measure on $\struct {X, \Sigma}$. 
Let $\size \mu$ be the variation of $\mu$. 
Then: 
:$\size {\map \mu A} \le \map {\size \mu} A$
for each $A \in \Sigma$. 
== Proof ==
Let $\tuple {\mu^+, \mu^-}$ be the Jordan decomposition of $\mu$.
Then: 
:$\mu = \mu^+ - \mu^-$
and:
:$\size \mu = \mu^+ + \mu^-$
We have: 
== Sources ==
* : $4.1$
Category:Signed Measures
Absolute Value of Simple Function is Simple Function

== Theorem ==
Let $\struct {X, \Sigma}$ be a measurable space.
Let $f: X \to \R$ be a simple function.
Then $\size f: X \to \R$, the absolute value of $f$, is also a simple function.
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : $8.7 \ \text{(vi)}$
Category:Simple Functions
Category:Absolute Value of Simple Function is Simple Function
Absolute Value of Simple Function is Simple Function/Proof 1

== Theorem ==
== Proof ==
By Sum of Positive and Negative Parts, we have:
:$\size f = f^+ + f^-$
We also have that Positive Part of Simple Function is Simple Function and Negative Part of Simple Function is Simple Function.
Hence $\size f$ is a pointwise sum of simple functions.
The result follows from Pointwise Sum of Simple Functions is Simple Function.
Category:Absolute Value of Simple Function is Simple Function
Absolute Value of Simple Function is Simple Function/Proof 2

== Theorem ==
== Proof ==
By Simple Function has Standard Representation, $f$ has a standard representation, say:
:$(1): \quad f = \ds \sum_{k \mathop = 0}^n a_k \chi_{E_k}$
Then, for all $x \in X$:
:$\map {\size f} x = \ds \size {\sum_{k \mathop = 0}^n a_k \map {\chi_{E_k} } x}$
by definition of pointwise absolute value.
The fact that $(1)$ forms a standard representation ensures that for every $x \in X$, precisely one $k$ has $x \in E_k$.
Now suppose that $x \in E_l$. 
Then $\map {\chi_{E_l} } x = 0$ $k \ne l$ by definition of characteristic function.
It follows that $\map {\size f} x = \size {a_l \cdot 1} = \size {a_l}$.
Now define $g: X \to \R$ by:
:$\map g x := \ds \sum_{k \mathop = 0}^n \size {a_k} \map {\chi_{E_k} } x$
By construction, $g$ is a simple function, and for $x \in E_l$:
:$\map g x = \size {a_l}$
Thus, since every $x$ is in $E_l$ for precisely one $l$, it has been shown that $g = \size f$.
As $g$ was a simple function, it follows that $\size f$ is also a simple function.
Category:Absolute Value of Simple Function is Simple Function
Absolute Value of Trigonometric Function

== Theorem ==
Let $\theta$ be an angle embedded in a Cartesian plane.
Let $\theta$ be such that the vertex of $\theta$ is located at the origin while one arm is coincident with the $x$-axis.
Let $\phi$ be the acute angle made by the other arm with the $x$-axis.
Let $f: \R \to \R$ be a trigonometric function.
Then $\size {\map f \theta}$ is equal to $\map f \phi$.
That is, a trigonometric function of an angle can be calculated by working out its quadrant to determine its sign, then using the equivalent value of what it is between $0$ and $90 \degrees$ to get its value.
== Proof ==
== Sources ==
* : $\text V$. Trigonometry: Angles larger than $90 \degrees$
Category:Trigonometric Functions
Absolute Value of Uniformly Convergent Product

== Theorem ==
Let $X$ be a compact topological space.
Let $\struct {\mathbb K, \norm {\,\cdot\,} }$ be a valued field.
Let $\sequence {f_n}$ be a sequence of continuous mappings $f_n: X \to \mathbb K$.
Let the infinite product $\ds \prod_{n \mathop = 1}^\infty f_n$ converge uniformly to $f$.
Then $\ds \prod_{n \mathop = 1}^\infty \norm {f_n}$ converges uniformly to $\norm f$.
== Proof ==
== Also see ==
* Absolute Value of Infinite Product
Category:Uniform Convergence
Category:Infinite Products
Absolute Value on Ordered Integral Domain is Strictly Positive except when Zero

== Theorem ==
Let $\struct {D, +, \times, \le}$ be an ordered integral domain.
For all $a \in D$, let $\size a$ denote the absolute value of $a$.
Then $\size a$ is (strictly) positive except when $a = 0$.
== Proof ==
Let $P$ be the (strict) positivity property on $D$.
Let $<$ be the (strict) total ordering defined on $D$ as:
:$a < b \iff a \le b \land a \ne b$
From the trichotomy law, exactly one of three possibilities holds for any $ a \in D$:
$(1): \quad \map P a$:
In this case $0 < a$ and so $\size a = a$.
So:
:$\map P a \implies \map P {\size a}$
$(2): \quad \map P {-a}$:
In this case $a < 0$ and so $\size a = -a$.
So:
:$\map P {-a} \implies \map P {\size a}$
$(3): \quad a = 0$:
In this case $\size a = a$.
So:
:$a = 0 \implies \size a = 0$.
Hence the result.
== Also see ==
* Absolute Value is Bounded Below by Zero
== Sources ==
* : Chapter $2$: Ordered and Well-Ordered Integral Domains: $\S 7$. Order
Category:Absolute Value Function
Absolutely Continuous Random Variable is Continuous

== Theorem ==
Let $\struct {\Omega, \Sigma, \Pr}$ be a probability space.
Let $X$ be an absolutely continuous random variable on $\struct {\Omega, \Sigma, \Pr}$. 
Then $X$ is a continuous random variable.
== Proof ==
Let $F_X$ be the cumulative distribution function for $X$. 
From the definition of an absolutely continuous random variable, we have: 
:$F_X$ is absolutely continuous.
From Absolutely Continuous Real Function is Continuous, we then have: 
:$F_X$ is continuous.
Then, from the definition of a continuous random variable, we have: 
:$X$ is a continuous random variable.
Category:Absolutely Continuous Random Variables
Category:Continuous Random Variables
Absolutely Continuous Real Function is Continuous

== Theorem ==
Let $I \subseteq \R$ be a real interval.
Let $f: I \to \R$ be an absolutely continuous real function.
Then $f$ is continuous.
== Proof ==
From Absolutely Continuous Real Function is Uniformly Continuous:
:$f$ is uniformly continuous.
Therefore, from Uniformly Continuous Real Function is Continuous:
:$f$ is continuous.
Category:Absolutely Continuous Real Functions
Category:Continuous Real Functions
Absolutely Continuous Real Function is Uniformly Continuous

== Theorem ==
Let $I \subseteq \R$ be a real interval.
Let $f: I \to \R$ be an absolutely continuous real function.
Then $f$ is uniformly continuous.
== Proof ==
Let $\epsilon$ be a positive real number.
Since $f$ is absolutely continuous, there exists real $\delta > 0$ such that for all collections of pairwise disjoint closed real intervals $\closedint {a_1} {b_1}, \dotsc, \closedint {a_n} {b_n} \subseteq I$ with: 
:$\ds \sum_{i \mathop = 1}^n \paren {b_i - a_i} < \delta$
we have: 
:$\ds \sum_{i \mathop = 1}^n \size {\map f {b_i} - \map f {a_i} } < \epsilon$
Consider specifically the case $n = 1$.
Suppose that $a \le x \le y \le b$ and:
:$0 \le y - x < \delta$
so that: 
:$\size {x - y} < \delta$
From the absolute continuity of $f$ we then have:
:$\size {\map f y - \map f x} < \epsilon$
so:
:$\size {\map f x - \map f y} < \epsilon$
Now suppose that $y < x$ and: 
:$\size {x - y} < \delta$
then: 
:$0 < x - y < \delta$ 
and: 
:$\size {\map f x - \map f y} < \epsilon$ 
So, in fact, for all $x, y \in \closedint a b$ with:
:$\size {x - y} < \delta$
we have:
:$\size {\map f x - \map f y} < \epsilon$
Since $\epsilon$ was arbitrary:
:$f$ is uniformly continuous.
Category:Absolutely Continuous Real Functions
Category:Uniformly Continuous Real Functions
Absolutely Convergent Complex Series/Examples/(z over (1-z))^n

== Example of Absolutely Convergent Complex Series ==
The complex series defined as:
:$\ds S = \sum_{n \mathop = 1}^\infty \paren {\dfrac z {1 - z} }^n$
is absolutely convergent, provided $\Re \paren z < \dfrac 1 2$.
== Proof ==
Suppose $S$ is absolutely convergent.
Then $\ds \sum_{n \mathop = 1}^\infty \cmod {\dfrac z {1 - z} }^n$ is convergent.
By Terms in Convergent Series Converge to Zero, this means that:
:$\lim_{n \mathop \to \infty} \cmod {\dfrac z {1 - z} }^n \to 0$
which means in turn that:
It remains to be shown that $S' := \ds \sum_{n \mathop = 1}^\infty \cmod {\dfrac z {1 - z} }^n$ is in fact a convergent series when $z < \dfrac 1 2$.
When $z < \dfrac 1 2$, we have that $\cmod {\dfrac z {1 - z} } < 1$, from above.
Let $w = \cmod {\dfrac z {1 - z} }$.
Then we have that:
The result follows.
== Sources ==
* : $\S 4.3$. Series: Example $\text{(iii)}$
Category:Examples of Absolutely Convergent Complex Series
Absolutely Convergent Generalized Sum Converges

== Theorem ==
Let $V$ be a Banach space.
Let $\norm {\, \cdot \,}$ denote the norm on $V$.
Let $d$ denote the corresponding induced metric.
Let $\family {v_i}_{i \mathop \in I}$ be an indexed subset of $V$ such that the generalized sum $\ds \sum \set {v_i: i \in I }$ converges absolutely.
Then the generalized sum $\ds \sum \set {v_i: i \in I}$ converges.
== Proof ==
The proof proceeds in two stages:
:$(1): \quad$ Finding a candidate $v \in V$ where the sum might converge to
:$(2): \quad$ Showing that the candidate is indeed sought limit.
That $\ds \sum \set {v_i: i \mathop \in I}$ converges absolutely means that $\ds \sum \set {\norm {v_i}: i \mathop \in I}$ converges.
Now, for all $n \in \N$, let $F_n \subseteq I$ be finite such that:
:$\ds \sum_{i \mathop \in G} \norm {v_i} > \sum \set {\norm {v_i}: i \mathop \in I} - 2^{-n}$
for all finite $G$ with $F_n \subseteq G \subseteq I$
It may be arranged that $n \ge m \implies F_m \subseteq F_n$ by passing over to $\ds F'_n = \bigcup_{m \mathop = 1}^n F_m$ if necessary.
Define:
:$\ds v_n = \sum_{i \mathop \in F_n} v_i$
Next, we need to show that the sequence $\sequence {v_n}_{n \mathop \in \N}$ is Cauchy.
So let $\epsilon > 0$, and let $N \in \N$ be such that:
:$2^{-N} < \epsilon$.
Then for $m \ge n \ge N$, we have:
Now to estimate this last quantity, observe:
Finally, by the defining property of $N$, as $n \ge N$:
:$2^{-n} < 2^{-N} < \epsilon$
Combining all of these estimates leads to:
:$\map d {v_m, v_n} < \epsilon$
It follows that $\sequence {v_n}_{n \mathop \in \N}$ is a Cauchy sequence.
As $V$ is a Banach space:
:$\ds \exists v \in V: \lim_{n \mathop \to \infty} v_n = v$
Having identified a candidate $v$ for the sum $\ds \sum \set {v_i: i \in I}$ to converge to, it remains to verify that this is indeed the case.
According to the definition of considered sum, the convergence is convergence of a net.
Next, Metric Induces Topology ensures that we can limit the choice of opens $U$ containing $v$ to open balls centered at $v$.
Now let $\epsilon > 0$ be arbitrary.
We want to find a finite $F \subseteq I$ such that:
:$\map d {\ds \sum_{i \mathop \in G} v_i, v} < \epsilon$
for all finite $G$ with $F \subseteq G \subseteq I$.
Now let $N \in \N$ such that:
:$\forall n \ge N: \map d {v_n, v} < \dfrac \epsilon 2$
with the $v_n$ as above.
By taking a larger $N$ if necessary, we need to ensure that $2^{-N} < \dfrac \epsilon 2$ holds as well.
We need to verify that the set $F_N$ defined above has sought properties.
So let $G$ be finite, such that:
:$F_N \subseteq G \subseteq I$
Then:
For the first of these terms, observe:
Note that:
:$2^{-N} < \dfrac \epsilon 2$
Combine these inequalities leads to:
:$\ds \map d {\sum_{i \mathop \in G} v_i, v} < \frac \epsilon 2 + \frac \epsilon 2 = \epsilon$
By definition of convergence of a net, it follows that:
:$\ds \sum \set {v_i: i \in I} = v$
== Also see ==
* Absolutely Convergent Series in Normed Vector Space is Convergent iff Space is Banach
Category:Absolute Convergence
Category:Generalized Sums
Category:Banach Spaces
Absolutely Convergent Generalized Sum Converges to Supremum

== 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 $I$.
Let the generalized sum $\ds \sum_{i \mathop \in I} v_i$ converge absolutely to $c \in \R$.
Then:
:$c = \sup \set{\ds \sum_{i \mathop \in F} \norm{v_i} : F \in \FF}$
== Proof ==
:
:$\exists E \in \FF : \ds \sum_{i \mathop \in E} \norm{v_i} > c$
Let:
:$0 < \epsilon < \ds \sum_{i \mathop \in F} \norm{v_i} - c$
Let $F \in \FF$.
Let $E' = F \cup E$.
We have:
Since $F$ was arbitrary, it follows:
:$\forall F \in \FF : \exists E' \in \FF : E' \supseteq F$ and $\size{\ds \sum_{i \mathop \in E'} \norm{v_i} - c} > \epsilon$
This contradicts the hypothesis that $\ds \sum_{i \mathop \in I} v_i$ converges absolutely to $c \in \R$.
Hence:
:$\forall E \in \FF : \ds \sum_{i \mathop \in E} \norm{v_i} \le c$
By definition of absolutely net convergence:
:$\forall \epsilon \in \R_{\mathop > 0}: \exists F \in \FF : \forall E \in \FF : E \supseteq F \leadsto \ds \sum_{i \mathop \in E} \norm{v_i} \in \hointl {c - \epsilon} c$
From Characterizing Property of Supremum of Subset of Real Numbers:
:$c = \sup \set{\ds \sum_{i \mathop \in F} \norm{v_i} : F \in \FF}$
Category:Banach Spaces
Category:Generalized Sums