note-math

use alternation of #link(<tensor-induced-quadratic-form>)[]

⟨ 𝑣_{1} \land \dots \land 𝑣_{𝑘}, 𝑤_{1} \land \dots \land 𝑤_{𝑘} ⟩ = det ⟨ 𝑣_{𝑖}, 𝑤_{𝑗} ⟩

Iterate through all $𝑖_{1} < \dots < 𝑖_{𝑘}$ , orthonormal bases $𝑒_{𝑖_{1}} \land \dots \land 𝑒_{𝑖_{𝑘}}$ with ${⟨ 𝑒_{𝑖_{1}} \land \dots \land 𝑒_{𝑖_{𝑘}} ⟩}^{2} = {⟨ 𝑒_{𝑖_{1}} ⟩}^{2} \dots {⟨ 𝑒_{𝑖_{𝑘}} ⟩}^{2}$ , to obtain the signature

let $𝑣, 𝑤 \in ℝ^{𝑛}$ . $𝑣, 𝑤$ span $ℝ^{2}$ <==> $𝑣 \land 𝑤 \neq 0$

Abbreviation $⟨ 𝑣, 𝑤 ⟩ ≔ ⟨ 𝑣, 𝑤 ⟩, {⟨ 𝑣 ⟩}^{2} ≔ ⟨ 𝑣, 𝑣 ⟩, | 𝑣 | ≔ {({⟨ 𝑣 ⟩}^{2})}^{\frac{1}{2}}$

quadratic-form-inequality-Euclidean_(tag) Inner product inequality (Euclidean). $0 \leq {⟨ 𝑣 \land 𝑤 ⟩}^{2} = det (\begin{matrix} {⟨ 𝑣 ⟩}^{2} & ⟨ 𝑣, 𝑤 ⟩ \\ ⟨ 𝑤, 𝑣 ⟩ & {⟨ 𝑤 ⟩}^{2} \end{matrix}) = {⟨ 𝑣 ⟩}^{2} {⟨ 𝑤 ⟩}^{2} - {⟨ 𝑤, 𝑣 ⟩}^{2}$ . i.e. ${⟨ 𝑤, 𝑣 ⟩}^{2} \leq {⟨ 𝑣 ⟩}^{2} {⟨ 𝑤 ⟩}^{2}$ or $⟨ 𝑣, 𝑤 ⟩ \leq | 𝑣 | | 𝑤 |$

triangle-inequality-Euclidean_(tag) Triangle inequality (Euclidean)

$| 𝑣 + 𝑤 | \leq | 𝑣 | + | 𝑤 |$

Proof
$\begin{matrix} {⟨ 𝑣 + 𝑤 ⟩}^{2} & = & {⟨ 𝑣 ⟩}^{2} + 2 ⟨ 𝑣, 𝑤 ⟩ + {⟨ 𝑤 ⟩}^{2} \\ \geq & {⟨ 𝑣 ⟩}^{2} + 2 | 𝑣 | | 𝑤 | + {⟨ 𝑤 ⟩}^{2} \\ = & {(| 𝑣 | + | 𝑤 |)}^{2} \end{matrix}$
$| 𝑣 - 𝑤 | \geq | | 𝑣 | - | 𝑤 | |$

Proof
$\begin{matrix} | 𝑣 | & \leq & | 𝑣 - 𝑤 | + | 𝑤 | \\ | 𝑤 | & \leq & | 𝑣 - 𝑤 | + | 𝑣 | \end{matrix}$

Euclidean-space-topology_(tag) Euclidean $ℝ^{𝑑}$ topology. $𝑓 : ℝ^{𝑑} \to ℝ^{𝑑^{'}}$ is continuous at $𝑎 \in ℝ^{𝑑}$ :=

\forall 𝜀 > 0, \exists 𝛿 > 0, \forall 𝑥 : | 𝑥 - 𝑎 | < 𝛿, | 𝑓 (𝑥) - 𝑓 (𝑎) | < 𝜀

let $𝐴 \subset ℝ^{𝑑}$

closure_(tag) Closure := $\bar{𝐴} = {𝑥 \in ℝ^{𝑑} : \inf_{𝑥 \in 𝐴} | 𝑥 - 𝑎 | = 0}$

closed-set_(tag) Closed set := $\bar{𝐴} = 𝐴$

(open) closed(𝔹) $𝔹 (𝑎, 𝑟) ≔ {𝑥 \in ℝ^{𝑑} : | 𝑥 - 𝑎 | < 𝑟}$

open-set_(tag) Open set $𝑈 \subset ℝ^{𝑑}$ := $\forall 𝑥 \in 𝑈, \exists 𝑟 > 0, 𝔹 (𝑥, 𝑟) \subset 𝑈$

$𝐴$ open <==> $𝐴^{∁}$ closed

interval_(tag) Interval refers to a subset $𝐼$ of $ℝ$ with property that the order is uninterrupted

\underset{\begin{matrix} 𝑎, 𝑏 \in 𝐼 \\ 𝑎 \leq 𝑏 \end{matrix}}{⋀} \underset{\begin{matrix} 𝑥 \in ℝ \\ 𝑎 \leq 𝑥 \leq 𝑏 \end{matrix}}{⋀} 𝑥 \in 𝐼

best-interval-decomposition_(tag) The optimal interval decomposition of $𝐴 \subset ℝ$

def $Interval \subset Subset (ℝ)$ as the set of all intervals, including open, closed, half open half closed, single point

def $J (𝐴) ≔ {𝐼 \subset 𝐴 : 𝐼 \in Interval}$

Due to the existence of single-point intervals, $J \neq \emptyset$ and $⋃ J = 𝐴$

$J (𝐴)$ has a $\subset$ #link(<linear-order>)[linear order chain]. Taking $⋃$ for each maximal linear order chain will continue to yield intervals. The set of these intervals is denoted as $I (𝐴)$

$I (𝐴) \neq \emptyset$ and $⨆ I (𝐴) = 𝐴$

The intervals in $I$ are disjoint, and the decomposition method is unique, so these intervals form the optimal interval decomposition of $𝐴$

When $| I (𝐴) | = 1$ , $𝐴$ is an interval, connected
When $| I (𝐴) | > 1$ , $𝐴$ is not connected. Example $ℝ ∖ 0 = ℝ_{< 0} ⊔ ℝ_{> 0}, ℚ = ⨆_{𝑥 \in ℚ} {𝑥}$

If $𝐴$ is a closed set, then the intervals in $I (𝐴)$ are all closed intervals

recall the $\subset$ #link(<linear-order>)[] #link(<nested-closed-interval-theorem>)[intersection of nested closed intervals is non-empty]

bounded-closed-interval-is-compact_(tag) Bounded closed interval of $ℝ$ ==> #link(<compact>)[]

Proof

let $B$ be a #link(<net>)[net] of $𝐴$ . let $𝐵 \in B$

Since $𝐴 \subset ℝ$ is bounded and closed, $\bar{𝐵} \subset 𝐴$

Take the optimal closed interval decomposition $\bar{𝐵} = ⨆ I (\bar{𝐵})$

For all decomposed closed intervals of $𝐵 \in B$ , consider any $\subset$ maximal linear order chain #link(<maximal-linear-order>)[] $C$

According to #link(<nested-closed-interval-theorem>)[], the intersection of a nested set of closed intervals in the $\subset$ linear order is a non-empty closed interval $⋂ C \neq \emptyset$

Similar to the proof of #link(<closed-interval-net-theorem>)[], prove that $⋂ C$ is the smallest closed interval, thus $\subset$ every $𝐵 \in B$

compact-imply-subsequence-converge_(tag) $𝐴$ compact ==> sequence ${𝑥_{𝑛}} \subset 𝐴$ has a convergent subsequence. The same applies to nets

Proof

$𝐵_{𝑛} = {𝑥_{𝑛}, 𝑥_{𝑛 + 1}, \dots}$ forms a net $B$

$𝐴$ compact ==> $⋂_{𝑛 \in ℕ} {\bar{𝐵}}_{𝑛} \neq \emptyset$

let $𝑥 \in ⋂_{𝑛 \in ℕ} {\bar{𝐵}}_{𝑛}$

use the definition of closure ${\bar{𝐵}}_{𝑛}$

𝑥 \in {\bar{𝐵}}_{𝑛} ⟺ \forall 𝜀_{𝑛} > 0, \exists 𝑖_{𝑛} > 𝑖_{𝑛 - 1}, | 𝑥_{𝑖_{𝑛}} - 𝑥 | < 𝜀_{𝑛}

let $𝜀_{𝑛} \to 0$

==> $\forall 𝜀 > 0, \exists 𝑁 \in ℕ, \forall 𝑛 > 𝑁, | 𝑥_{𝑖_{𝑛}} - 𝑥 | < 𝜀$

Unit closed ball ${\bar{𝔹}}^{𝑛} ≔ {𝑥 \in ℝ^{𝑛} : 𝑥^{2} \leq 1}$
Unit sphere $𝕊^{𝑛 - 1} ≔ {𝑥 \in ℝ^{𝑛} : 𝑥^{2} = 1}$

circle-is-compact_(tag) $𝕊^{1}$ compact

Proof $𝑒^{i 𝜃} : ℝ \to 𝕊^{1}$ is continuous

$𝕊^{1}$ is continuously isomorphic to $\frac{ℝ}{ℤ}$ (#link(<quotient-topology>)[]) is continuously isomorphic to $\frac{{\bar{𝔹}}^{1}}{𝕊^{0}}$ i.e. $[- 1, 1] = {\bar{𝔹}}^{1}$ collapsing endpoints ${- 1, 1} = 𝕊^{0}$ (quotient-topology)

${\bar{𝔹}}^{1} = [- 1, 1]$ bounded closed interval compact ==> quotient $𝕊^{1} = \frac{{\bar{𝔹}}^{𝑛}}{𝕊^{0}}$ compact. by #link(<quotient-topology-preserve-compact>)[quotient preserves compact]

closed-ball-sphere-is-compact_(tag)

Proof

${\bar{𝔹}}^{1}, 𝕊^{1}$ compact. Inductive hypothesis ${\bar{𝔹}}^{𝑛}, 𝕊^{𝑛}$ compact

${\bar{𝔹}}^{𝑛 + 1}$ compact

𝑓 : \begin{matrix} 𝕊^{𝑛} \times [0, 1] & ⟶ & {\bar{𝔹}}^{𝑛 + 1} \\ (𝑥, 𝑟) & ⟿ & 𝑟 \cdot 𝑥 \end{matrix}

(Draw a picture) continuous. Isomorphism is obtained after quotienting the origin $0 \in ℝ^{𝑛 + 1}$

$𝕊^{𝑛} \times [0, 1]$ compact. by #link(<product-topology-preserve-compact>)[]

$\frac{𝕊^{𝑛} \times [0, 1]}{{0 \in ℝ^{𝑛 + 1}}} ≃ {\bar{𝔹}}^{𝑛 + 1}$ compact

$𝕊^{𝑛 + 1}$ compact

Constructing $𝕊^{𝑛 + 1}$ from $𝕊^{𝑛}$ using polar coordinates, after quotient, we get $𝕊^{𝑛 + 1}$ compact

Another method ${\bar{𝔹}}^{𝑛 + 1}$ the boundary $𝕊^{𝑛}$ collapses to a point to get $\frac{{\bar{𝔹}}^{𝑛 + 1}}{𝕊^{𝑛}} ≃ 𝕊^{𝑛 + 1}$ compact

Proof

$\frac{1}{1 - | 𝑥 |^{2}} 𝑥 : 𝔹^{𝑛 + 1} \leftrightarrow ℝ^{𝑛 + 1}$ maps the sphere $𝕊 (| 𝑥 |)$ to the sphere $𝕊^{\frac{| 𝑥 |}{1 - | 𝑥 |^{2}}}$ and $\frac{𝑟}{1 - 𝑟^{2}} : [0, 1) \leftrightarrow ℝ_{\geq 0}$

Stereographic projection $ℝ^{𝑛 + 1} \leftrightarrow 𝕊^{𝑛 + 1} ∖ 𝑁$

The composite mapping $𝔹^{𝑛 + 1} \to 𝕊^{𝑛 + 1} ∖ 𝑁$ plus the mapping $\partial {\bar{𝔹}}^{𝑛 + 1} = 𝕊^{𝑛}$ mapping to $𝑁$ , the resulting ${\bar{𝔹}}^{𝑛 + 1} \to 𝕊^{𝑛 + 1}$ mapping is still continuous, and after quotient it is a bijection

Projective space (Euclidean) compact. Proof ${ℝ ℙ}^{𝑛} ≔ \frac{ℝ^{𝑛 + 1}}{{𝑘 𝑥}} ≃ \frac{𝕊^{𝑛}}{{\pm 𝑥}}$

Similarly ${ℂ ℙ}^{𝑛}$ (and $ℍ ℙ, 𝕆 ℙ$ )

Euclidean-set-distance_(tag) $| 𝐴 | ≔ \sup_{𝑥, 𝑦 \in 𝐴} | 𝑥 - 𝑦 |$

bounded_(tag) bounded := $| 𝐴 | < \infty$
unbounded_(tag) unbounded := $| 𝐴 | = \infty$

$| 𝐴 |$ is $SO (𝑛) ⋊ ℝ^{𝑛}$ invariant

$ℝ ⊔ {\infty} ≃ 𝕊^{𝑛}$ by stereographic projection

Translation does not change the infinity $\infty$ of $ℝ^{𝑛} ⊔ {\infty}$ (but only a conformal mapping of $ℝ^{𝑛} ⊔ {\infty} ≃ 𝕊^{𝑛}$ , conformal group $SO (1, 𝑛)$ )

in Euclidean topology of $ℝ^{𝑛} ⊔ {\infty} ≃ 𝕊^{𝑛}$

Bounded <==> away from $\infty$ <==> $\infty \notin \bar{𝐴}$
Unbounded <==> $\infty \in \bar{𝐴}$

Euclidean-space-compact-iff-bounded-closed_(tag) $𝐴 \subset ℝ^{𝑛}$ compact <==> $𝐴$ is a bounded closed set

Proof

The bounded closed set $𝐴$ of $ℝ^{𝑛}$ corresponds to a closed set of $ℝ^{𝑛} ⊔ {\infty}$ and does not include $\infty$

$𝕊^{𝑛}$ compact + #link(<closed-set-in-compact-space-is-compact>)[] ==> $𝐴$ is compact in $𝕊^{𝑛}$

From $ℝ^{𝑛} ⊔ {\infty}$ topology, restrict back to subspace $ℝ^{𝑛}$ topology + $𝐴 \subset ℝ^{𝑛}$

Get $𝐴$ compact

Closed set

let $𝑥 \in \bar{𝐴}$

$𝔹 (𝑥, 𝑟) \cap 𝐴$ forms a net of $𝐴$ . Note that it is possible that $𝑥 \notin 𝔹 (𝑥, 𝑟) \cap 𝐴$

compact ==> $\emptyset \neq ⋂_{𝑟 > 0} \bar{𝔹} (𝑥, 𝑟) \cap 𝐴 \subset 𝐴$
$⋂_{𝑟 > 0} \bar{𝔹} (𝑥, 𝑟) = 𝑥$

==> $\emptyset \neq {𝑥} \cap 𝐴 ⟹ 𝑥 \in 𝐴$

Bounded

The open ball of

ℝ^{𝑑}

does not contain

\infty

. The open ball family

{𝔹 (𝑥, 𝑟) \subset ℝ^{𝑛} : (𝑥 \in 𝐴) \land (𝑟 > 0)}

covers

𝐴

. Take #link(<compact-finite-open-cover>)[finite cover], still does not contain

\infty

let $B$ be net of $ℝ^{𝑛}$

nested-closed-set-theorem_(tag) The intersection of nested bounded closed sets of $ℝ^{𝑛}$ is non-empty. The intersection result is also a closed set. It can be understood as the minimal element of $\subset$ linear order chain nested closed sets

closed-net-theorem_(tag) The intersection of a net of bounded closed sets of $ℝ^{𝑛}$ is non-empty Proof

Map the closed set of

ℝ^{𝑛}

to the closed set of

ℝ^{𝑛} ⊔ {\infty} ≃ 𝕊^{𝑛}

𝕊^{𝑛}

is compact, so the intersection of nested closed sets or the intersection of a net of closed sets is non-empty. Boundedness makes it not converge to

\infty

limit-distance-vanish-net_(tag) := $lim_{𝐵 \in B} | 𝐵 | = 0$

or $\forall 𝜀 > 0, \exists 𝐵 \in B, | 𝐵 | < 𝜀$ . The tail of the net is bounded

A net can be composed of tails $𝐵_{𝑛} = {𝑥_{𝑛}, 𝑥_{𝑛 + 1}, \dots}$

Cauchy-completeness-Euclidean_(tag)

in $ℝ^{𝑛}$ , limit-distance-vanish net converges to a point

According to the closed set net theorem ==> let $𝑥 \in ⋂_{𝐵 \in B} \bar{𝐵} \neq \emptyset$

limit-distance-vanish $lim_{𝐵 \in B} | 𝐵 | = 0$ ==> $⋂_{𝐵 \in B} \bar{𝐵} = {𝑥}$

Some infinite-dimensional linear spaces e.g. #link(<Lebesgue-integrable>)[] $𝐿^{1}$ , bounded closed sets cannot be compact but still satisfy limit-distance-vanish net converging to a point

According to induction, finite summation is associative and commutative. But this does not guarantee infinite summation i.e.

let

Rearrangement $𝑓 : ℕ \leftrightarrow ℕ$
$𝑥_{𝑛} = 𝑎_{1} + \dots + 𝑎_{𝑛}$ converges to $𝑥$

Then $𝑦_{𝑛} = 𝑎_{𝑓 (1)} + \dots + 𝑎_{𝑓 (𝑛)}$ may not converge or converge to other value $𝑦 \neq 𝑥$

compare

$\sum \frac{1}{𝑛} = \infty$
$\sum {(- 1)}^{𝑛 + 1} \frac{1}{𝑛} = log (1 + 1) = log (2)$

Convergence (not ==>) Absolute convergence

let $𝑎_{𝑛}$ be a sequence $ℕ \to ℝ$

$\sum_{. . \infty} 𝑎_{𝑛}$ converges ==> $lim_{𝑛 \to \infty} 𝑎_{𝑛} = 0$

Proof $\forall 𝜀 > 0, \exists 𝑁 \in ℕ, \forall 𝑛 > 𝑁, | \sum_{𝑖 = 1 . . 𝑛} 𝑎_{𝑖} - 𝑎 | < \frac{𝜀}{2}$

==> By the triangle inequality
$| 𝑎_{𝑛} | = | \sum_{𝑖 = 1 . . 𝑛 + 1} 𝑎_{𝑖} - \sum_{𝑖 = 1 . . 𝑛} 𝑎_{𝑖} | \leq | \sum_{𝑖 = 1 . . 𝑛 + 1} 𝑎_{𝑖} - 𝑎 | + | \sum_{𝑖 = 1 . . 𝑛} 𝑎_{𝑖} - 𝑎 | < 𝜀$
$lim_{𝑛 \to \infty} 𝑎_{𝑛} \neq 0$ ==> $\sum_{. . \infty} 𝑎_{𝑛}$ does not converge

Any sequence $𝑥_{𝑛}$ can define $𝑎_{𝑛} = 𝑥_{𝑛} - 𝑥_{𝑛 - 1}$ such that $𝑥_{𝑛} = 𝑎_{1} + \dots + 𝑎_{𝑛}$

Rearrangement $𝑓 : ℕ \leftrightarrow ℕ$ does not change the tail behavior of the sequence $lim_{𝑛 \to \infty} 𝑎_{𝑓 (𝑛)} = lim_{𝑛 \to \infty} 𝑎_{𝑛}$

If $𝑎_{𝑛} \geq 0$ , $\sum 𝑎_{𝑛}$ rearrangement invariant

Proof

$\sum 𝑎_{𝑛} < \infty ⟹ \forall 𝜀 > 0, \exists 𝑁 \in ℕ, \forall 𝑚 > ℕ, \sum 𝑎_{𝑛} - 𝜀 < \sum_{𝑛 = 0 . . 𝑚} 𝑎_{𝑛} \leq \sum 𝑎_{𝑛}$

$𝑁^{'} ≔ \max {𝑓^{- 1} (1), \dots, 𝑓^{- 1} (𝑁)}$

==> ${1, \dots, 𝑁} \subset {𝑓 (1), \dots, 𝑓 (𝑁^{'})}$

==> $\forall 𝑀 > 𝑁^{'}, \sum 𝑎_{𝑛} - 𝜀 \leq \sum_{𝑛 = 0 . . 𝑁} 𝑎_{𝑛} \leq \sum_{𝑛 = 0 . . 𝑀} 𝑎_{𝑓 (𝑛)} \leq \sum 𝑎_{𝑛}$ (by $𝑎_{𝑛} \geq 0$ )

==> $lim_{𝑀 \to \infty} \sum_{𝑛 = 0 . . 𝑀} 𝑎_{𝑓 (𝑛)} = \sum 𝑎_{𝑛}$

def

\begin{matrix} 𝑎_{𝑛}^{+} & ≔ & switch (𝑎_{𝑛}) {\begin{matrix} \geq 0 \Rightarrow 𝑎_{𝑛} \\ < 0 \Rightarrow 0 \end{matrix} \\ 𝑎_{𝑛}^{-} & ≔ & switch (𝑎_{𝑛}) {\begin{matrix} \geq 0 \Rightarrow 0 \\ < 0 \Rightarrow - 𝑎_{𝑛} \end{matrix} \\ 𝑎_{𝑛}^{+}, 𝑎_{𝑛}^{-} & \geq & 0 \\ 𝑎_{𝑛} & = & 𝑎_{𝑛}^{+} - 𝑎_{𝑛}^{-} \end{matrix}

\sum | 𝑎_{𝑛} | < \infty ⟺ \sum 𝑎_{𝑛}^{+}, \sum 𝑎_{𝑛}^{-} < \infty

series-rearrangement-absolutely-convergence-real_(tag) Absolute convergence $\sum_{. . \infty} | 𝑎_{𝑛} |$ ==> $\sum_{. . \infty} 𝑎_{𝑛}$ converges and rearrangement invariant

Proof $𝑎_{𝑛} = 𝑎_{𝑛}^{+} - 𝑎_{𝑛}^{-}$ and use operation of convergent sequence

lim \sum_{. . 𝑁} 𝑎_{𝑛} = lim \sum_{. . 𝑁} 𝑎_{𝑛}^{+} - lim \sum_{. . 𝑁} 𝑎_{𝑛}^{-}

$\sum 𝑎_{𝑛}^{+} = \infty$ and $\sum 𝑎_{𝑛}^{-} < \infty$ ==> $\sum 𝑎_{𝑛} = + \infty$ and rearrangement invariant

Question The case of $𝑙^{2}$ norm ${(\sum | 𝑎_{𝑛} |^{2})}^{\frac{1}{2}}$ reduce to $𝑏_{𝑛} = | 𝑎_{𝑛} |^{2}$ ?

harmonic series $\sum \frac{1}{𝑛} = \infty$ vs $\sum \frac{1}{𝑛^{2}} = \frac{𝜋^{2}}{6}$ , say that, $𝑙^{1}$ convergence is closer to normal convergence. $𝑙^{2}$ convergence is more suitable for Fourier serise

The last possibility

series-rearrangement-real_(tag)

let $lim_{𝑛 \to \infty} 𝑎_{𝑛} = 0$ and $\sum 𝑎_{𝑛}^{+} = \sum 𝑎_{𝑛}^{-} = \infty$

$\exists 𝑓 : ℕ \leftrightarrow ℕ, \sum 𝑎_{𝑓 (𝑛)}$

Converges to $ℝ, + \infty, - \infty$
Does not converge to $ℝ, + \infty, - \infty$

Example

Convergent case $𝑎_{𝑛} = {(- 1)}^{𝑛 + 1} \frac{1}{𝑛}$
Divergent case $𝑎_{𝑛} = {(- 1)}^{𝑛}$

Proof

Converges to $𝐴 \in ℝ$

$𝑝_{1} ≔ \inf {𝑝 \in ℕ : 𝐴 < \sum_{. . 𝑝} 𝑎_{𝑛}^{+}}$ . Meaning: $𝑝_{1}$ is the smallest natural number that makes the positive summation greater than $𝐴$

$𝑞_{1} ≔ \inf {𝑞 \in ℕ : 𝐴 > \sum_{. . 𝑝_{1}} 𝑎_{𝑛}^{+} - \sum_{. . 𝑞} 𝑎_{𝑛}^{-}}$ . Meaning: $𝑞_{1}$ is the smallest natural number that makes the negative summation less than $𝐴$

$𝑝_{2} ≔ \inf {𝑝 \in ℕ : 𝐴 > \sum_{. . 𝑝} 𝑎_{𝑛}^{+} - \sum_{. . 𝑞_{1}} 𝑎_{𝑛}^{-}}$

And so on, exhaust all $𝑎_{𝑛}$

Rearrange $𝑎_{1}, 𝑎_{2}, \dots$ to

\begin{matrix} 𝑎_{1}^{+}, \dots, 𝑎_{𝑝_{1}}^{+} \\ - 𝑎_{1}^{-}, \dots, - 𝑎_{𝑞_{1}}^{-} \\ 𝑎_{𝑝_{1} + 1}^{+}, \dots, 𝑎_{𝑝_{2}}^{+} \\ ⋮ \end{matrix}

According to the definition of $𝑝_{𝑁 + 1}$

\begin{matrix} 0 & < & (\sum_{1}^{𝑝_{𝑁 + 1}} 𝑎_{𝑛}^{+} - \sum_{1}^{𝑞_{𝑁}} 𝑎_{𝑛}^{-}) - 𝐴 \\ \leq & (\sum_{1}^{𝑝_{𝑁 + 1}} 𝑎_{𝑛}^{+} - \sum_{1}^{𝑞_{𝑁}} 𝑎_{𝑛}^{-}) - (\sum_{1}^{𝑝_{𝑁 + 1} - 1} 𝑎_{𝑛}^{+} - \sum_{1}^{𝑞_{𝑁}} 𝑎_{𝑛}^{-}) \\ = & 𝑎_{𝑝_{𝑁 + 1}}^{+} \end{matrix}

According to the definition of $𝑞_{𝑁 + 1}$

0 < (\sum_{1}^{𝑝_{𝑁 + 1}} 𝑎_{𝑛}^{+} - \sum_{1}^{𝑞_{𝑁}, \dots, 𝑞_{𝑁 + 1} - 1} 𝑎_{𝑛}^{-}) - 𝐴 \leq 𝑎_{𝑝_{𝑁 + 1}}^{+}

And so on

- 𝑎_{𝑞_{𝑁 + 1}}^{-} \leq (\sum_{1}^{𝑝_{𝑁 + 1}, \dots, 𝑝_{𝑁 + 2} - 1} 𝑎_{𝑛}^{+} - \sum_{1}^{𝑞_{𝑁 + 1}} 𝑎_{𝑛}^{-}) - 𝐴 < 0

$lim_{𝑛 \to \infty} 𝑎_{𝑛} = 0 ⟹ lim_{𝑁 \to \infty} 𝑎_{𝑝_{𝑁}}^{+} = lim_{𝑁 \to \infty} 𝑎_{𝑞_{𝑁}}^{-} = 0$

==> $lim_{𝑁 \to \infty} \sum_{. . 𝑁} 𝑎_{𝑛} = 𝐴$

Converges to $+ \infty$

In the handling of $𝐴 \in ℝ$

Change $𝐴 < \sum_{1}^{\dots} 𝑎_{𝑛}^{+} - \sum_{1}^{\dots} 𝑎_{𝑛}^{-}$ to $𝑁 < \sum_{1}^{\dots} 𝑎_{𝑛}^{+} - \sum_{1}^{\dots} 𝑎_{𝑛}^{-}$

Change $𝐴 > \dots$ to $𝑁 > \dots$

Does not converge to $ℝ, + \infty, - \infty$

Change $𝐴 < \dots$ to $1 < \dots$

Change $𝐴 > \dots$ to $- 1 > \dots$

Series in $ℝ^{𝑑}$ that are rearrangement invariant are also absolutely convergent series

$\sum 𝑎_{𝑛}$ converges ==> $lim_{𝑛 \to \infty} 𝑎_{𝑛} = 0$

series-rearrangement-absolutely-convergence_(tag)

let $𝑎_{𝑛}$ be a $ℕ \to ℝ^{𝑑}$ sequence

$\sum | 𝑎_{𝑛} | < \infty$ ==> $\sum 𝑎_{𝑛}$ converges and is rearrangement invariant

Proof

$\sum 𝑎_{𝑛}$ converges. by using the triangle inequality $| \sum_{𝑁 + 1}^{\infty} 𝑎_{𝑛} | \leq \sum_{𝑁 + 1}^{\infty} | 𝑎_{𝑛} |$ and $ℝ^{𝑑}$ #link(<Cauchy-completeness-Euclidean>)[Cauchy sequence converges]
Rearrangement invariant

$\forall 𝜀 > 0, \exists 𝑁 \in ℕ, (| \sum_{. . 𝑁} 𝑎_{𝑛} - 𝐴 | < \frac{𝜀}{2}) \land (\sum_{𝑁 + 1}^{\infty} | 𝑎_{𝑛} | < \frac{𝜀}{2})$

$𝑁^{'} ≔ \max {𝑓^{- 1} (1), \dots, 𝑓^{- 1} (𝑁)}$

${1, \dots, 𝑁} \subset {𝑓 (1), \dots, 𝑓 (𝑁^{'})}$

$\forall 𝑀 > 𝑁^{'}, | \sum_{. . 𝑀} 𝑎_{𝑓 (𝑛)} - 𝐴 | \leq | \sum_{. . 𝑁} 𝑎_{𝑛} - 𝐴 | + \sum_{𝑁 + 1}^{\infty} | 𝑎_{𝑛} | < 𝜀$

Now consider the case where $\sum 𝑎_{𝑛}$ is not absolutely convergent

def $𝑉 ≔ {𝑥 \in ℝ^{𝑑} : \sum | ⟨ 𝑎_{𝑛}, 𝑥 ⟩ | < \infty}$

From the triangle inequality or the equivalence of $1$ -norm, $2$ -norm, $\infty$ -norm of $ℝ^{𝑑}$

$𝑉$ is a linear subspace
$𝑉 = ℝ^{𝑑} ⟹ \sum | 𝑎_{𝑛} | < \infty$

let $𝑉 \neq ℝ^{𝑑}$ . The $𝑉$ component of $\sum 𝑎_{𝑛}$ is absolutely convergent

Consider the $𝑉^{⟂}$ component of $\sum 𝑎_{𝑛}$

series-rearrangement_(tag)

let $𝑥 \in 𝑉^{⟂}$

$\sum {⟨ 𝑎_{𝑛}, 𝑥 ⟩}^{+} = \infty$ and $\sum {⟨ 𝑎_{𝑛}, 𝑥 ⟩}^{-} < \infty$ ==> $\sum 𝑎_{𝑛}$ converges to $\infty$ in the $𝑥$ component, rearrangement invariant
$\sum {⟨ 𝑎_{𝑛}, 𝑥 ⟩}^{+} = \sum {⟨ 𝑎_{𝑛}, 𝑥 ⟩}^{-} = \infty$ . $\sum 𝑎_{𝑛}$ is rearrangement unstable in the $𝑥$ component

The positive linear combination $𝑘 𝑎_{𝑛} + 𝑘^{'} 𝑎_{𝑛}^{'}$ with $sign (𝑘) = sign (𝑘^{'})$ of sequences with the same convergence pattern preserves their convergence pattern