Euclidean-space - note-math

use alternation of tensor-induced-quadratic-form

Iterate through all , orthonormal bases with , to obtain the signature

let . span <==>

Abbreviation

[quadratic-form-inequality-Euclidean] Inner product inequality (Euclidean). . i.e. or

[triangle-inequality-Euclidean] Triangle inequality (Euclidean)

Proof
Proof

[Euclidean-space-topology] Euclidean topology. is continuous at :=

let

[closure] Closure :=

[closed-set] Closed set :=

(open) closed(𝔹)

[open-set] Open set :=

open <==> closed

[interval] Interval refers to a subset of with property that the order is uninterrupted

[best-interval-decomposition] The optimal interval decomposition of

def as the set of all intervals, including open, closed, half open half closed, single point

def

Due to the existence of single-point intervals, and

has a linear order chain. Taking for each maximal linear order chain will continue to yield intervals. The set of these intervals is denoted as

and

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

When , is an interval, connected
When , is not connected. Example

If is a closed set, then the intervals in are all closed intervals

recall the linear-order intersection of nested closed intervals is non-empty

[bounded-closed-interval-is-compact] Bounded closed interval of ==> compact

Proof

let be a net of . let

Since is bounded and closed,

Take the optimal closed interval decomposition

For all decomposed closed intervals of , consider any maximal linear order chain maximal-linear-order

According to nested-closed-interval-theorem, the intersection of a nested set of closed intervals in the linear order is a non-empty closed interval

Similar to the proof of closed-interval-net-theorem, prove that is the smallest closed interval, thus every

[compact-imply-subsequence-converge] compact ==> sequence has a convergent subsequence. The same applies to nets

Proof

forms a net

compact ==>

let

use the definition of closure

let

==>

Unit closed ball
Unit sphere

[circle-is-compact] compact

Proof is continuous

is continuously isomorphic to (quotient-topology) is continuously isomorphic to i.e. collapsing endpoints (quotient-topology)

bounded closed interval compact ==> quotient compact. by quotient preserves compact

[closed-ball-sphere-is-compact]

Proof

compact. Inductive hypothesis compact

compact

(Draw a picture) continuous. Isomorphism is obtained after quotienting the origin

compact. by product-topology-preserve-compact

compact

compact

Constructing from using polar coordinates, after quotient, we get compact

Another method the boundary collapses to a point to get compact

Proof

maps the sphere to the sphere and

Stereographic projection

The composite mapping plus the mapping mapping to , the resulting mapping is still continuous, and after quotient it is a bijection

Projective space (Euclidean) compact. Proof

Similarly (and )

[Euclidean-set-distance]

[bounded] bounded :=
[unbounded] unbounded :=

is invariant

by stereographic projection

Translation does not change the infinity of (but only a conformal mapping of , conformal group )

in Euclidean topology of

Bounded <==> away from <==>
Unbounded <==>

[Euclidean-space-compact-iff-bounded-closed] compact <==> is a bounded closed set

Proof

The bounded closed set of corresponds to a closed set of and does not include

compact + closed-set-in-compact-space-is-compact ==> is compact in

From topology, restrict back to subspace topology +

Get compact

Closed set

let

forms a net of . Note that it is possible that

compact ==>

==>

Bounded

The open ball of does not contain . The open ball family covers . Take finite cover, still does not contain

let be net of

[nested-closed-set-theorem] The intersection of nested bounded closed sets of is non-empty. The intersection result is also a closed set. It can be understood as the minimal element of linear order chain nested closed sets

[closed-net-theorem] The intersection of a net of bounded closed sets of is non-empty Proof

Map the closed set of to the closed set of , 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

[limit-distance-vanish-net] :=

or . The tail of the net is bounded

A net can be composed of tails

[Cauchy-completeness-Euclidean]

in , limit-distance-vanish net converges to a point

According to the closed set net theorem ==> let

limit-distance-vanish ==>

Some infinite-dimensional linear spaces e.g. Lebesgue-integrable , 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
converges to

Then may not converge or converge to other value

compare

Convergence (not ==>) Absolute convergence

let be a sequence

converges ==>

Proof

==> By the triangle inequality
==> does not converge

Any sequence can define such that

Rearrangement does not change the tail behavior of the sequence

If , rearrangement invariant

Proof

==>

==> (by )

==>

def

[series-rearrangement-absolutely-convergence-real] Absolute convergence ==> converges and rearrangement invariant

Proof and use operation of convergent sequence

and ==> and rearrangement invariant

Question The case of norm reduce to ?

harmonic series vs , say that, convergence is closer to normal convergence. convergence is more suitable for Fourier serise

The last possibility

[series-rearrangement-real]

let and

Converges to
Does not converge to

Example

Convergent case
Divergent case

Proof

Converges to

. Meaning: is the smallest natural number that makes the positive summation greater than

. Meaning: is the smallest natural number that makes the negative summation less than

And so on, exhaust all

Rearrange to

According to the definition of

And so on

==>

Converges to

In the handling of

Change to

Does not converge to

Change to

Series in that are rearrangement invariant are also absolutely convergent series

converges ==>

[series-rearrangement-absolutely-convergence]

let be a sequence

==> converges and is rearrangement invariant

Proof

converges. by using the triangle inequality and Cauchy sequence converges
Rearrangement invariant

Now consider the case where is not absolutely convergent

def

From the triangle inequality or the equivalence of -norm, -norm, -norm of

is a linear subspace

let . The component of is absolutely convergent

Consider the component of

[series-rearrangement]

let

and ==> converges to in the component, rearrangement invariant
. is rearrangement unstable in the component

The positive linear combination with of sequences with the same convergence pattern preserves their convergence pattern