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
A more general inequality when should be . For simplicity, we temporarily do not use this more general assumption.

For the norm , the infimum of such that , or the supremum of over , is .

(Proof First, calculate an upper bound for , then prove that it is the supremum. Use differential techniques to prove that for , . Apply it to the components of the norm. Let , and convert to trying to calculate the maximum value of . Due to homogeneity, scaling does not affect the result. Assume . Use differential methods to calculate the maximum value of , obtaining the upper bound . Then, use the embedding in to show that this value can be achieved, so is the supremum. When , , which makes the normal triangle inequality invalid.)

[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. (Here we are not to define topology and no need to generalize to higher dimension, so not need to restrict to only open interval)

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

Assume is a bounded closed interval, and is a net of

Since is a closed set, the definition of is the same for the topologies of and

Since is bounded, we can define the non-empty infimum set and the supremum set

According to the property of nets (or using the corresponding interval net ), it can be proven that all numbers in are all numbers in

has an upper bound, has a lower bound, so we can take the infimum/supremum, and it satisfies

We prove that

Take , prove that

Proof

Define

because

is a closed set, so

Therefore

That is

Next, prove

For each , since is a net, there exists such that

Thus , so and

And , so

By the arbitrariness of selecting , we have being the upper bound of , thus , which means

Therefore

Since , then

Due to the arbitrary choice of , we have

Therefore

[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

Considering the translation invariance of infinity , use stereographic projection technique

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

bounded closed = compact ==> 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