Using the alternation of tensor-induced-quadratic-form
Iterate over all , orthonormal basis with , obtaining signature
let . Prop 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 for should be . For simplicity, this more general assumption is not used for now
For the norm
The infimum of , or
The supremum for , is
Proof
First compute an upper bound for , then prove it is the supremum
Use differential techniques to prove for . Using to reduce to a single variable, prove
Compute the maximum of . Due to homogeneity, scaling does not affect the result. Assume . Use differential method to compute the maximum of , which is .
For each norm component of , let . use
Sum these norm components
let . use
Using the embedding with to illustrate that the inequality can achieve equality , thus is the supremum.
When , , which causes the usual triangle inequality to fail.
[Euclidean-space-topology] Euclidean topology. is continuous at :=
let
[closure] Closure :=
[closed-set] Closed set :=
(open) ball
[open-set] Open set :=
open <==> closed
[interval] Interval refers to a subset of with property that the order is uninterrupted
[best-interval-decomposition] Optimal interval decomposition of
Def as the set of all intervals, including open, closed, half-open half-closed, and single point
No need to restrict to only open intervals, as this is not defining a topology, nor does it need to generalize to higher dimensions
Def , i.e., the set of all intervals among the subsets of
Due to the existence of single-point intervals, and
has linear order chains. Taking of each maximal linear order chain yields an interval. The set of these intervals is denoted
and
The intervals in are pairwise disjoint, and the decomposition is unique, thus these intervals constitute 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
[bounded-closed-interval-is-compact] A bounded closed interval of ==> compact
Proof
Assume is a bounded closed interval, is a net of
Since is closed, 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
By the property of the net (or using the corresponding interval net ), it can be proved that numbers in are numbers in
has an upper bound, has a lower bound, so we can take the infimum/supremum, and it satisfies
Prop
Take , prove
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 the selection of , is an upper bound of , thus , i.e.,
Hence
Since , we have
By the arbitrariness of the selection of , we have
Thus
By the arbitrariness of the selection of , is compact
[compact-imply-subsequence-converge] compact ==> sequence has a convergent subsequence. Similarly for nets
Proof
forms a net
compact ==>
let
let and
use the definition of closure
We can inductively choose such that is a subsequence. Proof . Choose such that and
==>
- The closed unit ball
- The 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)
is a bounded closed interval, hence compact ==> the quotient is compact. by quotient preserves compactness
[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
Using polar coordinates to construct from , after quotient, obtain compact
Another method boundary collapses to a point to get compact
Proof
. Sphere corresponds to sphere , then corresponds to radius
Stereographic projection
The composite map plus the map to yields a map that remains continuous; after quotienting , it becomes a bijection,
Projective space (Euclidean) compact. Proof
Similarly (and ) compact
[Euclidean-set-distance]
- [bounded] bounded :=
- [unbounded] unbounded :=
is invariant
The point at infinity is translation-invariant
by stereographic projection
in Euclidean topology of
- bounded <==> away from <==>
- unbounded <==>
[Euclidean-space-compact-iff-bounded-closed] compact <==> is bounded and closed
Proof
- <==
bounded closed set corresponds to a closed set in that does not include
compact + closed-set-in-compact-space-is-compact ==> is compact in
topology restricted back to subspace topology
obtain compact
- ==>
- closed set
let
forms a net of . Note that possibly
-
compact ==>
-
==>
i.e. closed
- bounded
[nested-closed-set-theorem] The intersection of a nested sequence of bounded closed sets in is nonempty. Its intersection is also a closed set, which can be understood as the minimal element of the โ linearly ordered chain of nested closed sets
[closed-net-theorem] The intersection of a net of bounded closed sets in is nonempty Proof
let be net of
[limit-distance-vanish-net] :=
or
The tail of a net is bounded
Sequences can form a net
[Cauchy-completeness-Euclidean]
in , a net with vanishing limit-distance converges to a point
bounded closed set = compact ==> let
limit-distance-vanish ==>
Some infinite-dimensional linear spaces e.g. Lebesgue-integrable , bounded closed sets do not imply compact, but still satisfy that a net with vanishing limit-distance converges to a point, due to the completeness of
By induction, finite summation is associative and commutative. But this does not guarantee it holds for infinite summation i.e.
let
- converges to
- Rearrangement
then may not converge or converge to another 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
Prop If , is invariant under rearrangement
Proof
==>
==> (by )
==>
def
[series-rearrangement-absolutely-convergence-real] Prop Absolute convergence ==> converges and is invariant under rearrangement
Proof and use the arithmetic operations of convergent sequences
and ==> and invariant under rearrangement
Question The case of norm reduces to ?
Harmonic series vs , say that, convergence is closer to normal convergence.
Final possibilities
[series-rearrangement-real]
let and
Prop
- Converges to
- Does not converge to
Example
Proof
- Converges to
. Meaning: is the smallest natural number such that the sum of positives exceeds
. Meaning: is the smallest natural number such that the negative sum is less than
And so on, exhausting all
Rearrange as
By the definition of
By the definition of
And so on
==>
- Converges to
In the treatment of
Change to
Change to
- Does not converge to
Change to
Change to
Similarly for
Prop A series in that is rearrangement invariant is also absolutely convergent
Prop converges ==>
[series-rearrangement-absolutely-convergence]
let be a sequence
==> converges and is rearrangement invariant
Proof
-
converges. By using the triangle inequality and the Cauchy sequence convergence in
-
Rearrangement invariant
Now consider the case where is not absolutely convergent
def
By the triangle inequality or the equivalence of -norm, -norm, -norm in
- is a linear subspace
let . The component of converges absolutely
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