aarjaneiro/planetmath huggingface dataset:
induced Alexandroff topology on a poset

induced Alexandroff topology on a poset

Let

(

X

,

≤

)

be a poset. For any

x

∈

X

define following subset:

(

-

∞

,

x

]

=

{

y

∈

X

|

y

≤

x

}

.

The

induced

Alexandroff

topology

τ

on

X

is defined as a topology generated by

{

(

-

∞

,

x

]

}

x

∈

X

.

Proposition

1.

(

X

,

τ

)

is a

T

0

, Alexandroff space.

Proof.

Let

x

,

y

∈

X

be such that

x

≠

y

. Note that this implies that

x

≰

y

or

y

≰

x

(because

≤

is antisymmetric). Therefore

x

∉

(

-

∞

,

y

]

or

y

∉

(

-

∞

,

x

]

. Thus

(

X

,

τ

)

is

T

0

.

Now in order to show that

(

X

,

τ

)

is Alexandroff it is enough to show that an arbitrary

intersection

of base sets is open. So assume that

{

x

i

}

i

∈

I

is a subset of

X

such that

A

=

⋂

i

∈

I

(

-

∞

,

x

i

]

≠

∅

and let

y

∈

A

. Then (since

≤

is

transitive

) it is clear that

(

-

∞

,

y

]

⊆

A

and thus

⋂

i

∈

I

(

-

∞

,

x

i

]

=

A

=

⋃

y

∈

A

(

-

∞

,

y

]

and therefore the intersection is open, which

completes

the proof.

□

Proposition 2.

Let

X

,

Y

be posets and

f

:

X

→

Y

a

function

. Then

f

preserves

order if and only if

f

is

continuous

in induced Alexandroff topologies.

Proof.

,,

⇒

” Assume that

f

preserves order and let

A

=

(

-

∞

,

y

]

⊆

Y

be an open base set. We wish to show that

f

-

1

⁢

(

A

)

is open in

X

. So take any

x

∈

f

-

1

⁢

(

A

)

. Now if

y

≤

x

, then

f

⁢

(

y

)

≤

f

⁢

(

x

)

(since

f

preserves order) and thus

f

⁢

(

y

)

∈

A

. Therefore

y

∈

f

-

1

⁢

(

A

)

. Since

y

was arbitrary we obtain that for any

x

∈

f

-

1

⁢

(

A

)

we have

(

-

∞

,

x

]

⊆

f

-

1

⁢

(

A

)

and thus

f

-

1

⁢

(

A

)

=

⋃

x

∈

f

-

1

⁢

(

A

)

(

-

∞

,

x

]

,

which implies that

f

-

1

⁢

(

A

)

is open.

,,

⇐

” Assume that

f

is continuous and let

y

≤

x

for some

x

,

y

∈

X

. Assume that

f

⁢

(

y

)

≰

f

⁢

(

x

)

. Let

A

=

(

-

∞

,

f

⁢

(

x

)

]

. Therefore

f

⁢

(

y

)

∉

A

, but

A

is open, so

f

-

1

⁢

(

A

)

is open (because

f

is continuous). Thus

(

-

∞

,

x

]

⊆

f

-

1

⁢

(

A

)

. But

y

≤

x

, so

y

∈

f

-

1

⁢

(

A

)

. But this implies that

f

⁢

(

y

)

∈

A

.

Contradiction

.

□

Title

induced Alexandroff topology on a poset

Canonical

name

InducedAlexandroffTopologyOnAPoset

Date of creation

2013-03-22 18:46:01

Last modified on

2013-03-22 18:46:01

Owner

joking (16130)

Last modified by

joking (16130)

Numerical id

4

Author

joking (16130)

Entry type

Derivation

Classification

msc 54A05

Generated on Sat Feb 10 11:12:56 2018 by

LaTeXML
induced partial order on an Alexandroff space

induced partial order on an Alexandroff space

Let

X

be a

T

0

, Alexandroff space. For

A

⊆

X

denote by

A

o

the

intersection

of all open

neighbourhoods

of

A

. Define a

relation

≤

on

X

as follows: for any

x

,

y

∈

X

we have

x

≤

y

if and only if

x

∈

{

y

}

o

. This relation will be called the

induced

partial order

on

X

.

Proposition

1.

(

X

,

≤

)

is a poset.

Proof.

Of course

x

∈

{

x

}

o

for any

x

∈

X

. Thus

≤

is

reflexive

.

Assume now that

x

≤

y

and

y

≤

x

for some

x

,

y

∈

X

. Assume that

x

≠

y

. Then, since

X

is a

T

0

space, there is an

open set

U

such that

x

∈

U

and

y

∉

U

or there is an open set

V

such that

y

∈

V

and

x

∉

V

. Both cases lead to

contradiction

, because we assumed that

x

∈

{

y

}

o

and

y

∈

{

x

}

o

. Thus every open neighbourhood of one

element

must also contain the other. Thus

≤

is antisymmetric.

Finally assume that

x

≤

y

and

y

≤

z

for some

x

,

y

,

z

∈

X

. Since

y

∈

{

z

}

o

, then

{

z

}

o

is an open neighbourhood of

y

and thus

{

y

}

o

⊆

{

z

}

o

. Therefore

x

∈

{

z

}

o

, so

≤

is

transitive

, which

completes

the proof.

□

Proposition 2.

Let

X

,

Y

be two,

T

0

, Alexandroff spaces and

f

:

X

→

Y

be a

function

. Then

f

is

continuous

if and only if

f

preserves the induced partial order.

Proof.

,,

⇒

” Assume that

f

is continuous and suppose that

x

,

y

∈

X

are such that

x

≤

y

. We wish to show that

f

⁢

(

x

)

≤

f

⁢

(

y

)

, so assume this is not the case. Let

A

=

{

f

⁢

(

y

)

}

o

. Then

f

⁢

(

x

)

∉

A

. But

A

is open, so

f

-

1

⁢

(

A

)

is also open (because we assumed that

f

is continuous). Furthermore

y

∈

f

-

1

⁢

(

A

)

and because

x

≤

y

, then

x

∈

f

-

1

⁢

(

A

)

, but this implies that

f

⁢

(

x

)

∈

A

. Contradiction.

,,

⇐

” Assume that

f

preserves the induced partial order and let

U

⊆

Y

be an open subset. Let

x

∈

U

. Then for any

y

≤

x

we have

f

⁢

(

y

)

≤

f

⁢

(

x

)

(because

f

preserves the induced partial order) and since

{

f

⁢

(

x

)

}

o

⊆

U

(because

U

is open and

{

f

⁢

(

x

)

}

o

is the smallest open neighbourhood of

f

⁢

(

x

)

) we have that

f

⁢

(

y

)

∈

U

. Thus

{

x

}

o

=

{

y

∈

X

|

y

≤

x

}

⊆

f

-

1

⁢

(

U

)

which implies that

f

-

1

⁢

(

U

)

is open because

f

-

1

⁢

(

A

)

contains a small neighbourhood of each point. This completes the proof.

□

Title

induced partial order on an Alexandroff space

Canonical name

InducedPartialOrderOnAnAlexandroffSpace

Date of creation

2013-03-22 18:45:55

Last modified on

2013-03-22 18:45:55

Owner

joking (16130)

Last modified by

joking (16130)

Numerical id

4

Author

joking (16130)

Entry type

Derivation

Classification

msc 54A05

Generated on Sat Feb 10 11:12:58 2018 by

LaTeXML
interior axioms

interior axioms

Let

S

be a set. Then an

interior operator

is a

function

:

∘

𝒫

(

S

)

→

𝒫

(

S

)

which

satisfies

the
following properties:

Axiom 1

.

S

∘

=

S

Axiom 2

.

For all

X

⊂

S

, one has

X

∘

⊆

S

.

Axiom 3

.

For all

X

⊂

S

, one has

(

X

∘

)

∘

=

X

∘

.

Axiom 4

.

For all

X

,

Y

⊂

S

, one has

(

X

∩

Y

)

∘

=

X

∘

∩

Y

∘

.

If

S

is a

topological space

, then the

operator

which assigns to
each set its

interior

satisfies these axioms.

Conversely

, given an
interior operator

∘

on a set

S

, the set

{

X

∘

∣

X

⊂

S

}

defines a

topology

on

S

in which

X

∘

is the
interior of

X

for any subset

X

of

S

. Thus, specifying an
interior operator on a set is

equivalent

to specifying a topology
on that set.

The

concepts

of interior operator and

closure operator

are closely
related.
Given an interior operator

∘

, one can
define a closure operator

c

by the condition

X

c

=

(

(

X

′

)

∘

)

′

and, given a closure operator

c

, one can
define an interior operator

∘

by the condition

X

∘

=

(

(

X

′

)

c

)

′

.

Title

interior axioms

Canonical name

InteriorAxioms

Date of creation

2013-03-22 16:30:37

Last modified on

2013-03-22 16:30:37

Owner

rspuzio (6075)

Last modified by

rspuzio (6075)

Numerical id

8

Author

rspuzio (6075)

Entry type

Definition

Classification

msc 54A05

Related topic

GaloisConnection

Defines

interior operator

Generated on Sat Feb 10 11:13:00 2018 by

LaTeXML
isolated

isolated

Let

X

be a

topological space

, let

S

⊂

X

, and let

x

∈

S

. The point

x

is said to be an

isolated

point of

S

if there exists an

open set

U

⊂

X

such that

U

∩

S

=

{

x

}

.

The set

S

is

isolated

or

discrete

if every point in

S

is an

isolated point

.

Title

isolated

Canonical name

Isolated

Date of creation

2013-03-22 12:05:59

Last modified on

2013-03-22 12:05:59

Owner

djao (24)

Last modified by

djao (24)

Numerical id

8

Author

djao (24)

Entry type

Definition

Classification

msc 54A05

Synonym

discrete set

Defines

isolated set

Defines

isolated point

Generated on Sat Feb 10 11:13:02 2018 by

LaTeXML
isomorphism

isomorphism

A

morphism

f

:

A

⟶

B

in a

category

is an

isomorphism

if there exists a morphism

f

-

1

:

B

⟶

A

which is its

inverse

. The objects

A

and

B

are

isomorphic

if there is an isomorphism between them.

A morphism which is both an isomorphism and an

endomorphism

is called an

automorphism

. The set of automorphisms of an object

A

is denoted

Aut

⁡

(

A

)

.

Examples:

•

In the

category of sets

and functions, a function

f

:

A

⟶

B

is an isomorphism if and only if it is

bijective

.

•

In the category of groups and

group homomorphisms

(or rings and

ring homomorphisms

), a

homomorphism

ϕ

:

G

⟶

H

is an isomorphism if it has an inverse map

ϕ

-

1

:

H

⟶

G

which is also a homomorphism.

•

In the category of vector spaces and

linear transformations

, a linear transformation is an isomorphism if and only if it is an

invertible linear transformation

.

•

In the category of topological spaces and

continuous maps

, a continuous map is an isomorphism if and only if it is a homeomorphism.

Title

isomorphism

Canonical

name

Isomorphism

Date of creation

2013-03-22 12:19:20

Last modified on

2013-03-22 12:19:20

Owner

djao (24)

Last modified by

djao (24)

Numerical id

6

Author

djao (24)

Entry type

Definition

Classification

msc 54A05

Classification

msc 15A04

Classification

msc 13A99

Classification

msc 20A05

Classification

msc 18A05

Defines

isomorphic

Defines

automorphism

Generated on Sat Feb 10 11:13:04 2018 by

LaTeXML
limit points of sequences

limit points of sequences

In a

topological space

X

, a point

x

is a

limit point

of the sequence

x

0

,

x

1

,

…

if, for every

open set

containing

x

, there are finitely many indices such that the corresponding elements of the sequence do not belong to the open set.

A point

x

is an

accumulation point

of the sequence

x

0

,

x

1

,

…

if, for every open set containing

x

, there are infinitely many indices such that the corresponding elements of the sequence belong to the open set.

It is worth noting that the set of limit points of a sequence can differ from the set of limit points of the set of elements of the sequence. Likewise the set of accumulation points of a sequence can differ from the set of accumulation points of the set of elements of the sequence.

Reference: L. A. Steen and J. A. Seebach, Jr. “Counterxamples in

Topology

” Dover Publishing 1970

Title

limit points of sequences

Canonical

name

LimitPointsOfSequences

Date of creation

2013-03-22 14:38:13

Last modified on

2013-03-22 14:38:13

Owner

rspuzio (6075)

Last modified by

rspuzio (6075)

Numerical id

7

Author

rspuzio (6075)

Entry type

Definition

Classification

msc 54A05

Defines

limit point of a sequence

Defines

limit point of the sequence

Defines

accumulation point of a sequence

Defines

accumulation point of the sequence

Generated on Sat Feb 10 11:13:06 2018 by

LaTeXML
neighborhood

neighborhood

The meaning of the word

neighborhood

in

topology

is not well standardized. For most authors, a

neighborhood

of a point

x

in a

topological space

X

is an

open subset

U

of

X

which contains

x

. If

X

is a

metric space

, then an open ball around

x

is one example of a neighborhood. Unless otherwise specified, this

definition

of neighborhood predominantes on this site.

More generally, a neighborhood of any subset

S

of

X

is defined to be an open set of

X

containing

S

.

Some authors use the word neighborhood to denote any subset

U

that contains an open subset containing

x

. This alternative usage has the advantage that it is easier to develop the theory of filters for topological spaces. At the other extreme, some analysis texts which deal only in metric spaces define a neighborhood to be an open ball around a point

x

.

Topologists tolerate this ambiguity because the most common usage, “

S

contains a neighborhood of

x

” is unaffected by the choice. In fact, almost any argument involving neighborhoods would be unaffected by shrinking a neighborhood to a smaller open set or to an open ball (in the context of metric spaces).

A

deleted neighborhood

of

x

is an open set of the form

U

∖

{

x

}

, where

U

is an open subset of

X

which contains

x

.

Title

neighborhood

Canonical

name

Neighborhood

Date of creation

2013-03-22 12:04:49

Last modified on

2013-03-22 12:04:49

Owner

djao (24)

Last modified by

djao (24)

Numerical id

13

Author

djao (24)

Entry type

Definition

Classification

msc 54A05

Synonym

neighbourhood

Related topic

TopologicalSpace

Related topic

NeighborhoodSystem

Related topic

Ball

Related topic

MetricSpace

Related topic

LocalBase

Defines

deleted neighborhood

Generated on Sat Feb 10 11:13:07 2018 by

LaTeXML
proof of characterization of connected compact metric spaces.

proof of characterization of connected compact metric spaces.

First we prove the right-hand arrow: if

X

is

connected

then the

property

stated in the

Theorem

holds. This

implication

is true in every

metric space

X

, without additional conditions.

Let us denote with

A

ε

the set of all points

z

∈

X

which can be joined to

x

with a

sequence

of points

p

1

,

…

,

p

n

with

p

1

=

x

,

p

n

=

z

and

d

⁢

(

p

i

,

p

i

+

1

)

<

ε

. If

z

∈

A

ε

then also

B

ε

⁢

(

z

)

⊂

A

ε

since given

w

∈

B

ε

⁢

(

z

)

we can simply add the point

p

n

+

1

=

w

to the sequence

p

1

,

…

,

p

n

.
This immediately shows that

A

ε

is an open subset of

X

. On the other hand we can show that

A

ε

is also closed. In fact suppose that

x

n

∈

A

ε

and

x

n

→

x

¯

∈

X

. Then there exists

k

such that

x

¯

∈

B

ε

⁢

(

x

k

)

and hence

x

¯

∈

A

ε

by the property stated above. Since both

A

ε

and its complementary set are open then,
being

X

connected, we conclude that

A

ε

is either empty or its complementary set is empty. Clearly

x

∈

A

ε

so we conclude that

A

ε

=

X

. Since this is true for all

ε

>

0

the first implication is proven.

Let us prove the reverse implication. Suppose by

contradiction

that

X

is not connected. This means that two non-empty open sets

A

,

B

exist such that

A

∪

B

=

X

and

A

∩

B

=

∅

. Since

A

is the complementary set of

B

and vice-versa, we know that

A

and

B

are closed too. Being

X

compact

we conclude that both

A

and

B

are compact sets.
We now claim that

δ

:=

inf

a

∈

A

,

b

∈

B

⁡

d

⁢

(

a

,

b

)

>

0

.

Suppose by contradiction that

δ

=

0

.
In this case by

definition

of

infimum

, there exist two sequences

a

k

∈

A

and

b

k

∈

B

such that

d

⁢

(

a

k

,

b

k

)

→

0

. Since

A

and

B

are compact, up to a

subsequence

we may and shall suppose that

a

k

→

a

∈

A

and

b

k

→

b

∈

B

. By the continuity of the distance function we conclude that

d

⁢

(

a

,

b

)

=

0

i.e.

a

=

b

which is in contradiction with the condition

A

∩

B

=

∅

. So the claim is proven.

As a

consequence

, given

ε

<

δ

it is not possible to join a point of

A

with a point of

B

. In fact in the sequence

p

1

,

…

,

p

n

there should exists two consecutive points

p

i

and

p

i

+

1

with

p

i

∈

A

and

p

i

+

1

∈

B

. By the previous observation we would conclude that

d

⁢

(

p

i

,

p

i

+

1

)

≥

δ

>

ε

.

Title

proof of

characterization

of connected compact metric spaces.

Canonical

name

ProofOfCharacterizationOfConnectedCompactMetricSpaces

Date of creation

2013-03-22 14:17:06

Last modified on

2013-03-22 14:17:06

Owner

paolini (1187)

Last modified by

paolini (1187)

Numerical id

4

Author

paolini (1187)

Entry type

Proof

Classification

msc 54A05

Generated on Sat Feb 10 11:13:12 2018 by

LaTeXML
sequence accumulating everywhere in [-1,ERROR(\tmspace)+.1667⁢e⁢m⁢1]

sequence accumulating everywhere in

[

-

1

,

\tmspace

+

.1667

⁢

e

⁢

m

⁢

1

]

We want to show that if

k

is an irrational number, then any

real number

of the

interval

[

-

1

,

1

]

is an

accumulation point

of the sequence

sin

⁡

2

⁢

k

⁢

π

,

sin

⁡

4

⁢

k

⁢

π

,

sin

⁡

6

⁢

k

⁢

π

,

…

(1)

In other words, the real numbers (1) come arbitrarily close to every number of the interval.

Proof.

Set on the perimeter of the

unit circle

, starting e.g. from the point

(

1

,

0

)

,  anticlockwise the points

P

0

,

P

1

,

P

2

,

…

(2)

with successive arc-distances

2

⁢

k

⁢

π

.  Since

k

is irrational,

2

⁢

k

⁢

π

and the length

2

⁢

π

of the perimeter are

incommensurable

.  Therefore no two of the points

P

i

coincide, whence we have an

infinite

sequence (2) of distinct points.  We can see that these points form an everywhere

dense set

on the perimeter, i.e. that an arbitrarily short arc contains always points of (2).

Let then

ε

be an arbitrary positive number.  Choose an integer

n

such that

2

⁢

π

n

<

ε

and the perimeter of the unit circle, starting from the point

(

1

,

0

)

,  into

n

equal arcs.  Each of the points

P

1

,

P

2

,

…

falls into one of these arcs, because the arcs

2

⁢

π

/

n

and

P

0

⁢

P

1

are incommensurable.  Thus, among the

n

+

1

first points

P

1

,

P

2

,

…

,

P

n

+

1

there must be at least two ones belonging to a same arc.  Let

P

μ

and

P

ν

(

μ

<

ν

) belong to the same arc.  Then the length

l

of the arc

P

μ

⁢

P

ν

is less than

ε

.  Starting from

P

μ

one comes to

P

ν

by moving on the perimeter

ν

-

μ

times in succession arcs with length

2

⁢

k

⁢

π

(when one has possibly to go around the perimeter several times).  Repeating that procedure, starting from the point

P

ν

, one comes to the point

P

ν

+

(

ν

-

μ

)

=

P

2

⁢

ν

-

μ

, and furthermore to

P

3

⁢

ν

-

2

⁢

μ

, to

P

4

⁢

ν

-

3

⁢

μ

, and so on.

The points

P

μ

,

P

ν

,

P

2

⁢

ν

-

μ

,

P

3

⁢

ν

-

2

⁢

μ

,

…

(3)

form on the perimeter a sequence of equidistant points, a

subsequence

of (2).  Since the arc-distance of successive points of (3) equals to

l

<

ε

, whence it is evident that any arc with length at least

ε

contains at least one of the points (3).  Consequently, the points (2) are everywhere dense on the perimeter of the unit circle.  Thus the same concerns their projections (

http://planetmath.org/ProjectionOfPoint

) on the

y

-axis, i.e. the sines (1) on the interval

[

-

1

,

1

]

.

Title

sequence accumulating everywhere in

[

-

1

,

\tmspace

+

.1667

⁢

e

⁢

m

⁢

1

]

Canonical

name

SequenceAccumulatingEverywhereIn11

Date of creation

2013-03-22 19:13:39

Last modified on

2013-03-22 19:13:39

Owner

pahio (2872)

Last modified by

pahio (2872)

Numerical id

7

Author

pahio (2872)

Entry type

Example

Classification

msc 54A05

Related topic

LissajousCurves

Generated on Sat Feb 10 11:13:14 2018 by

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spectral space

spectral space

A

topological space

is called

spectral

if

•

it is

compact

,

•

Kolmogorov (also called

T

0

(

http://planetmath.org/T0

)),

•

compactness

is preserved upon finite

intersection

of open compact sets, and

•

any nonempty

irreducible

subspace

of it contains a

generic point

In his thesis, Mel Hochster showed that for any

spectral space

there is

commutative

unitary ring

whose prime spectrum is

homeomorphic

to the spectral space.

References

1

M. Hochster

,
”

Prime Ideal

Structure

in

Commutative Rings

”,
Transactions of American Mathematical Society, Aug. 1969, vol. 142, p. 43-60

Title

spectral space

Canonical

name

SpectralSpace

Date of creation

2013-03-22 16:22:32

Last modified on

2013-03-22 16:22:32

Owner

jocaps (12118)

Last modified by

jocaps (12118)

Numerical id

9

Author

jocaps (12118)

Entry type

Definition

Classification

msc 54A05

Generated on Sat Feb 10 11:13:16 2018 by

LaTeXML
the category of T0 Alexandroff spaces is equivalent to the category of posets

the category of T0 Alexandroff spaces is equivalent to the category of posets

Let

𝒜

⁢

𝒯

be the category of all

T

0

, Alexandroff spaces and

continuous maps

between them. Furthermore let

𝒫

⁢

𝒪

⁢

𝒮

⁢

ℰ

⁢

𝒯

be the category of all posets and order preserving maps.

Theorem

.

The categories

𝒜

⁢

𝒯

and

𝒫

⁢

𝒪

⁢

𝒮

⁢

ℰ

⁢

𝒯

are

equivalent

.

Proof.

Consider two functors:

T

:

𝒜

⁢

𝒯

→

𝒫

⁢

𝒪

⁢

𝒮

⁢

ℰ

⁢

𝒯

;

S

:

𝒫

⁢

𝒪

⁢

𝒮

⁢

ℰ

⁢

𝒯

→

𝒜

⁢

𝒯

,

such that

T

⁢

(

X

,

τ

)

=

(

X

,

≤

)

, where

≤

is an

induced

partial order

on an Alexandroff space and

T

⁢

(

f

)

=

f

for continuous map. Analogously, let

S

⁢

(

X

,

≤

)

=

(

X

,

τ

)

, where

τ

is an induced Alexandroff

topology

on a poset and

S

⁢

(

f

)

=

f

for order preserving maps. One can easily show that

T

and

S

are well defined. Furthermore, it is easy to verify that

equalities

T

∘

S

=

1

𝒫

⁢

𝒪

⁢

𝒮

⁢

ℰ

⁢

𝒯

and

S

∘

T

=

1

𝒜

⁢

𝒯

hold, which

completes

the proof.

□

Remark.

Of course every finite topological space is Alexandroff, thus we have very nice ,,

interpretation

” of finite

T

0

spaces - finite posets (since functors

T

and

S

do not change set-theoretic properties of underlying sets such as finitness).

Title

the category of T0 Alexandroff spaces is equivalent to the category of posets

Canonical name

TheCategoryOfT0AlexandroffSpacesIsEquivalentToTheCategoryOfPosets

Date of creation

2013-03-22 18:46:04

Last modified on

2013-03-22 18:46:04

Owner

joking (16130)

Last modified by

joking (16130)

Numerical id

8

Author

joking (16130)

Entry type

Theorem

Classification

msc 54A05

Generated on Sat Feb 10 11:13:18 2018 by

LaTeXML