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test

1 :test:2010/03/20(土) 07:28:12
test

2 :test:2010/03/20(土) 07:29:53
白と黄色の本の問題解きます。

3 :132人目の素数さん:2010/03/20(土) 10:04:52
氏ねクズ

4 :test:2010/03/20(土) 20:29:57
>3 通報死マスタ/

5 :132人目の素数さん:2010/03/20(土) 21:25:05
http://at.yorku.ca/list/qa.htm#ta

6 :132人目の素数さん:2010/03/20(土) 21:55:54
幼稚園児レベルらしいいんだが、これ分かるか? おいらには分からんかった。

「66233のうち、23443が2999の場合、56890のうち22310はいくつになるか?」

7 :132人目の素数さん:2010/03/21(日) 01:21:47
When does the 0-dimensional Reflection Preserve Products?

Andrej Bauer

A Question about Zero-dimensional Spaces

We say that a topological space is zero-dimensional when the collection of its clopen (closed and open) subsets forms a basis for its topology.

For a topological space X, let z(X) be its zero-dimensional reflection, i.e., the same underlying set X but with the topology generated by the collection of the clopen subsets of X.

Question 1: Does z preserve products, i.e., is z(X ×Y) homeomorphic to z(X) ×z(Y)?

The question can be stated equivalently as follows.

Question 1': Is every clopen subset of X ×Y a union of the sets of the form U ×V, where U 'subset or equal' X is clopen and V 'subset or equal' Y is clopen?

8 :132人目の素数さん:2010/03/21(日) 01:32:14
The Sierpinski Triangle

Sonia Sabogal

The Sierpinski Triangle is obtained as the residual set remaining when
one begins with a triangle and applies the operations of dividing it
into four triangles, connecting the middle points of the triangle's
edges, and omitting the interior of the center one, then repeats this
operation on each of the surviving 3 triangles, then repeats again on
the surviving 9 triangles, and so on indefinitely. This space is
homeomorphic to the unique nonempty compact set K of the complex plane
that satisfies: K = w1(K) \cup w2(K) \cup w3(K) where w1, w2, w3 are
maps of the complex plane defined by: w1(z) = 0.5z, w2(z) = 0.5z+0.5
and w3(z) = 0.5z+0.5i (see W. J. Charatonik and A. Dilks; Topol. and
its Appl.; 55(1994), 215-238, Example 2.7).

Question: Is there a known topological characterization of Sierpinski triangle?

9 :132人目の素数さん:2010/03/21(日) 02:44:23
>>7
clopenって「閉かつ開」って意味なのか
初めて知った

10 :132人目の素数さん:2010/03/21(日) 12:51:26
Theorem 4.1.1: For a topological space (X, T), the following are true:
(i) φ and X are closed sets
(ii) Intersection of any number of closed sets is closed
(iii) Union of finitely many closed sets is closed.

φ<(X, T)ー>φ^c=closed
AUBU...<(X,T)->(AUBU...)^c=A^c&B^c&..=closed
A&B&...<(X,T)->(A&B&...)^c=A^cUB^cU..=closed

11 :132人目の素数さん:2010/03/21(日) 13:04:33
Theorem 4.2.1: In a topological space the following are euqivalent:
(i) G is open
(ii) G = G0

G<(X,T)->G<G<(X,T)->G<G0
G0<G
so G=G0

12 :132人目の素数さん:2010/03/21(日) 13:15:30
Theorem 4.2.2: In a topological space (X, T), the following are equivalent:
(i) K0 = U {G < K; G<T}
(ii) K0 is the largest open set contained in K.
(i)->K0<G<K->K0<K0,K0^c<!G<T
(i)->K0=UG<T->K0=open
G<S<K,S<T->S<U {G < K; G<T}=K0->S<K0

13 :132人目の素数さん:2010/03/22(月) 07:16:19
Theorem 4.2.4: The following are equivalent in a space (X, T)
(i) K~ = & {F > K; F is closed in X}
(ii) K~ is smallest closed set containing K.

(i)->F=F~=F+F'->K~=K+K'<! F->K'(>a)<!F->Na'&K=p<F=F+F'->Na'&F=!φ->a<F'!!
K~<F->b<!K~<F->K~=closed<F->K~(>K)&(F>K)<&(F>K)->b<K~!!->(ii)



14 :132人目の素数さん:2010/03/22(月) 10:20:57
Theorem 4.2.5: In (X, T) the following are equivalent
(i) F is closed
(ii) F = F~=F+F'
(iii) F' < F

(i)->F=F+F'>F'->(iii)->(i)
(ii)->F'<F->(iii),(iii)->F+F'<F<F+F'->(ii)
(i)=(iii)=(ii)

F=F&F'


15 :132人目の素数さん:2010/03/22(月) 10:59:34
Theorem 4.2.6: In a topological space (X, T), let Np denote the family of all neighbourhood of a
point p of X, then the following are true
(i) p belongs to each member of Np .
(ii) Every superset of a member of Np belongs to Np .
(iii) For every member N of Np there is another member G of Np , which is the neighbourhood
of each of its points.

(i)p<G<Np
(ii)V,U<Np->(V,U)<Np=UNp
(iii)N<Np->p<G<N<Np,G=open->G<Np and G<Ng,g<G

16 :132人目の素数さん:2010/03/22(月) 16:13:43
Theorem 4.4.1: A non-empty class of subsets of a set X is a base for a topology of X iff
(i) X = {B; B<B}
(ii) For any B1,B2<B B1&B2 is the union of members of B, i.e., if p <B1&B2, then there
exists B'<B such that p<B'<B<B1&B2.

(i)->φ,X<UB
(ii)->&T<(UB)&(UB)=U(B&B)=UB'

17 :132人目の素数さん:2010/03/22(月) 18:36:31
First Axiom of Countability. A topological space has a countable local base at each of its points.
Second Axiom of Countability. A topological space has a countable base.

Definition: A topological space is called first countable (respectively second countable) if it satisfies
the first (respectively second) axiom of countability.

Lindeloff’s Theorem: Every non-empty open set in a second countable space can be represented by a
countable union of basic open sets.

Corollary: Every open base of a second countable space has a countable subclass which is also an
open base.

Theorem 4.4.2: Every second countable space is separable.

18 :132人目の素数さん:2010/03/22(月) 19:09:59
Theorem 4.4.3: Every separable metric space is second countable.



19 :132人目の素数さん:2010/03/22(月) 19:42:13
Theorem 5.1.1: A topological space (X, T) is a T1 -space iff every singleton subset of X is closed.

x,y<X->x^c=open>y,y^c=open>x

20 :132人目の素数さん:2010/03/22(月) 19:44:59
Corollary: Every finite subset of a T1 -space is closed.

S=U{x}=Closed

21 :132人目の素数さん:2010/03/22(月) 19:51:00
Definition: A topological space (X, T ) is a called a T2 -space or a Hausdorff space if it satisfies the T2 -
axiom of separation.
Note the following then:
(a) Every Hausdorff space is a T1-space.
(b) Every metric space is a Hausdorff space.
(c) Every subspace of Hausdorff space is Hausdorff.
(d) The cartesian product of Hausdorff spaces is Hausdorff.
(e) The set {(x, x); x < X} is closed in X x X if X is Hausdorff.
(f) Every convergent sequence in a Hausdorff space converges to a unique limit.

22 : ◆27Tn7FHaVY :2010/03/23(火) 00:28:06
what is your aim?

23 :猫の嵐 ◆ghclfYsc82 :2010/03/23(火) 08:44:55
Mathematics is very good to be discussed here at 2chan.

--neko--


24 :132人目の素数さん:2010/03/25(木) 21:00:46
conquer the universe

Aimy high/

Behind the Old French expression that evolved into Amy is a more ancient Latin verb, amare, meaning ''to love.''

25 :132人目の素数さん:2010/05/06(木) 07:44:26
test

26 :132人目の素数さん:2010/06/22(火) 15:25:55
test

27 :132人目の素数さん:2010/06/23(水) 20:31:53
test

28 :132人目の素数さん:2010/06/24(木) 20:43:15


29 : p177219.amixcom.jp:2010/06/25(金) 03:41:53
tes

30 :フfusianasan:2010/06/25(金) 03:43:06
tes

31 :゙fusianasan:2010/06/25(金) 03:44:51
test

32 :132人目の素数さん:2010/06/28(月) 21:26:24
test

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