トップページ⇒math
32コメント8KB

test

■ このスレッドは過去ログ倉庫に格納されています
0001test2010/03/20(土) 07:28:12
test
0002test2010/03/20(土) 07:29:53
白と黄色の本の問題解きます。
0003１３２人目の素数さん2010/03/20(土) 10:04:52
氏ねクズ
0004test2010/03/20(土) 20:29:57
>3 通報死マスタ／
0005１３２人目の素数さん2010/03/20(土) 21:25:05
http://at.yorku.ca/list/qa.htm#ta
0006１３２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？」
0007１３２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?
0008１３２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?
0009１３２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
0010１３２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
0011１３２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
0012１３２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
0013１３２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)

0014１３２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'
0015１３２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
0016１３２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'
0017１３２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.
0018１３２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.

0019１３２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
0020１３２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
0021１３２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.
0022 ◆27Tn7FHaVY 2010/03/23(火) 00:28:06
what is your aim?
0023猫の嵐 ◆ghclfYsc82 2010/03/23(火) 08:44:55
Mathematics is very good to be discussed here at 2chan.

--neko--
0024１３２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.''
0025１３２人目の素数さん2010/05/06(木) 07:44:26
test
0026１３２人目の素数さん2010/06/22(火) 15:25:55
test
0027１３２人目の素数さん2010/06/23(水) 20:31:53
test
0028１３２人目の素数さん2010/06/24(木) 20:43:15
0029 p177219.amixcom.jp2010/06/25(金) 03:41:53
tes
0030ﾌfusianasan2010/06/25(金) 03:43:06
tes
0031ﾞfusianasan2010/06/25(金) 03:44:51
test
0032１３２人目の素数さん2010/06/28(月) 21:26:24
test
■ このスレッドは過去ログ倉庫に格納されています