开集名称，由何而来？集合

首先开机民本工程，我们考虑实数集合R一个集合：：

T = { ∅ ，I，R }

其中：字母“I”代表没有端的实数区间。

容易验证，拓扑T包含三个元素，而且确实满足拓扑的4项基本要求（公理），因而集合T是R上的一个简易拓扑，。

所以，I，作为拓扑T的元素，称为实数集合R中的“开区间”。

百度一下“无穷小微积分”，进入该站，下载“T？

opology”，打开文件，查找第10页，即可见到本文附件文字。

概念错误，越学越糊涂，

袁萌

附件：拓扑概念的正规定义

5. Denition of Topology and Basic Terminology.

Let’s start by repeating the denition of a topology on a set X:

Definition 5.1.

Let X be a set and let T be a family of subsets of T which satises the following four axioms:

(T1) The empty set ∅ is an element of T.

(T2) The set X is an element of T.

(T3) If Uj ∈T for each j ∈ J thenSj∈J Uj is in T.

(T4) If U1 and U2 are in T then U1 ∩U2 is in T.

then we say that T is a topology on the set X.

Here the axiom (T3) is referred to by saying T is closed under arbitrary unions and axiom (T4) is referred by saying T is closed under pairwise intersection.

（请注意！）

A set X together with a topology T on that set is called a topological space, more formally we will refer to this topological space as (X,T). If (X,T) is a topological space then the elements of T （再次请注意！！）are called open sets in X. Theorem 5.2

(T is closed under nite intersections.). Let T be a topology on a set X, and let U1,U2,...,Un be open sets in this topology where n is a positive integer. Then the intersection U1 ∩U2 ∩···∩Un is an open set.

Proof. We prove this statement by induction on n. If n = 1 then the collection of open sets has only one set U1 and the intersection is just U1 which is an open set. This shows that the statement is true when n = 1. Now suppose the statement is true for a positive integer n (this is the induction hypothesis). To complete the proof by induction we need to show that the statement is true for n+1. So suppose that U1,U2,...,Un+1 are open sets. Then we can write U1 ∩U2 ∩···∩Un+1 = V ∩Un+1 where V = U1 ∩U2 ∩···∩Un. Then V is an open set by the induction hypothesis, and V ∩Un+1 is open by axiom (T4). This shows that the statement is true for n+1 (that is, U1 ∩U2 ∩···∩Un+1 is open) and completes the proof by induction.

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If x is an element of X and U is an open set which contains x then we say that U is a neighborhood of x.

Theorem 5.3.

A subset V ⊆ X is an open set if and only if every element x ∈ V has a neighborhood that is contained in V .

Proof.
(⇒) Suppose that V
is an open set in X. Then for every element x ∈
V , V is a neighborhood of x and V ⊆ V .
(⇐) Suppose that V
is a subset of X and each element x ∈
V has a neighborhood Ux such that Ux ⊆ V . Then V
=Sx∈V Ux and
since each Ux is open then the union is open by axiom (T3). This
shows that V is an open set and completes the proof. [In the second
part of the previous proof, carefully explain the equality V
=Sx∈V Ux?] If T1
and T2 are two topologies on a set X and T2 ⊆ T1, then we say
that T1 is ner than T2, or that T2 is coarser than T1. If T2 ( T1
then T1 is strictly ner than T2. If neither of the
topologiesT1 andT2 are ner than the other then

we say that the two topologies are incomparable. We end this section by describing a number of examples of topological spaces.

Example 5.4.

For any set X let Tdiscrete = P(X) = {A | A ⊆ X}.

Clearly axioms (T1) and (T2) hold since ∅ and X are subsets of X. The union and the intersection of any collection of subsets of X is again a subset of X and this shows that axioms (T3) and (T4) hold as well. Therefore Tdiscrete forms a topology on X and this is called the discrete topology on X. So in the discrete topology, every subset of X is an open set. Therefore Tdiscrete is the largest possible topology on X, which is to say that the discrete topology is ner than every topology on X.

Example 5.5.

For any set X let

Ttrivial = {∅,X}. It is easily veried that Ttrivial is a topology on X and this is called the trivial topology on X. This topology contains only two open sets (assuming that X is a nonempty set). Therefore Ttrivial is the smallest possible topology on X, which is to say that every topology on X is ner than the trivial topology.

Example 5.6.

Just a reminder that for each positive integer n, Teuclid is a topology on Rn called the Euclidean topology. In particular, taking n = 1 gives the Euclidean topology Teuclid on the real line R, referred to briey as the Euclidean line. Here Teuclid = {U ⊂R| for each x ∈ U there is > 0 such that (x−,x + ) ⊆ U}. With this denition it is not hard to show that every open interval in R is an open set in the Euclidean topology. [Be sure you can write out an explanation for this.] But note carefully that there are open sets in the Euclidean topology that are not open intervals. For example, the union of two or more disjoint intervals (such as (−1,1) ∪ (√2,π)) will be an open set which is not an interval.

Example 5.7.

Consider the
collection of subsets T` of the real line R given by T` = {U
⊆R| for each
x ∈
U there is

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This set forms a topology T` on R called the lower limit topology on R. It is not hard to show that

(1) each nite half-open interval of the form [a,b) where a < b is an open set in the lower limit topology but not an open set in the Euclidean topology, while

(2) every open set in the Euclidean topology on R is open in the lower limit topology. Thus the lower limit topology on R is strictly ner than the Euclidean topology on R (that is Teuclid (T`).

Example 5.8.

Let X be a set and let x0 be an element of X. Dene T to be the collection of subsets of X consisting of X itself and all subsets of X which do not contain x0. Thus T = {X}∪{U ⊆ X | x0 / ∈ U} . This forms a topology on X which is called the excluded point topology.

Example 5.9.

Let X be any set and dene Tcofinite = {∅}∪{U ⊆ X | X−U is a nite set } = {∅}∪{X−F | F is a nite subset of X}. Then it is not hard to show thatTcofinite forms a topology on X, called the conite topology on X. (We will explain this using closed sets in the next section.) Note that if the set X itself is nite then every subset of X will be nite, and so the conite topology on a nite set X is the same as the discrete topology on X.