10.2 General topological spaces

中文

Given a set $X$, a topology on $X$ consists of a family of subsets $\mathcal{O}$, called open sets, which satisfy the following conditions:

(Top1) The empty set $\varnothing$ is open and the entire space $X$ is open, $\{\varnothing ,X\}\subset \mathcal{O}$

(Top2) If $U$ and $V$ are open sets then so is their intersection $U\cap V$

$$U\in \mathcal{O}\text{ and }V\in \mathcal{O}⟹U\cap V\in \mathcal{O}.$$

(Top3) If $\left\{V_i\mid i\in I\right\}$ is any family of open sets then their union ${\bigcup }_{i\in I}{V}_i$ is open.

Successive application of (Top2) implies that the intersection of any finite number of open sets is open, but $\mathcal{O}$ is not in general closed with respect to infinite intersections of open sets. On the other hand, $\mathcal{O}$ is closed with respect to arbitrary unions of open sets. The pair $(X,\mathcal{O})$ , where $\mathcal{O}$ is a topology on $X$, is called a topological space. We often refer simply to a topological space $X$ when the topology $\mathcal{O}$ is understood. The elements of the underlying space $X$ are normally referred to as points.

Example 10.1

Define $\mathcal{O}$ to be the collection of subsets $U$ of the real line $\mathbb{R}$ having the property that for every $x\in U$ there exists an open interval $(x-ϵ,x+ϵ)\subseteq U$ for some $ϵ>0$ . These sets agree with the definition of open sets given in Section 10.1. The empty set is assumed to belong to $\mathcal{O}$ by default, while the whole line $\mathbb{R}$ is evidently open since every point lies in an open interval. Thus (Top1) holds for the family $\mathcal{O}$. To prove (Top2) let $U$ and $V$ be open sets such that $U\cap V\ne \varnothing ,$ , the case where $U\cap V=\varnothing$ being trivial. For any $x\in U\cap V$ there exist positive numbers ${ϵ}_1$ and ${ϵ}_2$ such that

$$(x-{ϵ}_1,x+{ϵ}_1)\subseteq U\text{ and }(x-{ϵ}_2,x+{ϵ}_2)\subseteq V.$$

If $ϵ=\min ({ϵ}_1,{ϵ}_2)$ then $(x-ϵ,x+ϵ)\subseteq U\cap V$ , hence $U\cap V$ is an open set.

For (Top3), let $U$ be the union ofan arbitrary collection of open sets $\{U_i\mid i\in I\}$ If $x\in U,$ then $x\in U_j$ for some $j\in I$ and there exists $ϵ>0$ such that $(x-ϵ,x+ϵ)\subseteq U_j\subseteq U$ Hence $U$ is open and the family $\mathcal{O}$ forms a topology for $\mathbb{R}$. It is often referred to as the standard topology on $\mathbb{R}$. Any open interval $(a,b)$ where $a<b$ is an open set, for if $x\in (a,b)$ then $(x-ϵ,x+ϵ)\subset (a,b)$ for $\begin{array}{r}ϵ=\frac{1}{2}\min (x-a,b-x)\end{array}$ . A similar argument shows that the intervals may also be of semi-infinite extent, such as $(-\infty ,a)$ or $(b,\infty )$ Notice that infinite intersections of open sets do not generally result in an open set. For example, an isolated point $\{a\}$ is not an open set since it contains no finite open interval, yet it is the intersection of an infinite sequence of open intervals such as

$$(a-1,a+1),(a-\frac{1}{2},a+\frac{1}{2}),(a-\frac{1}{3},a+\frac{1}{3}),(a-\frac{1}{4},a+\frac{1}{4}),\dots$$

Similar arguments can be used to show that the open sets defined on ${\mathbb{R}}^2$ in Section 10.1 form a topology. Similarly, in ${\mathbb{R}}^n$ we define a topology where a set $U$ is said to be open if for every point $\mathbf{x}\in U$ there exists an open ball

$$B_r(\mathbf{x})=\{\mathbf{y}\in {\mathbb{R}}^2||\mathbf{y}-\mathbf{x}|<r\}\subset U,$$

where

$$|\mathbf{y}-\mathbf{x}|=\sqrt{(y_1-x_1{)}^2+(y_2-x_2{)}^2+\cdots +(y_n-x_n{)}^2}.$$

This topology will again be termed the standard topology on ${\mathbb{R}}^n$

Example 10.2

Consider the family ${\mathcal{O}}^{\prime }$ of all open intervals on $\mathbb{R}$ of the form $(-a,b)$ where $a,b>0;$ , together with the empty set. All these intervals contain the origin 0. It is not hard to show that (Top1)–(Top3) hold for this family and that $(X,{\mathcal{O}}^{\prime })$ is a topological space. This space is not very ‘nice’ in some of its properties. For example no two points $x$, $y\in \mathbb{R}$ lie in non-intersecting neighbourhoods. In a sense all points of the line are ‘arbitrarily close to each other in this topology.

A subset $V$ is called closed if its complement $X-V$ is open. The empty set and the whole space are clearly closed sets, since they are both open sets and are the complements of each other. The intersection of an arbitrary family of closed sets is closed, as it is the complement of a union of open sets. However, only finite unions of closed sets are closed in general.

Example 10.3

Every closed interval $[a,b]=\{x\mid a\le x\le b\}$ where $-\infty <a\le b<\infty$ is a closed set, as it is the complement of the open set $(-\infty ,a)\cup (b,\infty )$ . Every singleton set consisting of an isolated point $\{a\}\equiv [a,a]$ is closed. Closed intervals $[a,b]$ are not open sets since the end points $a$ or $b$ do not belong to any open interval included in [$a$, $b$].

If $A$ is any subset of $X$ the relative topology on $A$, or topology induced on $A$, is the topology whose open sets are

$${\mathcal{O}}_A=\{A\cap U|U\in \mathcal{O}\}.$$

Thus a set is open in the relative topology on $A$ iff it is the intersection of $A$ and an open set $U$ in $X$ (see Fig. 10.2). That these sets form a topology on $A$ follows from the following three facts:

1. $\varnothing \cap A=\varnothing ,X\cap A=A.$

2. $(U\cap A)\cap (V\cap A)=(U\cap V)\cap A.$

3. ${\bigcup }_{i\in I}(U_i\cap A)=\left({\bigcup }_{i\in I}{U}_i\right)\cap A.$

A subset $A$ of $X$ together with the relative topology ${\mathcal{O}}_A$ induced on it is called a subspace of $(X,\mathcal{O})$

Figure 10.2 Relative topology induced on a subset of a topological space

Example 10.4

The relative topology on the half-open interval $A=[0,1)\subset \mathbb{R}$ , induced on $A$ by the standard topology on $\mathbb{R}$, is the union of half-open intervals of the form $[0,a)$

where $0<a<1$ , and all intervals of the form $(a,b)$ where $0<a<b\le 1$ . Evidently some of the open sets in this topology are not open in $\mathbb{R}$.

Exercise

Show that if $A\subseteq X$ is an open set then all open sets in the relative topology on $A$ are open in $X$.

Exercise

If $A$ is a closed set, show that every closed set in the induced topology on $A$ is closed in $X$.

A point $x$ is said to be an accumulation point of a set $A$ if every open neighbourhood $U$ of $x$ contains points of $A$ other than $x$ itself, as shown in Fig. 10.3. What this means is that $x$ may or may not lie in $A$, but points of $A$ ‘cluster’ arbitrarily close to it (sometimes it is also called a cluster point of $A$). $A$ related concept is commonly applied to sequences of points $x_n\in X$ . We say that the sequence $x_n\in X$ converges to $x\in X$ or that $x$ is a limit point of $\{x_n\}$ , denoted $x_n\to x,$ , if for every open neighbourhood $U$ of $x$ there is an integer $N$ such that $x_n\in U$ for all $n\ge N$ . This differs from an accumulation point in that we could have $x_n=x$ for all $n>n_0$ for some $n_0$

Figure 10.3 Accumulation point of a set

The closure of any set $A$, denoted $\overline{A}$ , is the union of the set $A$ and all its accumulation points. The interior of $A$ is the union of all open sets $U\subseteq A,$ , denoted $A^o$ . The difference of these two sets, $b(A)=\overline{A}-A^o$ , is called the boundary of $A$.

Theorem 10.2 · Basic properties of closure, interior and boundary

The closure of any set $A$ is a closed set. The interior $A^o$ is the largest open set included in $A$. The boundary $b(A)$ is a closed set.

Proof

Let $x$ be any point not in $\overline{A}.$ . Since $x$ is not in $A$ and is not an accumulation point of $A$, it has an open neighbourhood $U_x$ not intersecting $A$. Furthermore, $U_x$ cannot contain any other accumulation point of $A$ else it would be an open neighbourhood of that point not intersecting $A$. Hence the complement $X-\overline{A}$ of the closure of $A$ is the union of the open sets $U_x$ . It is therefore itself an open set and its complement $\overline{A}$ is a closed set.

Since the interior $A^o$ is a union of open sets, it is an open set by (Top3). If $U$ is any open set such that $U\subseteq A$ then, by definition, $U\subseteq A^o$ . Thus $A^o$ is the largest open subset of $A$. Its complement is closed and the boundary $b(A)=\overline{A}\cap (X-A^o)$ is necessarily a closed set.

Exercise

Show that a set $A$ is closed if and only if it contains its boundary, $A\supseteq b(A)$

Exercise

A set $A$ is open if and only if $A\cap b(A)=\varnothing .$

Exercise

Show that all accumulation points of $A$ lie in the boundary $b(A)$

Exercise

Show that a point $x$ lies in the boundary of $A$ iff every neighbourhood of $x$ contains point both in $A$ and not in $A$.

Example 10.5

The closure of the open ball $B_a(\mathbf{x})\subset {\mathbb{R}}^n$ (see Example 10.1) is the closed ball

$$\overline{B_a}(\mathbf{x})=\{\mathbf{y}||\mathbf{y}-\mathbf{x}|\le a\}.$$

Since every open ball is an open set, it is its own interior, $B_a^o(\mathbf{x})=B_a(\mathbf{x})$ and its boundary is the $(n-1)$ -sphere of radius $a$, centre $x$,

$$b(B_a(\mathbf{x}))=S_a^{n-1}(\mathbf{x})=\{y\mid |\mathbf{y}-\mathbf{x}|=a\}.$$

Example 10.6

A set whose closure is the entire space $X$ is said to be dense in $X.$ . For example, since every real number has rational numbers arbitrarily close to it, the rational numbers $\mathbb{Q}$ are a countable set that is dense in the set of real numbers. In higher dimensions the situation is similar. The set of points with rational coordinates ${\mathbb{Q}}^n$ is a countable set that is dense in ${\mathbb{R}}^n$

Exercise

Show that $\mathbb{Q}$ is neither an open or closed set in $\mathbb{R}.$

Exercise

Show that ${\mathbb{Q}}^o=\varnothing$ and $b(\mathbb{Q})=\mathbb{R}$

It is sometimes possible to compare different topologies ${\mathcal{O}}_1$ and ${\mathcal{O}}_2$ on a set $X.$ . We say ${\mathcal{O}}_1$ is finer or stronger than ${\mathcal{O}}_2$ if ${\mathcal{O}}_1\supseteq {\mathcal{O}}_2$ . Essentially, ${\mathcal{O}}_1$ has more open sets than ${\mathcal{O}}_2$ In this case we also say that ${\mathcal{O}}_2$ is coarser or weaker than ${\mathcal{O}}_1$

Example 10.7

All topologies on a set $X$ lie somewhere between two extremes, the discrete and indiscrete topologies. The indiscrete or trivial topology consists simply of the empty set and the whole space itself, ${\mathcal{O}}_1=\{\varnothing ,X\}$ . It is the coarsest possible topology on $X-$ if $\mathcal{O}$ is any other topology then ${\mathcal{O}}_1\subseteq \mathcal{O}$ by (Top1). The discrete topology consists of all subsets of $X,{\mathcal{O}}_2=2^X$ . This topology is the finest possible topology on $X,$ , since it includes all other topologies ${\mathcal{O}}_2\supseteq \mathcal{O}$ . For both topologies (Top1)–(Top3) are trivial to verify.

Given a set $X,$ , and an arbitrary collection of subsets $\mathcal{U}$, we can ask for the weakest topology $\mathcal{O}(\mathcal{U})$ containing $\mathcal{U}$. This topology is the intersection of all topologies that contain $\mathcal{U}$ and is called the topology generated by $\mathcal{U}$. It is analogous to the concept of the vector subspace $L(M)$ generated by an arbitrary subset $M$ of a vector space $V$ (see Section 3.5).

A constructive way of defining $\mathcal{O}(\mathcal{U})$ is the following. Firstly, adjoin the empty set and the entire space $X$ to $\mathcal{U}$ if they are not already in it. Next, extend $\mathcal{U}$ to a family $\hat{\mathcal{U}}$ consisting of all finite intersections $U_1\cap U_2\cap \cdots \cap U_n$ of sets $U_i\in \mathcal{U}\cup \{\varnothing ,X\}$ . Finally, the set $\mathcal{O}(\mathcal{U})$ consisting of arbitrary unions of sets from $\hat{\mathcal{U}}$ forms a topology. To prove (Top2),

$$\bigcup _{i\in I}\left(\bigcap _{a=1}^{n_i}{U}_{ia}\right)\cap \bigcup _{j\in J}\left(\bigcap _{b=1}^{n_j}{V}_{jb}\right)=\bigcup _{i\in I}\bigcup _{j\in J}\left(U_{i1}\cap \cdots \cap U_{i{n}_i}\cap V_{j1}\cap \cdots \cap V_{j{n}_j}\right).$$

Property (Top3) follows immediately from the construction.

Example 10.8

On the real line $\mathbb{R}$, the family of all open intervals generates the standard topology since every open set is a union of open sets of the form $(x-ϵ,x+ϵ)$ . Similarly, the standard topology on ${\mathbb{R}}^2$ is generated by the set of open balls

$$\mathcal{U}=\{B_a(\mathbf{r})|a>0,\mathbf{r}=(x,y)\in {\mathbb{R}}^2\}.$$

To prove this statement we must show that every set that is an intersection of two open balls $B_a(\mathbf{r})$ and $B_b({\mathbf{r}}^{\prime })$ is a union of open balls from $\mathcal{U}$. If $\mathbf{x}\in B_a(\mathbf{r})$ , let $ϵ<a$ be such that $B_{ϵ}(\mathbf{x})\subset B_a(\mathbf{r})$ . Similarly if $\mathbf{x}\in B_b({\mathbf{r}}^{\prime })$ , let ${ϵ}^{\prime }<a$ be such that $B_{{ϵ}^{\prime }}(\mathbf{x})\subset B_b({\mathbf{r}}^{\prime })$ . Hence, if $\mathbf{x}\in B_a(\mathbf{r})\cap B_b({\mathbf{r}}^{\prime })$ then $B_{{ϵ}^{\prime \prime }}(\mathbf{x})\subset B_a(\mathbf{r})\cap B_b({\mathbf{r}}^{\prime })$ where ${ϵ}^{\prime \prime }=\min (ϵ,{ϵ}^{\prime })$ . The proof easily generalizes to intersections of any finite number of open balls. Hence the standard topology of ${\mathbb{R}}^2$ is generated by the set of all open balls. The extension to ${\mathbb{R}}^n$ is straightforward.

Exercise

Show that the discrete topology on $X$ is generated by the family of all singleton sets $\{x\}$ where $x\in X$

A set $A$ is said to be a neighbourhood of $x\in X$ if there exists an open set $U$ such that $x\in U\subset A$ . If $A$ itself is open it is called an open neighbourhood of $x$. A topological space $X$ is said to be first countable if every point $x\in X$ has a countable collection $U_1(x),U_2(x),\dots$ . of open neighbourhoods of $x$ such that every open neighbourhood $U$ of $x$ includes one of these neighbourhoods $U\supset U_n(x)$ . A stronger condition is the following: a topological space $(X,\mathcal{O})$ is said to be second countable or separable if there exists a countable set $U_1,U_2,U_3,...$ . that generates the topology of $X$.

Example 10.9

The standard topology of the Euclidean plane ${\mathbb{R}}^2$ is separable, since it is generated by the set of all rational open balls,

$${\mathcal{B}}_{\text{ rat }}=\left\{B_a(\mathbf{r})\mid a>0\in \mathbb{Q},\mathbf{r}=(x,y)\text{ s.t. }x,y\in \mathbb{Q}\right\}.$$

The set $B_{\mathrm{rat}}$ is countable as it can be put in one-to-one correspondence with a subset of ${\mathbb{Q}}^3$ Since the rational numbers are dense in the real numbers, every point $x$ ofan open set $U$ lies in a rational open ball. Thus every open set is a union of rational open balls. By a similar argument to that used in Example 10.8 it is straightforward to prove that the intersection of two sets from $B_{\mathrm{rat}}$ is a union of rational open balls. Hence ${\mathbb{R}}^2$ is separable. Similarly, all spaces ${\mathbb{R}}^n$ where $n\ge 1$ are separable.

Let $X$ and $Y$ be two topological spaces. Theorem 10.1 motivates the following definition: a function $f:X\to Y$ is said to be continuous if the inverse image $f^{-1}(U)$ of every open set $U$ in $Y$ is open in $X$. If $f$ is one-to-one and its inverse $f^{-1}:Y\to X$ is continuous, the function is called a homeomorphism and the topological spaces $X$ and $Y$ are said to be homeomorphic or topologically equivalent, written $X\cong Y$ . The main task of topology is to find topological invariants – properties that are preserved under homeomorphisms. They may be real numbers, algebraic structures such as groups or vector spaces constructed from the topological space, or specific properties such as compactness and connectedness. The ultimate goal is to find a set oftopological invariants that characterize a topological space. In the language of category theory, Section 1.7, continuous functions are the morphisms of the category whose objects are topological spaces, and homeomorphisms are the isomorphism of this category.

Example 10.10

Let $f:X\to Y$ be a continuous function between topological spaces. If the topology on $X$ is discrete then every function $f$ is continuous, for no matter what the topology on $Y$, every inverse image set $f^{-1}(U)$ is open in $X$. Similarly if the topology on $Y$ is indiscrete than the function $f$ is always continuous since the only inverse images in $X$ of open sets are $f^{-1}(\varnothing )=\varnothing$ and $f^{-1}(Y)=X$ , which are always open sets by (Top1).

Problems

Problem 10.1

Give an example in ${\mathbb{R}}^2$ of each of the following:

(a) A family of open sets whose intersection is a closed set that is not open.

(b) A family of closed sets whose union is an open set that is not closed.

(c) A set that is neither open nor closed.

(d) A countable dense set.

(e) A sequence of continuous functions $f_n:{\mathbb{R}}^2\to \mathbb{R}$ whose limit is a discontinuous function

Problem 10.2

If $\mathcal{U}$ generates the topology on $X$ show that $\{A\cap U\mid U\in \mathcal{U}\}$ generates the relative topology on $A$.

Problem 10.3

Let $X$ be a topological space and $A\subset B\subset X$ . If $B$ is given the relative topology, show that the relative topology induced on $A$ by $B$ is identical to the relative topology induced on it by $X$.

Problem 10.4

Show that for any subsets $U$, $V$ of a topological space $\overline{U\cup V}=\overline{U}\cup \overline{V}.$ . Is it true that $\overline{U\cap V}=\overline{U}\cap \overline{V}?$ What corresponding statements hold for the interior and boundaries of unions and intersections of sets?

Problem 10.5

If $A$ is a dense set in a topological space $X$ and $U\subseteq X$ is open, show that $U\subseteq \overline{A\cap U}$

Problem 10.6

Show that a map $f:X\to Y$ between two topological spaces $X$ and $Y$ is continuous if and only if $f(\overline{U})\subseteq \overline{f(U)}$ for all sets $U\subseteq X.$ . Show that $f$ is a homeomorphism only if $f(\overline{U})=\overline{f(U)}$ for all sets $U\subseteq X$

Problem 10.7

Show the following:

(a) In the trivial topology, every sequence $x_n$ converges to every point of the space $x\in X.$

(b) In $R^2$ the family of open sets consisting of all open balls centred on the origin $B_r(0)$ is a topology. Any sequence ${\mathbf{x}}_n\to \mathbf{x}$ converges to all points on the circle of radius $|\mathbf{x}|$ centred on the origin.

(c) If $C$ is a closed set of a topological space $X$ it contains all limit points of sequences $x_n\in C$

(d) Let $f:X\to Y$ be a continuous function between topological spaces $X$ and $Y$. If $x_n\to x$ is any convergent sequence in $X$ then $f(x_n)\to f(x)$ in $Y$.

Problem 10.8

If $W$, $X$ and $Y$ are topological spaces and the functions $f:W\to X,g:X\to Y$ are both continuous, show that the function $h=g\circ f:W\to Y$ is continuous.