Topological Preliminaries

2.3 Definitions Let $X$ be a topological space, as defined in Sec. 1.2.

(a) A set $E \subset X$ is closed if its complement $E^c$ is open. (Hence $\emptyset$ and $X$ are closed, finite unions of closed sets are closed, and arbitrary intersections of closed sets are closed.)

(b) The closure $\bar{E}$ of a set $E \subset X$ is the smallest closed set in $X$ which contains $E$ . (The following argument proves the existence of $\bar{E}$ : The collection $\Omega$ of all closed subsets of $X$ which contain $E$ is not empty, since $X \in \Omega$ ; let $\bar{E}$ be the intersection of all members of $\Omega$ .)

(c) A set $K \subset X$ is compact if every open cover of $K$ contains a finite subcover. More explicitly, the requirement is that if $\{V_\alpha\}$ is a collection of open sets whose union contains $K$ , then the union of some finite subcollection of $\{V_\alpha\}$ also contains $K$ .
In particular, if $X$ is itself compact, then $X$ is called a compact space.

(d) A neighborhood of a point $p \in X$ is any open subset of $X$ which contains $p$ . (The use of this term is not quite standardized; some use

“neighborhood of $p$ ” for any set which contains an open set containing $p$ .)

(e) $X$ is a Hausdorff space if the following is true: If $p \in X$ , $q \in X$ , and $p \neq q$ , then $p$ has a neighborhood $U$ and $q$ has a neighborhood $V$ such that $U \cap V = \emptyset$ .

(f) $X$ is locally compact if every point of $X$ has a neighborhood whose closure is compact.

Obviously, every compact space is locally compact.

We recall the Heine-Borel theorem: The compact subsets of a euclidean space $R^n$ are precisely those that are closed and bounded ([26],† Theorem 2.41). From this it follows easily that $R^n$ is a locally compact Hausdorff space. Also, every metric space is a Hausdorff space.

2.4 Theorem Suppose $K$ is compact and $F$ is closed, in a topological space $X$ . If $F \subset K$ , then $F$ is compact.

PROOF If $\{V_\alpha\}$ is an open cover of $F$ and $W = F^c$ , then $W \cup \bigcup_\alpha V_\alpha$ covers $X$ ; hence there is a finite collection $\{V_{\alpha_i}\}$ such that

Then $F \subset V_{\alpha_1} \cup \cdots \cup V_{\alpha_n}$ . //////

Corollary If $A \subset B$ and if $B$ has compact closure, so does $A$ .

2.5 Theorem Suppose $X$ is a Hausdorff space, $K \subset X$ , $K$ is compact, and $p \in K^c$ . Then there are open sets $U$ and $W$ such that $p \in U$ , $K \subset W$ , and $U \cap W = \emptyset$ .

PROOF If $q \in K$ , the Hausdorff separation axiom implies the existence of disjoint open sets $U_q$ and $V_q$ , such that $p \in U_q$ and $q \in V_q$ . Since $K$ is compact, there are points $q_1, \dots, q_n \in K$ such that

Our requirements are then satisfied by the sets

Corollaries

(a) Compact subsets of Hausdorff spaces are closed.

(b) If $F$ is closed and $K$ is compact in a Hausdorff space, then $F \cap K$ is compact.

Corollary (b) follows from (a) and Theorem 2.4.

† Numbers in brackets refer to the Bibliography.

2.6 Theorem If $\{K_\alpha\}$ is a collection of compact subsets of a Hausdorff space and if $\bigcap_\alpha K_\alpha = \emptyset$ , then some finite subcollection of $\{K_\alpha\}$ also has empty intersection.

PROOF Put $V_\alpha = K_\alpha^c$ . Fix a member $K_1$ of $\{K_\alpha\}$ . Since no point of $K_1$ belongs to every $K_\alpha$ , $\{V_\alpha\}$ is an open cover of $K_1$ . Hence $K_1 \subset V_{\alpha_1} \cup \cdots \cup V_{\alpha_n}$ for some finite collection $\{V_{\alpha_i}\}$ . This implies that

2.7 Theorem Suppose $U$ is open in a locally compact Hausdorff space $X$ , $K \subset U$ , and $K$ is compact. Then there is an open set $V$ with compact closure such that

PROOF Since every point of $K$ has a neighborhood with compact closure, and since $K$ is covered by the union of finitely many of these neighborhoods, $K$ lies in an open set $G$ with compact closure. If $U = X$ , take $V = G$ .

Otherwise, let $C$ be the complement of $U$ . Theorem 2.5 shows that to each $p \in C$ there corresponds an open set $W_p$ such that $K \subset W_p$ and $p \notin \bar{W}_p$ . Hence $\{C \cap \bar{G} \cap \bar{W}_p\}$ , where $p$ ranges over $C$ , is a collection of compact sets with empty intersection. By Theorem 2.6 there are points $p_1, \dots, p_n \in C$ such that

The set

then has the required properties, since

2.8 Definition Let $f$ be a real (or extended-real) function on a topological space. If

is open for every real $\alpha$ , $f$ is said to be lower semicontinuous. If

is open for every real $\alpha$ , $f$ is said to be upper semicontinuous.

A real function is obviously continuous if and only if it is both upper and lower semicontinuous.

The simplest examples of semicontinuity are furnished by characteristic functions:

(a) Characteristic functions of open sets are lower semicontinuous.
(b) Characteristic functions of closed sets are upper semicontinuous.

The following property is an almost immediate consequence of the definitions:

(c) The supremum of any collection of lower semicontinuous functions is lower semicontinuous. The infimum of any collection of upper semicontinuous functions is upper semicontinuous.

2.9 Definition The support of a complex function $f$ on a topological space $X$ is the closure of the set

The collection of all continuous complex functions on $X$ whose support is compact is denoted by $C_c(X)$ .

Observe that $C_c(X)$ is a vector space. This is due to two facts:

(a) The support of $f+g$ lies in the union of the support of $f$ and the support of $g$ , and any finite union of compact sets is compact.
(b) The sum of two continuous complex functions is continuous, as are scalar multiples of continuous functions.

(Statement and proof of Theorem 1.8 hold verbatim if “measurable function” is replaced by “continuous function,” “measurable space” by “topological space”; take $\Phi(s, t) = s + t$ , or $\Phi(s, t) = st$ , to prove that sums and products of continuous functions are continuous.)

2.10 Theorem Let $X$ and $Y$ be topological spaces, and let $f: X \to Y$ be continuous. If $K$ is a compact subset of $X$ , then $f(K)$ is compact.

PROOF If $\{V_\alpha\}$ is an open cover of $f(K)$ , then $\{f^{-1}(V_\alpha)\}$ is an open cover of $K$ , hence $K \subset f^{-1}(V_{\alpha_1}) \cup \cdots \cup f^{-1}(V_{\alpha_n})$ for some $\alpha_1, \dots, \alpha_n$ , and therefore $f(K) \subset V_{\alpha_1} \cup \cdots \cup V_{\alpha_n}$ . ////

Corollary The range of any $f \in C_c(X)$ is a compact subset of the complex plane.

In fact, if $K$ is the support of $f \in C_c(X)$ , then $f(X) \subset f(K) \cup \{0\}$ . If $X$ is not compact, then $0 \in f(X)$ , but $0$ need not lie in $f(K)$ , as is seen by easy examples.

2.11 Notation In this chapter the following conventions will be used. The notation

will mean that $K$ is a compact subset of $X$ , that $f \in C_c(X)$ , that $0 \le f(x) \le 1$ for all $x \in X$ , and that $f(x) = 1$ for all $x \in K$ . The notation

will mean that $V$ is open, that $f \in C_c(X)$ , $0 \le f \le 1$ , and that the support of $f$ lies in $V$ . The notation

will be used to indicate that both (1) and (2) hold.

2.12 Urysohn’s Lemma Suppose $X$ is a locally compact Hausdorff space, $V$ is open in $X$ , $K \subset V$ , and $K$ is compact. Then there exists an $f \in C_c(X)$ , such that

In terms of characteristic functions, the conclusion asserts the existence of a continuous function $f$ which satisfies the inequalities $\chi_K \le f \le \chi_V$ . Note that it is easy to find semicontinuous functions which do this; examples are $\chi_K$ and $\chi_V$ .

PROOF Put $r_1 = 0$ , $r_2 = 1$ , and let $r_3, r_4, r_5, \dots$ be an enumeration of the rationals in $(0, 1)$ . By Theorem 2.7, we can find open sets $V_0$ and then $V_1$ such that $\bar{V}_0$ is compact and

Suppose $n \ge 2$ and $V_{r_1}, \dots, V_{r_n}$ have been chosen in such a manner that $r_i < r_j$ implies $\bar{V}_{r_j} \subset V_{r_i}$ . Then one of the numbers $r_1, \dots, r_n$ , say $r_i$ , will be the largest one which is smaller than $r_{n+1}$ , and another, say $r_j$ , will be the smallest one larger than $r_{n+1}$ . Using Theorem 2.7 again, we can find $V_{r_{n+1}}$ so that

Continuing, we obtain a collection $\{V_r\}$ of open sets, one for every rational $r \in [0, 1]$ , with the following properties: $K \subset V_1$ , $\bar{V}_0 \subset V$ , each $\bar{V}_r$ is compact, and

Define

and

The remarks following Definition 2.8 show that $f$ is lower semicontinuous and that $g$ is upper semicontinuous. It is clear that $0 \le f \le 1$ , that

$f(x) = 1$ if $x \in K$ , and that $f$ has its support in $\bar{V}_0$ . The proof will be completed by showing that $f = g$ .

The inequality $f_r(x) > g_s(x)$ is possible only if $r > s$ , $x \in V_r$ , and $x \notin \bar{V}_s$ . But $r > s$ implies $V_r \subset V_s$ . Hence $f_r \le g_s$ for all $r$ and $s$ , so $f \le g$ .

Suppose $f(x) < g(x)$ for some $x$ . Then there are rationals $r$ and $s$ such that $f(x) < r < s < g(x)$ . Since $f(x) < r$ , we have $x \notin V_r$ ; since $g(x) > s$ , we have $x \in \bar{V}_s$ . By (3), this is a contradiction. Hence $f = g$ . ////

2.13 Theorem Suppose $V_1, \dots, V_n$ are open subsets of a locally compact Hausdorff space $X$ , $K$ is compact, and

Then there exist functions $h_i < V_i$ ( $i = 1, \dots, n$ ) such that

Because of (1), the collection $\{h_1, \dots, h_n\}$ is called a partition of unity on $K$ , subordinate to the cover $\{V_1, \dots, V_n\}$ .

PROOF By Theorem 2.7, each $x \in K$ has a neighborhood $W_x$ with compact closure $\bar{W}_x \subset V_i$ for some $i$ (depending on $x$ ). There are points $x_1, \dots, x_m$ such that $\bar{W}_{x_1} \cup \cdots \cup \bar{W}_{x_m} \supset K$ . If $1 \le i \le n$ , let $H_i$ be the union of those $\bar{W}_{x_j}$ which lie in $V_i$ . By Urysohn’s lemma, there are functions $g_i$ such that $H_i < g_i < V_i$ . Define

Then $h_i < V_i$ . It is easily verified, by induction, that

Since $K \subset H_1 \cup \cdots \cup H_n$ , at least one $g_i(x) = 1$ at each point $x \in K$ ; hence (3) shows that (1) holds. ////