A MAPPING THEOREM FOR METRIC SPACES

BY V. K. BALACHANDRAIg

1. Introduction. It is well known (see for example [2; 252, 66.2]) that ifR is a compact metric space, and f a closed continuous map of R onto a topo-logical space R*, then R* f(R) is also a metric space. Since Whyburn hasproved [3; 73, Theorems 9, 10] that normality and strong (= perfect) separabilityare preserved by a closed continuous map f, provided the image space R* isassumed locally separable, it follows (by Urysohn’s metrization theorem) thatin the theorem stated in the beginning the hypothesis "R is a compact metricspace" can be replaced by "R is a separable metric space" (provided R* isassumed locally separable). The question arises whether the "separability"restriction on R can also be eliminated; i.e., whether u closed continuous (locallyseparable) image R* of any metric space R is a metric space.

In this note we show that the answer is in the affirmative for all closed con-tinuous maps which are also open. The problem in the general case, however,remains unanswered.

In the proof of our result, it is interesting to note that we do not make use ofany "metrization theorems" as in the case of the previously referred result.Rather we obtain explicitly a metric for the image space R* by making use ofthe well-known Hausdorff distance function d(X, Y) for closed sets of the originalspace R.

2. Definitions and preliminary remarks. As usual a map f is called closed(open) if it transforms every closed (open) set into a closed (open) set of theimage.

In this paper we use the term topological space for any set R satisfying thethree standard neighborhood axioms of Hausdorff. (Hence no separationpostulates are to be assumed unless explicitly stated.)A topological space R in which single points are closed will be called T1 A

topological space R satisfying the 1-st (2-nd) axiom of countability will becalled locally (strongly) separable [2; 74, 21]. In a locally separable space R,if {Gn} (n 1, 2, ...) is a countable fundamental family of open neighborhoodsof a point p, then this can be replaced, whenever necessary, by the countablefundamental family {Hn} where H I:=l Gi which has the additionalproperty of being a decreasing family.A sequence of points {p} in a topological spaco R is said to converge to p (in

symbols, p -- p) if every open neighborhood of p contains all but a finite numlerof the p. It follows from the definition that if pn - p and (p,) is a subsequenceof (p), then pn, p.

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462 V. K. BALACHANDRAI

3. Lemmas.

LEMMA 1. An open continuous image R* f(R) of a locally separable spaceR is locally separable.

Proof. For, if p* R* and p* f(p), then since R is locally separable, thereis a countable fundamental family of open neighborhoods {Gn} for p; since f isopen, G* f(Gn) is open. Again if G* is any open neighborhood of p*, f-1 (G*)is an open neighborhood of p (by continuity of f) and therefore contains a Gso that G* f f-1 (G*) f(Gn) G*. Hence {G*} forms a countable funda-mental family of open neighborhoods for p*. That is, R* is locally separable.

LEMMA 2. In a locally separable space R, x X (X denotes the closure of X)if and only if there exists a sequence (xn) of points x, X (n 1, with x ---> x.

Proof. Choose for x a decreasing fundamental family of open neighborhoods{Hn} (n 1,2, ...),H1 H2 .... (If there are only a finite number ofdistinctH,sayH1, ,Hm,thensetH Hmforn > m.) IfxX, thenH. X 0 (n 1,2, ...). IfxH Xthenxn--.x. On the other hand,if Xn X with x X, then clearly Hn X 0 (n 1, 2, ...) or x X.

LEMMA 3. Let f be an open map of a locally separable space R onto a topologicalf-space R* f (R). Then, if y* y* in R* and y (y*), we can choose a

f-point y (y*) (n 1, 2, ...), such that y ---> y in R.

Proof. Let H H2 be a decreasing fundamental family of openneighborhoods of y. If any H. (j 1, 2, ...) be disjoint with an infinity off- (y*), say f-’ (y) (i 1, 2, ...), thenf(H) y 0 (i 1, 2, ...).Since f is open and H. is an open neighborhood of y, f(H) is an open neighbor-hood of y* and hence contains all the y for sufficiently large i (since y --* y*).This contradiction shows that every H. must meet all f-l(y*n) from a certainstage onwards. So for each j there is a least integer n; such that H. f- (Yn)0 for n >_ nWe shall now show how to choose the sequence (y). For each integer n ni

choose as y any point in f- (y*); for n n, choose as yn any point in H f-(y,*). Next for arbitrary n > nl, choose as y any point in H f- (y), wherethe integer m (depending on n) is determined as follows" set m n if Hn f-(y) 0; otherwise set m k, the largest integer n such that H f-1 (y) 0(then n/ > n __>. n). Thereby we obtain a sequence (yn) which is such that(each) Hi contains all y for n _> Max [j, n.]. It follows that y -, y, completingthe proof.

DEFINITION. If X, Y are non-null closed subsets of a metric space R withdistance function d the Hausdorff distance d(X, Y) is defined to be the greaterof the two numbers

sup p(x Y) and sup p(y X)xeX

A IIAPPING THEORENI FOR METRIC SPACES 463

(p(x, Y) denotes the minimum distance of x from Y, i.e. p(x, Y) infr d(x, y))1; 106, VIII.LEMMA 4. If d (X, Xn) "--+ O, then for each x , X we can choose a sequence

(xn), x, , X, (n 1, 2, ...) such that x, x.

Proof. Let e > 0 be fixed. By definition of p(x, X,) there exists xn Xnwithd(x, Xn) < p(x, Xn) -+- e/n (n 1,2, ...). Sincep(x, Xn) <_ d (X, Xn)and d (X, X,) --+ 0, it follows from the above inequality, that d (x, x,) -+ 0.

4. The theorem. Letf be an open, closed and continuous map of a metric spaceR onto a topological space R*. Then R* is a metric space.

Proof. By Lemma 1 it results that R* is locally separable. Again sinceR is T1 and f is closed, R* is T1. Now we may clearly assume the metric d of Rto be bounded; introduce in (the set) R* a metric d* as follows" d*(x*, y*) d(X, Y), where X f-(x*), Y f-(y*). (Since R* is T1 and f is continuous,X and Y are closed in R.)We shall now show that the given topology of R* is the same as the one

induced by the metric d*. Since R* is locally separable it suffices, in view ofLemma 2, to show that a sequence x* --+ x* in the given topology of R*, if and

x*) -+ 0 as n --+only if, x* -+ x* under the metric d*, i.e.,First let x* -+ x* in R*. If possible, suppose that d* (x*, x*) q- 0; then, for

a certain e > 0, d*(x*,, x*) d(X,, X) > e, for an infinity of values of n (X,f-(x*), X f-l(x*)). This means that for an infinity of values of n either(1) p(y X) > e, (y X) or (2) p(x, X) > e, (xn X). By replacing (x*) by asuitable subsequence (which converges also to x*) we can suppose without lossof generality that one of these possibilities happens for all values of n.

Casei. Letp(y,,X,) > e>O(sothatXX) (n 1,2, ...).If the set Y {y} is totally bounded, then (y) contains a Cauchy sub-

sequence (yn,) [2; 236, 63.2], which can be so chosen that d(y,,, y,)(for any m’, n’). Then for any fixed m’ and all n’

(A)p(y,, X,,,) >_ p(y, X,,) d(y, y,,,)

But, since x (= f(X,)) form u subsequence (x2) of (x*) and x*, -+ x*, it resultsthat x --+ x*. Therefore by Lemma 3, points x, Xn, can be so selected thatx, -+ ym, (for any fixed m’) which contradicts inequality (A).

If the set Y {y} is not totally bounded, there exists some subsequence(y,) and an e > 0 with d(y,, y,) > 2e (for any m’, n’ such that m’ n’).Since f is open, x* --+ x*, and y, X f-l(x*) (n 1, 2, ...), for each n’ wecan certainly choose (cf. Lemma 3) an n" such that

464 V. K. BALACI-IADRAI

Then

d(x,,,, x,,,) >_ d(y,,, y,,) d(y,, x,,,) d(y,, x,,,)

(m" n")>2 2 2 =Hence X" {xn,, is closed in R, and therefore f(X") is closed in R* (since f isclosed). On the other hand, since, f(X") {x*,, }, x* -- x* (and so alsox*,, --. x*), and x*,, x* (since X,,, X), f(X") cannot be closed, so that wehave again a contradiction.The assumption p(y, X,) > e is therefore untenable.Caseii. Letp(x,X) > > 0 (xX). If we writeX’ {x} then it is

clear that X’ X O. Since f is closed, continuous f(X’) f(X’) [2; 323, 36.5]whence f(X’).f(X) O. On the other hand f(X) x* belongs to f(X’)since f(X’) {x*} and x* -- x*. This contradiction shows that the assumptionp(xn, X) > e is also untenable. Hence d*(x*, x*) ---. O.

Conversely, if d*(x*, x*) d(X,, X) 0 and x X, then points x X canbe found such that x --* x (Lemma 4). It follows by continuity of f, x*, --* x*.Thus, x*, -- x* in R*, if and only if, d*(x*, x*) O, which completes the proofof the theorem.

REFERENCES

1. C. KURATOWSKI, Topologie I, second edition, Warsaw, 1948.2. n. VAIDYANATHASWAMY, Treatise on set-topology, Madras, 1947.3. G. T. WHYBURN, Open and closed mappings, Duke Mathematical Journal, vol. 17(1950),

pp. 69-74.

MADRAS UNIVERSITY.