ALMOST CONVEX METRICS AND PEANO COMPACl " IFICATIONS

Let (X,d) denote a locally connected, connected separable metric space. We say the X is S-metrizable provided there is a topologically equivalent metric ρ on X such that (X,ρ) has Property S, i.e., for any ϵ>0, X is the union of finitely many connected sets of ρ-diameter less than ϵ. It is well-known that S-metrizable spaces are locally connected and that if ρ is a Property S metric for X, then the usual metric completion (X˜,ρ˜) of (X,ρ) is a compact, locally connected, connected metric space; i.e., (X˜,ρ˜) is a Peano compactification of (X,ρ). In an earlier paper, the author conjectured that if a space (X,d) has a Peano compactification, then it must be S-metrizable. In this paper, that conjecture is shown to be false; however, the connected spaces which have Peano compactificatons are shown to be exactly those having a totally bounded, almost convex metric. Several related results are given.


INTRODUCTION.
Throughout this note let (X,d) denote a metric space.We say that d is convex R.F.DICMAN, JR.   provided that, for any pair x,yX, there is zgX such that d(x,z) d(z,y) d(x,y)/2.
It is Tmos ovex if, for x,yeX and O, there is zX such that Id(x,z)-d(x,y)/2 a and Id(z,y)-d(x,y)/21 We say that X is S-metizae provided there is a topologically equivalent met- ric on X such that (X,$) has Property S, i.e., for any 0, X is the union of finitely many connected sets of 0-diameter less than .It is well-known that S- metrizable spaces are locally connected and that if is a Property S metric for X, then the usual metric completion (X,) of (X,) is a compact, locally connected, con- nected metric space, i.e., (,) is a Peano compactification of (X,) [3, p. 154].
It is a famous result of R. H. Bing that any continuous curve P (i.e., a compact, locally connected, connected metric space) can be assigned a convex metric [I].
In an earlier paper [4], the author conjectured that, if X is locally connected and if X has a Peano compactification, then X is S-metrizable.In this paper we show, by example, that this conjecture is false; however, we do obtain a characterization of such spaces in terms of the existence of a totally bounded, almost convex metric for X.We also obtain several related results characterizing totally bounded (Smetrizable, almost convex) metrics. 2.
THEOREM 2.1.A connected metric space (X,d) has a Peano compactification if and only if it has a topologically equivalent totally bounded, almost convex metric.
PROOF.The necessity.Let (P,r) be a Peano compactification of X, i.e., P is a continuous curve and X is a dense subset of P. By R. H. Bing's result, there exists an equivalent metric for P such that is convex.It then follows that e X is totally bounded and almost convex; cf.[i, Thm.i0].
The Sufficiency.Let r be an almost convex, totally bounded metric for X.Let (,) be the usual metric completion of (X,r).We will argue that (,) is a Peano compactification of (X,r).Clearly, X is compact since r is totallv bounded.
Furthermore, r is a convex metric for X" let x,yX.Since r is almost convex, there exists a sequences Xl,X2,...,yl,y 2 and Zl,Z 2 in X sch that r(x ,z r(x ,yn)/212 -n and Ir(z Since r is totally bounded, without loss of generality, we may assume that each of the sequences Xl,X2,...,yl,Y2,..., and Zl,Z2,... is Cauchy in X.Then by the com- pleteness of (X,r) it follows that lim x x and lim Yn Y" Furthermore if n nn-> lim z z, r(x,z) r(z,y) r(x,y)/2 since r is continuous.Thus r is convex and n n-oo complete.It follows from Theorem 3.1 of [5] that the spheres S~(x,) of X are con- r nected sets.This implies that X is locally connected and this completes the proof.EXAMPLE 2.1.Let P be the square {(x,y)2 0 x,y i} in the plane.For nq, let Ln {(i/n,y): 0 y I} and let L 0 {(O,y): 0 v I}.Set X P\n0 Ln.Then P is a Peano compactification of X; however, X is not S-metrizable.
Suppose 0 is an S-metric for X and let A {(x,l): 0 of X\(A u B) have limit points in each of A and B. Thus, any collection of con- nected sets of 0-diameter less than s/3 that covers a component C has at least one n such connected subset lying entirely in C This implies that p is not an S-metric n for X; however, if d is the relative metric on X inherited from the usual metric on P, d is almost convex and totally bounded. 3.

RELATED RESULTS.
A compatible normal sequence in a space Z is a sequence LI,U 2 of open covers of Z such that Un+1 star-refines Un for n 1,2 and so, for any xsZ, {St(X,Un ): n 1,2 is a neighborhood base for x [5].
[6, Prop.23.4]A To-space is metrizable if and only if it has a compatible normal sequence.
COROLLARY 3.1.A metric space X is totally bounded if and only if X has a com- patible normal sequence UI,U2,... where each Un is a finite cover of X.
PROOF.Suppose (X,d) is totally bounded.It follows from the total boundedness of (X,d) that there is a finite open cover U1 of X such that 6d(U) 1/3 for all Us[l 1 where $d(U) sup{d(x,y): x,yeU}, the d-diameter of U. Since U1 is finite, there is -2 a Lebesgue number el 3 such that, if d(x,y) el' then x and y be in some member of LI I. Again, by the total boundedness of (X,d), there is a finite open cover V 1 of A and B are compact and hence 0(A,B) e > O. Now the components CI, C 2,