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Normal Nonmetrizable Moore Space from Continuum Hypothesis Or Proc. Natt Acad. Sci. USA Vol. 79, pp. 1371-1372, February 1982 Mathematics Normal nonmetrizable Moore space from continuum hypothesis or nonexistence of inner models with measurable cardinals (general topology/large cardinals/relative consistency results) WILLIAM G. FLEISSNER Department of Mathematics and Statistics, University of Pittsburgh, Pittsburgh, Pennsylvania 15260 Communicated by R. H. Bing, October 5, 1981 ABSTRACT Assuming the continuum hypothesis, a normal than co that satisfy [1]; for example K = sup{w, 2w, 22 , . }. nonmetrizable Moore space is constructed. This answers a ques- If[2] or [3] fails for such a cardinal, then there is an inner model tion raised by F. B. Jones in 1931, using an axiom well known at with a measurable cardinal, because the covering lemma would that time. For the construction, a consequence of the continuum fail with respect to the core model; see ref. 6. hypothesis that also follows from the nonexistence of an inner We now construct the space. Let F be the set of functions model with a measurable cardinal is used. Hence, it is shown that from w to E. For n E w, set n = {fln:f E F} and set I = to prove the consistency of the statement that all normal Moore U on. For o-E E, set [C] = {fE F: aoCf. spaces are metrizable one must assume the consistency of the net statement that measurable cardinals exist. Let Z be the family of subsets, Z of z satisfying card Z ' K and for some n E w, Z C In- By [2], we can list Z as {Za: a In 1931, Jones (1) asked whether all normal Moore spaces were < K+}. For ar EC , define or* to be the greatest ordinal in range metrizable. Towards this end, he proved that, assuming o. For p E: set I2P = {a- o-* < p*} and Z(p) = {Za: a < 28°< 2k', separable Moore spaces are metrizable. Bing (2) p*}. By induction on n EE W, we define, for o- E Xn and m E showed that a Moore space is metrizable if and only if it is col- co, the sets A(om) to satisfy lectionwise normal. The most important results after Bing are: (i) assuming Martin's axiom plus not continuum hypothesis, card A(am) = Km [4] there is a normal nonmetrizable Moore space (3); (ii) assuming Gidel's axiom ofconstructibility, normal Moore spaces are col- U A(o,m) = Z(of n (U Sk) [5] lectionwise Hausdorff (4); (iii) assuming the product measure mew k<n extension axiom, all normal Moore spaces are metrizable (5). Nyikos (5) called this last result a provisional solution to the if a, C a,' and m '< m', then A(o-,m) C A(o-',m'). [6] normal Moore space problem because a strongly compact car- Let Qk be the set of triples (gp, r) satisfying p, T EC Sk [7] dinal was used in the consistency proof. In this paper, a normal nonmetrizable Moore space is con- for all i < k - 1, p(i) < r(i) < p(i + 1) < r(i + 1) [8] structed, not from a new axiom, but from the familiar continuum g is a function from Z(p*) to {0, 1}. hypothesis. Moreover, the solution ofthe normal Moore space [9] problem is completed by showing that a large cardinal assump- Set Q = U Qk. For or E= En, set B(of) = [a] U G(o), where G(of) kEw tion is necessary to show the consistency of all normal Moore is the set of triples E Q satisfying spaces being metrizable. These two things are done simulta- (g,p,T) neously by constructing a normal nonmetrizable Moore space either o C p or or C T [10] from an axiom that is implied by the continuum hypothesis and whose failure implies the existence of inner models with mea- for all i < n, p(O)i = (°) [11] surable (and larger) cardinals. if ZE A(a,n) and Z C Em, I begin by setting out in detail the hypotheses assumed be- yond the Zermelo-Frankel axioms for set theory with choice. then g(Zfln m) = 1 iffom E Z. [12] Let Hyp be the axiom that asserts that for some K: The point set of our space, X, is F U Q. A basis for the to- there is an increasing sequence K,,, n E cl, of cardinals pology on X is {B(u): o E Y} U {{q}: q E Q}. It is easy to see that X is T1 and regular. cofinal in K such that for all n E C), 2Kn < K, [1] Set G. = {B(o): o- E Y} U {{q}: q E Qk where k > n}. [13] +K= 2 [21 K For every infinite subset U of U Gn, if x E U, then U is a there is stationary subset, E, of K+ such that for all nnw 8 E E, there is an increasing sequence 85, [3a] basis at x. Hence X is a metacompact Moore space. To show that i E co, cofinal in 8, X is normal, it suffices to separate, for each n E clo and each Z C the sets = U o-E and = F- Define , -' In Hz Z} Kz Hz. and for all 3 < K+, E n , is not stationary in /. [3b] {[ao: C = {y E K ifp* < Y, K = The continuum hypothesis implies Hyp; let CL and let E be then Z n Yp = Z,, for some a < y}. [14] the set oflimit ordinals less than wl. There are cardinals greater For 3 e- K+, define y(f3) to be the least element of C greater The publication costs ofthis article were defrayed in part by page charge than /3. For a E E'+3, define j(u) - n + 3 to satisfy payment. This article must therefore be hereby marked "advertise- ment" in accordance with 18 U. S. C. §1734 solely to indicate this fact. if y(cr(n)) < or(n + 2), then Z fl Xo=(n) E A(oj(cr)) [15] 1371 Downloaded by guest on September 25, 2021 1372 Mathematics: Fleissner Proc. Natl. Acad. Sci. USA 79 (1982) if y(o(n + 1)) = y(v(n + 1)), [16] tively choose {a.: a< K} C S' such that aji) < co3(j) iff i < j or i = j and a < /3. Enumerate A(oa,n) as {Z (a,8): 8 < then for some i < max{j(co), j(l')}, o-(o)i $ v(o)i K}. For a < 3 < K there is g such that (g,o,,oq) EE B(0a) n B(o,) (We use [3b] to obtain [16]). Set W<, = U {B(p): or C p E unless [12] fails. Thus for each pair a,,B there are 8, 7, and m Xj(o)}. To separate Hz and Kz, it will suffice to show that if oc such that Z(a,8) = Z(p,,r) n X0I'm and either orjlm E Z(a, 8) v E on+3(Jni E Z, and vIn § Z, then W(o) n w(v) = 0. and o-rm § Z(/38r) or oqjm 0 Z(a,8) and oim EE Z(P,{, Be- Aiming for a contradiction, assume that (g,p,r) E= W(o) n W(v), cause K > (2Kn)+, we can apply the Erdos-Rado theorem, K where (0) < v(O). If y(v(n)) < or(n + 2), we get a contradiction (3)2. to quickly get a contradiction. from [12]. Otherwise y(o-(n + 1)) = y(v(n + 1)), and we get a contradiction from and I thank P. J. Nyikos, T. C. Przymusinski, and M. E. Rudin for sug- [8], [16], [11]. gestions that simplified the presentation of this example. I also thank The following notion breaks the demonstration that X is not Frank Tall and Bill Weiss for organizing, Erindale College for hosting, metrizable into several short combinatorial arguments. and the Canadian Mathematical Society for funding the valuable and Definition: A subset, S, of in is stafull if for all cr E S and stimulating conference during which some ofthis work was done. This j < n, {r(j): olj v E S} is stationary. work was supported in part by National Science Foundation Grant For each 8 E E, set Y8 = ff E F: f(O) = 8}. Because {Y8: 8 MCS-79-01848. E E} is a discrete family of closed sets, if X were metrizable, there would be a disjoint family {U8: 8 E E} of open sets with 1. Jones, F. B. (1937) BulL Am. Math. Soc. 47, 671-677. C Aiming for a contradiction, assume that {U&: 8 E E} 2. Bing, R. H. (1951) Can. J. Math. 3, 175-186. Ys Us. 3. Tall, F. D. (1969) Dissertation (Univ. of Wisconsin, Madison, is such a family. Let T = {o- E X: B(of) U,(0)}. For some n E w, WI). T f in contains a stafull set, S. (Assume not, then inductively 4. Fleissner, W. G. (1974) Proc. Am. Math. Soc. 46, 294-298. define yji such that y § UY(o)). Refine S to a stafull subset S' 5. Nyikos, P. J. (1980) Proc. Am. Math. Soc. 78, 429-435. such that for all u,, v E S and i < n, o(O)i = v(O),. Next, induc- 6. Dodd, A. & Jensen, R. B. (1981) Ann. Math Log. 20, 43-75. Downloaded by guest on September 25, 2021.
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