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On monotony of у- dimension

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ROCZNIKI POLSKIEGO TOWARZYSTWA MATEMATYCZNEGO Seria I : PRACE MATEMATYCZNE X I I (1968)

ANNALES SOCIETATIS MATHEMATICAE POLONAE Series I : COMMENTATIONES MATHEMATICAE X I I (1968)

E. Du d a (Wrocław)

On monotony of у- dimension

There are several different notions of dimension, two of a few more important are: Menger’s “small” inductive dimension “ind” [4] and Cech’s “great” inductive dimension “Ind” [1]. Both are defined by in­ duction; it has recently been shown [2] that one can assign to any sequence у = {yx, y2, ...) consisting of 0’s and l ’s a new notion of dimen­ sion, called y-dimension and denoted “у-dim”, in such a way that, vaguely speaking, in a consecutive step of calculating y-dimX for a topological space X one follows “ind” if у* = 0 and “Ind” if y< = 1.

It is known ([3], p. 26) that in the class of all topological spaces “ind” is monotone, i.e., if Y a X , then ind Y < in d X The same proof (with obvious modifications) works to the effect in the same class that “Ind” is monotone with respect to closed subsets, i.e., if Y — Y czX , then

Ind Y ^ IndX. Combining these two results we infer that

Th e o r e m 1. In the class of all topological spaces y-dimension

is monotone with respect to closed subsets for every sequence у of 0's and l ’s.

However, using an example of Tychonoff [5] one can prove ([3], p. 154; see also [6]) that in general “Ind” is not monotone. It is the purpose of the present paper to show that

Th e o r e m 2 . In the class of all topological spaces y-dimension is monotone if and only if у = (0, 0, ...).

It is unknown, however, whether this theorem is still true if we restrict ourselves to the class of normal spaces or to the class of metric spaces only.

Let X be a topological space and {xx, x2, . . . , Xk} a sequence of fc distinct points not belonging to X. By Xk we shall mean the space

X k == {xx, X2, . .. , Xk)■ w X

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20 К. D u d a

bourhoods of an x in X are of the form {xx, . .. , xk} w U, where U a X

is an open neighbourhood of x in X. Obviously, X is a closed subset of X k.

Le m m a. Let X be a topological space. I f у = (yx, y2, . ..) is a sequence consisting of 0’s and l ’s such that yx — y2 = ... = yk — 0, where Jc ^ 1,

then

y-dimX& = hĄ-{yk+l, yk+2, ...)-dim X.

P roof. Intending to calculate y-diuiX* by a direct application of the definition of у-dimension we have to consider a triple sequence of the form: a point p xeX k, its neighbourhood U1 <= X k, the boundary F x of TJX) a point p2gFi, its neighbourhood U2 <= F x, the boundary F2 of U2 taken relative to F xm, a point p se F 2, and so on. If we proceed care­ fully, that is if we take each time p t = xx and Ui = {Xi} , i = 1, 2, . .. , Tc, we come — as one can easily check — to F k — X. But if we do not follow this device, then we may come to another F k — Z — Z <=. X. Anyway, if we do not impose upon Z any restrictions (except for that of being an admissible set in calculation of у-dimension), then

(1) [y-dimXfc = fc + sup(y*+1, y*+2, ...)-dim Z, where supremum is taken over all admissible Z’s.

F k being a closed subset of X , by Theorem 1 we have (2) (y*+i , 7k+2, • • .)-dimZ < (yk + 1, yk + 2, .. .)-dimX

for each admissible Z. But obviously X is admissible, and therefore formulas (1) and (2) imply our lemma.

We may now turn to the proof of Theorem 2. We need to construct for each у = (yx, y2, .. •) Ф (0, 0, ...) a topological space A together with a subspace В a A such that y-dimX < y-dimB. As our construction will be based upon Tychonoff’s example, let us recall it here:

Let о be the least countable ordinal and let [0, со] = {n: 0 < n < со} be provided with the order topology. Similarly, let Q be the least uncount­ able ordinal and [0, Q~\ — {a: 0 < a < Q], with the order topology. Let S denote the topological product of [0, со] and [0, £?], and let T be the complement in 8 of the single point (co,i3).

Pr o o f of Theorem 2. It is known ([3], p. 155) that for any compact space О, if ind<7 = 0, then also IndO = 0. Since 8 is both compact and ind 8 = 0, then Ind$ = 0. However, T is not normal and therefore IndT Ф 0.

How let у = (ух, y2? •••) be a sequence of 0’s and l ’s. If yx = 1, it suffices to take 8 itself together with T , because then

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On monotony of y-dimension 21 If y x = y 2 = . . . = y k = 0 and 7fc+ 1 = 1 , where к > 1 , then it suffices to take 8k together with Tk, because Tk is a subspace of 8k and, by lemma and (3),

y-dim$* = Tc + (yk+lf yk+2, ...)-dim# = k,

y-d\mTk = k + (yk+i , Ук+2, ...)-dimT > к.

R eferen ces

[1] E . C ech, 8 m la dimension des espaces parfaitement normaux, Bull. Int.

Acad. Boheme 33 (1932), pp. 38-55.

[2] B . D u d a, On notions of dimension, Bull. Acad. Polon. Sci. Ser. Sci. Math.

Astronom. Phys. 13 (1965), pp. 777-780.

[3] W. H n re w icz and H. W a llm a n , Dimension theory, Princeton 1948.

[4] K. M en ger, Dimensionstheorie, Leipzig und Berlin 1928.

[5] A. T y c h o n o ff, tfber die topologische Erweiterung von Eaiimen, Math. Ann.

102 (1930), pp. 544-661.

[6] N. Y e d e n is s o ff, Bemarques sm la dimension des espaces topologiques,

Учение Записки Московского Университета 30 (1939), pp. 131-140.

INSTYTUT MATEMATYCZNY P O L S K IE J AKADEM II NAUK

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