数理解析研究所講究録
1531 巻 2007 年 31-40

31Borel classes dimensions

リンショーピン大学

ヴィタリ チャティルコ (Vitalij Chatyrko)
Department of Mathematics

Link\"oping University
服部泰直 (Yasunao Hattori)

島根大学総合理工学部

Faculty of Science and Engineering
Shimane University

1

Introduction and results.

The classes of topological spaces are assumed to be

1. non-empty (we suppose that at least the empty spaceand$\emptyset$

is a member),

2. monotone with respect to closed subsets.
The letter is used to denote a such class and the following classes ofspaces satisfiy the conditions 1 and 2 above.
$\mathcal{P}$$\bullet$$\bullet$$\bullet$$\bullet$The class of compact metrizable spaces$\mathcal{K}$The class of a-compact metrizable spaces

.
$S$

.

The class of completely metrizable spaces C.

The class of separable completely metrizable spaces$C_{0}$.

32
Let $X$ be a space and $A,$ disjoint subsets of $X$ . We recall that a closed
set $C\subset X$ is said to be a partition between and in $X$ if there are disjoint$B\subset V$
open subsets and of $X$ such thatand $C=X\backslash (U\cup V)$ .
In [4] Lelek introduced the small inductive dimension modulo a class ,P-ind, which is a natural generalization of well known dimension functions
such as the small inductive dimension ind and the small inductive compact.
ness degree
$B$

$A$

$U$

$V$

$B$

$A\subset U,$$P$$\mathrm{c}\mathrm{m}\mathrm{p}$Definition 1.1 Let $X$ be a regular -space and a class of spaces. Then
we define the small inductive dimension modulo a class $P,$ -ind $X$ , of $X$ asfollows.
$P$$T_{1}$$P$(i)$\mathcal{P}$-ind $X=-1$ iff $X\in P$ .

(ii) For a natural number $P$ -ind $X\leq n$ if for any point $x\in X$ and any
there exists a partition between
closed subset of $X$ with$P$
$X$
and in such that -ind $C<n$ .
$n,$

$A$

$C$

$x\not\in A$

$x$

$A$

The small inductive dimension modulo a class $P$ has a natural transfinite
extension.

Definition 1.2 Let $X$ be a regular -space and either an ordinal number
or the integer-l. Then the small transfinite inductive dimension modulo ,-trind $X$ , of $X$ is defined as follows.
$T_{1}$$\alpha$$P$$\prime p$(i) $P$-trind $X=-1$ iff $X\in P$ ;(ii) $P$-trind $X\leq\alpha$ if for any point $x\in X$ and any closed subset of $X$
there exists a partition between and in $X$ such thatwith
$P$ -trind $C<\alpha$ .
$A$

$C$

$x\not\in A$(iii) $P$-trind $X=\alpha$ if $P$-trind $X\leq\alpha$ and;$x$$\mathcal{P}$$A$-trind $X>\beta$ for any ordinal$\beta<\alpha$(iv)$\mathcal{P}$-trind $X=\infty$ if -trind $X>\alpha$ for any ordinal a.
$\prime p$

We notice the following.
$\bullet$$\{\emptyset\}$-trind$X=\mathrm{t}\mathrm{r}\mathrm{i}\mathrm{n}\mathrm{d}X,$. the small transfinite dimension.

$\mathrm{i}.\mathrm{e}_{)}$

33$\bullet$$\bullet$$\bullet$(and -trind-indnite) compactness degree.
$\mathcal{K}$$X=\mathrm{c}\mathrm{m}\mathrm{p}X$$\mathcal{K}$(and -trind-indcompleteness degree.

$C$

If

$C$

$X=\mathrm{i}\mathrm{c}\mathrm{d}X$$P_{2}\subset P_{1}$trcmp, then$X\leq \mathrm{t}\mathrm{r}\mathrm{i}\mathrm{n}\mathrm{d}X$$X=\mathrm{t}\mathrm{r}\mathrm{c}\mathrm{m}\mathrm{p}X$$X=\mathrm{t}\mathrm{r}\mathrm{i}\mathrm{c}\mathrm{d}X$), i.e., the small (transfi-), i.e., the small (transfinite)-trind $X\leq P_{2}$ -trind $X$ ; in particular, tricd $X\leq$holds.

$P_{1}$

Here, we shall consider on the absolute Borel classes. For each ordinalandbe the absolute additive class and thenumber , letabsolute m..ultiplicative classe a, respectively. Further, $A(a)\cap \mathcal{M}(a)$ is said
to be the absolute ambiguous class and we write $AB=\cup\{A_{\alpha} : \alpha<\omega_{1}\}$ .
We notice that the absolute Borel classes in the universe of metrizable spaces
satisfy the conditions 1 and 2.
Recall that in the universe of separable metrizable spaces, we have the$A(\alpha)$$\alpha$$a$$\mathcal{M}(a)$$a$following.
$\bullet A(0)=\{\emptyset\}$$\bullet \mathcal{M}(0)=\mathcal{K}$$\bullet A(1)=S$.

.

$\bullet \mathcal{M}(1)=C_{0}$$\bullet$.

.

A diagram of the hierarchy of absolute Borel classes:$A(1)=S$$C_{/}$$A(2)$...

$\mathrm{C}_{/}$$\backslash ^{\mathrm{O}}$$A(2)\cap \mathcal{M}(2)$$\{\emptyset\}\subseteq \mathcal{K}\subseteq A(1)\cap \mathcal{M}(1)$$\backslash \sigma\backslash$$\backslash \mathrm{o}_{\mathcal{M}(1)=C_{0}}O_{/}$$\mathcal{M}(2)$We have a trivial example which shows the difference between trind andicd
and cmptrcmp: The Hilbert cube
has trind-ind ) $=-1$ . Furthermore, E. Pol constructed the following example.
$\mathrm{I}\mathrm{I}^{\infty}$$S$$\mathrm{I}\mathrm{I}^{\infty}$$\mathrm{I}\mathrm{I}^{\infty}=\infty$$\mathrm{I}\mathrm{I}^{\infty}(=$$\mathrm{I}\mathrm{I}^{\infty}=$

34

Example 1.1 (E. Pol, [5]) There exists a -compact, completely metrizable space $P$ such that trcmp $P=\infty$ (i.e., trind $P=$ trcmp $P=$ oo and.
tricd $P=S$ -trind$\sigma$$P=A(1)\cap \mathcal{M}(1)- \mathrm{t}\mathrm{r}\mathrm{i}\mathrm{n}\mathrm{d}P=-1)$

Thus, we may ask whether we can generalize Pol’s example to every

ordinal number$a<\omega_{1}$.

It is well know that the small compactness degree cmp is related to an
extension property, i.e., de Groot proved that a separable metrizable space$X$ is rim-compact (i.e., cmp $X\leq 0$ ) iff $X$ has a metric compactificationsuch that $\dim(\mathrm{Y}-X)\leq 0$ . Connect with this theorem, we introduce othertwo dimension-like functions.
$\mathrm{Y}$Definition 1.3 Let $P$ be a class of spaces. We recall that a separable metrizable space is a $P$ -hull (resp. $P$ -kemel) of a separable metrizable space $X$ if). Then the small transfinite P-deficiency,(resp.
and$P$-trdef $X$ , and the smalltransfinite $P$ -surplus, $P$-trsur $X$ , of a separable
metrizable space $X$ are defined by$\mathrm{Y}$$X\subset \mathrm{Y}$$\mathrm{Y}\in P$$P$-trdef $X= \min$ {( -def$\mathcal{P}$$P$$\mathrm{Y}\subset X$( -sur$\mathcal{P}$:$X= \min\{\mathrm{i}\mathrm{n}\mathrm{d}(\mathrm{Y}\backslash X)$-trsur $X= \min$ {:$\mathrm{t}\mathrm{r}\mathrm{i}\mathrm{n}\mathrm{d}(Y\backslash X)$$\mathrm{t}\mathrm{r}\mathrm{i}\mathrm{n}\mathrm{d}(X\backslash Y)$$X= \min\{\mathrm{i}\mathrm{n}\mathrm{d}(X\backslash \mathrm{Y})$It is clear that the functions

$P$

:

$Y$

$\mathcal{P}$is an$\mathrm{Y}$:is an -hull of $X$ },$Y$$\mathrm{Y}$$\mathcal{P}$is an-hull of $X\}$ ),$\prime p$-kernel of $X$ },is an -kernel of $X\}$ ).
$P$-trsur are transfinite exten-sur, respectively, which are discussed in, then-trdef $X$ and-trdef-trdef and$\mathcal{P}$sions of the functions $P$-def and $P$[1]. It is also clear that if P2-trsur $X\leq P_{2}$ -trsur $X$ .
Recall also that for the function -def is the well known compact deficiency def. We will denote the transfinite extension -trdef of the compact
deficiency def by trdef.
$\subset P_{1}$$P_{1}$$X\leq \mathcal{P}_{2}$$P_{1}$$\mathcal{K}$$\mathcal{K}$Facts (cf. [1]). Let $X$ be a separable metrizable space andnumber. Then we have the following.

$a$an ordinal

35-sur $X$ holds and the-ind
1. If $a=0$ , thenandconverse of the inequalities do not hold. (We notice that$X.$
$X$
-def $X=$ def ) We also notice-ind $X=$ cmp andso-ind $X=A(\mathrm{O})$ -sur $X$ trivially holds
and hencethat.
-def $X$ can not be defined if
and$X\leq \mathcal{M}(\mathrm{O})- \mathrm{d}\mathrm{e}\mathrm{f}X\leq \mathcal{M}(\mathrm{O})$$\Lambda t(\mathrm{O})$$\mathcal{M}(0)=\mathcal{K}$$\mathcal{M}(0)$$\mathcal{M}(0)$$A(\mathrm{O})=\{\emptyset\}$$A(\mathrm{O})$$X\neq\emptyset$$A(\mathrm{O})$2. If $a=1$ , then $A(1)$ -ind $X\leq A(1)$ -def $X=A(1)$ -sur $X$ and-sur $X$ hold. The converses of the inind $X=\mathcal{M}(1)$ -defequalities above do not hold. (We notice that $A(1)=S$ and)-ind
and so$\mathcal{M}(1)-$$X\leq \mathcal{M}(1)$$\mathcal{M}(1)=C_{0}$$X=\mathrm{i}\mathrm{c}\mathrm{d}X.$$\mathcal{M}(1)$

3. If
ind

$\alpha\geq 2$, then$A(\alpha)- \mathrm{i}\mathrm{n}\mathrm{d}X=A(a)- \mathrm{d}\mathrm{e}\mathrm{f}X=A(a)$$X=\mathcal{M}(a)- \mathrm{d}\mathrm{e}\mathrm{f}X=\mathcal{M}(a)$-sur $X$ and$\mathcal{M}(a)-$-sur $X$ hold.

-def $X=\mathcal{M}(a)-$M. Charalambous [2] showed that the equality
ind $X$ can not be extended to the transfinite dimension for the case of $\alpha=1$ .
$\mathcal{M}(\alpha)$

Example 1.2 (M. Charalambous, [2]) There exists a separable metrizand tricdC $(=$
able space $C$ such that C-trdefC
tricd $X$ holds for every-trind $C$ ) $=\infty$ . (We notice that -trdef
separable metrizable space.)$(=\mathcal{M}(1)- \mathrm{t}\mathrm{r}\mathrm{d}\mathrm{e}\mathrm{f}C)=\omega_{0}$$\leq$$C_{0}$$\mathcal{M}(1)$Thus, it seems to be natural that we ask whether for each ordinal number

there exits a separable metrizable space $X$ such that $M(a)$ -trdef $X=$
and $A(a)$ -trind $X=\infty$ .
-trdef-trind $X=\infty$ orandConnect with the questions above, we have the following.

$\alpha<\omega_{1}$$\omega_{0}$$X=\omega_{0}$$A(\alpha)$$\mathcal{M}(a)$Theorem 1.1 Let a be any ordinal with $1\leq a<\omega_{1}$ .
andspaces(1) There exist separable met$r\cdot izable$$(a)fX_{\alpha},$ $f$$(b)\mathcal{M}(a)$$\mathcal{M}(\alpha)$$Y_{\alpha}$,$fZ_{\alpha}\leq\omega_{0\mathrm{z}}$-trind $X_{\alpha}=-1$ and-trind$A(\alpha)$-trind$\mathcal{K}$$X_{\alpha}=\infty$

such that-trsur;(and hence $A(a)\cap$$X_{\alpha}=\infty)_{i}$$X_{\alpha}=\infty$$\mathcal{M}(\alpha)$-trind$Y_{\alpha}=\infty$);

and-trind$A(\alpha+1)\cap \mathcal{M}(a+1)$ -trind $Z_{\alpha}=-1$ .

$(d)\mathcal{M}(a)$$Z_{\alpha}$$\mathrm{Y}_{\alpha}$

where.f is either trdef or$(c)A(a)$ -trind $Y_{\alpha}=-1$ andtrind$X_{\alpha},$-trind$Z_{\alpha}=A(\alpha)$$Z_{\alpha}=\infty$(and hence$A(a)\cap \mathcal{M}(\alpha)-$36such that(2) There does not exist a separable metrizable space.
-trindand-trindtrind$A(\alpha)-$$W_{\alpha}$$W_{\alpha}\neq\infty,$$A(a)\cap \mathcal{M}(\alpha)$$W_{\alpha}\neq\infty$$\mathcal{M}(\alpha)$$W_{\alpha}=\infty$

Theorem 1.2 There exists a separable metrizable space $X$ with trdef $X=$and
we have B-trindX
such that for each-trsur.
orwhere $B=A(\alpha),$B-trdefX$=\infty$$1\leq\alpha<\omega_{1}$$X=\omega_{0}$$\mathcal{K}$$\mathcal{M}(\alpha)$$=B- \mathrm{t}\mathrm{r}\mathrm{s}\mathrm{u}\mathrm{r}X=\omega_{0f}$$A(a)\cap \mathcal{M}(\alpha)$

Remark 1.1 By Thereoms 1:1 and 1.2, it follows that the equalities-ind $X$ and $A(a)$ -sur $X=A(a)$ -ind $X$ can not be extended toin Theorem
andtransfinite-dimensional cases. For the spaces
1.1, we additionally have that$\mathcal{M}(\alpha)-$$\mathrm{d}\mathrm{e}\mathrm{f}X=\mathcal{M}(a)$$X_{\alpha},$-trdef$X_{\alpha}=A(\alpha)$-trsur$\bullet$$\mathcal{M}(\alpha)$$\bullet$$\mathcal{M}(\alpha)- \mathrm{t}\mathrm{r}\mathrm{d}\mathrm{e}\mathrm{f}Y_{\alpha}=\mathcal{M}(\alpha)$$Y_{\alpha}=-1$-trdef$\omega_{0}$$\mathrm{Y}_{|alpha}$$Z_{\alpha}$;$Z_{\alpha}=A(\alpha)$-trsur $X_{\alpha}=A(a)$ -trsur$Z_{\alpha}=$.

We refer the readers to the books [1], [3] and [7] for the dimensions modulo
classes, dimension theory and the theory of Borel sets, respectively.

2

Outline of proofs.

All classes of topological spaces considered here are additionally assumed tobe finitely additive. We will follow some idea of E. Pol [5]. Let be a classif for everyof topological spaces. A space $X$ is said to have the propertyof pairs of disjoint compact subsets of $X$ there exist
sequence.
partitions between and in $X$ and an integer $N$ such thatis closed hereditary.
It is evident that the property.
We have two propositions on the property$P$$(*)_{P}$$\{(A_{i}, B_{i})\}_{i=1}^{\infty}$$L_{i}$$A_{i}$$\bigcap_{i=1}^{N}L_{i}\in P$$B_{i}$$(*)_{P}$$(*)_{\mathcal{P}}$Proposition 2.1 If a space $X$ is covered by a finite family of closed sets suchthen $X$ also possesses
that each element of this cover possesses property
this property.
$(*)_{P}$Proposition 2.2 Let $X$ be a space.
.
property$(*)_{\mathcal{P}}$

If

$P$

-trind $X\neq\infty$ then $X$ possesses

37

Let

be the Hilbert cube anda subspace of the unit interval II. For each $n\geq 1$ we denotethe subset {: $x_{j}=0$ for $j\geq n+1$ } ofby . For each $n\geq 1$, we putand each $i=1,$$\mathrm{I}\mathrm{I}^{\infty}=\{(x_{i})’.

0\leq x_{i}\leq 1, i=1,2, \ldots\}$$Z= \{0, \frac{1}{2}, \frac{1}{3}, \ldots\}$$(x_{i})\in \mathrm{I}\mathrm{I}^{\infty}$$\ldots,$$\mathrm{I}\mathrm{I}^{\infty}$$\mathrm{I}\mathrm{I}^{n}$$n$$A_{i}^{n}=\{(x_{i})\in \mathrm{I}\mathrm{I}^{n}\subset \mathrm{I}\mathrm{I}^{\infty} :Choose for each $n\geq 1$ a subset$B_{i}^{n}=\{(x_{i})\in \mathrm{I}^{n}\subset \mathrm{I}\mathrm{I}^{\infty} :x_{i}=0\},$$E_{n}$

of

x_{i}=1\}$.

and put$\mathrm{I}\mathrm{I}^{n}$.

(1)$X=( \{0\}\cross \mathrm{I}\mathrm{I}^{\infty})\cup(\bigcup_{n=1}^{\infty}\{\frac{1}{n}\}\cross E_{n})$Furthermore, we put. It is clear that$Y$
is compact, andis a subspace of the topological sum. Thus, trind. Observe also that trind. Hence$\mathrm{Y}=(\{0\}\cross \mathrm{I}\mathrm{I}^{\infty})\mathrm{U}(\bigcup_{n=1}^{\infty}\{\frac{1}{n}\}\cross \mathrm{I}\mathrm{I}^{n})$$X\subset Y\subset Z\cross \mathrm{I}\mathrm{I}^{\infty},$$(Y\backslash X)\leq\omega_{0}$$\oplus_{n=1}^{\infty}\mathrm{I}\mathrm{I}^{n}$$\omega_{0}$$Y\backslash X$

trdef$(X\backslash (\{0\}\cross \mathrm{I}\mathrm{I}^{\infty}))\leq$$X\leq\omega_{0}$and$\mathcal{K}$-trsur$X\leq\omega_{0}$.

(2)Lemma 2.1 If for each $m\geq 1$ there exists an integer $k(m)\geq m+1$ such
$i\leq m$ ,that for any $n\geq k(m)$ and any pari,ition
andinbetween, then $P$ -trind $X=\infty$ .
we have$A_{i}^{n}$$L_{i}^{n}$$B_{i}^{n}$$\mathrm{I}\mathrm{I}^{n},$$E_{n} \cap\bigcap_{i=1}^{m}L_{i}^{n}\not\in P$

Proof. By Proposition 2.2, it suffices to show that $X$ does not have the. For each $i=1,2,$
property
be a partition between compact
let
:
and
sets
$N$
$N$
$x_{i}=1\}$ We shall show thatfor every natural number . Let
betweenbe a natural number. For each $i>1$
let us consider a partition$-$,
andin such that $L_{i}=L_{i}\cap X$ . Note that for every there exists
a natural number
such that for any $n\geq ni$. Let a fixed
is a partition between
inand
integer with $n \geq\max\{n_{1}, \ldots, n_{N}, k(N)\}$ . Then
, and $C\not\in P$ by theis a closed subset ofassumption. So.
$(*)_{P}$$L_{i}$$\ldots$$A_{i}=\{(0, (x_{j}))\in\{\mathrm{O}\}\cross \mathrm{I}\mathrm{I}^{\infty} :x_{i}=0\}$$B_{i}=\{(0, (x_{j}))\in\{0\}\cross \mathrm{I}\mathrm{I}^{\infty}$$\bigcap_{i=1}^{N}L_{i}\not\in P$$L_{i}’$$A_{i}$$B_{i}$$i$$\mathrm{Y}$$n_{i}\geq 2$$L_{i}^{n}=L_{i}^{J} \cap(\{\frac{1}{n}\}\cross \mathrm{I}\mathrm{I}^{n})$$\{\frac{1}{\tau\iota}\}\cross A_{i}^{n}$$\{\frac{1}{n}\}\cross B_{i}^{n}$$\{\frac{1}{n}\}\cross \mathrm{I}\mathrm{I}^{n}$$n$$C=( \bigcap_{i=1}^{N}L_{i}^{n})\cap(\{\frac{1}{n}\}\cross E_{n})=$$( \bigcap_{i=1}^{N}L_{i})\cap(\{\frac{1}{n}\}\cross E_{n})$$\bigcap_{i=1}^{N}L_{i}$$\bigcap_{i=1}^{N}L_{i}\not\in P$

We shall also use the following.

Lemma 2.2 ([8, Lemma 5. ) Let
be partitions between the opposite
and $1\leq p<n$ . Thenandin , wherefacesmeeting the, there is a continuum
for any $k\neq i_{j},j=1,$
faces and .
$\mathit{2}J$$A_{i_{j}}^{n}$$B_{i_{j}}^{n}$$\ldots,p$$A_{k}^{n}$$B_{k}^{n}$$L_{i_{j}}$$1\leq i_{1}<i_{2}\ldots<i_{\mathrm{p}}\leq n$$\mathrm{I}\mathrm{I}^{n}$$C \subset\bigcap_{j=1}^{p}L_{i_{j}}$38be an ordinal number withandof II such that

Lemma 2.3 Let$sub_{\mathit{8}}etsQ_{\alpha},$$P_{\alpha}$$a$$1\leq\alpha<\omega_{1}$. Then there exist$D_{\alpha}$1.

$Q_{\alpha}\in A(a)-\mathcal{M}(a)_{l}$2.

$P_{\alpha}\in \mathcal{M}(\alpha)-A(a)$,3. $D_{\alpha}\in A(a+1)\cap \mathcal{M}(\alpha+1)-(A(a)\cup \mathcal{M}(\alpha))$ .
Proof of Theorem 1.1. (1) We shall prove for$\mathrm{Y}_{\alpha}$

only. We put$\mathrm{Y}_{\alpha}=(\{0\}\cross \mathrm{I}\mathrm{I}^{\infty})\cup(\bigcup_{n=2}^{\infty}\{\frac{1}{n}\}\cross\pi_{n}^{-1}(Q_{\alpha})))$:
II be
is the subspace II described in Lemma 2.3 andwhere
the projection onto the n-th factor. By the construction of , it is clear, and. Since the-trsurthat
absolute Borel classes are preserved under perfect preimages, it follows thatand hence $A(a)$ -trind $Y_{\alpha}=-1$ . Now, it. Thus,. To apply Lemma 2.1, for every-trindsuffices to show thatnatural number $m$ let $k(m)=m+1$ . For each $n\geq k(m)$ and each $i\leq n$
in . By Lemmea 2.2, there exsits
andbe a partition between
let
. Let
and
a continuum such that$\pi_{n}^{G}=\pi|C$ : $Carrow$ II be the restriction of the projection
over . Then
. Sinceis, it follows thatandclosed set of. Thus, it follows from Lemma 2.1 that
. This completes the proof.
trind(2) The second part of Theorem 1.1 is a direct consequence of tbe following proposition.
$\mathrm{I}\mathrm{I}^{n}arrow$$Q_{\alpha}$$\pi_{n}$$Y_{\alpha}$$\mathcal{M}(\alpha)$$\mathcal{M}(a)- \mathrm{t}\mathrm{r}\mathrm{d}\mathrm{e}\mathrm{f}Y_{\alpha}\leq \mathrm{t}\mathrm{r}\mathrm{d}\mathrm{e}\mathrm{f}Y_{\alpha}\leq\omega_{0}$$\pi_{n}^{-1}(Q_{\alpha})\in A(a)$$\mathrm{Y}_{\alpha}\leq\omega_{0}$$Y_{\alpha}\in A(a)$$Y_{\alpha}=\infty$$\mathcal{M}(\alpha)$$\mathrm{I}\mathrm{I}^{n}$$A_{i}^{n}$$L_{i}^{n}$$C$$B_{i}^{n}$$C \subset\bigcap_{i=1}^{n}L_{i}^{n}$$C\cap A_{i}^{n}\neq\emptyset\neq C\cap B_{i}^{n}$$\pi_{n}$$C \cap\pi_{n}^{-1}(Q_{\alpha})=(\pi_{n}^{C})^{-1}(Q_{\alpha})\subset\bigcap_{i=1}^{n}L_{i}^{n}\cap\pi_{n}^{-1}(Q_{\alpha})$$\bigcap_{i=1}^{n}L_{i}^{n}\cap\pi_{n}^{-1}(Q_{\alpha})$$C$$C\cap\pi_{n}^{-1}(Q_{\alpha})$$(\pi_{n}^{C})^{-1}(Q_{\alpha})\not\in \mathcal{M}(a)$$\bigcap_{i=1}^{n}L_{i}^{n}\cap\pi_{n}^{-1}(Q_{\alpha})\not\in \mathcal{M}(a)$$\mathcal{M}(\alpha)-$$\mathrm{Y}_{\alpha}=\infty$

Proposition 2.3 Let $X$ be a separable metrizable space with $A(a)$ -trind $X\leq$. Then-trindand$\mu_{1}$$\mathcal{M}(a)$$A(a)\cap \mathcal{M}(\alpha)$$X\leq\mu_{2}$-trind$X=$

Proof. The proposition can be proved by a standard transfinite induction
on $\nu=\max\{\mu_{1}, \mu_{2}\}$ .

Connect with Proposition 2.1, we ask the following question.

39such that
Question 2.1 Does there exsit a separable metrizable space-trind-trind $X_{\alpha}> \max$ { $A(a)$ -trind} for each ordinal number ? In particular, does there exist a separable metrizable space$X$ such that-ind $X=1$ and -ind $X=S$-trind $X=0$ ?
$X_{\alpha}$$A(\alpha)\cap \mathcal{M}(a)$$X_{\alpha},$$\mathcal{M}(\alpha)$$X_{\alpha}$$a$$C_{0}\cap S$$C_{0}$

Recall from M.G. Charalambous ([2]) that we call a subset of a space$X$ a Bemstein set if $|A\cap B|=|(X\backslash A)\cap B|=c$ for every uncountable Borel
set of $X$ , where denotes the casrdinality of the continuum. It is known
that every uncountable completely metrizable space $X$ has countably many
disjoint Berstein sets. We notice that $A\not\in AB$ for every Berstein set of an
uncountable completely metrizable space $X$ .
$A$

$B$

$c$

$A$

Proof of Theorem 1.2. Let $F$ be a Berstein set of II. We put $X=$
. Then, we can show that $X$ is the desired
space by an argument similar to Theorem 1.1.
$( \{0\}\cross \mathrm{I}\mathrm{I}^{\infty})\cup(\bigcup_{n=1}^{\infty}\{\frac{1}{n}\}\cross\pi_{n}^{-1}(F))$

Connect with Theorem 1.1, we may ask the following question.
Question 2.2 For each ordinal numbers and with $1\leq a<\omega_{1}$ and
which
and
do there exist separable metrizable spaces
satisfy the folowing conditions?
$a$$\beta$$0\leq\beta<\omega_{1}$$X_{\alpha,\beta}$

1. $A(a)$ -trind$X_{\alpha},\rho=\beta$$Y_{\alpha,\beta}$,

2.

$\mathcal{M}(a)$-trind3.

$\mathcal{M}(a)$-trind $X_{\alpha,\beta}=A(a)$ -trind$\mathrm{Y}_{\alpha,\beta}=\beta$, and$\mathrm{Y}_{\alpha,\beta}=-1$.

References[1] J.M. Aarts, T. Nishiura, Dimension and Extensions, North-Holland,
Amsterdam 1993
[2] M.G. Charalambous, On transfinite inductive dimension and deficiency modulo a class P, Topol. Appl. 81 (1997) 123-135[3] R. Engelking, Theory of dimensions, finite and infinite, Heldermann
Verlag, Lemgo, 1995.

40[4] A. Lelek, Dimension and mappings of spaces with finite deficiency,
Colloquium Mathematicum 12 (1964) 221-227[5] E. Pol, The Baire-category method in some compact extension problems, Pacific J. Math. 122: 1 (1986) 197-210[6] R. Pol, On transfinite inductive compactness degree, Colloq. Math.
53 (1987), 57-61.

[7] S.M. Srivastava, A Course on Borel Sets, Springer Verlag, New York,

1998[8] L.R. Rubin, R.M. Schori, J. J. Walsh, New dimension-theory techniques for constructing infinite-dimensional examples, General Topol.
Appl. 10 (1979) 93-102(V.A. Chatyrko)
Department of Mathematics, Link\"oping University, 58183 Linkeping, Swe-den.
vitja@mai.

$\mathrm{l}\mathrm{i}\mathrm{u}$.se(Y. Hattori)
Department of Mathematics, Shimane University, Matsue, Shimane, 6908504 Japanhattori@riko.shimane-u.ac.jp