数理解析研究所講究録
1531 巻 2007 年 63-67

63

HAUSDORFF HYPERSPACES OF EUCLIDEAN SPACES
AND THEIR DENSE SUBSPACES
Wieslaw Kubis’
Instytut Matematyki, Akademia Swigtokrzyska
Kielce, Poland
シフェントクシスカ学院

筑波大学数学系

酒井克郎 (Katsuro Sakai)

Institute of Mathematics, University of Tsukuba
Here, we introduce the results obtained in the paper [11] and related problems. We
consider metric spaces and their hyperspaces endowed with the Hausdorff metric.
Specifically, given a metric space, we shall denote by Cld(X) and Bd(X)the hyperspaces consisting of all nonempty closed sets and of all nonempty boundedclosed sets in $X$ respectively and we denote by
the Hausdorff metric, whichis infinite-valued on Cld(X) if $X$ is unbounded. When $X$ is compact, the spaceCld(X)is equal to the hyperspace $\exp(X)$ of all nonempty compact
sets with the Vietoris topology. Even if $X$ is noncompact, on the space $\exp(X)$ , theHausdorff metric topology coincides with the Vietoris topology. However, in case $X$
is noncompact, these topologies are very different on the spaces Cld(X) and Bd(X).
Vietoris hyperspaces $\exp(X)$ have been studied extensively for many years. Amongthe known results, let us mention the theorem of Curtis and Schori [8] (cf. [13,Chapter 8]), saying that $\exp(X)$ is homeomorphic tothe Hilbert cube-1,
if and only if $X$ is a Peano continuum, that is, it is compact, connected andlocally connected. Later, Curtis [7] characterized non-compact metric spaces $X$ forwhich $\exp(X)$ is homeomorphic to the Hilbert cube minus a pointor the pseudo-interiorof . In particular,
ishomeomorphic to. For more information concerning Vietoris hyperspaces, werefer to the book of Ilanes and Nadler [10].
It is well known that the hyperspace $\exp(X)$ is an ANR
if and only if $X$ islocally connected (and connected). On the other hand, it is proved in [6] that thespace Bd(X) is an ANR
whenever the metric on $X$ is almost convex, that is,$X=\langle X, d\rangle$$d_{H}$$(=\mathrm{B}\mathrm{d}(X))$$(\cong)$$[$$\mathrm{Q}=$$1]^{\omega}$$\mathrm{Q}\backslash \mathrm{O}(=\mathrm{Q}\backslash \{0\})$$\mathrm{s}=(-1,1)^{\omega}$$\mathrm{Q}^{1}$$\mathrm{B}\mathrm{d}(\mathbb{R}^{m})=\exp(\mathbb{R}^{m})$$\mathrm{Q}\backslash \mathrm{O}$$(\mathrm{A}\mathrm{R})$$(\mathrm{A}\mathrm{R})$$\cdot$1It is well known that is homeomorphic to the separable Hilbert space$\mathrm{s}$$\ell_{2}$.

64for every $\alpha>0,$ $\beta>0$ and for every $y\in X$ such that $d(x, y)<\alpha+\beta$ , there exists$z\in X$ with $d(x, z)<\alpha$ and $d(z, y)<\beta$ . This condition was further weakened in[12], which has turned out to be actually a necessary and sufficient one by Banakhand Voytsitskyy [3]. In the last paper, several equivalent conditions are given, which
are too technical to mention them here. We refer to [3] for the details. On the other
hand, Cld(X) is not connected whenever $X$ is a metric space which is not totally
many components.
bounded. For example, Cld(R) has. Thenis denoted byThe completion of a metric space, via the isometric embedding
Bd(X) can be identified with the subspace of, having in mind this identi. Thus we shall often write
is the completion of Bd(X). By such a reason, we also
fication. In this case,. For each $0\leq k<m$ , letconsider a dense subspace of a metric space$x,$$2^{\aleph_{0}}$$\tilde{X}=\langle\tilde{X}, d\rangle$$X=\langle X, d\rangle$$\mathrm{B}\mathrm{d}(\tilde{X})$$\mathrm{B}\mathrm{d}(X)\subseteq \mathrm{B}\mathrm{d}(\tilde{X})$$A\vdash\neq \mathrm{c}1_{\overline{X}}A$$\mathrm{B}\mathrm{d}(\tilde{X})$$D$$\nu_{k}^{m}=${$X=\langle X, d\rangle$$x=(x_{i})_{i=1}^{m}\in \mathbb{R}^{m}$:$x_{i}\in \mathbb{R}\backslash \mathbb{Q}$except for at most$k$many },$i$of $\dim\leq k$ .
which is the universal space for completely metrizable subspaces in,is homeomorphic to the -dimensional N\"obeling space
In case $2k+1<m,$which is the universal space for all separable completely metrizable spaces. Note$\mathbb{R}^{m}$$k$$\nu_{k}^{m}$that$\nu_{0}^{m}=(\mathbb{R}\backslash \mathbb{Q})^{m}\cong \mathbb{R}\backslash \mathbb{Q}$

Theorem 1. Suppose$\nu_{k}^{2k+1}$.

$\langle m, k\rangle=\langle 1,0\rangle$

or $0\leq k<m-1$ . Then,$\langle \mathrm{B}\mathrm{d}(\mathbb{R}^{m}), \mathrm{B}\mathrm{d}(\nu_{k}^{m})\rangle\cong\langle \mathrm{Q}\backslash 0, \mathrm{s}\backslash 0\rangle$

Consequently,$\mathrm{B}\mathrm{d}(\nu_{k}^{m})\cong\ell_{2}$.

.

This can be derived from the following:such that
set inTheorem 2. Let $D$ be a dense
and in case $m>1$ it is assumed that $D=p[D]\cross \mathbb{R}$ , where
projection onto the first $m-1$ coordinates. Then,$\mathbb{R}^{m}$$G_{\delta}$$\mathbb{R}^{m}\backslash D$

is also dense inis the$\mathbb{R}^{m}$$p:\mathbb{R}^{m}arrow \mathbb{R}^{m-1}$.

$\langle \mathrm{B}\mathrm{d}(\mathbb{R}^{m}), \mathrm{B}\mathrm{d}(D)\rangle\cong\langle \mathrm{Q}\backslash 0,\mathrm{s}\backslash \mathrm{O}\rangle$

Question 1. In case $m>1$ , under the only assumption that, is the pairis also dense in
set and? In particular, is the pair
phic to?
$\mathbb{R}^{m}\backslash D$$G_{\delta}$$\langle \mathrm{Q}\backslash \mathrm{O}, \mathrm{s}\backslash \mathrm{O}\rangle$$\mathbb{R}^{m}$is a dense
homeomorhomeomorphic to$D\subseteq \mathbb{R}^{m}$$\langle \mathrm{B}\mathrm{d}(\mathbb{R}^{m}), \mathrm{B}\mathrm{d}(D)\rangle$$\langle \mathrm{B}\mathrm{d}(\mathbb{R}^{m}), \mathrm{B}\mathrm{d}(\nu_{m-1}^{m})\rangle$$\langle \mathrm{Q}\backslash 0, \mathrm{s}\backslash 0\rangle$

We also consider the following dense subspaces of Bd(X):
Nwd(X) –all nowhere dense closed sets;
Perf(X) –all perfect sets;2$\bullet$$\bullet$2I.e., completely metrizable closed sets which are dense in itself.

65

Cantor(X) –all compact sets homeomorphic to the Cantor set.
, we can also consider the following subspace:In case$\bullet$$X=\mathbb{R}^{m}$$\bullet$$\mathfrak{R}(\mathbb{R}^{m})$–all closed sets of the Lebesgue measure zero.

For these spaces, we have the following:

Theorem 3. Let $F$ be one of the following subspaces of$\mathrm{N}\mathrm{w}\mathrm{d}(\mathbb{R}^{m}),$Then,$\langle$$\mathrm{B}\mathrm{d}(\mathbb{R}^{m}),$$\mathrm{P}\mathrm{e}\mathrm{r}\mathrm{f}(\mathrm{R}^{m}),$$F)\cong\langle \mathrm{Q}\backslash 0, \mathrm{s}\backslash \mathrm{O}\rangle$$\mathrm{C}\mathrm{a}\mathrm{n}\mathrm{t}\mathrm{o}\mathrm{r}(\mathrm{R}^{m})$, hence$F\cong P_{2}$and$\mathrm{B}\mathrm{d}(\mathbb{R}^{m})$$\mathfrak{R}(\mathbb{R}^{m})$:.

.

To prove Theorems 2 and 3 above, we adopt the characterization of the pseudoboundaryof the Hilbert cube , see [5].
We also study the space Cld(R). It is very different from the hyperspace exp(R). It
many components, Bd(R) is the only separable
is not hard to see that Cld(R) has
one and any other component has weight . Applying Torutczyk’s Characterization of Hilbert space [14] (cf. [15]), we can prove$\mathrm{Q}$$\mathrm{Q}\backslash \mathrm{s}$$2^{\mathrm{N}_{0}}$$2^{\aleph_{0}}$be a nonseparable component of Cld(R) which does not contain$(-\infty, 0]$ . Then.

Theorem 4. Let$\mathbb{R}_{f}[0, +\infty),$$\mathcal{H}$$\mathcal{H}\cong\ell_{2}(2^{\aleph_{0}})$

Question 2. Does Theorem 4 hold even if
Question 3. For $m>1$ , is$\mathcal{H}$contains$\mathrm{C}\mathrm{l}\mathrm{d}(\mathbb{R}^{m})\backslash \mathrm{B}\mathrm{d}(\mathbb{R}^{m})$an$\mathbb{R},$$\ell_{2}(2^{\aleph_{0}})$$[0, \infty)$or$(-\infty, 0]$?

-manifold?

Now, we consider the subspacesof Cld(R).
, Nwd(R), Perf(R) and
Similarly to Bd(R), it can be shown that those complements are -sets in Cld(R).
-manifold and is a -set
Due to Negligibility Theorem ([1], [9]), if $M$ is an
in $M$ then $M\backslash A\cong M$ . Thus, the following follows from Theorem 4:$\mathrm{C}\mathrm{l}\mathrm{d}(\mathbb{R}\backslash \mathbb{Q})$$\mathfrak{R}(\mathbb{R})$$\mathrm{Z}_{\sigma}$$\ell_{2}(2^{\mathrm{N}_{0}})$$\mathrm{Z}_{\sigma}$be a nonseparable component of Cld(R) which does not contain$[0, +\infty),$ $(-\infty, 0]$ . Then, the following spaces are homeomorphic to:

Corollary 5. Let$\mathbb{R},$$A$$\mathcal{H}$$\ell_{2}(2^{\aleph_{0}})$$\mathcal{H}\cap \mathfrak{R}(\mathbb{R}),$$\mathcal{H}\cap \mathrm{N}\mathrm{w}\mathrm{d}(\mathbb{R}),$$\mathcal{H}\cap \mathrm{P}\mathrm{e}\mathrm{r}\mathrm{f}(\mathbb{R})$and$\mathcal{H}\cap \mathrm{C}\mathrm{l}\mathrm{d}(\mathbb{R}\backslash \mathbb{Q})$.

Borel classes. Given a metric space X, , let
be its completion. Then, thehyperspaceis the completion of the hyperspace Bd(X). Concerning Borel
classes of hyperspaces, the following are also shown in the paper [11]:(1) Bd(X) is
if $X$ is a-compact.
in(2) Bd(X) isin
if $X$ is Polish.3
3I.e., separable and completely metrizable$\langle$$\mathrm{B}\mathrm{d}(\tilde{X})$$F_{\sigma\delta}$$G_{\delta}$$\mathrm{B}\mathrm{d}(\tilde{X})$$\mathrm{B}\mathrm{d}(\tilde{X})$$d\rangle$$\langle\tilde{X}, d\rangle$

66(3) Bd(X) is Polish for every Polish space $X$ in which bounded sets are totallybounded.
in Bd(X) for every separable metric space $X$ .
(4) Nwd(X) is(5) Perf(X) isin Bd(X) if $X$ is separable and locally compact.
in Bd(X) for every Polish space $X$ .
(6) Perf(X) is(7) Bd(X) is analytic for every analytic metric space $X$ in which bounded sets
are totally bounded.
$G_{\delta}$$G_{\delta}$$F_{\sigma\delta}$

which admits the metricFix a dense set $X$ in a separable Banach spaceinduced from the norm of . Then X, is an almost convex metric space and. In case $X$ is , the space
therefore by a result of [6] the space Bd(X) is an
Bd(X) is completely metrizable by (2). If additionally $E$ is finite-dimensional then.
Bd(X) is Polish by (3). In case $X$ is -compact, by (1), Bd(X) is absolutely
$E$

$E$

$d$

$d\rangle$$\langle$$\mathrm{A}\mathrm{R}$$G_{\delta}$$F_{\sigma\delta}$$\sigma$

Remarks. Recently, Banakh and Voytsitskyy [4] proved that the space Cld(X)
if and only if $X$ is a completely metrizable(resp. Bd(X)) is homeomorphic to
nowhere locally compact metric space such that each (resp. bounded) subset of $X$
is totally bounded and the completion of $X$ is connected and locally connected.
$\ell_{2}$$\tilde{X}$

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