CHINEN, NAOTSUGU

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TitleISOMETRIES ON THE SYMMETRIC PRODUCTS OF THEEUCLIDEAN SPACES WITH USUAL METRICS (Thepresent situation of set-theoretic and geometric topology and itsprospects)Author(s)CHINEN, NAOTSUGUCitation数理解析研究所講究録 (2014), 1884: 7-12Issue Date2014-04URL http://hdl.handle.net/2433/195696RightType Departmental Bulletin PaperTextversionpublisherKyoto UniversityISOMETRIES ON THE SYMMETRIC PRODUCTS OF THEEUCLIDEAN SPACES WITH USUAL METRICS(NAOTSUGU CHINEN)(NATIONAL DEFENSE ACADEMY OF JAPAN)1. INTRODUCTIONAs an interesting construction in topology, Borsuk and Ulam [3] introducedthe n-th symmetric product of a metric space $(X, d)$ , denoted by $F_{n}(X)$ . Namely$F_{n}(X)$ is the space of non-empty finite subsets of $X$ with at most $n$ elementsendowed with the Hausdorff metric $d_{H}$ , i.e., $F_{n}(X)=\{A\subset X|1\leq|A|\leq n\}$ and$d_{H}(A, B)= \inf\{\epsilon|A\subset B_{d}(B, \epsilon)$ and $B \subset B_{d}(A, \epsilon)\}=\max\{d(a, B),$ $d(b, A)a\in$$A,$ $b\in B\}$ for any $A,$ $B\in F_{n}(X)$ (see [10, p.6]).For the symmetric products of $\mathbb{R}$ , it is known that $F_{2}(\mathbb{R})\approx \mathbb{R}\cross[0, \infty)$ and$F_{3}(\mathbb{R})\approx \mathbb{R}^{3}$ (see Section 3). It was proved in [3] that $F_{n}(I)$ is homeomorphic to$I^{n}$ (written $F_{n}(I)\approx I^{n}$ ) if and only if $1\leq n\leq 3$ , and that for $n\geq 4,$ $F_{n}(I)$ cannot be embedded into $\mathbb{R}^{n}$ , where $I=[0,1]$ has the usual metric. Thus, for $n\geq 4,$$F_{n}(\mathbb{R})\not\simeq \mathbb{R}^{n}$ . Molski [12] showed that $F_{2}(I^{2})\approx I^{4}$ , and that for $n\geq 3$ neither$F_{n}(I^{2})$ nor $F_{2}(I^{n})$ can be embedded into $\mathbb{R}^{2n}$ . Thus, for $n\geq 3,$ $F_{n}(\mathbb{R}^{2})$ pt $\mathbb{R}^{2n}$ and$F_{2}(\mathbb{R}^{n})\not\simeq \mathbb{R}^{2n}.$Turning toward the symmetric product $F_{n}(\mathbb{S}^{1})$ of the circle $\mathbb{S}^{1}$ , Chinen andKoyama [9] prove that for $n\in \mathbb{N}$ , both $F_{2n-1}(\mathbb{S}^{1})$ and $F_{2n}(\mathbb{S}^{1})$ have the samehomotopy type of the $(2n-1)$-sphere $\mathbb{S}^{2n-1}$ . In [7] Bott corrected Borsuk’sstatement [4] and showed that $F_{3}(\mathbb{S}^{1})\approx \mathbb{S}^{3}$ . In [9], another proof of it is given.For a metric space $(X, d)$ , we denote by $Isom_{d}(X)$ (Isom(X) for short) thegroup of all isometries from $X$ into itself, i.e., $\phi$ : $Xarrow X\in Isom_{d}(X)$ if $\phi$is a bijection satisfying that $d(x, x’)=d(\phi(x), \phi(x’))$ for any $x,$ $x’\in X$ . Let$n\in \mathbb{N}$ . Every isometry $\phi$ : $Xarrow X$ induces an isometry $\chi_{(n)}(\phi)$ : $(F_{n}(X), d_{H})arrow$$(F_{n}(X), d_{H})$ defined by $\chi_{(n)}(\phi)(A)\overline{\neg}\phi(A)$ for each $A\in F_{n}(X)$ . Thus, thereexists a natural monomorphism $\chi_{(n)}$ : $Isom_{d}(X)arrow Isom_{d_{H}}(F_{n}(X))$ . It is clearthat $\chi_{(n)}$ : $Isom_{d}(X)arrow Isom_{d_{H}}(F_{n}(X))$ is an isomorphism if and only if $\chi_{(n)}$ isan epimorphism, i.e., for every $\Phi\in Isom_{d_{H}}(F_{n}(X))$ there exists $\phi\in Isom_{d}(X)$such that $\Phi=\chi_{(n)}(\phi)$ .In this paper, it is of interest to know whether $\chi_{(n)}$ : $Isom_{d}(X)$ $arrow$$Isom_{d_{H}}(F_{n}(X))$ is an isomorphism for a metric space $(X, d)$ . Recently, Borovikovaand Ibragimov [5] prove that $(F_{3}(\mathbb{R}), d_{H})$ is bi-Lipschitz equivalent to $(\mathbb{R}^{3}, d)$ andthat $\chi_{(3)}$ : $Isom_{d}(\mathbb{R})arrow Isom_{d_{H}}(F_{3}(\mathbb{R}))$ is an isomorphism, where $\mathbb{R}$ has the usualmetric $d$ . The following result is a generalization of the result above and theaffirmative answer to [6, p.60, Conjecture 2.1].1884 2014 7-12 7Theorem 1.1. Let $l\in \mathbb{N}$ and let $X=\mathbb{R}^{l}$ or $X=\mathbb{S}^{l}$ with the usual metric $d.$Then $\chi_{(n)}$ : $Isom_{d}(X)arrow Isom_{d_{H}}(F_{n}(X))$ is an isomorphism for each $n\in \mathbb{N}.$In Section 4, we give the main ideas of proof of Theorem 1.1. In Example5.2 below, we present a compact metric space ($X$, d) such that $\chi_{(n)}(Isom_{d}(X))\neq$$Isom_{d_{H}}(F_{n}(X))$ for all $n\geq 2$ , i.e., $\chi_{(n)}$ : $Isom_{d}(X)arrow Isom_{d_{H}}(F_{n}(X))$ is not anisomorphism. And, in Section 3, we provide another proof of [5, Theorem 6]. Itsproof is based on the proof of [11, Lemma 2.4].2. PRELIMINARIESNotation 2.1. Let denote the set of all natural numbers and real numbers by $\mathbb{N}$ and$\mathbb{R}$ , respectively. Let $d$ be the usual metric on $\mathbb{R}^{l}$ , i.e., $d(x, y)= \{\sum_{i=1}^{l}(x_{i}-y_{i})^{2}\}^{1/2}$for any $x=(x_{1}, \ldots, x_{l}),$ $y=(y_{1}, \ldots, y_{l})\in \mathbb{R}^{l}$ . Write $\mathbb{S}^{l}=\{x=(x_{1}, \ldots, x_{l+1})\in$$\mathbb{R}^{l+1}|\sum_{i=1}^{l+1}x_{i}^{2}=1\}$ with the length metric $d$ . Denote the identity map from $X$into itself by $id_{X}.$Definition 2.2. Let $(X, d)$ be a metric space, let $x\in X$ , let $Y,$ $Z$ be subsetsof $X$ and let $\epsilon>0$ . Set $d(Y, Z)= \inf\{d(y, x)|y\in Y, z\in Z\}$ , and $B_{d}(Y, \epsilon)=$$\{x\in X|d(x, Y)\leq\epsilon\}$ . If $Y=\{y\}$ , for simplicity of notation, we write $B_{d}(y, \epsilon)=$$B_{d}(Y, \epsilon)$ and $S_{d}(y, \epsilon)=S_{d}(Y, \epsilon)$ .For $n\in \mathbb{N}$ , the n-th symmetric product of $X$ is defined by$F_{n}(X)=\{A\subset X|1\leq|A|\leq n\},$where $|A|$ is the cardinality of $A$ . Write $F_{(m)}(X)=\{A\in 2^{X}||A|=m\}$ for each$m\in \mathbb{N}$ . Let Isom$(X, Y)=\{\phi\in Isom(X)|\phi(y)=y$ for each $y\in Y\}$ for $Y\subset X.$Set $r(A)= \min\{\{1\}\cup\{d(a, a’)|a, a’\in A, a\neq a’\}\}$ for each $A\in F_{n}(X)$ .3. A METRIC SPACE IS $BI$-LIPSCHITZ EQUIVALENT TO THE SYMMETRICPRODUCT OF $\mathbb{R}$In this section, we give another proof of [5, Theorem 6] which is based on theproof of [11, Lemma 2.4].Definition 3.1. Let $n\in \mathbb{N}$ . Set $F_{n}^{*}(I)=\{A\in F_{n}(I)|0,1\in A\}$ . It is knownthat $F_{2}^{*}(I)=\{\{0,1\}\},$ $F_{3}^{*}(I)=\{\{0, t, 1\}|0\leq t\leq 1\}\approx \mathbb{S}^{1}$ , and, $F_{4}^{*}(I)=$$\{\{0, s, t, 1\}|0\leq s\leq t\leq 1\}$ is homeomorphic to the dance hat (see [16]). Ingeneral, $F_{2n}^{*}(I)$ is contractible but not collapsible, and $F_{2n+1}^{*}(I)$ has the samehomotopy type of $\mathbb{S}^{2n+1}$ . In [1], it is called the spaces $F_{2n}^{*}(I),$ $n\geq 2$ , higherdimensional dunce hats (see [1]).8Definition 3.2 ([11]). Let ($X$, d) be a metric space with diam $X\leq 2$ . SetCon$e^{}$ $(X)=Xx[0, \infty)/(X\cross\{O\})$ , is said to be the open cone over $X$ , with themetric $d_{C}([(x_{1}, t_{1})],$ $[(x_{2}, t_{2})])=|t_{1}-t_{2}|+ \min\{t_{1}, t_{2}\}\cdot d(x_{1}, x_{2})$ .Definition 3.3. Let $f$ : $(X, d)arrow(Y, d’)$ be a map. The map $f$ is said to beLipschitz $(bi-$Lipschitz, respectively) if there exists $L>0$ such that$d’(f(x_{1}), f(x_{2}))\leq Ld(x_{1}, x_{2})$$(L^{-1}d(x_{1}, x_{2})\leq d’(f(x_{1}),$ $f(x_{2}))\leq Ld(x_{1}, x_{2})$ , respectively)for any $x_{1},$ $x_{2}\in X$ . ( $X$, d) is said to be $bi$-Lipschitz equivalent to $(Y, d’)$ if thereexists a surjective bi-Lipschitz map from ($X$ , d) to $(Y, d’)$ .Theorem 3.4 ([11]). Let $n\in \mathbb{N}$ with $n\geq 2$ . Then $(F_{n}(\mathbb{R}), d_{H})$ is $bi$-Lipschitzequivalent to $(\mathbb{R}\cross Cone^{o}(F_{n}^{*}(I)), \rho)$ , where $\rho=\sqrt{d^{2}+(d_{H})_{C}^{2}}.$Sketch of Proof. Let $Z= \{A\in F_{n}(\mathbb{R})|\min A=0\}$ . For every $A\in Z$ there existsthe unique $E\in F_{n}^{*}(I)$ such that $A=tE$, where $t= \max A.$Stepl: $(F_{n}(\mathbb{R}), d_{H})$ is bi-Lipschitz equivalent to $(\mathbb{R}\cross Z, \rho_{1})$ , where $\rho_{1}$ $=$$\sqrt{d^{2}+(d_{H})^{2}}$ . In fact, we can show the following.Stepl.1: $A$ map $f$ : $F_{n}( \mathbb{R})arrow \mathbb{R}\cross Z:A\mapsto(\min A, A-\min A)$ is $\sqrt{5}$-Lipschitz.Stepl.2: $A$ map $f^{-1}$ : $\mathbb{R}\cross Zarrow F_{n}(\mathbb{R}):(b, A)\mapsto A+b$ is 2-Lipschitz.Step2: $(Z, d_{H})$ is bi-Lipschitz equivalent to (Cone $(F_{n}^{*}(I)),$ $(d_{H})_{C}$). In fact, wecan show the following.Step2.1: $A$ map $g:Zarrow Cone^{o}(F_{n}^{*}(I))$ : $tE\mapsto[(E, t)]$ is 1-Lipschitz.Step2.2: $A$ map $g^{-1}:Cone^{o}(F_{n}^{*}(I))arrow Z:[(E, t)]\mapsto tE$ is 3-Lipschitz.By the above, $(id_{\mathbb{R}}\cross g)\circ f$ : $F_{n}(\mathbb{R})arrow \mathbb{R}\cross Zarrow \mathbb{R}\cross Cone^{o}(F_{n}^{*}(I))$ is a bi-Lipschitzequivalence. $\square$Corollary 3.5. $(F_{2}(\mathbb{R}), d_{H})$ is $bi$-Lipschitz equivalent to $(\mathbb{R}\cross[0, \infty), d)$ .Proof. By Definition 3.1, $F_{2}^{*}(I)$ is one point, thus (Cone $(F_{2}^{*}(I)),$ $(d_{H})_{C}$ ) is corre-sponding to $([0, \infty), d)$ . By Theorem 3.4, $(F_{2}(\mathbb{R}), d_{H})$ is bi-Lipschitz equivalentto $(\mathbb{R}\cross[0, \infty), d)$ . $\square$The following result is first proved in [5, Theorem 6]. We give another proofby use of Theorem 3.4.Corollary 3.6 ([5]). $(F_{3}(\mathbb{R}), d_{H})$ is $bi$-Lipschitz equivalent to $(\mathbb{R}^{3}, d)$ .Sketch of Proof. We note $F_{3}^{*}(If)=\{\{0, t, 1\}|0\leq t\leq 1\}\approx \mathbb{S}^{1}$Stepl: We can show that (Cone $(F_{n}^{*}(I)),$ $(d_{H})_{C}$ ) is bi-Lipschitz equivalent to(Cone $(\mathbb{S}^{1}),$ $(d|_{\mathbb{S}^{1}})_{C}$ ).Step2: We can show that $(\mathbb{R}^{2}, d)$ is bi-Lipschitz equivalent to (Cone $(\mathbb{S}^{1}),$ $(d|_{S^{1}})_{C}$ ).By Theorem 3.4, $(F_{3}(\mathbb{R}), d_{H})$ is bi-Lipschitz equivalent to $(\mathbb{R}^{3}, d)$ . $\square$9Remark 3.7. We note that $F_{2}(\mathbb{R}^{2})\approx \mathbb{R}^{4}$ . Indeed, we can define a homeomorphism$h:F_{2}(\mathbb{R}^{2})arrow \mathbb{R}^{2}\cross Cone^{o}(\mathbb{S}^{1}/x\sim-x)(\approx \mathbb{R}^{4})$ by$h(A)=\{\begin{array}{ll}(m(A), [(\frac{2(A-m(A))}{diamA}, diam A)]) if diam A\neq 0( m(A) , the cone point) if diam A=0,\end{array}$where $m(A)=a$ if $A=\{a\}$ and $m(A)=(a+a’)/2$ if $A=\{a, a’\}$ . In general, wesee that $F_{2}(\mathbb{R}^{\iota})\approx \mathbb{R}^{l}\cross$ Cone$O(\mathbb{S}^{l-1}/x\sim-x)$ for each $l\in \mathbb{N}.$4. ISOMETRIESLemma 4.1. Let $n\in \mathbb{N}$ and let ($X$, d) be a metric space such that(1) $\Phi|_{F_{1}(X)}\in$ Isom$(F_{1}(X))$ for each $\Phi\in$ Isom$(F_{n}(X))$ , and that(2) Isom$(F_{n}(X), F_{1}(X))=\{id_{F_{n}(X)}\}.$Then, $\chi_{(n)}$ : Isom$(X)arrow$ Isom$(F_{n}(X))$ is an isomorphism.Proof. Let $\Phi\in$ Isom$(F_{n}(X))$ and let $A_{x}=\{x\}\in F_{1}(X)$ for each $x\in X.$By assumption, $\Phi|_{F_{1}(X)}\in$ Isom$(F_{1}(X))$ . Denote $\Phi(A_{x})\in F_{1}(X)$ by $\{\phi(x)\}$for each $x\in X$ . Then, $\phi$ : $Xarrow X$ : $x\mapsto\phi(x)$ is an isometry. Set$\Phi’=\chi_{(n)}(\phi^{-1})0\Phi\in$ Isom$(F_{n}(X))$ . We claim that $\Phi’|_{F_{1}(X)}=$ id $|_{F_{1}(X)}$ . In-deed, $\Phi|_{F_{1}(X)}=\chi_{(n)}(\phi)|_{F_{1}(X)}$ and $\chi_{(n)}(\phi^{-1})=(\chi_{(n)}(\phi))^{-1}$ By assumption, wehave that $\Phi’=id_{F_{n}(X)}$ , therefore, $\Phi=\chi_{(n)}(\phi)$ , which completes the proof. $\square$Definition 4.2. Let ($X$, d) be a metric space, let $n\in \mathbb{N}$ , let $\epsilon>0$ and let$A\in F_{n}(X)$ . Define$D_{n}(A, \epsilon)=\sup\{k\in \mathbb{N}|A_{1}, \ldots, A_{k}\in S_{d_{H}}(A, \epsilon), d_{H}(A_{i}, A_{j})=2\epsilon(i\neq j)\}\in \mathbb{N}\cup\{\infty\}.$Lemma 4.3. Let $l,$ $n\in \mathbb{N}$ , let $X=\mathbb{R}^{l}$ or $X=\mathbb{S}^{l}$ and let $\Phi\in$ Isom$(F_{n}(X))$ .Then, $\Phi|_{F_{1}(X)}\in$ Isom$(F_{1}(X))$ .Sketch of Proof. Let $n\in \mathbb{N}$ with $n\geq 2.$Stepl: Let $A=\{a_{1}\}\in F_{1}(X)$ and let $\epsilon>0$ with $\epsilon<r(A)$ . We can show that$D_{n}(A, \epsilon)=3.$Step2: Let $m\in \mathbb{N}$ with $m\geq 2$ , let $A=\{a_{1}, \ldots, a_{m}\}\in F_{(m)}(X)$ and let $\epsilon>0$with $\epsilon<r(A)/5$ . We can show that $D_{n}(A, \epsilon)>3.$Let $\Phi\in$ Isom$(F_{n}(X))$ and let $A\in F_{n}(X)$ . From the definition of $D_{n}(A, \epsilon)$ ,we obtain $D_{n}(A, \epsilon)=D_{n}(\Phi(A), \epsilon)$ for each $0< \epsilon<\min\{r(A), r(\Phi(A))\}$ . Bythe above, we see that $A\in F_{1}(X)$ if and only if $\Phi(A)\in F_{1}(X)$ . Therefore,$\Phi|_{F_{1}(X)}\in$ Isom$(F_{1}(X))$ . $\square$Lemma 4.4. Let $l,$ $n\in \mathbb{N}$ . Then, Isom$(F_{n}(\mathbb{R}^{\iota}), F_{1}(\mathbb{R}^{l}))=\{id_{F_{n}(\mathbb{R}^{l})}\}.$10Sketch of Proof.Stepl: Let $l,$ $n\in \mathbb{N}$ and let $\Phi\in Isom(F_{n}(\mathbb{R}^{\iota}), F_{1}(\mathbb{R}^{l}))$ . Then, $\Phi|_{F_{2}(\mathbb{R}^{l})}=id_{F_{2}(\mathbb{R}^{l})}.$Step2: Let $n\in \mathbb{N}$ with $n\geq 2$ and let $\Phi\in$ Isom$(F_{n}(\mathbb{R}^{l}), F_{1}(\mathbb{R}^{l}))$ and let $A\in$$F_{(m)}(\mathbb{R}^{\iota})$ . We can show that $\Phi(A)\subset A$ . If similar arguments apply to$\Phi(A)$ and $\Phi^{-1}$ , we obtain $A=\Phi^{-1}(\Phi(A))\subset\Phi(A)$ , therefore, $A=\Phi(A)$ .$\square$Lemma 4.5. Let $l,$ $n\in \mathbb{N}$ . Then Isom$(F_{n}(\mathbb{S}^{\iota});^{F_{1}(\mathbb{S}^{\iota}))}=\{id_{F_{n}(\mathbb{S}^{l})}\}.$Proof. Let $\Phi\in$ Isom$(F_{n}(\mathbb{S}^{l}), F_{1}(\mathbb{S}^{l})),$ $m\in \mathbb{N}$ with $2\leq m\leq n$ and let $A\in F_{(m)}(\mathbb{S}^{\iota})$ .We show that $A=\Phi(A)$ . Let $a\in A$ and let $a’\in \mathbb{S}^{\iota}$ be the anti-point of $a$ . Since$d_{H}(\{a’\}, \Phi(A))=d_{H}(\Phi(\{a’\}), \Phi(A))=d_{H}(\{a’\}, A)=\pi$ , we have $a\in\Phi(A)$ ,therefore, $A\subset\Phi(A)$ . If similar arguments apply to $\Phi(A)$ and $\Phi^{-1}$ , we obtain$\Phi(A)\subset\Phi^{-1}(\Phi(A))=A$ , therefore, $A=\Phi(A)$ , which completes the proof. $\square$The proof of Theorem 1.1. By Lemmas 4.3, 4.4 and 4.5, the conditions in Lemma4.1 hold for $(X, d)$ , which completes the proof. a5. QUESTIONSQuestion 5.1. Let 1, $n\in \mathbb{N}$ with $n\geq 2$ . When ($X$ , d) is a following space, $is$$\chi_{n}$ : $Isom_{d}(X)arrow Isom_{d_{H}}(F_{n}(X))$ an isomorp hism 2(1) $X=\mathbb{R}^{l}$ has a metric $d_{\infty}$ , where $d_{\infty}(x, y)= \max\{|x_{i}-y_{i}||i=1, \ldots, l\}$for any $x=(x_{1}, \ldots, x_{l}),$ $y=(y_{1}, \ldots, y_{l})\in X.$(2) $X$ is a convex subset of $\mathbb{R}^{l}.$(3) $X$ is an $\mathbb{R}$-tree (see [2] for $\mathbb{R}$ -trees).(4) $X$ is the hyperbolic $l$ -space (see [8] for the hyperbolic $l$ -space).Example 5.2. Let $n,$ $m\in \mathbb{N}$ with $2\leq n\leq m$ and let ($X$ , d) be an $m$-pointsdiscrete metric space satisfying that $d(x, x’)=1$ whenever $x\neq x’$ . Then, $F_{n}(X)$is a discrete metric space such that $d_{H}(A, A’)=1$ for any $A,$ $A’\in F_{n}(X)$ with$A\neq A’$ . Thus, Isom$(X)|=|X|!<|F_{n}(X)|!=$ Isom$(F_{n}(X))|$ , therefore, $\chi_{(n)}$ :$Isom_{d}(X)arrow Isom_{d_{H}}(F_{n}(X))$ is not an isomorphism.This drives us to the following question as the generalization of Theorem 1.1,Question 5.3. Let ($X$ , d) be a connected metric space. Then, is $\chi_{(n)}$ : Isom$(X)arrow$$Isom(F_{n}(X))$ an isomorphism?$\mathbb{S}^{3}?$Question 5 $.4$ . It is known that $F_{3}(\mathbb{S}^{1})\approx \mathbb{S}^{3}$ . Is $F_{3}(\mathbb{S}^{1})bi$-Lipschitz equivalent to11REFERENCES[1] R. N. Andersen, M. M. Marjanovi\v{c}, and R. M. Schori, Symmetric products and higher-dimensional dunce hats, Topology Proc. 18 (1993), 7-17.[2] M. Bestvina, $\mathbb{R}$ -trees in topology, geometry, and group theory, Handbook of geometrictopology, 55-91, North-Holland, Amsterdam, 2002.[3] K. Borsuk and S. Ulam, On symmetric products of topological spaces, Bull. A. M. S.37(1931), 875-882.[4] K. Borsuk, On the third symmetric potency of the circumference, Ifund. Math. 36 (1949),236-244.[5] M. Borovikova and Z. Ibragimov, The third symmetric product of $\mathbb{R}$ , Comput. MethodsFunct. Theory 9 (2009), 255-268.[6] M. Borovikova, Z. Ibragimov and H. Yousefi, Symmetric products of the real line, J. Anal.18 (2010), 53-67.[7] R. Bott, On the third symmetric potency of $S_{1}$ , Fund. Math. 39 (1952), 264-268.[8] M. R. Bridson and A. Haefliger, Metric spaces of non-positive curvature, Grundlehren derMathematischen Wissenschaften, 319. Springer-Verlag, Berlin, 1999.[9] N. Chinen and A. Koyama, On the symmetric hyperspace of the circle, Topology Appl. 157(2010), 2613-2621.[10] A. Illanes and S. B. Nadler, Jr., Hyperspaces, Marcel Dekker, 1999.[11] L. V. Kovalev, Symmetric products of the line; embeddings and retractions, arXiv:1212.$1698v1$ [math.MG] 7 Dec 2012.$[12]$ R. Molski, On symmetric product, Fund. Math. 44 (1957), 165-170.[13] H. R. Morton, Symmetric product of the circle, Proc. Camb. Phil. Soc. 63 (1967), 349-352.[14] R. M. Schori, Hyperspaces and symmetric products of topological spaces, Fund. Math. 39(1968), 77-87.[15] W. Wu, Note sur les produits essentiels sym\’etriques des espaces topologiques, C. R. Acad.Sci. Paris 224 (1947), 1139-1141.[16] E. C. Zeeman, On the dunce hat, Topology 39 (1964), 341-358.DEPARTMENT OF MATHEMATICS, NATIONAL DEFENSE ACADEMY OF JAPAN, YOKOSUKA239-8686, JAPAN$E$-mail address: naochin@nda.ac.jp12

ISOMETRIES ON THE SYMMETRIC PRODUCTS OF THE EUCLIDEAN SPACES WITH USUAL METRICS (The present situation of set-theoretic and geometric topology and its prospects)

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ISOMETRIES ON THE SYMMETRIC PRODUCTS OF THE EUCLIDEAN SPACES WITH USUAL METRICS (The present situation of set-theoretic and geometric topology and its prospects)

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