by

Following [1] we say that a subset $X$$X$XX$X$ of the $n$$n$nn$n$-dimensional real Euclidean space $R^n$$R^n$R^(n)R^{n}$R^n$ is $k$$k$kk$k$-independent ( $0⩽k⩽n-1$$0⩽k⩽n-1$0 <= k <= n-10 \leqslant k \leqslant n-1$0⩽k⩽n-1$ ) if any $k+2$$k+2$k+2k+2$k+2$ distinct points of that subset are linearly independent. ( ${}^1$${}^1$^(1){ }^{1}${}^1$ )

In what follows the homeomorphic image of the set $\{(x^1,\dots ,x^m)$$\left\{\left(x^1,\dots ,x^m\right)\right.${(x^(1),dots,x^(m)):}\left\{\left(x^{1}, \ldots, x^{m}\right)\right.$\{(x^1,\dots ,x^m)$ : $\sum {\left(x^i\right)}^2<1\}$$\left.\sum {\left(x^i\right)}^2<1\right\}${: sum(x^(i))^(2) < 1}\left.\sum\left(x^{i}\right)^{2}<1\right\}$\sum {\left(x^i\right)}^2<1\}$ in $R^m$$R^m$R^(m)R^{m}$R^m$ will be said to be an open $m$$m$mm$m$-cell; the homeomorphic image of the set $\{(x^1,\dots ,x^m):\sum {\left(x^i\right)}^2=1\}$$\left\{\left(x^1,\dots ,x^m\right):\sum {\left(x^i\right)}^2=1\right\}${(x^(1),dots,x^(m)):sum(x^(i))^(2)=1}\left\{\left(x^{1}, \ldots, x^{m}\right): \sum\left(x^{i}\right)^{2}=1\right\}$\{(x^1,\dots ,x^m):\sum {\left(x^i\right)}^2=1\}$ will be said to be an $m-1$$m-1$m-1m-1$m-1$-sphere.
K. Borsuk [1] has proved the following imbedding theorem concerning $k$$k$kk$k$-independent sets:

If $X$$X$XX$X$ is a compact $k$$k$kk$k$-independent set in $R^n$$R^n$R^(n)R^{n}$R^n$ and if $N$$N$NN$N$ is an open subset in $X$$X$XX$X$ containing $k$$k$kk$k$ distinct points, then $X\mathrm{\setminus }N$$X\mathrm{\setminus }N$X\\NX \backslash N$X\mathrm{\setminus }N$ is homeomorphic with a subset of $R^{n-k}$$R^{n-k}$R^(n-k)R^{n-k}$R^{n-k}$.

In [6], p. 503 and in [4], another notion of $k$$k$kk$k$-independence is applied, which is useful in applications in the approximation theory and which will be called in the sequel $k$$k$kk$k$-vectorial-independence.

The subset $X$$X$XX$X$ of $R^n$$R^n$R^(n)R^{n}$R^n$ will be said to be $k$$k$kk$k$-vectorial-independent if for any $k$$k$kk$k$ of its distinct points $x_1,\dots ,x_k$$x_1,\dots ,x_k$x_(1),dots,x_(k)x_{1}, \ldots, x_{k}$x_1,\dots ,x_k$ the vectors $\overrightarrow{O{x}_1},\dots ,\overrightarrow{O{x}_k}$$\overrightarrow{O{x}_1},\dots ,\overrightarrow{O{x}_k}$vec(Ox_(1)),dots, vec(Ox_(k))\overrightarrow{O x_{1}}, \ldots, \overrightarrow{O x_{k}}$\overrightarrow{O{x}_1},\dots ,\overrightarrow{O{x}_k}$, where $O$$O$OO$O$ is the origin in $R^n$$R^n$R^(n)R^{n}$R^n$, are linearly independent.

Observation 1. A $k$$k$kk$k$-vectorial-independent subset $X$$X$XX$X$ in $R^n$$R^n$R^(n)R^{n}$R^n$ is $k-2$$k-2$k-2k-2$k-2$-independent in the sense of [1].

Indeed, if $x_1,\dots ,x_k$$x_1,\dots ,x_k$x_(1),dots,x_(k)x_{1}, \ldots, x_{k}$x_1,\dots ,x_k$ are $k$$k$kk$k$ distinct points in $X$$X$XX$X$, then they cannot be contained in any $k-2$$k-2$k-2k-2$k-2$-dimensional hyperplane $H^{k-2}$$H^{k-2}$H^(k-2)H^{k-2}$H^{k-2}$, because such a hyperplane generates a $k-1$$k-1$k-1k-1$k-1$-dimensional subspace (i.e. a $k-1$$k-1$k-1k-1$k-1$-dimensional hyperplane passing through the origin), and if $x_1,\dots ,x_k$$x_1,\dots ,x_k$x_(1),dots,x_(k)x_{1}, \ldots, x_{k}$x_1,\dots ,x_k$ were in $H^{k-2}$$H^{k-2}$H^(k-2)H^{k-2}$H^{k-2}$, the vectors $\overrightarrow{O{x}_1},\dots ,\overrightarrow{O{x}_k}$$\overrightarrow{O{x}_1},\dots ,\overrightarrow{O{x}_k}$vec(Ox_(1)),dots, vec(Ox_(k))\overrightarrow{O x_{1}}, \ldots, \overrightarrow{O x_{k}}$\overrightarrow{O{x}_1},\dots ,\overrightarrow{O{x}_k}$ would be linearly dependent, being in $R^{k-1}$$R^{k-1}$R^(k-1)R^{k-1}$R^{k-1}$.

Observation 2. If $X$$X$XX$X$ is a $k$$k$kk$k$-independent subset in $R^n$$R^n$R^(n)R^{n}$R^n$, then it may be considered a $k+2$$k+2$k+2k+2$k+2$-vectorial-independent subset in $R^{n+1}$$R^{n+1}$R^(n+1)R^{n+1}$R^{n+1}$ if we consider $R^n$$R^n$R^(n)R^{n}$R^n$ as a hyperplane $H^n$$H^n$H^(n)H^{n}$H^n$ in $R^{n+1}$$R^{n+1}$R^(n+1)R^{n+1}$R^{n+1}$ not passing through the origin.

Indeed, consider $k+2$$k+2$k+2k+2$k+2$ distinct points $x_1,\dots ,x_{k+2}$$x_1,\dots ,x_{k+2}$x_(1),dots,x_(k+2)x_{1}, \ldots, x_{k+2}$x_1,\dots ,x_{k+2}$ in $X$$X$XX$X$ in $H^n$$H^n$H^(n)H^{n}$H^n$. The vectors $\overrightarrow{O{x}_1},\dots ,\overrightarrow{O{x}_{k+2}}$$\overrightarrow{O{x}_1},\dots ,\overrightarrow{O{x}_{k+2}}$vec(Ox_(1)),dots, vec(Ox_(k+2))\overrightarrow{O x_{1}}, \ldots, \overrightarrow{O x_{k+2}}$\overrightarrow{O{x}_1},\dots ,\overrightarrow{O{x}_{k+2}}$ in $R^{n+1}$$R^{n+1}$R^(n+1)R^{n+1}$R^{n+1}$ are linearly independent. If they were linearly dependent, there would exist a subspace $R^{k+1}$$R^{k+1}$R^(k+1)R^{k+1}$R^{k+1}$ in $R^{n+1}$$R^{n+1}$R^(n+1)R^{n+1}$R^{n+1}$ containing them, which would intersect $H^n$$H^n$H^(n)H^{n}$H^n$ in a $k$$k$kk$k$-dimensional hyperplane $H^k$$H^k$H^(k)H^{k}$H^k$ containing the points $x_1,\dots ,x_{k+2}$$x_1,\dots ,x_{k+2}$x_(1),dots,x_(k+2)x_{1}, \ldots, x_{k+2}$x_1,\dots ,x_{k+2}$, which is a contradiction.

It was conjectured by A. M. Gleason (see [8]) that the $k$$k$kk$k$-independent compact subset $X$$X$XX$X$ in $R^n$$R^n$R^(n)R^{n}$R^n$ is homeomorphic with a subset of $S^{n-k}$$S^{n-k}$S^(n-k)S^{n-k}$S^{n-k}$, the $n-k$$n-k$n-kn-k$n-k$-sphere. Investigations about this imbedding conjecture were announced by C. T. Yang in [8], but we have not been able to obtain any information about his results.

In Theorem 2 of [4] it was proved that if $X$$X$XX$X$ is a compact $k$$k$kk$k$-vectorialindependent set in $R^n,N$$R^n,N$R^(n),NR^{n}, N$R^n,N$ is open in $X$$X$XX$X$ and contains $k-2$$k-2$k-2k-2$k-2$ distinct points, then $X\mathrm{\setminus }N$$X\mathrm{\setminus }N$X\\NX \backslash N$X\mathrm{\setminus }N$ is homeomorphic with a subset of $R^{n-k+1}$$R^{n-k+1}$R^(n-k+1)R^{n-k+1}$R^{n-k+1}$, which is an analogue of the imbedding theorem of $K$$K$KK$K$. Borsuk for $k$$k$kk$k$-vectorial-independent sets. By a similar reformulation of the conjecture of A. M. Gleason, we obtain:

If $X$$X$XX$X$ is a $k$$k$kk$k$-vectorial-independent compact subset of $R^n$$R^n$R^(n)R^{n}$R^n$, then it is homeomorphic with a subset of $S^{n-k+1}$$S^{n-k+1}$S^(n-k+1)S^{n-k+1}$S^{n-k+1}$.

Making use of our Observation 2 above, we can see that this conjecture implies the conjecture of Gleason. Our conjecture for $k=n$$k=n$k=nk=n$k=n$ is the well-known theorem of J. Mairhuber [3] in the approximation theory. For $k=1$$k=1$k=1k=1$k=1$ it is obviously true, and for $k=2$$k=2$k=2k=2$k=2$ an imbedding of $X$$X$XX$X$ into a proper subset of $S^{n-1}$$S^{n-1}$S^(n-1)S^{n-1}$S^{n-1}$ may be realised by the radial projection with respect to the origin of $R^n$$R^n$R^(n)R^{n}$R^n$ into the geometrical sphere with its centre at the origin.

Let $X$$X$XX$X$ be a compact $k$$k$kk$k$-independent set in $R^n$$R^n$R^(n)R^{n}$R^n$ containing an $m$$m$mm$m$-cell. Then, as has been proved by S. S. Ryškov [5], the following inequality is valid: $\left({}^2\right)$$\left({}^2\right)$(^(2))\left({ }^{2}\right)$\left({}^2\right)$

Suppose now that the compact $k$$k$kk$k$-vectorial-independent subset $X$$X$XX$X$ in $R^n$$R^n$R^(n)R^{n}$R^n$ is of dimension $m$$m$mm$m$. Then there exists a closed subset $X_0$$X_0$X_(0)X_{0}$X_0$ of $X$$X$XX$X$ of the same dimension $m$$m$mm$m$ and an $n-1$$n-1$n-1n-1$n-1$-hyperplane $H^{n-1}$$H^{n-1}$H^(n-1)H^{n-1}$H^{n-1}$ which has the property of separating strictly the origin $O$$O$OO$O$ and the subset $X_0$$X_0$X_(0)X_{0}$X_0$. Denote by $X_0^{\prime }$$X_0^{\prime }$X_(0)^(')X_{0}^{\prime}$X_0^{\prime }$ the radial projection with respect to 0 of $X_0$$X_0$X_(0)X_{0}$X_0$ into $H^{n-1}$$H^{n-1}$H^(n-1)H^{n-1}$H^{n-1}$. Obviously, $X_0^{\prime }$$X_0^{\prime }$X_(0)^(')X_{0}^{\prime}$X_0^{\prime }$ is a $k$$k$kk$k$-vectorial-independent homeomorphic image of $X_0$$X_0$X_(0)X_{0}$X_0$, and therefore
$\left({}^2\right)$$\left({}^2\right)$(^(2))\left({ }^{2}\right)$\left({}^2\right)$ In [5], Ryškov defines the so-called $k$$k$kk$k$-regular sets as being in fact $k$$k$kk$k$-independent in the sense of [1], and has announced his inequality for these sets. But all the reasonings in the text are valid for $k$$k$kk$k$-l-independent sets. In Uspehi Mat. Nauk 15 (6) (1960), pp. 125-132, the definition of the $k$$k$kk$k$-regular sets is changed in this sense. Our inequality follows from the inequality of Ryskov applied to $k$$k$kk$k$-independent sets.
from Observation 1 it follows that $X_0^{\prime }$$X_0^{\prime }$X_(0)^(')X_{0}^{\prime}$X_0^{\prime }$ is a $k-2$$k-2$k-2k-2$k-2$-independent subset of $H^{n-1}$$H^{n-1}$H^(n-1)H^{n-1}$H^{n-1}$. Applying the inequality of S. S. Ryškov we conclude that

The present note aims at giving a proof of our conjecture in the particular case where the $k$$k$kk$k$-vectorial-independent set $X$$X$XX$X$ in $R^n$$R^n$R^(n)R^{n}$R^n$ contains an $n-k+1$$n-k+1$n-k+1n-k+1$n-k+1$-cell. More precisely, we shall prove the following.

THEOREM. Let $X$$X$XX$X$ be a compact subset of $R^n,n⩾2$$R^n,n⩾2$R^(n),n >= 2R^{n}, n \geqslant 2$R^n,n⩾2$, which is $k$$k$kk$k$-vectorialindependent and contains an $n-k+1$$n-k+1$n-k+1n-k+1$n-k+1$-cell. Then $X$$X$XX$X$ is homeomorphic with a subset of $S^{n-k+1}$$S^{n-k+1}$S^(n-k+1)S^{n-k+1}$S^{n-k+1}$.

The inequality (*) restricts $k$$k$kk$k$ in this case to $k⩽3$$k⩽3$k <= 3k \leqslant 3$k⩽3$ or $k=n$$k=n$k=nk=n$k=n$. As we have observed above, in the case of $k=2$$k=2$k=2k=2$k=2$ the proof is simple, and in the case of $k=n$$k=n$k=nk=n$k=n$ it is known. Therefore only the case $k=3$$k=3$k=3k=3$k=3$ will be considered, and to justify our theorem in this case, we observe that each geometrical sphere $S^{n-2}$$S^{n-2}$S^(n-2)S^{n-2}$S^{n-2}$ in a hyperplane $H^{n-1}$$H^{n-1}$H^(n-1)H^{n-1}$H^{n-1}$ in $R^n$$R^n$R^(n)R^{n}$R^n$ not passing through the origin is a 3 -vectorial-independent set in $R^n$$R^n$R^(n)R^{n}$R^n$ containing $n-2$$n-2$n-2n-2$n-2$-cells.

In the proof given here we apply a method utilised by I. J. Schoenberg and C. T. Yang in [7] for proving the theorem of J. Mairhuber. An important moment in the proof is the employing of the following theorem of M. Brown [2]:

If $h$$h$hh$h$ is a homeomorphic imbedding of $S^{n-1}\times I$$S^{n-1}\times I$S^(n-1)xx IS^{n-1} \times I$S^{n-1}\times I$ into $S^n$$S^n$S^(n)S^{n}$S^n$, then the closure of either complementary domain of $h({\mathbb{S}}^{n-1}\times \{1/2\})$$h\left({\mathbb{S}}^{n-1}\times \{1/2\}\right)$h(S^(n-1)xx{1//2})h\left(\mathbb{S}^{n-1} \times\{1 / 2\}\right)$h({\mathbb{S}}^{n-1}\times \{1/2\})$ in ${\mathbb{S}}^n$${\mathbb{S}}^n$S^(n)\mathbb{S}^{n}${\mathbb{S}}^n$ is a closed $n$$n$nn$n$-cell. (Here $I=[0,1]$$I=[0,1]$I=[0,1]I=[0,1]$I=[0,1]$.)

We begin with a lemma:
Lemma. Let $X$$X$XX$X$ be a compact Hausdorff space having the following properties:
(i) $X$$X$XX$X$ contains an open $n$$n$nn$n$-cell $Q$$Q$QQ$Q$ as an open subset;
(ii) if $N$$N$NN$N$ is a non-empty open subset of $X$$X$XX$X$, then $X\mathrm{\setminus }N$$X\mathrm{\setminus }N$X\\NX \backslash N$X\mathrm{\setminus }N$ may be imbedded in a proper subset of $S^n$$S^n$S^(n)S^{n}$S^n$.

Then $X$$X$XX$X$ is homeomorphic with a subset of $S^n$$S^n$S^(n)S^{n}$S^n$.
Proof. If $X$$X$XX$X$ is not connected, the proof is immediate.
Suppose that $X$$X$XX$X$ is connected. Let $A$$A$AA$A$ be an annulus, that is to say the homeomorphic image of the set $S^{n-1}\times I$$S^{n-1}\times I$S^(n-1)xx IS^{n-1} \times I$S^{n-1}\times I$, which is contained in the $n$$n$nn$n$-cell $Q$$Q$QQ$Q$. Since $Q$$Q$QQ$Q$ is open in $X,A$$X,A$X,AX, A$X,A$ separates $X$$X$XX$X$ and so does ${\sigma }^{n-1}$${\sigma }^{n-1}$sigma^(n-1)\sigma^{n-1}${\sigma }^{n-1}$, the image in $A$$A$AA$A$ of $S^{n-1}\times \{1/2\}$$S^{n-1}\times \{1/2\}$S^(n-1)xx{1//2}S^{n-1} \times\{1 / 2\}$S^{n-1}\times \{1/2\}$, i.e. $X\mathrm{\setminus }A=Y_1\cup Y_2,X\mathrm{\setminus }{\sigma }^{n-1}=V_1\cup V_2$$X\mathrm{\setminus }A=Y_1\cup Y_2,X\mathrm{\setminus }{\sigma }^{n-1}=V_1\cup V_2$X\\A=Y_(1)uuY_(2),X\\sigma^(n-1)=V_(1)uuV_(2)X \backslash A=Y_{1} \cup Y_{2}, X \backslash \sigma^{n-1}=V_{1} \cup V_{2}$X\mathrm{\setminus }A=Y_1\cup Y_2,X\mathrm{\setminus }{\sigma }^{n-1}=V_1\cup V_2$, where $Y_1,Y_2$$Y_1,Y_2$Y_(1),Y_(2)Y_{1}, Y_{2}$Y_1,Y_2$, and respectively $V_1,V_2$$V_1,V_2$V_(1),V_(2)V_{1}, V_{2}$V_1,V_2$ are non-empty open disjoint subsets of $X$$X$XX$X$. Suppose that $Y_1\subset V_1,Y_2\subset V_2$$Y_1\subset V_1,Y_2\subset V_2$Y_(1)subV_(1),Y_(2)subV_(2)Y_{1} \subset V_{1}, Y_{2} \subset V_{2}$Y_1\subset V_1,Y_2\subset V_2$ and introduce the notations: $B_1=V_1\cup {\sigma }^{n-1},B_2=V_2\cup {\sigma }^{n-1}$$B_1=V_1\cup {\sigma }^{n-1},B_2=V_2\cup {\sigma }^{n-1}$B_(1)=V_(1)uusigma^(n-1),B_(2)=V_(2)uusigma^(n-1)B_{1}=V_{1} \cup \sigma^{n-1}, B_{2}=V_{2} \cup \sigma^{n-1}$B_1=V_1\cup {\sigma }^{n-1},B_2=V_2\cup {\sigma }^{n-1}$. The sets $B_1$$B_1$B_(1)B_{1}$B_1$ and $B_2$$B_2$B_(2)B_{2}$B_2$ are both 'connected and are not separated by ${\sigma }^{n-1}$${\sigma }^{n-1}$sigma^(n-1)\sigma^{n-1}${\sigma }^{n-1}$. Denote by $f$$f$ff$f$ and by $g$$g$gg$g$ the homeomorphisms of $X\mathrm{\setminus }{Y}_2$$X\mathrm{\setminus }{Y}_2$X\\Y_(2)X \backslash Y_{2}$X\mathrm{\setminus }{Y}_2$ and, respectively, of $X\mathrm{\setminus }{Y}_1$$X\mathrm{\setminus }{Y}_1$X\\Y_(1)X \backslash Y_{1}$X\mathrm{\setminus }{Y}_1$ in $S^n$$S^n$S^(n)S^{n}$S^n$, which exist according to (ii). Since the above sets both contain $A$$A$AA$A$, from the theorem of M . Brown [2]
it follows that the complementary domains of $f{\sigma }^{n-1}$$f{\sigma }^{n-1}$fsigma^(n-1)f \sigma^{n-1}$f{\sigma }^{n-1}$ and $g{\sigma }^{n-1}$$g{\sigma }^{n-1}$gsigma^(n-1)g \sigma^{n-1}$g{\sigma }^{n-1}$ in $S^n$$S^n$S^(n)S^{n}$S^n$ are open $n$$n$nn$n$-cells. Thus we may suppose that $f$$f$ff$f$ and $g$$g$gg$g$ are homeomorphisms which both transform ${\sigma }^{n-1}$${\sigma }^{n-1}$sigma^(n-1)\sigma^{n-1}${\sigma }^{n-1}$ in the equator $E$$E$EE$E$ of $S^n$$S^n$S^(n)S^{n}$S^n$ and by which $B_1$$B_1$B_(1)B_{1}$B_1$ is mapped into the north hemisphere and $B_2$$B_2$B_(2)B_{2}$B_2$ into the south hemisphere of $S^n$$S^n$S^(n)S^{n}$S^n$ (suppose that $S^n$$S^n$S^(n)S^{n}$S^n$ is a geometrical sphere). Consider the following homeomorphism of $E$$E$EE$E$ onto itself: $l=g\circ f^{-1}\mid E$$l=g\circ f^{-1}\mid E$l=g@f^(-1)∣El=g \circ f^{-1} \mid E$l=g\circ f^{-1}\mid E$. Let $h$$h$hh$h$ be an extension of the homeomorphism $l$$l$ll$l$ to a homeomorphism of the whole north hemisphere onto itself. Then $h\circ f$$h\circ f$h@fh \circ f$h\circ f$ will be a homeomorphism of $B_1$$B_1$B_(1)B_{1}$B_1$ into the north hemisphere carrying ${\sigma }^{n-1}$${\sigma }^{n-1}$sigma^(n-1)\sigma^{n-1}${\sigma }^{n-1}$ onto $E$$E$EE$E$. Consider the mapping

$\phi$$\phi$varphi\varphi$\phi$ is a well-defined mapping which is one-to-one and continuous. To prove its continuity, let $U\subset \phi X$$U\subset \phi X$U sub varphi XU \subset \varphi X$U\subset \phi X$ be open. If $U\cap E=\mathrm{\varnothing }$$U\cap E=\mathrm{\varnothing }$U nn E=O/U \cap E=\emptyset$U\cap E=\mathrm{\varnothing }$, then ${\phi }^{-1}U$${\phi }^{-1}U$varphi^(-1)U\varphi^{-1} U${\phi }^{-1}U$ is open according to the continuity of $h\circ f$$h\circ f$h@fh \circ f$h\circ f$ and $g$$g$gg$g$. Suppose $U\cap E\ne \varnothing$$U\cap E\ne \varnothing$U nn E!=O/U \cap E \neq \varnothing$U\cap E\ne \varnothing$. Then the sets $h\circ f{B}_1\cap U$$h\circ f{B}_1\cap U$h@fB_(1)nn Uh \circ f B_{1} \cap U$h\circ f{B}_1\cap U$ and $g{B}_2\cap U$$g{B}_2\cap U$gB_(2)nn Ug B_{2} \cap U$g{B}_2\cap U$ are open in the relative topology of $h\circ f{B}_1$$h\circ f{B}_1$h@fB_(1)h \circ f B_{1}$h\circ f{B}_1$ and $g{B}_2$$g{B}_2$gB_(2)g B_{2}$g{B}_2$ respectively. Therefore the sets

and

are open in the relative topology of $B_1$$B_1$B_(1)B_{1}$B_1$ and $B_2$$B_2$B_(2)B_{2}$B_2$, respectively. Let $G_1$$G_1$G_(1)G_{1}$G_1$ and $G_2$$G_2$G_(2)G_{2}$G_2$ be open sets in $X$$X$XX$X$ such that $G_1\cap B_1=W_1,G_2\cap B_2=W_2$$G_1\cap B_1=W_1,G_2\cap B_2=W_2$G_(1)nnB_(1)=W_(1),G_(2)nnB_(2)=W_(2)G_{1} \cap B_{1}=W_{1}, G_{2} \cap B_{2}=W_{2}$G_1\cap B_1=W_1,G_2\cap B_2=W_2$. Then the sets $G_1\cup V_2$$G_1\cup V_2$G_(1)uuV_(2)G_{1} \cup V_{2}$G_1\cup V_2$ and $G_2\cup V_1$$G_2\cup V_1$G_(2)uuV_(1)G_{2} \cup V_{1}$G_2\cup V_1$ are open in $X$$X$XX$X$ and

But $W_1\cup W_2={\phi }^{-1}U$$W_1\cup W_2={\phi }^{-1}U$W_(1)uuW_(2)=varphi^(-1)UW_{1} \cup W_{2}=\varphi^{-1} U$W_1\cup W_2={\phi }^{-1}U$, which completes the proof of the continuity of $\phi$$\phi$varphi\varphi$\phi$. From the compactness of $X$$X$XX$X$ it follows that $\phi$$\phi$varphi\varphi$\phi$ is a homeomorphic imbedding of $X$$X$XX$X$ into $S^n$$S^n$S^(n)S^{n}$S^n$.

Proof of the theorem. If $X$$X$XX$X$ in the theorem contains an $n-2$$n-2$n-2n-2$n-2$-cell (remember that only the case $k=3$$k=3$k=3k=3$k=3$ is considered), then it contains an open $n-2$$n-2$n-2n-2$n-2$ - cell as an open set. Indeed, suppose that $Q$$Q$QQ$Q$ is an open $n-2$$n-2$n-2n-2$n-2$-cell in $X$$X$XX$X$ such that $X\mathrm{\setminus }\overline{Q}\ne \mathrm{\varnothing }$$X\mathrm{\setminus }\overline{Q}\ne \mathrm{\varnothing }$X\\ bar(Q)!=O/X \backslash \bar{Q} \neq \emptyset$X\mathrm{\setminus }\overline{Q}\ne \mathrm{\varnothing }$. According to Theorem 2 in [4], $Q$$Q$QQ$Q$ is open in any closed proper subset in $X$$X$XX$X$ in which it is contained. Then $Q$$Q$QQ$Q$ is open in $X$$X$XX$X$ according to the normality of this space. From Theorem 2 in [4] it also follows that $X\mathrm{\setminus }N$$X\mathrm{\setminus }N$X\\NX \backslash N$X\mathrm{\setminus }N$ may be topologically imbedded into a proper subset of $S^{n-2}$$S^{n-2}$S^(n-2)S^{n-2}$S^{n-2}$ for any non-empty, open subset $N$$N$NN$N$ in $X$$X$XX$X$. It follows that all the conditions of the lemma are satisfied and therefore $X$$X$XX$X$ is homeomorphic with a subset of $S^{n-2}$$S^{n-2}$S^(n-2)S^{n-2}$S^{n-2}$.

[1] K. Borsuk, On the $k$$k$kk$k$-independent sets of the Euclidean space and of the Hilbert space, Bull. Acad. Sci. Pol. 5 (1957), pp. 351-356.
[2] M. Brown, A proof of the generalised Schoenflies theorem, Bull. Amer. Math. Soc. 66 (1960), pp. 74-76.
[3] J. C. Mairhuber, On Haar's theorem concerning Chebyshev approximation problems having unique solutions, Proc. Amer. Math. Soc. 7 (1956), pp. 609-615.
[4] A. B. Németh, Homeomorphe projections of $k$$k$kk$k$-independent sets and Chebyshev subspaces of finite dimensional Chebyshev spaces, Mathematica 9 (1967), pp. 325-333.
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THE COMPUTING INSTITUTE OF THE ACADEMY OF ROUMANIA Cluj

Reçu par la Rédaction le 5. 5. 1968