About an imbedding conjecture for k-independent sets

Abstract

Authors

A.B. Nemeth
Institutul de Calcul

Keywords

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Paper coordinates

A.B. Németh, About an imbedding conjecture for k-independent sets. Fund. Math. 67 1970 203–207.

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Journal

Fundamenta Mathematicae

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Institute of Mathematics Polish Academy of Sciences

DOI
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0016-2736

 

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1730-6329

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About an imbedding conjecture for k k kkk-independent sets

by

A. B. Németh (Cluj)

Following [1] we say that a subset X X XXX of the n n nnn-dimensional real Euclidean space R n R n R^(n)R^{n}Rn is k k kkk-independent ( 0 ⩽ k ⩽ n − 1 0 ⩽ k ⩽ n − 1 0 <= k <= n-10 \leqslant k \leqslant n-10⩽k⩽n−1 ) if any k + 2 k + 2 k+2k+2k+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 , … , x m ) x 1 , … , x m {(x^(1),dots,x^(m)):}\left\{\left(x^{1}, \ldots, x^{m}\right)\right.{(x1,…,xm) : ∑ ( x i ) 2 < 1 } ∑ x i 2 < 1 {: sum(x^(i))^(2) < 1}\left.\sum\left(x^{i}\right)^{2}<1\right\}∑(xi)2<1} in R m R m R^(m)R^{m}Rm will be said to be an open m m mmm-cell; the homeomorphic image of the set { ( x 1 , … , x m ) : ∑ ( x i ) 2 = 1 } x 1 , … , x m : ∑ x i 2 = 1 {(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\}{(x1,…,xm):∑(xi)2=1} will be said to be an m − 1 m − 1 m-1m-1m−1-sphere.
K. Borsuk [1] has proved the following imbedding theorem concerning k k kkk-independent sets:
If X X XXX is a compact k k kkk-independent set in R n R n R^(n)R^{n}Rn and if N N NNN is an open subset in X X XXX containing k k kkk distinct points, then X ∖ N X ∖ N X\\NX \backslash NX∖N is homeomorphic with a subset of R n − k R n − k R^(n-k)R^{n-k}Rn−k.
In [6], p. 503 and in [4], another notion of k k kkk-independence is applied, which is useful in applications in the approximation theory and which will be called in the sequel k k kkk-vectorial-independence.
The subset X X XXX of R n R n R^(n)R^{n}Rn will be said to be k k kkk-vectorial-independent if for any k k kkk of its distinct points x 1 , … , x k x 1 , … , x k x_(1),dots,x_(k)x_{1}, \ldots, x_{k}x1,…,xk the vectors O x 1 → , … , O x k → O x 1 → , … , O x k → vec(Ox_(1)),dots, vec(Ox_(k))\overrightarrow{O x_{1}}, \ldots, \overrightarrow{O x_{k}}Ox1→,…,Oxk→, where O O OOO is the origin in R n R n R^(n)R^{n}Rn, are linearly independent.
Observation 1. A k k kkk-vectorial-independent subset X X XXX in R n R n R^(n)R^{n}Rn is k − 2 k − 2 k-2k-2k−2-independent in the sense of [1].
Indeed, if x 1 , … , x k x 1 , … , x k x_(1),dots,x_(k)x_{1}, \ldots, x_{k}x1,…,xk are k k kkk distinct points in X X XXX, then they cannot be contained in any k − 2 k − 2 k-2k-2k−2-dimensional hyperplane H k − 2 H k − 2 H^(k-2)H^{k-2}Hk−2, because such a hyperplane generates a k − 1 k − 1 k-1k-1k−1-dimensional subspace (i.e. a k − 1 k − 1 k-1k-1k−1-dimensional hyperplane passing through the origin), and if x 1 , … , x k x 1 , … , x k x_(1),dots,x_(k)x_{1}, \ldots, x_{k}x1,…,xk were in H k − 2 H k − 2 H^(k-2)H^{k-2}Hk−2, the vectors O x 1 → , … , O x k → O x 1 → , … , O x k → vec(Ox_(1)),dots, vec(Ox_(k))\overrightarrow{O x_{1}}, \ldots, \overrightarrow{O x_{k}}Ox1→,…,Oxk→ would be linearly dependent, being in R k − 1 R k − 1 R^(k-1)R^{k-1}Rk−1.
Observation 2. If X X XXX is a k k kkk-independent subset in R n R n R^(n)R^{n}Rn, then it may be considered a k + 2 k + 2 k+2k+2k+2-vectorial-independent subset in R n + 1 R n + 1 R^(n+1)R^{n+1}Rn+1 if we consider R n R n R^(n)R^{n}Rn as a hyperplane H n H n H^(n)H^{n}Hn in R n + 1 R n + 1 R^(n+1)R^{n+1}Rn+1 not passing through the origin.
Indeed, consider k + 2 k + 2 k+2k+2k+2 distinct points x 1 , … , x k + 2 x 1 , … , x k + 2 x_(1),dots,x_(k+2)x_{1}, \ldots, x_{k+2}x1,…,xk+2 in X X XXX in H n H n H^(n)H^{n}Hn. The vectors O x 1 → , … , O x k + 2 → O x 1 → , … , O x k + 2 → vec(Ox_(1)),dots, vec(Ox_(k+2))\overrightarrow{O x_{1}}, \ldots, \overrightarrow{O x_{k+2}}Ox1→,…,Oxk+2→ in R n + 1 R n + 1 R^(n+1)R^{n+1}Rn+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}Rk+1 in R n + 1 R n + 1 R^(n+1)R^{n+1}Rn+1 containing them, which would intersect H n H n H^(n)H^{n}Hn in a k k kkk-dimensional hyperplane H k H k H^(k)H^{k}Hk containing the points x 1 , … , x k + 2 x 1 , … , x k + 2 x_(1),dots,x_(k+2)x_{1}, \ldots, x_{k+2}x1,…,xk+2, which is a contradiction.
It was conjectured by A. M. Gleason (see [8]) that the k k kkk-independent compact subset X X XXX in R n R n R^(n)R^{n}Rn is homeomorphic with a subset of S n − k S n − k S^(n-k)S^{n-k}Sn−k, the n − k n − k n-kn-kn−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 XXX is a compact k k kkk-vectorialindependent set in R n , N R n , N R^(n),NR^{n}, NRn,N is open in X X XXX and contains k − 2 k − 2 k-2k-2k−2 distinct points, then X ∖ N X ∖ N X\\NX \backslash NX∖N is homeomorphic with a subset of R n − k + 1 R n − k + 1 R^(n-k+1)R^{n-k+1}Rn−k+1, which is an analogue of the imbedding theorem of K K KKK. Borsuk for k k kkk-vectorial-independent sets. By a similar reformulation of the conjecture of A. M. Gleason, we obtain:
If X X XXX is a k k kkk-vectorial-independent compact subset of R n R n R^(n)R^{n}Rn, then it is homeomorphic with a subset of S n − k + 1 S n − k + 1 S^(n-k+1)S^{n-k+1}Sn−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=nk=n is the well-known theorem of J. Mairhuber [3] in the approximation theory. For k = 1 k = 1 k=1k=1k=1 it is obviously true, and for k = 2 k = 2 k=2k=2k=2 an imbedding of X X XXX into a proper subset of S n − 1 S n − 1 S^(n-1)S^{n-1}Sn−1 may be realised by the radial projection with respect to the origin of R n R n R^(n)R^{n}Rn into the geometrical sphere with its centre at the origin.
Let X X XXX be a compact k k kkk-independent set in R n R n R^(n)R^{n}Rn containing an m m mmm-cell. Then, as has been proved by S. S. Ryškov [5], the following inequality is valid: ( 2 ) 2 (^(2))\left({ }^{2}\right)(2)
[ k + 2 2 ] m + [ k + 1 2 ] ⩽ n . k + 2 2 m + k + 1 2 ⩽ n . [(k+2)/(2)]m+[(k+1)/(2)] <= n.\left[\frac{k+2}{2}\right] m+\left[\frac{k+1}{2}\right] \leqslant n .[k+22]m+[k+12]⩽n.
Suppose now that the compact k k kkk-vectorial-independent subset X X XXX in R n R n R^(n)R^{n}Rn is of dimension m m mmm. Then there exists a closed subset X 0 X 0 X_(0)X_{0}X0 of X X XXX of the same dimension m m mmm and an n − 1 n − 1 n-1n-1n−1-hyperplane H n − 1 H n − 1 H^(n-1)H^{n-1}Hn−1 which has the property of separating strictly the origin O O OOO and the subset X 0 X 0 X_(0)X_{0}X0. Denote by X 0 ′ X 0 ′ X_(0)^(')X_{0}^{\prime}X0′ the radial projection with respect to 0 of X 0 X 0 X_(0)X_{0}X0 into H n − 1 H n − 1 H^(n-1)H^{n-1}Hn−1. Obviously, X 0 ′ X 0 ′ X_(0)^(')X_{0}^{\prime}X0′ is a k k kkk-vectorial-independent homeomorphic image of X 0 X 0 X_(0)X_{0}X0, and therefore
( 2 ) 2 (^(2))\left({ }^{2}\right)(2) In [5], Ryškov defines the so-called k k kkk-regular sets as being in fact k k kkk-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 kkk-l-independent sets. In Uspehi Mat. Nauk 15 (6) (1960), pp. 125-132, the definition of the k k kkk-regular sets is changed in this sense. Our inequality follows from the inequality of Ryskov applied to k k kkk-independent sets.
from Observation 1 it follows that X 0 ′ X 0 ′ X_(0)^(')X_{0}^{\prime}X0′ is a k − 2 k − 2 k-2k-2k−2-independent subset of H n − 1 H n − 1 H^(n-1)H^{n-1}Hn−1. Applying the inequality of S. S. Ryškov we conclude that
[ k 2 ] m + [ k − 1 2 ] ⩽ n − 1 k 2 m + k − 1 2 ⩽ n − 1 [(k)/(2)]m+[(k-1)/(2)] <= n-1\left[\frac{k}{2}\right] m+\left[\frac{k-1}{2}\right] \leqslant n-1[k2]m+[k−12]⩽n−1
The present note aims at giving a proof of our conjecture in the particular case where the k k kkk-vectorial-independent set X X XXX in R n R n R^(n)R^{n}Rn contains an n − k + 1 n − k + 1 n-k+1n-k+1n−k+1-cell. More precisely, we shall prove the following.
THEOREM. Let X X XXX be a compact subset of R n , n ⩾ 2 R n , n ⩾ 2 R^(n),n >= 2R^{n}, n \geqslant 2Rn,n⩾2, which is k k kkk-vectorialindependent and contains an n − k + 1 n − k + 1 n-k+1n-k+1n−k+1-cell. Then X X XXX is homeomorphic with a subset of S n − k + 1 S n − k + 1 S^(n-k+1)S^{n-k+1}Sn−k+1.
The inequality (*) restricts k k kkk in this case to k ⩽ 3 k ⩽ 3 k <= 3k \leqslant 3k⩽3 or k = n k = n k=nk=nk=n. As we have observed above, in the case of k = 2 k = 2 k=2k=2k=2 the proof is simple, and in the case of k = n k = n k=nk=nk=n it is known. Therefore only the case k = 3 k = 3 k=3k=3k=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}Sn−2 in a hyperplane H n − 1 H n − 1 H^(n-1)H^{n-1}Hn−1 in R n R n R^(n)R^{n}Rn not passing through the origin is a 3 -vectorial-independent set in R n R n R^(n)R^{n}Rn containing n − 2 n − 2 n-2n-2n−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 hhh is a homeomorphic imbedding of S n − 1 × I S n − 1 × I S^(n-1)xx IS^{n-1} \times ISn−1×I into S n S n S^(n)S^{n}Sn, then the closure of either complementary domain of h ( S n − 1 × { 1 / 2 } ) h S n − 1 × { 1 / 2 } h(S^(n-1)xx{1//2})h\left(\mathbb{S}^{n-1} \times\{1 / 2\}\right)h(Sn−1×{1/2}) in S n S n S^(n)\mathbb{S}^{n}Sn is a closed n n nnn-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 XXX be a compact Hausdorff space having the following properties:
(i) X X XXX contains an open n n nnn-cell Q Q QQQ as an open subset;
(ii) if N N NNN is a non-empty open subset of X X XXX, then X ∖ N X ∖ N X\\NX \backslash NX∖N may be imbedded in a proper subset of S n S n S^(n)S^{n}Sn.
Then X X XXX is homeomorphic with a subset of S n S n S^(n)S^{n}Sn.
Proof. If X X XXX is not connected, the proof is immediate.
Suppose that X X XXX is connected. Let A A AAA be an annulus, that is to say the homeomorphic image of the set S n − 1 × I S n − 1 × I S^(n-1)xx IS^{n-1} \times ISn−1×I, which is contained in the n n nnn-cell Q Q QQQ. Since Q Q QQQ is open in X , A X , A X,AX, AX,A separates X X XXX and so does σ n − 1 σ n − 1 sigma^(n-1)\sigma^{n-1}σn−1, the image in A A AAA of S n − 1 × { 1 / 2 } S n − 1 × { 1 / 2 } S^(n-1)xx{1//2}S^{n-1} \times\{1 / 2\}Sn−1×{1/2}, i.e. X ∖ A = Y 1 ∪ Y 2 , X ∖ σ n − 1 = V 1 ∪ V 2 X ∖ A = Y 1 ∪ Y 2 , X ∖ σ n − 1 = V 1 ∪ 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∖A=Y1∪Y2,X∖σn−1=V1∪V2, where Y 1 , Y 2 Y 1 , Y 2 Y_(1),Y_(2)Y_{1}, Y_{2}Y1,Y2, and respectively V 1 , V 2 V 1 , V 2 V_(1),V_(2)V_{1}, V_{2}V1,V2 are non-empty open disjoint subsets of X X XXX. Suppose that Y 1 ⊂ V 1 , Y 2 ⊂ V 2 Y 1 ⊂ V 1 , Y 2 ⊂ V 2 Y_(1)subV_(1),Y_(2)subV_(2)Y_{1} \subset V_{1}, Y_{2} \subset V_{2}Y1⊂V1,Y2⊂V2 and introduce the notations: B 1 = V 1 ∪ σ n − 1 , B 2 = V 2 ∪ σ n − 1 B 1 = V 1 ∪ σ n − 1 , B 2 = V 2 ∪ σ 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}B1=V1∪σn−1,B2=V2∪σn−1. The sets B 1 B 1 B_(1)B_{1}B1 and B 2 B 2 B_(2)B_{2}B2 are both 'connected and are not separated by σ n − 1 σ n − 1 sigma^(n-1)\sigma^{n-1}σn−1. Denote by f f fff and by g g ggg the homeomorphisms of X ∖ Y 2 X ∖ Y 2 X\\Y_(2)X \backslash Y_{2}X∖Y2 and, respectively, of X ∖ Y 1 X ∖ Y 1 X\\Y_(1)X \backslash Y_{1}X∖Y1 in S n S n S^(n)S^{n}Sn, which exist according to (ii). Since the above sets both contain A A AAA, from the theorem of M . Brown [2]
it follows that the complementary domains of f σ n − 1 f σ n − 1 fsigma^(n-1)f \sigma^{n-1}fσn−1 and g σ n − 1 g σ n − 1 gsigma^(n-1)g \sigma^{n-1}gσn−1 in S n S n S^(n)S^{n}Sn are open n n nnn-cells. Thus we may suppose that f f fff and g g ggg are homeomorphisms which both transform σ n − 1 σ n − 1 sigma^(n-1)\sigma^{n-1}σn−1 in the equator E E EEE of S n S n S^(n)S^{n}Sn and by which B 1 B 1 B_(1)B_{1}B1 is mapped into the north hemisphere and B 2 B 2 B_(2)B_{2}B2 into the south hemisphere of S n S n S^(n)S^{n}Sn (suppose that S n S n S^(n)S^{n}Sn is a geometrical sphere). Consider the following homeomorphism of E E EEE onto itself: l = g ∘ f − 1 ∣ E l = g ∘ f − 1 ∣ E l=g@f^(-1)∣El=g \circ f^{-1} \mid El=g∘f−1∣E. Let h h hhh be an extension of the homeomorphism l l lll to a homeomorphism of the whole north hemisphere onto itself. Then h ∘ f h ∘ f h@fh \circ fh∘f will be a homeomorphism of B 1 B 1 B_(1)B_{1}B1 into the north hemisphere carrying σ n − 1 σ n − 1 sigma^(n-1)\sigma^{n-1}σn−1 onto E E EEE. Consider the mapping
φ x = { h ∘ f x for x in B 1 , g x for x in B 2 . φ x = h ∘ f x       for       x  in  B 1 , g x       for       x  in  B 2 . varphi x={[h@fx," for ",x" in "B_(1)","],[gx," for ",x" in "B_(2).]:}\varphi x=\left\{\begin{array}{lll} h \circ f x & \text { for } & x \text { in } B_{1}, \\ g x & \text { for } & x \text { in } B_{2} . \end{array}\right.φx={h∘fx for x in B1,gx for x in B2.
φ φ varphi\varphiφ is a well-defined mapping which is one-to-one and continuous. To prove its continuity, let U ⊂ φ X U ⊂ φ X U sub varphi XU \subset \varphi XU⊂φX be open. If U ∩ E = ∅ U ∩ E = ∅ U nn E=O/U \cap E=\emptysetU∩E=∅, then φ − 1 U φ − 1 U varphi^(-1)U\varphi^{-1} Uφ−1U is open according to the continuity of h ∘ f h ∘ f h@fh \circ fh∘f and g g ggg. Suppose U ∩ E ≠ ∅ U ∩ E ≠ ∅ U nn E!=O/U \cap E \neq \varnothingU∩E≠∅. Then the sets h ∘ f B 1 ∩ U h ∘ f B 1 ∩ U h@fB_(1)nn Uh \circ f B_{1} \cap Uh∘fB1∩U and g B 2 ∩ U g B 2 ∩ U gB_(2)nn Ug B_{2} \cap UgB2∩U are open in the relative topology of h ∘ f B 1 h ∘ f B 1 h@fB_(1)h \circ f B_{1}h∘fB1 and g B 2 g B 2 gB_(2)g B_{2}gB2 respectively. Therefore the sets
W 1 = φ − 1 ( h ∘ f B 1 ∩ U ) = ( h ∘ f ) − 1 ( h ∘ f B 1 ∩ U ) W 1 = φ − 1 h ∘ f B 1 ∩ U = ( h ∘ f ) − 1 h ∘ f B 1 ∩ U W_(1)=varphi^(-1)(h@fB_(1)nn U)=(h@f)^(-1)(h@fB_(1)nn U)W_{1}=\varphi^{-1}\left(h \circ f B_{1} \cap U\right)=(h \circ f)^{-1}\left(h \circ f B_{1} \cap U\right)W1=φ−1(h∘fB1∩U)=(h∘f)−1(h∘fB1∩U)
and
W 2 = φ − 1 ( g B 2 ∩ U ) = g − 1 ( g B 2 ∩ U ) W 2 = φ − 1 g B 2 ∩ U = g − 1 g B 2 ∩ U W_(2)=varphi^(-1)(gB_(2)nn U)=g^(-1)(gB_(2)nn U)W_{2}=\varphi^{-1}\left(g B_{2} \cap U\right)=g^{-1}\left(g B_{2} \cap U\right)W2=φ−1(gB2∩U)=g−1(gB2∩U)
are open in the relative topology of B 1 B 1 B_(1)B_{1}B1 and B 2 B 2 B_(2)B_{2}B2, respectively. Let G 1 G 1 G_(1)G_{1}G1 and G 2 G 2 G_(2)G_{2}G2 be open sets in X X XXX such that G 1 ∩ B 1 = W 1 , G 2 ∩ B 2 = W 2 G 1 ∩ B 1 = W 1 , G 2 ∩ 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}G1∩B1=W1,G2∩B2=W2. Then the sets G 1 ∪ V 2 G 1 ∪ V 2 G_(1)uuV_(2)G_{1} \cup V_{2}G1∪V2 and G 2 ∪ V 1 G 2 ∪ V 1 G_(2)uuV_(1)G_{2} \cup V_{1}G2∪V1 are open in X X XXX and
W 1 ∪ W 2 = ( G 1 ∪ V 2 ) ∩ ( G 2 ∪ V 1 ) . W 1 ∪ W 2 = G 1 ∪ V 2 ∩ G 2 ∪ V 1 . W_(1)uuW_(2)=(G_(1)uuV_(2))nn(G_(2)uuV_(1)).W_{1} \cup W_{2}=\left(G_{1} \cup V_{2}\right) \cap\left(G_{2} \cup V_{1}\right) .W1∪W2=(G1∪V2)∩(G2∪V1).
But W 1 ∪ W 2 = φ − 1 U W 1 ∪ W 2 = φ − 1 U W_(1)uuW_(2)=varphi^(-1)UW_{1} \cup W_{2}=\varphi^{-1} UW1∪W2=φ−1U, which completes the proof of the continuity of φ φ varphi\varphiφ. From the compactness of X X XXX it follows that φ φ varphi\varphiφ is a homeomorphic imbedding of X X XXX into S n S n S^(n)S^{n}Sn.
Proof of the theorem. If X X XXX in the theorem contains an n − 2 n − 2 n-2n-2n−2-cell (remember that only the case k = 3 k = 3 k=3k=3k=3 is considered), then it contains an open n − 2 n − 2 n-2n-2n−2 - cell as an open set. Indeed, suppose that Q Q QQQ is an open n − 2 n − 2 n-2n-2n−2-cell in X X XXX such that X ∖ Q ¯ ≠ ∅ X ∖ Q ¯ ≠ ∅ X\\ bar(Q)!=O/X \backslash \bar{Q} \neq \emptysetX∖Q¯≠∅. According to Theorem 2 in [4], Q Q QQQ is open in any closed proper subset in X X XXX in which it is contained. Then Q Q QQQ is open in X X XXX according to the normality of this space. From Theorem 2 in [4] it also follows that X ∖ N X ∖ N X\\NX \backslash NX∖N may be topologically imbedded into a proper subset of S n − 2 S n − 2 S^(n-2)S^{n-2}Sn−2 for any non-empty, open subset N N NNN in X X XXX. It follows that all the conditions of the lemma are satisfied and therefore X X XXX is homeomorphic with a subset of S n − 2 S n − 2 S^(n-2)S^{n-2}Sn−2.

References

[1] K. Borsuk, On the k k kkk-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 kkk-independent sets and Chebyshev subspaces of finite dimensional Chebyshev spaces, Mathematica 9 (1967), pp. 325-333.
[5] S. S. Ryškov (С. С. Рышков), О к-регулярных вложениях, Докл. Акад. Наук. СССР 127 (1959), pp. 272-273.
[6] Yu. A. Šaškin (Ю. А. Шашкин), Систем Коровкина в пространствах непрерывных функций, Изв. Акад. Наук. СССР 26 (1962), рр. 495-512.
[7] I. J. Schoenberg and C. T. Yang, On the unicity of solutions of problems of best approximation, Ann. Math. Pura Appl. 54, pp. 1-12.
[8] C. T. Yang, On non-orientable closed surfaces in Euclidean spaces, Canad, J. Math. 14 (1962), pp. 660-668.
THE COMPUTING INSTITUTE OF THE ACADEMY OF ROUMANIA Cluj
Reçu par la Rédaction le 5. 5. 1968

  1. ( 1 1 ^(1){ }^{1}1 ) For the sake of simplicity, the affine space and the vectorial Euclidean space of dimension n n nnn are denoted by the same symbol R n R n R^(n)R^{n}Rn.
1970

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