Nonequality of dimensions for metric spaces
Transactions of the American Mathematical Society
Introduction. There are three classical set-theoretic notions of dimension; they are [2, p. 153]: Small inductive dimension ( = Menger-Urysohn dimension), denoted by ind such that ind (S) = -1 if S is empty, ind (S) = n if for every point p e S and open set U containing p there is an open set V satisfying p e V <=■ U, ind (boundary of V) S n-l, and ind (S)=n if ind (S)fZn but ind (S) = n-1 is not true. Large inductive dimension (due to Urysohn), denoted by Ind such that Ind (S) = -1 if S is
... S) = -1 if S is empty, Ind (S) á n if for every closed set Ce S and open set U containing C there is an open set V satisfying Cc v= U, Ind (boundary of V) = n-1, and Ind (S) = n if Ind (S)^n but Ind (S)Sn-l is not true. Covering dimension (=Lebesgue covering dimension), denoted by dim such that dim (S) = -1 if S is empty, dim (S) S n if every finite open cover of S has a finite open refining cover of order ^n+1, that is, no point of S belongs to more than n -I-1 members of the refinement, and dim (S)=n if dim (S) ^ n but dim (S) ^ n -1 is not true. It is well known that for separable metric space S ind (S) = Ind (S) = dim (S). Recently Katetov showed that for any metric space S Ind (S) = dim (S). However, the question, is ind (S)=dim (S) for all metric spaces, remained open. We shall answer this question in the negative. (For a comprehensive account of the preceding material the reader is invited to consult P. Aleksandrov's paper [1, pp. 1-4].) We shall prove, in succeeding sections, the following statement. Theorem. There is a complete metric space A such that ind (A)=0 but dim (A) = 1.