Abstract. It is shown that refinable maps preserve weak infinite-dimension, but not strong infinite-dimension.The purpose of this note is to show that refinable maps preserve weak infinitedimension, but not strong infinite-dimension. Under a refinable map between compacta, the domain and image must have the same finite-dimension or must both have infinite-dimension (see [4, Theorem 1.8(4); 6, Theorem I, 16]).The term compactum is used to mean a compact metric space. A map /: X -Y between compacta is said to be an e-mapping, e > 0, if diam f~x(y) < e for each y G Y. If x and y are points of a metric space, d(x, y) denotes the distance from x to y. A map r: X -» Y between compacta is refinable [2] if for each e > 0 there is a surjective e-mapping/: X -» Y such thatSuch a map / is called an e-refinement of r. A space X is weakly infinite-dimensional if for each countable family {(A¡, B¡)\i = 1,2,3,...} of pairs of disjoint closed sets in X there are partitions S¡ between A¡ and Bi with D°l, Sx■ -0. A space X is strongly infinite-dimensional if X is not weakly infinite-dimensional. A space X is countable-dimensional if X = U°l, Xi with dim X¡ < 0 for each i. If X is countable-dimensional, then X is weakly infinite-dimensional (see [3, II2F p. 16
]).We need the following.Lemma [5, Lemma 1]. Let f be a map from a compactum X to an ANR A and e > 0. Then there is a positive number Ô > 0 such that if g is any S-mapping from X onto any compactum Y, then there is a map A: Y -> A such that d(f, hg) < e.By using the lemma, we show the following theorem. Theorem 1. Every refinable map preserves weak infinite-dimension. In other words, there is no refinable map from a weakly infinite-dimensional compactum to a strongly infinite-dimensional compactum.