Abstract: We shall find the following necessary and sufficient conditions: I. the space is metrizable (cf. Nagata [1], Smirnof [2]), II. the space is strongly 0-dimensional. Property II means that any two closed disjoint sets in the space can be separated (by the empty set). We shall prove furthermore that the conditions I and II are equivalent to the following topological properties: the space is a Hausdorff space having a 0-dimensional NS-base. We call an open base of the space an NS-base, if it is the sum of a countable number of locally finite families (a family of open sets is locally finite, if any point of the space is contained in an open set which intersects at most a finite number of sets of the family). If the sets of this base are both open and closed, we call it a 0-dimensional NS-base. In a metric space M a non-archimedean metric can therefore be introduced (in a topologically equivalent way) if and only if M is strongly 0-dimensional. This settles a problem raised by A. F. Monna [4] some years ago. The question remains unsolved as to whether the condition of strong 0-dimensionality may be replaced by a weaker form of 0-dimensionality (any point and closed set, mutually disjoint, can be separated). Of course the answer is positive in the case of separable metric spaces, for both notions are then equivalent (cf. [3, p. 15]). However, it seems probable to me that these notions are not equivalent in general metric spaces (cf. [3, appendix], for the case of more general topological spaces). Does there exist a (weakly) 0-dimensional metric space in which two certain disjoint closed sets cannot be separated? In such a space it would be impossible to introduce a non-archimedean metric.