Non-archimedean metrics in topology
Non-archimedean metrics in topology
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拓扑中的非阿基米德度量
DOI：
10.1090/s0002-9939-1956-0080905-8
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发表时间：
1956
期刊：
影响因子：
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通讯作者：
J. Groot
中科院分区：
文献类型：
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作者：
J. Groot
我们将找到以下充要条件：一。空间是可度量化的（参见Nagata [1]，Smirnof [2]），II.空间是强0维的。性质II意味着空间中任何两个不相交的闭集合都可以被（空集）分开。我们将进一步证明条件I和II等价于以下拓扑性质：空间是具有0维NS-基的Hausdorff空间。我们称空间的一个开基为NS-基，如果它是可数个局部有限族的和（一个开集族是局部有限的，如果空间的任何一点包含在一个开集中，该开集最多与该族的有限个集合相交）。如果这个基的集合既开又闭，我们称它为0维NS-基。因此，在度量空间M中，可以引入非阿基米德度量（以拓扑等价的方式）当且仅当M是强0维的。这解决了A. F. [14]前几天。关于强0维的条件是否可以被较弱形式的0维（任何点和闭集，相互不相交，可以分离）取代的问题仍然没有解决。当然，在可分度量空间的情况下，答案是肯定的，因为这两个概念是等价的（参见。[3，第15页]）。然而，在我看来，这些概念在一般的度量空间中是不等价的（参见。[3，附录]，对于更一般的拓扑空间的情况）。是否存在一个（弱）0维度量空间，其中两个特定的不相交闭集不能分离？在这样的空间中，不可能引入非阿基米德度量。
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.