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This paper gives a negative answer to the following question posed by A. D. Wallace. Does each totally nonaposyndetic,2 bicompact, connected, Hausdorff space contain a weak cut point?3 Conditions which assure the existence of weak cut points in totally nonaposyndetic continua are given in [1] and [2]. Example 1 gives a totally disconnected, perfect, first-countable, bicompact, Hausdorff space K, of cardinal c, such that (the closure of) each subset of cardinal less than c is nowhere dense. The space K will be used in describing the main example. EXAMPLE 1. Let I be the space of all functions f from the natural numbers into [0, 1]. Let I have the order topology relative to the lexicographical order.4 Let K be the set of all f in I such that, if f(i)-=, then eitherf(j)=0, for j>i, orf(j)=1, forj>i. EXAMPLE 2. Let z be the first ordinal that is preceded by c ordinals. Let S consist of four long lines (of "length" c), put together as in Figure 1, with the connecting points a. and a'. Let K' be a copy of K and, for each point or subset A of K, let A' denote the corresponding point or subset of K'. Let k1, k2, . *, k=,... , for x <z, be a most economical well-ordering of K. Let K" be the union of K and K' with ki and k' identified. For y <z, let K, be the closure of { kx I x?< y }. Let Kz=K. Let H be the decomposition space of K" XS, the nondegenerate elements of which are {(k, ay)I kKy } and {(k', a )j k'IKY }, for Y ? z. Figure 2 will aid in visualizing H. It does not indicate the identifications that prevent cutting by the vertices. The intervals in the figure represent copies of S. H is a totally nonaposyndetic, bicompact, Hausdorff space that

A totally nonaposyndetic, compact, Hausdorff space with no cut point