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Communicated by B. MondIn this note we define a rather pathological connectedness property of Hausdorff spaces which is stronger than ordinary connectedness. We obtain a few basic properties of such spaces and derive a method for constructing them. It turns out that countable Hausdorff spaces having the connectedness property are as easily constructed as uncountable ones. Hence we have still another method for constructing countable connected Hausdorff spaces.DEFINITION. The subset K of the space X saturates X if K is infinite and for each open set U in X, K-U is finite. The space X is said to be saturated if X is a point or some subset of X saturates X. THEOREM 1. If U is an open subset of the saturated space X, then U is connected.PROOF. Suppose K is a subset of X that saturates X and C/is an open subset of X such that U is not connected. Then U is the union of two disjoint closed sets R and S. U-S is open, U-S £ R, and hence K-R is finite. Similarly, K-S is finite. But K = (K-R) u {K-S).COROLLARY. Every saturated space is connected. THEOREM If x and y are two points of the saturated space X and £f is an open cover of X, then there is a simple chain of elements of Sf from x to y with at most three links.PROOF. Suppose x e Ue Sf andy e We £f. Since l / u (fis connected, there is a point z such that zeU nW. Let V be an element of Sf containing z, and form the simple chain from the sets U, V, and W.

A dendritic space is a connected space in which every two points are separated by a third point. In this paper we describe a very natural method for obtaining a dendritic compactification of any connected space for which a dendritic compactification exists. The method is an extension of the familiar process of compactifying E 1 by adjoining-oo and + oo. In what follows, an arc is a Hausdorff continuum with only two noncut points. A ray is an arc minus one of its noncut points. The space X is semi-locally connected at the point p if each open set containing p contains an open set V containing p such that X-V has at most finitely many components. LEMMA. If the space X is arcwise connected but is not semilocally connected at the point p, then there exists an open set U containing p such that if V is an open set containing p and lying in U, then X-V has infinitely many components that intersect both V and X-U.

Communicated by G. SzekeresBy an arc we mean a nondegenerate simply ordered set which is compact and connected in its order topology. A space X is rigid if the only homeomorphism of X onto itself is the identity map. Examples of rigid totally disconnected compact ordered spaces may be found in [1], [2], and [3]. It is the purpose of this note to prove the existence of an arc such that no two of its subarcs are homeomorphic. The proof makes use of arcs of large cardinality and the technique of inverse limit spaces.In what follows, we use the convention that each ordinal is the set of all ordinals preceding it and that the cardinals are certain ones of the ordinals. If M is a set, the cardinal of M is denoted by | M |. co denotes the least infinite cardinal, and Q denotes the least uncountable cardinal. For each regular cardinal X > ft), let L x denote the cross product X x [0,1) ordered lexicographically and let A x = L x u {X} with the order on L x extended to A k in such a way that each point of L x is less than X. L a has sometimes been called the long line. We call any arc order isomorphic to A x a long arc of order X. If a is an infinite regular cardinal and X is a set, then by an a-sequence in X we mean a map of a into X. If X is a topological space and x is a point of X, then the weight of X at x, denoted by w(x), is the least cardinal a such that X has a base at x of cardinal a. Note that if X is an arc, x is an interior point of X, and there exist an increasing asequence converging to x and a decreasing j3-sequence converging to x, then w(x) = sup {a,/?}. A long arc X also has the important property that for each point x of X different from the right end point, there exists a decreasing cosequence converging to x. We shall need the following lemma, which is probably well known.