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.