A new characterization of trees

It is proved that a continuum is a tree if and only if for each pair of nondegenerate subcontinua K and L with K C L, it follows that K contains a cutpoint of L. A connected Hausdorff space X is said to be tree-like [1] if each pair of distinct points can be separated by a third point, i.e., if x, y E X and x $ y then there exists z E X so that X {z} = A U B where A and B are disjoint open sets, x E A and y e B. A tree is a compact tree-like space. It is well known [5] that a continuum is a dendrite if and only if it is a metrizable tree. It is a theorem of R. L. Moore [3] that a metrizable continuum X is a dendrite if and only if each nondegenerate subcontinuum of X contains uncountably many cutpoints of X. In the setting of Hausdorff continua this theorem does not generalize to yield a characterization of trees. (For a counterexample, see [4, Example 9].) Moore's theorem is now sixty years old, and trees have been intensively studied for over forty years and are quite well understood. Therefore, it is surprising that a suitable variation of Moore's theorem, permitting a characterization of trees, has so far eluded discovery. In this note we obtain such a variation. A point z of a connected space X is a cutpoint of X provided X {z} is not connected. A continuum is a compact connected Hausdorff space. A continuum X is hereditarily unicoherent if, for each pair of subcontinua A and B it follows that A n B is connected. An ordered continuum is a nondegenerate continuum with exactly two non-cutpoints. Finally, a continuum X is said to satisfy property T provided, for each pair of nondegenerate subcontinua K and L with K C L, it follows that K contains a cutpoint of L. THEOREM 1. A continuum is a tree if and only if it satisfies the property T. PROOF. Suppose X is a tree and that K and L are nondegenerate subcontinua with K c L. Let x and y be elements of K. Since L is a tree there exists z E L such that L {z} = L1 U L2, a separation, with x E L1 and y E L2. But K is a subcontinuum of L, so z E K. Hence K contains a cutpoint of L. Now assume X is a continuum satisfying property T. The proof that X is a tree will be dividied into 4 parts. 1. X is hereditarily unicoherent. For suppose there are subcontinua A and B of X such that A n B = P U Q where P and Q are nonempty separated sets. Then Received by the editors June 12, 1987 and, in revised form, October 9, 1987. 1980 Mathematics Subject Classification (1985 Revzsion). Primary 54F20, 54F50, 54F65; Secondary 54F55.