Let (X, T, n) be a topological transformation group, where X is a Hausdorff continuum. We will say that X is irreducibly T-invariant if no proper subcontinuum of X is ^-invariant. Wallace, [6], has shown that if T is abelian and X is irreducibly ^-invariant, then X has no cut point; he then asked if this statement remains true if "abelian" is replaced by "compact". In this paper we answer this question in the affirmative, and prove a related result when T satisfies a recursive property.A general problem due to Wallace is as follows: Assume T leaves an endpoint of X fixed. Under what conditions on X and T does T have another fixed point? This problem has been investigated by Wallace [8], Wang, [5], Chu, [1], and Gray, [3,4]. In Theorem 2, we show that if X is locally connected, and T is generated by a compact subgroup and a connected subgroup, then T has another fixed point.Departing from [7] slightly, we call a subcontinuum C of X a universal subcontinuum (USC) if given a subcontinuum D of X, D n C is a continuum. The intersection of arbitrarily many USC is again a USC. If X-x = U u V, where U and V are non-empty separated subsets of X (hereafter referred to as a "separation of X-x"), then U u {x} is a USC. The property of being a USC is topological. The proofs of these statements are to be found in [7].The terminology pertaining to transformation groups is taken from [2].
THEOREM 1. Let (X, T, n) be a topological transformation group where X is a Hausdorff continuum and one of the following conditions is satisfied:(i) T is compact, (ii) X is locally connected and T is pointwise regularly almost periodic.
If X is irreducibly T-invariant, then X contains no cut point.PROOF. Assume that X is irreducibly T-invariant. We make the following observations:1. If x e X, no proper USC of X contains the orbit Tx of x.For otherwise the intersection, D, of all USC which contains Tx is a proper subcontinuum of X containing Tx. We easily verify that that D is T-invariant, and hence X is not irreducibly T-invariant. 310