Discontinuous Closed Darboux Functions

Harvey Rosen

Abstract:

A function is Darboux (closed) if f(C) is connected (closed) for each connected (closed) subset C of X. Closed Darboux functions that are discontinuous are constructed. One such function is from the unit interval [0,1] onto a continuum Y.

Although a closed Darboux function does not have to be continuous, it does if X and Y are Euclidean
spaces. This is due to H. Pawlak's result in [1] that a closed function is continuous if and only if the image of each segment is connected. It follows that each closed Darboux function is continuous, where I denotes the topological space [0,1] with the usual topology . In [2], H. Pawlak and R. J. Pawlak give three ways to construct discontinuous closed Darboux functions . We answer some of their open problems here.

A topological space (X,T) is paracompact if each open cover of X has a locally finite open refinement. A function is nowhere constant at x if f(U) is nondegenerate for each open neighborhood U of x. Let H denote the Hashimoto-type topology on [0,1] generated by the base and A is countable . H is finer than , but ([0,1], H) is not paracompact. According to Theorem 1 of [2], every closed Darboux (and therefore continuous) function considered as a function is closed Darboux and discontinuous at each point where it is nowhere constant. It is asked if versions of this result still hold when H is replaced by any topology T finer than or ``close to compact'' like paracompact.

Proof: Let be a sequence in -converging to a point p such that -converges to f(p) and for . Let be a sequence of disjoint closed intervals not containing f(p) such that has center and radius . denotes the collection of all sequences of closed intervals such that has center and radius with . Define T to be the topology with base consisting of along with all sets of the form , where and .

Since is closed and T is finer than , then is closed and ( [0,1],T) is Hausdorff. Let J be any subinterval of [0,1]. Then f(J) is a point or an interval because is Darboux . Suppose f(J) is an interval. If , then f(J) is T-connected. If then is either (1) an interval or (2) the union of disjoint intervals and . For case (1), is T-connected because relative T-open sets in are relative -open sets in , too. Since f(p) is in the T-closure of is T-connected. For case (2), f(p) is in the T-closure of the T-connected sets and , and so is T-connected. This shows is a Darboux function. By construction, it is discontinuous at p.

We show ([0,1], T) is paracompact. Let be a T-open cover of [0,1]. We may assume . If , then has a finite subcover of [0,1]. So suppose for some V in , where and . Pick so has radius with , and choose . Whenever for , we let int for . But whenever for where and , we let int for . Then for each is a -open cover of having a finite subcover because is -compact, and W misses each member of . Let and meets . has a finite subcover of [0,1] - U, which is -compact. Therefore is a locally finite T-open refinement of . Finally, any paracompact Hausdorff space, like ([0,1],T), is normal.

If and the space Y is no longer an interval, then we can choose Y to be compact as the next example shows. We cannot choose it to be both compact and Hausdorff. For, if is closed Darboux and discontinuous and Y is compact and Hausdorff, then Urysohn's lemma would ensure there exists a continuous function such that is discontinuous besides being closed and Darboux.
According to [1], this is impossible.

Proof: Let Q denote the set of rational numbers in [0,1] with the relative topology from , and for let be the one-point compactification of Q [3]. Define by

Then f is discontinuous because when is open in Y but is not -open in [0,1].

Suppose . Then and Suppose F is a -closed subset of [0,1]. Then F is -compact. Suppose . Then f(F) = F and F is a -compact and -closed subset of Q. Consequently Q - f(F) is open in Y, and so f(F) is closed in Y. Now suppose . Then , which is closed in Y because is a -open subset of Q. This shows f is a closed function.

Let K be a connected subset of [0,1]. We show f(K) is connected. We may suppose K is an interval instead of a point. Since . Assume , a separation. A and B are disjoint sets open in and suppose . There is an open set U in Y such that . A is a subset of Y - U, which is a -closed and -compact subset of Q because Therefore Y - U and hence A is nowhere dense in Q. But for some open subset V of Y. Since implies is a -open subset of Q. Therefore is somewhere dense in Q, a contradiction. This shows f(K) is a connected set and f is a Darboux function.

A connected topological space is said to have an exploding point a with respect to a point if is a component of and there exist disjoint open sets U and V with and . Theorem 2 of [2] states that if has an exploding point a with respect to and is a dense compact connected subspace of , then there exists a closed Darboux function which is discontinuous at . It is asked whether the compactness of the subspace X of the explosion set can be weakened or when can be a connected Alexandroff compactification of a connected locally compact space X. Figure 1 illustrates that both situations, minus local compactness, can occur as in the following theorem. The picture shows a fan consisting of line segments emanating from the same endpoint a and limiting on a line segment L whose other endpoint is . denotes the half of L which contains a. Then let and , the one-point compactification of X.

Proof: Let U be an open neighborhood of in such that cl , and let F = (cl cl , which is closed in X. X is dense in because X is not compact. Therefore is connected and . Since X is completely regular, there exists a continuous function such that and . Define

by Then is continuous, but f is discontinuous at because .

We claim f is a closed function. Suppose K is closed in . Since K-U is closed in is compact. If , then K is a compact subset of X, and so is compact and therefore closed. But if , which is compact and therefore closed.

For the sake of completeness, we show here that f is a Darboux function in the same fashion as in [2]. Suppose C is connected. If , then and so is connected. If and , then because a is an exploding point of with respect to . Therefore there exists bd , and so f(p) = 1. We show f(C) = [0,1] to see it is connected. Assume there exists such that . Let , and . Then and is a separation, contrary to C being connected.

In Theorem 3 of [2], Pawlak and Pawlak extend a homeomorphism to a closed Darboux discontinuous function. They show that for a nondegenerate locally connected metrizable continuum X and , there exist a locally connected continuum and a locally connected, connected paracompact space each having X as a subspace such that every homeomorphism can be extended to a closed Darboux function discontinuous at . They ask how close to compact can be chosen. We show can actually be compact.

Proof: Let and let be a sequence of distinct elements of X different from a and converging to a. Define , and define a topology on generated by the neighborhood system for , where denotes the -neighborhood of x in X. Define , and define a topology on generated by the following neighborhood system:

By construction and are locally connected, connected, and compact.

Define a function by

Choose a sequence in X such that in X. Then in and in because is an open neighborhood of containing no . Therefore is discontinuous at .

Let C be a connected subset of . Then whenever . Therefore , where denotes the projection defined by if and if . Since is continuous, is connected. Then is connected because each is a connected subset of . This shows f is Darboux .

That is closed follows from the facts that is one-to-one and .