Exercise 19, § 11
General Topology, I, 158
It is set in small type as supplementary.
Let X be a locally compact space. It can be shown that X can be identified with a dense open subspace of the universal compact space constructed in § 9, Exercise 25 (cf. Chapter IX, § 1, Exercise 8). Suppose also that X is connected and locally connected.
a) Show that, for every compact subset K of X, the set of components of which are not relatively compact is finite [cf. Exercise 14 b)]. Deduce that a point of belongs to the closure of at most one of the components of [if () are the non-compact components of , and the compact set of points of which lie in the closure of , construct a compact space whose underlying set is the sum of X and the sets , such that X is identified with a dense open subspace of this space].
b) Let be the following equivalence relation on : "for every compact subset K of X, x and y lie in the closure of the same non-compact component of ". Let R be the equivalence relation on whose equivalence classes are the classes mod and the subsets consisting of a single point of X. Show that R is Hausdorff and that X may be identified with a dense open subspace of the compact space ; show that in the compact space every point is the intersection of its neighbourhoods which are both open and closed. The points of are called the ends of the space X.
c) Let be a compact space such that may be identified with a dense open subspace of and such that in the compact space every point is the intersection of its neighbourhoods which are both open and closed (*). Show that there is a continuous mapping of into which induces the identity mapping on . [If is the union of two disjoint non-empty closed sets and , show that there exist two disjoint open subsets in such that and such that is compact].
d) Show that if is -compact, then has a countable base.
e) The real line has two ends, and () has one. Give an example of a connected, locally connected, locally compact subspace of for which the set of ends has the power of the continuum.*
Solution
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