Exercise 14, § 2
General Topology, IV
Bourbaki marks this one as harder than the rest.
Let be a connected Hausdorff topological space not consisting of a single point, and let denote the diagonal of .
a) Show that there is no partition of into two open sets such that and . [Begin by remarking that and are connected for each (use Exercise 4 a) of Chapter I, § 11); deduce that, if , we have , and show that this implies that, for each , one or the other of the sets , is empty; hence show that one or the other of the sets is empty, which is contrary to hypothesis.] Deduce that either is connected or else has exactly two components such that .
b) A linear ordering on is said to be compatible with the topology of if is coarser than . Deduce from a) that there exists such a linear ordering on if and only if is not connected, and that there are then exactly two such linear orderings. [With the notation of a), show that the order relation must be either or .] Show that the intervals, for these orderings, are connected in the topology , and that they are the only connected subsets of with respect to .
c) Show that if is either locally connected or locally compact in the topology , and if there exists a linear ordering on compatible with , then we must have . [If is locally connected, use b). If is locally compact, show that every compact neighbourhood of a point is also a neighbourhood of with respect to , by considering the frontier of the neighbourhood].
d) Let be the subspace of consisting of and all pairs such that and . Show that there exists a linear ordering on which is compatible with the topology induced on by the topology of , but that this topology is strictly finer than . \ Give an example of a connected Hausdorff space , no point of which has a fundamental system of connected neighbourhoods, and on which there is a linear ordering compatible with the topology of [cf. Chapter I, § 11, Exercise 2 b)].
Solution
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