By Milgram R. (ed.)

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Uik , then this will not cover (0, 1). For, the union 48 4. Topological Properties of such a ﬁnite collection will again be Ui , where i is the maximum of i1 , . . , ik . And Ui omits some points from (0, 1), such as 1/i. Hence no ﬁnite reﬁnement of this cover is itself a cover of (0, 1) and so (0, 1) is not compact. 23 can be very easily generalized to prove the following result. 27 If f : S → T is a continuous map and S is compact, then the image of f is compact. That is to say, the image of a compact space under a continuous map is compact.

Prove that if T is a Hausdorﬀ space and x1 , . . , xn is a ﬁnite list of distinct points in T , then there are open sets U1 , . . , Un each containing one, and only one, of the points x1 , . . , xn . 9. If the set of integers were given the indiscrete topology, would it be connected? Compact? Hausdorﬀ? 10. 37, do actually form a topology. 5 Deconstructionist Topology In this chapter we consider ways of relating a new space to other spaces which are more familiar. 1. When this is not the case, it is often possible to express the new space in terms of other, perhaps more familiar, spaces.

This line does not pass through 0), then we can proceed as usual. If, however, the line from x to y passes through 0, then we need to replace one of the end points. To do this, take any point in Rn which is not on the straight line through x and y. This point will either lie in U or in V . If it is in U , then we replace x by this new point and, if it is in V , then we replace y by it. Either way we end up with two points, one in U and one in V , such that the straight line between the two points does not pass through 0, so is contained in Rn − {0}.