By John B. Conway

ISBN-10: 3319023675

ISBN-13: 9783319023670

ISBN-10: 3319023683

ISBN-13: 9783319023687

This textbook in element set topology is geared toward an upper-undergraduate viewers. Its mild velocity could be worthwhile to scholars who're nonetheless studying to jot down proofs. necessities comprise calculus and not less than one semester of research, the place the coed has been adequately uncovered to the guidelines of simple set conception similar to subsets, unions, intersections, and capabilities, in addition to convergence and different topological notions within the genuine line. Appendices are integrated to bridge the distance among this new fabric and fabric present in an research direction. Metric areas are one of many extra common topological areas utilized in different parts and are accordingly brought within the first bankruptcy and emphasised during the textual content. This additionally conforms to the technique of the booklet firstly the actual and paintings towards the extra common. bankruptcy 2 defines and develops summary topological areas, with metric areas because the resource of notion, and with a spotlight on Hausdorff areas. the ultimate bankruptcy concentrates on non-stop real-valued services, culminating in a improvement of paracompact spaces.

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**Example text**

We can easily see that every ﬁnite subset of X is compact, but ﬁnding nontrivial examples of compact sets requires us to ﬁrst prove some results. 2. Let (X, d) be a metric space. (a) If K is a compact subset of X, then K is closed and bounded. (b) If K is compact and F is a closed set contained in K, then F is compact. (c) The continuous image of a compact subset is a compact subset. Proof. (a) If x ∈ / K, then for each z in K let rz , sz > 0 such that B(z; rz ) ∩ B(x; sz ) = ∅. Now {B(z; rz ) : z ∈ K} is an open cover of K.

If K satisﬁes (c) and G is an open cover of K, then there is an r > 0 such that for each x in K there is a G in G such that B(x; r) ⊆ G. Let G be an open cover of K, and suppose the claim is false; thus, for every n ≥ 1 there is an xn in K such that B(xn ; n−1 ) is not contained in any set G in G. By (c), there is an x in K and a subsequence {xnk } such that xnk → x. Since G is a cover, there is a G in G such that x ∈ G; choose a positive such that B(x; ) ⊆ G. Let nk > 2 −1 such that xnk ∈ B(x; /2).

B) Let A be a nonempty relatively open and closed subset of E. Since A = ∅, there is some integer N with A ∩ EN = ∅. But A ∩ EN is both relatively open and closed in EN , so EN ⊆ A by the connectedness of EN . By hypothesis, EN −1 ∩ EN = ∅, so EN −1 ∩ A = ∅, and it follows that EN −1 ⊆ A. Continuing, we get that En ⊆ A for 1 ≤ n ≤ N . Since EN ∩ EN +1 = ∅, similar arguments show that EN +1 ⊆ A. Continuing, we get that En ⊆ A for all n ≥ 1. That is, E = A, and so E is connected. 7. The union of two intersecting connected subsets of a metric space is connected.

### A Course in Point Set Topology by John B. Conway

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