By John B. Conway

This textbook in element set topology is aimed toward an upper-undergraduate viewers. Its light velocity may be valuable to scholars who're nonetheless studying to write down proofs. must haves contain calculus and at the very least one semester of research, the place the coed has been accurately uncovered to the guidelines of simple set concept similar to subsets, unions, intersections, and capabilities, in addition to convergence and different topological notions within the genuine line. Appendices are incorporated to bridge the distance among this new fabric and fabric present in an research path. Metric areas are one of many extra regularly occurring topological areas utilized in different components and are consequently brought within the first bankruptcy and emphasised in the course of the textual content. This additionally conforms to the method of the e-book 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 features, culminating in a improvement of paracompact areas.

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**Sample text**

Define a metric d on X by letting d agree with di on Xi ; and when x ∈ Xi , y ∈ Xj , and i ̸= j, then d(x, y) = 1. ) (a) Show that d is indeed a metric on X. (b) Show that {Xi : i ∈ I} is the collection of components of X and each of these components is an open subset of X. (c) Show that (X, d) is separable if and only if I is a countable set. 6(a) and obtain a theorem whose conclusion is that E = {Ei : i ∈ I} is connected. (8) If E is a subset of Rq , x, y ∈ E, and ϵ > 0 such that there is an ϵ-chain from x to y, show that for any ϵ′ with 0 < ϵ′ < ϵ there is an ϵ′ -chain from x to y.

Gn } is a finite cover of K. Hence K is compact. 26 1. Metric Spaces (d) implies (c). Assume {xn } is a sequence of distinct points in K. By (d), {xn } has a limit point; since K is closed, that limit point must be in K. 7(a), but we must manufacture an actual subsequence of the original sequence. This takes a little bit of care and eﬀort, which we leave to the interested reader. (c) implies (d). If S is an infinite subset, then S has a sequence of distinct points {xn }; by (c), there is a subsequence {xnk } that converges to some point x.

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. Since K is compact, there are points z1 , . . , zn in K such that n K ⊆ k=1 B(zk ; rzk ). Let s = min{szk : 1 ≤ k ≤ n}. Note that B(x; s) ∩ K = ∅; in fact, if there is a y in B(x; s) ∩ K, then there is a k such that y ∈ B(x; s)∩B(zk ; rzk ) ⊆ B(x; szk )∩B(zk ; rzk ), which contradicts the choice of the numbers szk and rzk . Therefore, B(x; s) ⊆ X\K. Since x was arbitrary, this says that X\K is open.