By Steven G. Krantz

Do not get me incorrect - Krantz is nice yet this is often primarily child Rudin - with out the proofs - that's kind of like a bar with no beer.

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**Sample text**

LiS)} U (4/5. 615} is an open covering of the interval J = [0. 1]. However, not all the elements V are actually needed to cover J. In fact (-115. 7. 7 It is the special property displayed in this example that distinguishes compact sets from the point of view of topology. 6 If C is an open covering of a set S and if V is another open covering of S such that each element of V is also an element ofC. then we call Va subcovering ofC. We call V afinite subcovering if V has just finitely many elements.

U (0). Then the points 1,1/2,1/3, ... are isolated points of T The point 0 is nol isolated. Every element of T is a 0 boundary point, and there are no others. 1 Observe that the interior points of a set S are elements of S- by their very definition. Also isolated points of S are elements of S. However. a boundary point of S mayor may not be an element of S. 2 Accumulation Points Let S be a subset of IR. In particular, x is an accumulation point of S if it is the limit of an eventually nonconstant sequence in S.

1 Let S c JR be a set. Then S is closed ifand only if each Cauchy sequence has a limit that is also an element of S. 3 The set E = [-2. 3) c JR is of course closed. If {a j I is any Cauchy sequence in E, then the sequence will have a limit in E - since the endpoints are included in the set, there is no possibility for the sequence to converge to an exterior point. 0 Note that it follows from the completeness of the real numbers that any Cauchy sequence whatever will have a limit in JR. The main point of this proposition is that.