119 A point a ∈ A which is not a limit point of A is called an isolated point of A. Observe that a point a ∈ R is a limit point of a set D if for all δ > 0 the set (B(a,δ ) \{a}) ∩ D is nonempty. ■ Example 5.1.5 (a) Let A =[0,1). Then a = 0 is a limit point of A and b = 1 is also a limit point of A. In fact, any point of the interval [0,1] is a limit point of A. The set [0,1) has no isolated points. (b) Let A = Z. Then A does not have any limit points. Every element of Z is an isolated point of Z. (c) Let A = {1/n : n ∈ N}. Then a = 0 is the only limit point of A. All elements of A are isolated points. (d) Let A = Q. Then every real number is a limit point of A. This follows directly from the density of Q (see Theorem 1.6.3 and also Exercise 2.1.8). ■ Example 5.1.6 If G is an open subset of R then every point of G is a limit point of G. In fact, more is true. If G is open and a ∈ G, then a is a limit point of G\{a}. Indeed, let δ > 0 be such that B(a;δ ) ⊂ G. Then (G \{a}) ∩ B(a;δ )=(a− δ ,a) ∪ (a,a + δ ) and therefore it is nonempty. The next result is useful when studying sequences and limit points. Theorem 5.1.6 Suppose A is an infnite set. Then there exists a one-to-one function f : N → A. Proof: Let A be an infnite set. We defne f as follows. Choose any element a1 ∈ A and set f (1)= a1. Now the set A \{a1} is again infnite because otherwise A = {a1}∪ (A \{a1}) would be the union of two fnite sets and hence fnite. So we may choose a2 ∈ A with a2 ̸ = a1 and we defne f (2)= a2 2. Having defned f (1),..., f (k), we choose ak+1 ∈ A such that ak+1 ∈ A \{a1,...,ak} and defne f (k + 1)= ak+1 (such an ak+1 exists because A \{a1,...,ak} is infnite and, so, nonempty). The function f so defned clearly has the desired properties. □ To paraphrase, the previous theorem says that in every infnite set we can fnd a sequence made up of all distinct points. The following theorem is a variation of the Bolzano-Weierstrass theorem. Theorem 5.1.7 Any infnite bounded subset of R has at least one limit point. Proof: Let A be an infnite subset of R and let {an} be a sequence of A such that am ≠ an for m ≠ n (see Theorem 5.1.6). Since {an} is bounded, by the Bolzano-Weierstrass theorem (Theorem 2.4.1), it has a convergent subsequence {ank }. Set b = limk →∞ ank . Given δ > 0, there exists K ∈ N such that ∈ B(b;δ ) for k ≥ K. Since the set {ank : k ≥ K} is infnite, it follows that at least one ank with ank k ≥ K is different from b. This shows that for every δ > 0, the set (B(b;δ ) \{b}) ∩ A is nonempty and thus b is a limit point of A. □ The defnitions and results given below provide the framework for discussing convergence within subsets of R. 2This fact relies on a basic axiom of set theory called the Axiom of Choice. See [Lay13] for more details.
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