119 A point a∈Awhich is not a limit point of Ais called anisolated point of A. Observe that a point a∈Ris a limit point of a set Dif for all δ >0 the set (B(a,δ)\ {a})∩Dis nonempty. ■ Example 5.1.5 (a) Let A= [0,1). Then a=0 is a limit point of Aand 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 Adoes not have any limit points. Every element of Zis 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 Aare 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 Gis an open subset of Rthen every point of Gis a limit point of G. In fact, more is true. If Gis open anda∈G, thenais 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 Ais an infinite set. Then there exists a one-to-one function f : N→A. Proof: Let Abe an infinite set. We define f as follows. Choose any element a1 ∈Aand set f (1) =a1. Now the set A\ {a1}is again infinite because otherwise A={a1}∪(A\ {a1}) would be the union of two finite sets and hence finite. So we may choose a2 ∈Awith a2̸ =a1 and we define f (2) =a2 2. Having defined f (1), . . . , f (k), we choose ak+1 ∈A such that ak+1 ∈A\ {a1, . . . ,ak} and define f (k+1) =ak+1 (such an ak+1 exists because A\ {a1, . . . ,ak} is infinite and, so, nonempty). The function f so defined clearly has the desired properties. □ To paraphrase, the previous theorem says that in every infinite set we can find a sequence made up of all distinct points. The following theorem is a variation of the Bolzano-Weierstrass theorem. Theorem 5.1.7 Any infinite bounded subset of Rhas at least one limit point. Proof: Let Abe an infinite subset of Rand let {an}be a sequence of Asuch that am̸=an for m̸=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∈Nsuch that ank ∈B(b;δ) for k ≥K. Since the set {ank : k ≥K} is infinite, it follows that at least one ank with k ≥Kis different fromb. This shows that for every δ >0, the set (B(b;δ)\ {b})∩Ais nonempty and thus bis a limit point of A. □ The definitions 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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