Kenneth Kuttler

2.3. SEQUENCES AND CAUCHY SEQUENCES 41

Example 2.2.12 Consider A = B(x,δ ) , an open ball in Fp. Then every point of B(x,δ ) isa limit point of A.

If z ∈ B(x,δ ) , consider z+ 1k (x− z)≡ wk for k ∈ N. Then

∥wk−x∥=∥∥∥∥z+

1k(x− z)−x

∥∥∥∥=

∥∥∥∥(1− 1k

)z−(

1− 1k

)x∥∥∥∥= k−1

k∥z−x∥< δ

and also ∥wk− z∥ ≤ 1k ∥x− z∥< δ/k so wk→ z. Furthermore, the wk are distinct. Thus z

is a limit point of A as claimed. This is because every ball containing z contains infinitelymany of the wk and since they are all distinct, they can’t all be equal to z.

In a general metric space, peculiar things can occur. In particular, you can have anonempty open set which has no limit points at all.

Example 2.2.13 Let Ω ̸= /0 and define for x,y ∈ Ω, d (x,y) = 0 if x = y and d (x,y) = 1if x ̸= y. Then you can show that this is a perfectly good metric space on Ω. However,every set is both open and closed. There are also no limit points for any nonempty set sinceB(x,1/2) = {x}. You should consider why every set is both open and closed.

Next is the definition of what is meant by the closure of a set.

Definition 2.2.14 Let A be a nonempty subset of X for X a metric space. Then A isdefined to be the intersection of all closed sets which contain A. This is called the closureof A. Note X is one such closed set which contains A.

Lemma 2.2.15 Let A be a nonempty set in X . Then A is a closed set and

A = A∪A′

where A′ denotes the set of limit points of A.

Proof: First of all, denote by C the set of closed sets which contain A. Then defineA≡ ∩C . This is a closed set from Theorem 2.2.11.

The interesting part is the next claim. First note that from the definition, A ⊆ A so ifx ∈ A, then x ∈ A. Now consider y ∈ A′ but y /∈ A. If y /∈ A, a closed set, then there existsB(y,r)⊆ AC

. Thus y cannot be a limit point of A, a contradiction. Therefore, A∪A′ ⊆ ANext suppose x ∈ A and suppose x /∈ A. Is x ∈ A′? If not, then there is r > 0 such that

B(x,r)∩A = /0. But then B(x,r)C is a closed set containing A so from the definition, it alsocontains A which is contrary to the assertion that x ∈ A. Hence if x /∈ A, then x ∈ A′ and soA∪A′ ⊇ A ■

2.3 Sequences and Cauchy SequencesIt was discussed above what is meant by convergence of a sequence in a metric space. Itwas shown that if a sequence converges, then so does every subsequence. The context inthis section will be a metric space (X ,d).

Of course the converse does not hold. Consider ak = (−1)k it has a subsequence con-verging to 1 but the sequence does not converge. However, if you have a Cauchy sequence,defined next, then convergence of a subsequence does imply convergence of the Cauchysequence. This is a very important observation.

2.3. SEQUENCES AND CAUCHY SEQUENCES 41Example 2.2.12 Consider A = B(x, 5), an open ball in F?. Then every point of B(x, 5) isa limit point of A.If z € B(x, 5), consider z+} (x—z) = wy for k € N. Then1||w, —x|| = ate (x—z)—x1 1 k-1-|(-)-(-D4-Seaeand also ||w, —2z|| < ¢||x—z|| < 6/k so wy + z. Furthermore, the w; are distinct. Thus zis a limit point of A as claimed. This is because every ball containing z contains infinitelymany of the w, and since they are all distinct, they can’t all be equal to z.In a general metric space, peculiar things can occur. In particular, you can have anonempty open set which has no limit points at all.Example 2.2.13 Let Q 4 @ and define for x,y € Q, d(x,y) = 0 ifx=y and d(x,y) = 1if x # y. Then you can show that this is a perfectly good metric space on Q. However,every set is both open and closed. There are also no limit points for any nonempty set sinceB(x, 1/2) = {x}. You should consider why every set is both open and closed.Next is the definition of what is meant by the closure of a set.Definition 2.2.14 Let A be a nonempty subset of X for X a metric space. Then A isdefined to be the intersection of all closed sets which contain A. This is called the closureof A. Note X is one such closed set which contains A.Lemma 2.2.15 Let A be a nonempty set in X. Then A is a closed set andA=AUA'where A’ denotes the set of limit points of A.Proof: First of all, denote by @ the set of closed sets which contain A. Then defineA=N@. This is a closed set from Theorem 2.2.11.The interesting part is the next claim. First note that from the definition, A C A so ifx € A, then x € A. Now consider y € A’ but y ¢ A. If y ¢ A, a closed set, then there existsB(y,r) C AC . Thus y cannot be a limit point of A, a contradiction. Therefore, AUA’ C ANext suppose x € A and suppose x ¢ A. Is x € A’? If not, then there is r > 0 such thatB(x,r) MA =. But then B(x,r)° is a closed set containing A so from the definition, it alsocontains A which is contrary to the assertion that x € A. Hence if x ¢ A, then x € A’ and soAUA'DA2.3. Sequences and Cauchy SequencesIt was discussed above what is meant by convergence of a sequence in a metric space. Itwas shown that if a sequence converges, then so does every subsequence. The context inthis section will be a metric space (X,d).Of course the converse does not hold. Consider a, = (—1 )k it has a subsequence con-verging to | but the sequence does not converge. However, if you have a Cauchy sequence,defined next, then convergence of a subsequence does imply convergence of the Cauchysequence. This is a very important observation.