MTH 533,
Homework §9.1
1. Find the limit of each of the following vector sequences.
1 2k 2 − k + 1
,
a) xk =
k k 2 + 2k − 1
1
b) xk = 1, sin πk, cos
k
√
1/k 1
2
c) xk = k − k + k, k ,
k
*2. Prove Theorem 9.6.
Proof. We want to show that a sequence { xk } in Rn is Cauchy if and only if it converges.
(=⇒) If a sequence { xk } in Rn is Cauchy, then given ε = 1, there exists N such that k ≥ N =⇒
kxk − xN k ≤ 1. Choose r := max { kx1 − xN k , · · · , kxN −1 − xN k , 1 } , then { xk } ⊂ Br (xN), i.e.,
{ xk } is bounded. By the Bolzano-Weierstrass theorem, { xk } has a convergent subsequence xkj ,
say xkj → a as j → ∞. Let ε > o. Since { xk } is Cauchy, there exists N1 such that n, m ≥ N1 =⇒
kxn − xm k ≤ ε. Since xkj → a as j → ∞, there exists N2 such that kj ≥ N2 =⇒ xkj − a ≤ ε.
Choose N > max { N1 , N2 }. Then for n ≥ N , kxn − ak ≤ kxn − xN k + kxN − ak < 2ε, which
implies that { xk } converges to a.
(⇐=) We proved this in class.
*3. A subset E of Rn is said to be sequentially compact if and only if every sequence xk ∈ E has a
convergent subsequence whose limit belongs to E. Prove that every closed ball in Rn is sequentially
compact.
Proof. Since a closed ball B is bounded,
by the Bolzano-Weierstrass theorem, any sequence { xk } ⊂
B has a convergent subsequence xkj , say xkj → a as j → ∞. Since a is a limit point of B and
B is closed, a ∈ B.
*4.
a) Let E be a subset of Rn . A point a ∈ Rn is called a cluster point of E if E ∩ Br (a) contains
infinitely many points for every r > 0. Prove that a is a cluster point of E if and only if for each
r > 0, E ∩ Br (a)\ { a } is nonempty.
Proof. (=⇒) Let r > 0. Since E ∩Br (a) contains infinitely many points, E ∩Br (a)\ { a } contains
infinitely many points.
(⇐=) Suppose that a is not a cluster point of E. Then there exists r > 0 so that E ∩ Br (a)\ { a }
contains only finitely many points, say x1 , · · · , xk . Choose r0 = min { kx1 − ak , · · · , kxk − ak },
then E ∩ Br0 (a)\ { a } = ∅, a contradiction.
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b) Prove that every bounded infinite subset of Rn has at least one cluster point.
Proof. Let Ebe a bounded infinite subset of Rn . Choose { xk } ⊂ E such that the cardinality
of { xk } is infinite.
By the Bolzano-Weierstrass theorem, the sequence { xk } has a convergent
subsequence xkj , say xkj → a as j → ∞. Then for each r > 0, E ∩ Br (a)\ { a } is nonempty,
which implies that a is a cluster points of E.
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