The Real And Complex Number Systems

Remark: The exercise is useful in the theory of Topological Entorpy.
Infinite Series And Infinite Products
Sequences
8.1 (a) Given a real-valed sequence a n bounded above, let u n supa k : k ≥ n.
Then u n ↘ and hence U lim n→ u n is either finite or −. Prove that
U lim n→
sup a n lim n→
supa k : k ≥ n.
Proof: It is clear that u n ↘ and hence U lim n→ u n is either finite or −.
If U −, then given any M 0, there exists a positive integer N such that as n ≥ N,
we have
u n ≤−M
which implies that, as n ≥ N, a n ≤−M. So, lim n→ a n −. Thatis,a n is not bounded
below. In addition, if a n has a finite limit supreior, say a. Thengiven 0, and given
m 0, there exists an integer n m such that
a n a −
which contradicts to lim n→ a n −. From above results, we obtain
U lim n→
sup a n
in the case of U −.
If U is finite, then given 0, there exists a positive integer N such that as n ≥ N, we
have
U ≤ u n U .
So, as n ≥ N, u n U which implies that, as n ≥ N, a n U . In addition, given
′ 0, and m 0, there exists an integer n m,
U − ′ a n
by U ≤ u n supa k : k ≥ n if n ≥ N. From above results, we obtain
U lim n→
sup a n
in the case of U is finite.
(b)Similarly, if a n is bounded below, prove that
V lim n→
inf a n lim n→
infa k : k ≥ n.
Proof: Since the proof is similar to (a), we omit it.
If U and V are finite, show that:
(c) There exists a subsequence of a n which converges to U and a subsequence which
converges to V.
Proof: SinceU lim sup n→ a n by (a), then
(i) Given 0, there exists a positive integer N such that as n ≥ N, wehave
a n U .
(ii) Given 0, and m 0, there exists an integer Pm m,
U − a Pm .
Hence, a Pm is a convergent subsequence of a n with limit U.
Similarly for the case of V.