Department of Mathematics, University of Houston
Math 3333 - David Blecher
Homework 2 Solutions.
12.3 (g). The set S={n
n+1 :n∈N}has no maximum, and sup(S) = 1. To prove this, first notice that
1 is an upper bound for S(since n<n+ 1, so dividing by n+ 1 gives n
n+1 <1, for all n∈N).
To show that 1 is the least upper bound for Swe argue by contradiction. So suppose that
M < 1, and that Mis an upper bound for S. Thus n
n+1 ≤M, ∀n∈N. However,
n
n+ 1 ≤M⇔n≤M(n+ 1) ⇔n≤Mn +M⇔n(1 −M)≤M⇔n≤M
1−M.
Note that 1 −M > 0 since M < 1. Since the last few lines hold ∀n∈N, we have shown that
M
1−Mis an upper bound for N, which contradicts the Archimidean property. This contradiction
shows that 1 is the least upper bound for S.
To see that Shas no maximum, we can use a principle which says that if a set has a
maximum, then that maximum is also the supremum. In our case the supremum is 1, which
is not in S, so it cannot be the maximum.
12.6 If Sis bounded, then it is bounded above, and hence by the completeness axiom it has a least
upper bound or supremum. To show that this supremum is unique, suppose that M1and M2
are two suprema of S. Then either M1< M2,M2< M1, or M1=M2. If M1< M2then M2
is not a least upper bound, because M1is an upper bound which is smaller. This contradicts
the fact that M2is a least upper bound. So we cannot have M1< M2. A similar argument
shows that we cannot have M2< M1. Thus M1=M2. (4 points)
12.7 (a) First suppose that k= 0. In this case, k S ={kx :x∈S}={0}which has supremum 0.
On the other hand, 0 sup(S) = 0. So sup(kS) = ksup(S) if k= 0. [2 points]
Next suppose that k > 0. If x∈Sthen x≤sup(S). Thus kx ≤ksup(S) [1 point]. It follows
that ksup(S) is an upper bound for kS ={kx :x∈S}. Thus sup(kS)≤ksup(S) [2 points].
On the other hand, if x∈S, then kx ≤sup(kS). Dividing by kwe have x≤1
ksup(kS) [1
point]. Thus sup(S)≤1
ksup(kS) [1 point]. Multiplying by kwe have ksup(S)≤sup(kS) [1
point]. So now we have sup(kS) = ksup(S) [1 point].
There were bonus points for this Homework. I will try put up the point distribution later.
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