Past exam of the mathematics course of the University of Cambridge / 2024 / ia / Paper 4 / 7E / b

Table of contents
- i b
  - Solution i
- ii b
  - Solution ii
- iii b
  - Solution iii

i

Solution

Yes. Put

α = sup S

and

β = sup T

. Every

s + t \leq α + β

, so this is an upper bound. Given

ε > 0

, choose

s \in S

with

s > α - ε /2

and

t \in T

with

t > β - ε /2

. Then

s + t > α + β - ε

. Hence

sup (S + T) = sup S + sup T .

ii

Solution

No. Take

S = T = {- 2, - 1}

. Then

sup S = sup T = - 1, (sup S) (sup T) = 1,

but

ST = {1, 2, 4}

and therefore

sup (ST) = 4

.

iii

Solution

No, even when every element of $S$ is positive. Take

S = {\frac{1}{2}}, T = {1, 2} .

Then

S^{T} = {1/2, 1/4}

, so

sup (S^{T}) = \frac{1}{2}, (sup S)^{s u p T} = (\frac{1}{2})^{2} = \frac{1}{4} .

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