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(i) S is a subset of R and β ∈ R. Let T = {s + β : s ∈ S }. Show that if Sup S exists, then so does sup T. Moreover, Sup T = β + Sup S.

(ii) Using what you have proved in (i), show that if x ∈ R, x ≠ 0 and S = {gx : g ∈ Z }, then Sup S does not exist.

Here is my proof for (i)

Let the upper bound of S be called m. Then m > s for all s ∈ S. By property of real numbers, we can say that m + β > s+ β for some β ∈ R. Thus, we see that every s+ β ∈ T is an upper bound of T. T is bounded upwards. Our hypothesis implies that S has a least upper bound. Let Sup S = s' => if x < s', then x is not an upper bound of S. By property of real numbers, if x + β < s' + β, then x + β is not an upper bound of T. T also has a least upper bound, which is Sup T = s' + β => Sup T = β + Sup S as desired.

I got to (ii) and somehow am not seeing the light on where to start. Any pointers?

Juniper Leaf
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1 Answers1

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Hint: $S$ is translation of itself by $x$: $T=\{s+x:s\in S\}=S$. Therefore, if the supremum of $S$ exists, $\sup S=\sup S+x$ and $x=0$.

Arash
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