Set Addition with the Reals, The Suprema Lemma

We'll need some properties of Q,N to be able to determine this suprema. So let's get to those:

c+A

Let A⊆R be nonempty and bounded above, and let c∈R. Define the set c+A by:

c+A={c+a:a∈A}

Then sup(c+A)=c+sup(A).

Proof
For (i), set s=sup(A). We see that a≤s for all a∈A consequently, but then a+c≤c+s=c+sup(A) as required.

For (ii), let b be an arbitrary upper bound for c+A, so then c+a≤b for all a∈A. Then a≤b−c for all a∈A. So then b−c is an upper bound for A. Since s is the least upper bound for A, then we have it that b−c≤s, thus b≤c+s=c+sup(A), completing (ii).

Therefore, sup(c+A)=c+sup(A).
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