supremum Proof by contradiction .

Question

(The second part of the solution is confusing me)

The Question:

A is a bounded subset of ℝ

B = {b|b = 2a + 3, a ∈ A}

Prove that Sup(B) = 2Sup(A) + 3

Part 1 of solution:

We get 2a + 3 ≤ 2Sup(A) + 3

So B is bounded and Sup(B) ≤ 2Sup(A) + 3

Part 2 of solution

We then want to show Sup(B) ≥ 2Sup(A) + 3

As then we have equality:

The answer uses proof by contradiction and states that:

Assuming that Sup(B) ≥ 2Sup(A) + 3 is not true, then we have Sup(B) < 2Sup(A) + 3

Therefore: we have Sup(A)₂ = [Sup(B)- 3]/2 which is an upper bound of A less than Sup(A)…a contradiction.

My confusion

What I don’t don’t understand is: Yes Sup(A)₂ is less than Sup(A) but how do we know Sup(A)₂ is an upper bound of A?

Thanks!

Answer 1

Try the following result: Let A and C sets, we define $A+C=\{a+c: a\in A, c\in C\}$, then $\sup(A+C)=\sup(A)+\sup(C)$. Now let $C=\{3\}$ and $B=A+C$