How Do Transformations Affect the Supremum of a Set?

[Calu]

Homework Statement

8. Let ##A## be a non-empty subset of ##R## which is bounded above. Define
##B = \{x ∈ R : x − 1 ∈ A\}##, ##C = \{x ∈ R : (x + 1)/2 ∈ A\}.##
Prove that sup B = 1 + sup A, sup C = 2 sup A − 1.
The attempt at a solution

Note that ##sup A## exists. Let ##x ∈ B##; then ##x − 1 ∈ A##, and ##x − 1 ≤ sup A##. We have that ##x ≤ sup A + 1##, from which we deduce that ##B## is bounded above and ##sup B ≤ sup A + 1##. Now suppose that ##m## is an upper bound for ##B##. For ##x ∈ A##, ##x + 1 ∈ B## and ##x + 1 ≤ m##. It follows that ##x ≤ m − 1##, from which we deduce that ##m − 1## is an upper bound for ##A## and ##sup A ≤ m − 1##. Now ##sup A + 1 ≤ m##. Combining the above, ##sup B = sup A + 1##.

I understand this solution up until we reach ##sup B = sup A + 1##. I understand that they both must be less than or equal to ##m##, as ##m## is an upper bound for ##B## and an upper bound for ##A + 1##. However I don't understand how we know that they are exactly equal to each other. Could anyone help me out please?

 

[verty]

##m## is arbitrary, it can be any upper bound, even the least upper bound.

 

[Fredrik]

8. Let ##A## be a non-empty subset of ##R## which is bounded above. Define
##B = {x ∈ R : x − 1 ∈ A}##, ##C = {x ∈ R : (x + 1)/2 ∈ A}.##
Prove that sup B = 1 + sup A, sup C = 2 sup A − 1.

LaTeX tip: \{x\in\mathbb R:x-1\in A\}

 

[micromass]

LaTeX tip: \{x\in\mathbb R:x-1\in A\}

I fixed the LaTeX in the OP

 

[HallsofIvy]

First, since A is bounded above, so is B and so B has a supremum. Let [itex]\alpha[/itex] be the supremum of A.

1) Suppose [itex]x> \alpha+ 1[/itex]. Then [itex]x- 1>\alpha[/itex]. Since [itex]\alpha[/itex] is an upper bound on A x- 1 is not in A. Therefore, x= (x- 1)+ 1 is not in B. That is [itex]\alpha+ 1[/itex] is an upper bound on B.

2) Suppose there exist y, an upper bound on B, with [itex]y< \alpha+ 1[/itex]. Then if [itex]y< x< \alpha+ 1[/itex] x is not in B. But then [itex]x- 1< \alpha[/itex] is not in A which contradicts the fact that [itex]\alpha[/itex] is the least upper bound on A.