# Proof of Supremum of a Set

1. May 26, 2014

### Calu

The problem statement, all variables and given/known data

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?

Last edited by a moderator: May 26, 2014
2. May 26, 2014

### verty

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

Last edited: May 26, 2014
3. May 26, 2014

### Fredrik

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

4. May 26, 2014

### micromass

Staff Emeritus
I fixed the LaTeX in the OP

5. May 26, 2014

### HallsofIvy

Staff Emeritus
First, since A is bounded above, so is B and so B has a supremum. Let $\alpha$ be the supremum of A.

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

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