What is the Proof of sup(A)+1 as the Least Upper Bound for B=A+1?

[ak123456]

Homework Statement

Let A \inR be a non-empty,bounded set .Define
B=A+1=\left\{a+1:a\inA\right\} Prove that sup(B) =sup(A)+1

Homework Equations

The Attempt at a Solution

let a\inA b\inB s\inR because B=A+1=\left\{a+1:a\inA\right\}
so b=a+1 \forallb\inB s>= b \rightarrow s>=a+1 , s>=a so sup(B) =sup(A)+1
i thought that is too simple in my way , any other ways?

 

[jgens]

From the definition of A and B, if \mathrm{sup}(A) = \alpha then we clearly have that \alpha + 1 \geq a + 1 = b. This proves that \mathrm{sup}(A) + 1 is an upperbound for B. Now can you prove that this number must be the least upper bound?

 

[ak123456]

From the definition of A and B, if \mathrm{sup}(A) = \alpha then we clearly have that \alpha + 1 \geq a + 1 = b. This proves that \mathrm{sup}(A) + 1 is an upperbound for B. Now can you prove that this number must be the least upper bound?

i see ,thx