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

[ak123456]

Homework Statement

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

Homework Equations

The Attempt at a Solution

let a$$\in$$A b$$\in$$B s$$\in$$R because B=A+1=$$\left\{$$a+1:a$$\in$$A$$\right\}$$
so b=a+1 $$\forall$$b$$\in$$B 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