Thread: proof dealing with supremum and bounds

1. proof dealing with supremum and bounds

I'm having trouble understanding how to prove this problem. I need some help.

Let x Real number and Let S be a non empty subset of real numbers that is bounded above. We define a new set x + S, by x+S={x+s: s S}

a) Prove that x +S is bounded above
b) Prove that x + sup(S) is an upper bound of x + S. Conclude that the sup(x+ S) ≤ x + sup(S)
C) Prove that x + sup(S) = sup(S) = sup(x+S)

Thanks for any help provided.

2. Originally Posted by tuyt6444
I'm having trouble understanding how to prove this problem. I need some help.

Let x Real number and Let S be a non empty subset of real numbers that is bounded above. We define a new set x + S, by x+S={x+s: s S}

a) Prove that x +S is bounded above
Suppose p is an upper bound for S. Then $\displaystyle y\le p$ for all y in S. What can you say about y+ S?

b) Prove that x + sup(S) is an upper bound of x + S. Conclude that the sup(x+ S) ≤ x + sup(S)
C) Prove that x + sup(S) = sup(S) = sup(x+S)
This last statement is clearly untrue. Do you really have that "sup(S)" in the middle?

Thanks for any help provided.

3. Sorry I made a mistake on that last one the correct problem is

(c) prove that x+sup(S) = sup(x+S)