# Thread: Prove that an infimum equals a supremum in different sets

1. ## Prove that an infimum equals a supremum in different sets

Suppose that A is contained in the set of all real numbers and is bounded above. Let U be the set of upper bounds, i.e., U={x is an element of the real numbers: x>=a for all a elements of A}. Prove that U is bounded below and that inf U=sup A. (Note: U is not the empty set by the assumption that A is bounded above.)

Thanks for the help!

2. ## Re: Prove that an infimum equals a supremum in different sets

Originally Posted by lovesmath
Suppose that A is contained in the set of all real numbers and is bounded above. Let U be the set of upper bounds, i.e., U={x is an element of the real numbers: x>=a for all a elements of A}. Prove that U is bounded below and that inf U=sup A. (Note: U is not the empty set by the assumption that A is bounded above.)
First we need to say that $\displaystyle A$ is not empty, therefore $\displaystyle U$ is bounded below.
Let $\displaystyle \alpha=\sup(A)$. You need to show that $\displaystyle \alpha$ is a lower bound for $\displaystyle U$.
Then show that $\displaystyle \alpha$ is the greatest lower bound of $\displaystyle U$.

### how to prove supremum and infimum

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