Question

Let A be non empty sebset of real number which is bounded above let -A be the set of all real number -x where x belong A show that sup(A)= -inf(-A)

Answer 1

Answer on Question #56674 – Math – Real Analysis

Let $A$ be non empty subset of real numbers which is bounded above. Let $-A$ be the set of all real numbers $-x$, where $-x$ belong $A$. Show that $\sup(A) = -\inf(-A)$.

Solution

A is a non empty subset of real number which is bounded above, so $\sup (A)$ exists. By definition, for all $x\in A$, we have that $\sup (A)\geq x$, so $-\sup (A)\leq -x$ for all $x\in A$.

- A is non empty subset of real numbers which is bounded below, so $\inf (-A)$ exists. So we have that $\inf (-A)\leq y$ for all $y\in -A$, then $-\inf (-A)\geq y$ for all $y\in -A$, so $-\inf (-A)\geq x$ for all $x\in A$. $-\inf (-A)$ is a upper bound of $A$.

By definition, $x$ is a supremum if $x$ is an upper bound of $A$ and for any other upper bound $y$ of $A$ $x \leq y$. From these definitions and from what was proven above it is clear that $-\inf(A) \geq \sup(-A)$ and $-\sup(A) \leq \inf(-A)$. Combining these two inequalities gives $\sup(A) = -\inf(-A)$, as required.

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