Let A be a nonempty set of real numbers which is bounded below. Let -A be the set of all numbers -x, where x ∈ A . Prove that inf A = − sup ( − A )

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Let A be a nonempty set of real numbers which is bounded below. Let -A be the set of all numbers -x, where   x  ∈  A . Prove that  inf  A  =  −  sup  (  −  A  )

soyafh · Accepted Answer

Step 1  inf  A  =  −  sup  (  −  A  )Rudin&#039;s book gives definitions of the concepts involved, and I would stick close to what those definitions say.A  is bounded below, i.e. it has a lower bound x. That means   ∀  a  ∈  A  ,     x  ≤  a . Consequently   ∀  a  ∈  A  ,     −  x  ≥  −  a .  ∀  b  ∈  −  A     ∃  a  ∈  A     b  =  −  a , hence   ∀  b  ∈  −  A  ,     −  x  ≥  b . Thus −x is an upper bound of −A.Thus we have proved that for every lower bound x of A, -x is an upper bound of -A. In particular -inf A is an upper bound of -A. In order to show that -inf A is the smallest upper bound of -A, one must show that no number less than -inf A is an upper bound of -A. Suppose   c  &amp;lt;  −  inf  A . Then   −  c  &amp;gt;  inf  A . Since -c is greater than the largest lower bound of A, -c is not a lower bound of A. Hence for some   a  ∈  A ,   a  &amp;lt;  −  c , and so   −  a  &amp;gt;  c . Since   −  a  ∈  −  A, we have a member of -A that is greater than c, so c is not an upper bound of -A.

Dana Chung · Accepted Answer

Step 1Let A be a non-empty set of real numbers that is bounded below. By definition of bounded below, we may choose   α  ∈      R   such that   α  ≤  x for every   x  ∈  A . This implies that   −  x  ≤  −  α for every   x  ∈  A so that   −  α is an upper bound for -A. Thus, -A is a non-empty set of real numbers that is bounded above and, therefore, has a supremum, say   β , by the axiom of completeness.We must show that   −  β is the infimum of A. First, note   β is an upper bound for -A (by definition of supremum) or   β  ≥  −  x for every   x  ∈  A . Thus,   −  β  ≤  x for every   x  ∈  A and   −  β is a lower bound for A. Next, we must show that   β is the greatest lower bound of A. Thus, assume that   β  &amp;lt;  γ. Then,   −  γ  &amp;lt;  −  β so (since   β is the supremum of -A), there is some   x  ∈  A with   −  γ  &amp;lt;  x  &amp;lt;  −  β . Therefore,   β  &amp;lt;  −  x  &amp;lt;  γ  with   x  ∈  A so that   γ cannot be a lower bound of -A.To understand why these particular details are written out in grotesque detail, I would consider the material that you likely just learned. If you are trying to show that an infimum can be defined in terms of a supremum, then you have likely just learned these concepts, as well as concepts like upper and lower bounds. So I think you&#039;ve really got to refer quite explicitly to those definitions. By contrast, I used the order properties, like   x  &amp;lt;  y    ⟹    −  y  &amp;lt;  −  x without specific reference since that&#039;s probably at least a little bit in the past.

Let A be a nonempty set of real numbers which is bounded below. Let -A be the set of all numbers -x, where x in A. Prove that inf A=- sup (-A)

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