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CRN 11378: HW 1

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'''Problem 2.''' Let A be a non-empty set of real numbers that is bounded from above. Show: If s and t both are suprema of A, then s=t. (This shows that a set can have at most one supremum.)
 
'''Problem 2.''' Let A be a non-empty set of real numbers that is bounded from above. Show: If s and t both are suprema of A, then s=t. (This shows that a set can have at most one supremum.)
  
'''Problem 3.''' Suppose that every non-empty set that is bounded form above has a supremum. Show that then every non-empty set that is bounded form below has an infimum.
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'''Problem 3.''' Suppose that every non-empty set that is bounded from above has a supremum. Show that then every non-empty set that is bounded form below has an infimum.
  
 
'''Problem 4.''' A real number m is called maximum of the set AR, if mA and ma for all aA.
 
'''Problem 4.''' A real number m is called maximum of the set AR, if mA and ma for all aA.

Latest revision as of 14:01, 9 October 2019

Problem 1. Let A and B be two non-empty sets that are bounded from above. Show: If sup, then B contains an element that is an upper bound of A.

Problem 2. Let A be a non-empty set of real numbers that is bounded from above. Show: If s and t both are suprema of A, then s=t. (This shows that a set can have at most one supremum.)

Problem 3. Suppose that every non-empty set that is bounded from above has a supremum. Show that then every non-empty set that is bounded form below has an infimum.

Problem 4. A real number m is called maximum of the set A\subseteq\mathbb{R}, if m\in A and m\geq a for all a\in A.

  1. Show: If m is the maximum of A, then m is also the supremum of A.
  2. Let A=\{x\in\mathbb{Q}\ |\ x^2\leq 5\}. Show that A is bounded from above, but that A has no maximum.

Problem 5. Show that the Nested Interval Property together with the Archimedean Principle implies the Axiom of Completeness.

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