CRN 11378: HW 1
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− | '''Problem 1.''' If supA<supB, show that B contains an element that is an upper bound of A. | + | '''Problem 1.''' Let A and B be two non-empty sets that are bounded from above. Show: If supA<supB, show that 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. (Suprema are unique.) | '''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. (Suprema are unique.) |
Revision as of 00:07, 3 September 2019
Problem 1. Let A and B be two non-empty sets that are bounded from above. Show: If supA<supB, show that 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. (Suprema are unique.)
Problem 3. Suppose 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.
Problem 4. A real number m is called maximum of the set A⊆R, if m∈A and m≥a for all a∈A.
- Show: If m is the maximum of A, then m is also the supremum of A.
- Let A={x∈Q | x2≤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.