CRN 11982: HW 1
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'''Problem 3.''' 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.''' 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 4.''' A real number $m | + | '''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. | # 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. | # 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''. | '''Problem 5.''' Show that the ''Nested Interval Property'' together with the ''Archimedean Principle'' implies the ''Axiom of Completeness''. |
Latest revision as of 09:34, 27 August 2014
Problem 1. Exercise 1.3.2.
Problem 2. Exercise 1.3.3(a)(b).
Problem 3. 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 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.