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 | + | '''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⊆R, if m∈A and m≥a for all a∈A. | '''Problem 4.''' A real number m is called maximum of the set A⊆R, if m∈A and m≥a for all a∈A. |
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.
- Show: If m is the maximum of A, then m is also the supremum of A.
- 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.