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 $A\subseteq\mathbb{R}$, 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 $A$. | # 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. | # 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''. | '''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 $A\subseteq\mathbb{R}$, 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.
