CRN 11378: HW 1
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| − | '''Problem 1.''' Let $A$ and $B$ be two non-empty sets that are bounded from above. Show: If $\sup A < \sup B$, | + | '''Problem 1.''' Let $A$ and $B$ be two non-empty sets that are bounded from above. Show: If $\sup A < \sup B$, 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 $s=t$. ( | + | '''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 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 $A\subseteq\mathbb{R}$, if $m\in A$ and $m\geq a$ for all $a\in A$. | '''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 $a\in A$. | ||
Latest revision as of 13:01, 9 October 2019
Problem 1. Let $A$ and $B$ be two non-empty sets that are bounded from above. Show: If $\sup A < \sup B$, 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.
