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Latest revision as of 22:21, 14 October 2013

Problem 1. We say that $m$ is the maximum of a set $A$ if $m\in A$ and $m\geq a$ for all $a\in A$.

Suppose a set $A$ of real numbers has a maximum, call it $m$. Show that $m$ is also the supremum of $A$.

Problem 2. Let $A=\{x\in\mathbb{Q}\ |\ x^2\leq 3\}$. Show that A is bounded from above, but that $A$ has no maximum.

@@ Line 4: / Line 4: @@
 '''Problem 2.''' Let $A=\{x\in\mathbb{Q}\ |\ x^2\leq 3\}$. Show that A is bounded from above, but that $A$ has no maximum.
-'''Problem 3.''' Let $A$ and $B$ be two sets of real numbers.
-#Assume that $\emptyset\neq A\subseteq B$ and that $B$ is bounded from above. Show that $A$ is bounded from above, and moreover that $\sup A\leq \sup B$.
-#State a similar theorem for sets that are bounded from below. (No proof required.)
-'''Problem 4.'''
-A set is called bounded if it is bounded from above and bounded from  below. Assume that every non-empty bounded set of real numbers has both an infimum and a supremum. Show that this implies the Completeness Axiom for the Real Numbers.
-Hint: This is not as easy as it may look at first glance.
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