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HelmutKnaust at 04:21, 15 October 2013

2013-10-15T04:21:11Z

HelmutKnaust at 04:11, 15 October 2013

2013-10-15T04:11:48Z

HelmutKnaust: Created page with "'''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 th..."

2013-09-02T22:41:03Z

Created page with "'''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 th..."

New page

'''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.

'''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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	Hint: This is not as easy as it may look at first glance.		Hint: This is not as easy as it may look at first glance.
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		+	<script type="text/javascript" src="http://www.wolfram.com/cdf-player/plugin/v2.1/cdfplugin.js"></script>
		+	<script type="text/javascript">
		+	var cdf = new cdfplugin();
		+	cdf.embed('http://helmut.knaust.info/NB/Bungee.cdf', 777, 943);
		+	</script>

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 '''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.
-−
-<script type="text/javascript" src="http://www.wolfram.com/cdf-player/plugin/v2.1/cdfplugin.js"></script>
-<script type="text/javascript">
-var cdf = new cdfplugin();
-cdf.embed('http://helmut.knaust.info/NB/Bungee.cdf', 777, 943);
-</script>