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		<title>CRN 10459: HW 3 - Revision history</title>
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		<updated>2026-04-21T04:03:59Z</updated>
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		<id>http://helmut.knaust.info/mediawiki/index.php?title=CRN_10459:_HW_3&amp;diff=5307&amp;oldid=prev</id>
		<title>HelmutKnaust: Created page with &quot;'''Problem 11.''' Suppose $(a_n)$ is a Cauchy sequence, and that $(b_n)$ is a sequence satisfying $\lim_{n\to\infty} |a_n-b_n|=0$. Show that $(b_n)$ is a Cauchy sequence.  '''...&quot;</title>
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				<updated>2025-09-30T15:37:11Z</updated>
		
		<summary type="html">&lt;p&gt;Created page with &amp;quot;&amp;#039;&amp;#039;&amp;#039;Problem 11.&amp;#039;&amp;#039;&amp;#039; Suppose $(a_n)$ is a Cauchy sequence, and that $(b_n)$ is a sequence satisfying $\lim_{n\to\infty} |a_n-b_n|=0$. Show that $(b_n)$ is a Cauchy sequence.  &amp;#039;&amp;#039;&amp;#039;...&amp;quot;&lt;/p&gt;
&lt;p&gt;&lt;b&gt;New page&lt;/b&gt;&lt;/p&gt;&lt;div&gt;'''Problem 11.''' Suppose $(a_n)$ is a Cauchy sequence, and that $(b_n)$ is a sequence satisfying $\lim_{n\to\infty} |a_n-b_n|=0$. Show that $(b_n)$ is a Cauchy sequence.&lt;br /&gt;
&lt;br /&gt;
'''Problem 12.''' Suppose $(a_n)$ is a '''bounded''' sequence such that all of its '''converging''' subsequences converge to the same limit, say $L$. Show that $(a_n)$ converges to $L$ as well.&lt;br /&gt;
&lt;br /&gt;
'''Problem 13.''' Consider the following two properties:&lt;br /&gt;
# Every non-empty set that is bounded from above has a supremum.&lt;br /&gt;
# Every Cauchy sequence converges. &lt;br /&gt;
Show that (2)$\Rightarrow$(1). ((1)$\Rightarrow$(2) was done in class, via the Bolzano-Weierstrass Theorem.)&lt;br /&gt;
&lt;br /&gt;
'''Problem 14.''' &lt;br /&gt;
# Show: If $x$ is an accumulation point of $A\cup B$, then $x$ is an accumulation point of $A$, or $x$ is an accumulation point of $B$ (or both). &lt;br /&gt;
# Does the result also hold for a countably infinite collection of sets? Give a proof, or provide a counterexample.&lt;br /&gt;
&lt;br /&gt;
'''Problem 15.''' Find all accumulation points of the set &lt;br /&gt;
\[\left\{\frac{1}{m}+\frac{1}{n}\ |\ m,n\in\mathbb{N}\right\}\]&lt;br /&gt;
Remember that $A=B\  \Leftrightarrow\  (A\subseteq B)\wedge (B\subseteq A)$.&lt;/div&gt;</summary>
		<author><name>HelmutKnaust</name></author>	</entry>

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