CRN 11982: HW 6

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(Created page with "'''Problem 26.''' Exercise 3.3.3 '''Problem 27.''' Exercise 3.3.7 (b,c,e) '''Problem 28.''' Exercise 3.3.9 (b,f) '''Problem 29.''' Exercise 4.2.9 '''Problem 30.''' A set $...")
 
 
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'''Problem 26.''' Exercise 3.3.3
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'''Problem 26.''' Exercise 3.3.1
  
'''Problem 27.''' Exercise 3.3.7 (b,c,e)
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'''Problem 27.''' Exercise 3.3.9 (b,e)
  
'''Problem 28.''' Exercise 3.3.9 (b,f)
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'''Problem 28.''' Exercise 3.3.10
  
'''Problem 29.''' Exercise 4.2.9
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'''Problem 29.''' A set $X$ is called ''LP-compact'', if every infinite subset of $X$ has a limit point belonging to $X$.
 
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'''Problem 30.''' A set $X$ is called ''LP-compact'', if every infinite subset $A$ of $X$ has a limit point belonging to $X$.
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Show that a set is LP-compact if and only if it is closed and bounded.
 
Show that a set is LP-compact if and only if it is closed and bounded.
  
Hint: Show first that every bounded infinite set has a limit point.
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* Hint: Recall that every bounded infinite set has a limit point.
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'''Problem 30.''' Show that every non-empty open set is the countable union of pairwise disjoint open intervals: Let $O$ be a non-empty open set of real numbers. Then there are open intervals $I_n$, $n\in \mathbb{N}$, such that
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* $\displaystyle O=\bigcup_{n\in\mathbb{N}} I_n$, and
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* for all $m,n\in\mathbb{N}$:  $I_m\cap I_n=\emptyset$ or $I_m=I_n$.

Latest revision as of 20:18, 28 October 2014

Problem 26. Exercise 3.3.1

Problem 27. Exercise 3.3.9 (b,e)

Problem 28. Exercise 3.3.10

Problem 29. A set $X$ is called LP-compact, if every infinite subset of $X$ has a limit point belonging to $X$.

Show that a set is LP-compact if and only if it is closed and bounded.

  • Hint: Recall that every bounded infinite set has a limit point.

Problem 30. Show that every non-empty open set is the countable union of pairwise disjoint open intervals: Let $O$ be a non-empty open set of real numbers. Then there are open intervals $I_n$, $n\in \mathbb{N}$, such that

  • $\displaystyle O=\bigcup_{n\in\mathbb{N}} I_n$, and
  • for all $m,n\in\mathbb{N}$: $I_m\cap I_n=\emptyset$ or $I_m=I_n$.
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