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My question is for each of the following, to prove or disprove (by giving a counter example):

  • Every cover of a closed interval by closed intervals has a finite subcover.
  • Every cover of an open interval by closed intervals has a finite subcover.
  • Every cover of an open interval by open intervals has a finite subcover.

Now we actually haven't talked about covers and subcovers in class, buy from reading it would seem it's just as a set of sets whose union include all elements of the set being covered, and it's subsets.

Doesn't this mean that I can disprove all of them with pretty much the same example?

for The first I can choose $[0,1]$ and the cover of $[0,0]$ and $[\frac{1}{n}, 1]$ for $n\in \mathbb{N}$

for The second I can choose $(0,1)$ and the cover of $[\frac{1}{n}, 1-\frac{1}{n}]$ for $n\in \mathbb{N}$

for The third I can choose $(0,1)$ and the cover of $(\frac{1}{n}, 1)$ for $n\in \mathbb{N}$

Is this true or am I understanding it wrong?

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    $\begingroup$ Your counterexample for the first case doesn't quite work as written since $[0,0] = \{0\}$ is not really an interval. It might be worth asking your prof. how interval is defined in your class to check whether that counterexample is valid. $\endgroup$ – Mnifldz Apr 8 '15 at 20:05
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for 1st case consider the counterexample... consider closed interval $[-1,1]$...consider cover as $[-1,0]$ and $[1/n,1]$ for $n\in \mathbb{N}$...

rest of two are ok.

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  • $\begingroup$ I see, thanks!! $\endgroup$ – guest Apr 8 '15 at 21:13

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