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This might be extremely obvious but a proof such as this one show a "cover" of [0,2] to be:

the collection of open sets: $\{ (\frac{1}{n},2) \mid n \text{ is a positive integer} \}$

This is nice and gives us an open cover but when does it include the point 1, or 2? Does it not need to include these points to cover [0,2]? Right now I am only seeing it cover (0,2) or am I seeing something wrong?

Basically I am trying to prove that the interval

[a,b] $= \{x \mid a < x < b\}$ for $a,b,x$ all rationals

is not compact. So I would like to use the set:

$\{(a, b-1/n) \mid n \text{ is positive}\}$

To create my open cover with no finite subset but it just doesn't include the endpoints.. unless it does which this proof seems to be indicating. Could you let me know

  • If you can cover closed intervals like this
  • Any suggestions on my intended problem would be appreciated.
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  • $\begingroup$ That set wouldn't cover [0,2]. The points 1 and 2 aren't in the alleged cover. Any closed interval in the reals is compact, so your attempt to prove it isn't will fail. $\endgroup$ – zibadawa timmy Nov 7 '13 at 6:15
  • $\begingroup$ Okay, would it be fine for me to keep my idea but just union the set with something like {x | 3a/4 < x < 5a/4} and {y | 3b/4 < y < 5y/4} to get the end points in there? $\endgroup$ – Alex Nov 7 '13 at 6:20
  • $\begingroup$ The set of rationals in any non-degenerate interval is not compact. $\endgroup$ – dfeuer Nov 7 '13 at 6:47
  • $\begingroup$ @dfeuer: :-) You’ve had some sleep now. $\endgroup$ – Brian M. Scott Nov 7 '13 at 11:02
  • $\begingroup$ A union (even an infinite union) of open sets is still an open set. Thus, no union of open intervals contained in a closed interval can ever cover the closed interval. $\endgroup$ – robjohn Jan 16 '15 at 10:45
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No, the set described does not cover $M=[0,2]\cap\Bbb Q$. There are two ways to look at compactness of a subset $S$ of a topological space, which turn out to be equivalent (as you should prove to yourself): one is compactness of $S$ when considered under the subspace topology, and the other is the statement that any set of open sets in the space whose union includes $S$ as a subset (also called an open cover of $S$) has a finite subset whose union also includes $S$ as a subset. Thus you don't need to think about open sets that lie in $S$ but rather more broadly.

Consider the following open cover of $M$: $$\mathcal A=\{(-1,\sqrt 2)\}\cup\{(\sqrt 2+1/x,3)\mid x\in \Bbb R_+\}.$$

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take an increasing sequence of irrational numbers $y_{k}$ such that $0< y_{k}< y<2$, for all $k$, where $y$ is an irrational number and the cover will be $(y_{k},y_{k+1})\cup[0,y_{1})\cup(y,2]$.

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