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I want to use the definition of a compact set to determine whether the following set $\Omega$ is compact (Clearly, $\Omega$ is not compact due to the Heine-Borel theorem).

Let $\Omega = [1, 3)$ and let $\mathscr{F} = \{A_n \, : \, n \in \mathbb{N} \}$ where $A_n = \left(0, 3-\frac{1}{n} \right)$. Then the family $\mathscr{F}$ is an open cover for $\Omega$, because

$$\Omega \, \subseteq \, \bigcup_{n \in \mathbb{N}} \, \mathscr{F} \, = \, \left(0,3\right)$$

However, if we let $\mathscr{G} = \{A_{n_1}, \dotsc , A_{n_k}\}$ be any finite subfamily of $\mathscr{F}$, and if $m = \mathrm{max}\{n_1, \dotsc, n_k\}$, then

$$\bigcup \, \mathscr{G} \, = \, A_{n_1} \cup \cdots \cup A_{n_k} \, = \, \left(0, 3-\frac{1}{m}\right)$$

Since we can always find a rational number $q$ where $3-\frac{1}{m} < q < 3$, it follows that although $q \in \Omega$, we have $q \notin \bigcup \mathscr{G}$. Therefore, $\Omega \not \subseteq \bigcup \mathscr{G}$, so $\Omega$ is not compact.

If the above scratchings are correct, I was wondering if anyone could show me how to prove that if $x \in \Omega$, then $x \in \bigcup \, \mathscr{F}$.

Thank you.

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    $\begingroup$ If $x \in [1,3)$ then $x<3$ and hence there is some $n$ such that $x < 3-{1 \over n}$ from which we get $x \in A_n$. $\endgroup$
    – copper.hat
    Nov 12, 2015 at 21:35
  • $\begingroup$ Didn't you already concede tha t$\bigcup\mathscr F=(0,3)$? $\endgroup$ Nov 12, 2015 at 21:36
  • $\begingroup$ Your first displayed line asserts just that and more. It won't build character to prove that $[1,3)\subseteq (0,3)$ ;/ $\endgroup$
    – BrianO
    Nov 12, 2015 at 21:43

2 Answers 2

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The above is correct.

To see that it is indeed a cover, let $x \in [1,3)$. As $x < 3$, we know that $3 - x > 0$ and so for large enough $N$ we know that $\frac{1}{N} < 3 -x$, or $x < 3 - \frac{1}{N}$, so that $x \in \left[1,3-\frac{1}{N}\right)$.

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  • $\begingroup$ Oops: your very last phrase should be "so that $x\in[1,3-\frac 1 N)$". $\endgroup$
    – BrianO
    Nov 12, 2015 at 21:46
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If $x\in\Omega$, then $x<3$. If $x<2$, we are done. Otherwise, let $n>\frac{1}{3-x}$. Then $x<3-\frac{1}{n}$.

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