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.
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 $x \in \left[1,3-\frac{1}{N}\right)$.