# Showing the $[1, 2) \cup (3,4]$ is Not Compact Using Definition of Open Cover

So, when I see the set $[1, 2) \cup (3,4]$, I know that it is not compact by the Heine-Borel (Since it is not closed). However, I am struggling to come up with open covers for such a set. How do you think about finding an open cover?

The answer given is $(0, 4-\frac{1}{n})$, for $n \in \Bbb N$ which makes sense to me, however, I am not understanding how this is chosen.

For example, would $(1+\frac{1}{n}, 5)$ for $n \in \Bbb N$ also be an acceptable open cover?

Similarly, for trying to show that $\Bbb N$ is not compact, the answer given for an open cover is $(-n, n)$, but I feel than $(0,n)$ is also an open cover that would work.

Is the choice of open cover particularly important?

Thanks, AP

• $\{(1,4-\frac{1}{n})\}_n$ is not an open cover of $[1,2)\cup(3,4]$, as it doesn't contain $1$ or $4$. – carmichael561 Jul 16 '17 at 22:43
• @carimichael edited to reflect that answer given is $(0, 4-\frac{1}{n})$, but that still would not include $4$ – user345 Jul 16 '17 at 22:49
• I suggest something along the lines of $\{(1-\frac{1}{n},2-\frac{1}{n})\cup(3+\frac{1}{n},4+\frac{1}{n})\}_{n\geq 2}$. – carmichael561 Jul 16 '17 at 22:50

It should consists of open sets, and their union should include the set $X= [0,1) \cup (3,4]$, and no finitely many of them should cover this set. The first is why the given answer and your modification don't work: $1$ is not in the union of the second (and neither is $4$ for the first one), so they're not even covers of $X$. The sets $(0,n)$ also work for $\mathbb{N}$ assuming $0 \notin \mathbb{N}$) otherwise $0$ is not covered, and you need to add (-1,2)$as well, e.g. A cover that does work for$X$:$U_n = (0,2-\frac{1}{n+1}) \cup (3+\frac{1}{n+1}, 5)$. Any$x \in [1,2)$is covered eventually by some$U_n$and the same can be said for any$x \in (3,4]$. But take finitely many and consider the largest indexed$U_n$among them... • Thanks for pointing out that the given answer was not a correct cover. I was struggling to understand what a cover was, but this seems to make more sense now. I was also assuming that$0 \notin \mathbb{N}$for the second example – user345 Jul 16 '17 at 23:21 Let$S_n=(-\infty,2-2^{-n})\cup (2+2^{-n},\infty)$for$n\in \mathbb N.$Then$\{S_n: n\in \mathbb N\}$is an open cover of$[1,2)\cup (3,4]$with no finite subcover. Observe that for any$S\subset \mathbb R$where$S\ne \overline S,$we may take$p\in \overline S$\$S$and let $$C=\{\mathbb R \;\backslash \;[p-r,p+r]: r>0\}.$$ Then$C$is an open cover of$S$because$\cup C=\mathbb R$\$\{p\}\supset S.$Now if$D$is a finite subset of$C$then$\cup D\subset \mathbb R$\$[p-r,p+r]$for some$r>0.$But$p\in \overline S$so $$\phi \ne S\cap (p-r,p+r)\subset S \cap [p-r,p+r].$$ So$D$is not a cover of$S.\$