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

To start with the last one: yes the choice is very important.

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...

  • $\begingroup$ 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 $\endgroup$ – 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 $S.$


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