0
$\begingroup$

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

$\endgroup$
  • 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
2
$\begingroup$

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...

$\endgroup$
  • $\begingroup$ 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
0
$\begingroup$

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 $S.$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.