3
$\begingroup$

I have to

Show that every open set in $\mathbb{R}$ contains a closed set

My proof is:

Case 1) Take an open interval $(a,b).$ Either there exists an $x \in (a,b)$ such that $a < x < b$ or not. So, if $a < x < b$, $\{x\}$ is a closed set in $(a,b).$ If there doesn't exist such an $x,$ then, the interior of $(a,b)$ is the empty set, which is a closed set (it is also an open set, but that is not relevant).

Case 2) If the open interval is infinite, I could also pick a singleton set contained in it.

I particularily think that it could be incorrect to say that "If there doesn't exist such an $x,$ then, the interior of $(a,b)$ is the empty set," but I am not sure.

And also, judging by the complexity of another answer to the same question, Does every open interval of $\mathbb{R}$ contain a closed interval?, I beleive that my proof is missing some reasonings and understanding about what I need to prove, and that I am wrong about the whole thing.

$\endgroup$
  • 9
    $\begingroup$ Every open set contains the emptyset, which is closed. :) $\endgroup$ – Cornman Dec 30 '17 at 4:32
  • 1
    $\begingroup$ Your case 1 argument assumes without saying it that every open set includes an interval. This is true for all open sets except the empty set. You should be asking that $(a,b)$ is inside the open set. You don't need case 2 because even if the open set includes an infinite interval it includes a finite interval and you can go back to case 1. $\endgroup$ – Ross Millikan Dec 30 '17 at 5:05
  • 1
    $\begingroup$ Going off cornman's comment your proof would at the very least have to include a case for the empyset, which is both closed and open (called clopen sometimes). Observing that the empty set contains itself settles that part. As cornman suggested, this could be a one sentence proof by using the empty set as the closed set for all of them $\endgroup$ – David Reed Dec 30 '17 at 5:12
4
$\begingroup$

Let $U\subset \mathbb{R}$ an open set. If $U=\emptyset$, is closed and we finish. If not, let $x\in U$ and, because is open, exist $\epsilon>0$ such that $I=(x-\epsilon,x+\epsilon)\subset U$. Now, consider the closed interval $[x-\frac{\epsilon}{2},x+\frac{\epsilon}{2}]\subset I \subset U$ and we finish.

$\endgroup$
4
$\begingroup$

Every open set contains the empty set, which is closed.

$\endgroup$
2
$\begingroup$

Every set in $\mathbb{R}$ contains a closed set.

Every non-empty set in $\mathbb{R}$ contains a non-empty closed set, just take the singleton of some point it contains.

$\endgroup$

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.