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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.

Trux
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    Every open set contains the emptyset, which is closed. :) – Cornman Dec 30 '17 at 04:32
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    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. – Ross Millikan Dec 30 '17 at 05:05
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    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 – David Reed Dec 30 '17 at 05:12

3 Answers3

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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.

Martín Vacas Vignolo
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Every open set contains the empty set, which is closed.

HeMan
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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.

Gaston Burrull
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