# Every open interval has an open subinterval

I want to prove the following propsotion:

Let $I\subset \mathbb{R}$ be an open interval. If $b\in I$, there exist $a$ and $c$ in I with $a < b < c$.

I think that we should give a proof by contradiction:

My first attempt is:

The proposition can be written in the language of symbolic logic,

$\forall~ b\in I~~~~ \exists~ a,c\in I~~~~a<b<c$

The opposite of this statement is:

$\exists~ b\in I~~~~ \forall~ a,c\in I~~~~(a\geq b)\vee (b\geq c)$

How can i find a contradiction from here?

My second attempt is:

Let $I=(m,n)$. The following propositions are true:

$\forall~ b\in I~~~~ \exists~ \epsilon>0~~~~b>n-\epsilon$

and

$\forall~ b\in I~~~~ \exists~ \delta>0~~~~b<m+\delta$

Therefore, if we choose $a=n-\epsilon$ and $c=m+\delta$, we find $a<b<c$. But i can't show $n-\epsilon\in I$ and $m+\delta\in I$.

• Perhaps you also wish to show that $a,c\in I?$ – Alekos Robotis Oct 17 '17 at 14:23
• This follows trivially from the fact that an open interval has neither a least member nor a greatest member. In other words the thing you want to prove is a definition of an open interval. – Paramanand Singh Oct 17 '17 at 16:26
• @ParamanandSingh I try to write this fact formally. – user315531 Oct 17 '17 at 16:29
• You did not get my point. Definitions are not proved. They are taken as basis to prove various properties about the things which have been defined. – Paramanand Singh Oct 17 '17 at 16:31
• In that case it is even more important that you define the term "open interval" in some other manner without using least/greatest member idea. And if you have one such definition do provide it. And then one has to begin with that definition and prove the proposition you state. The accepted answer here use the definition based on least member and greatest member and thus does not constitute a proof. – Paramanand Singh Oct 18 '17 at 5:21

Let's assume $\exists~ b\in I~~\forall~ a,c\in I:~~ (a\geq b)\lor (c\leq b)$. Then, self-evidentally, $$\exists~ b\in I~~\forall~ a,c\in I\setminus \left\{ {b}\right\}:~~ (a> b)\lor (c<b).$$But because $a,c \in I\setminus \left\{ {b}\right\}$ is symmetric, it also follows that $$\exists~ b\in I~~\forall~ a,c\in I\setminus \left\{ {b}\right\}:~~ (a<b)\lor (c>b).$$ In order for both statements to hold, it must be $$\exists~ b\in I~~\forall~ a,c\in I\setminus \left\{ {b}\right\}:~~ (a,c>b)\lor (a,c<b).$$ Now if there were for all $b\in I$ two pairs $(a',c'), (a'',c'')\in I^2$ with $a',c'>b$ but $a'',c''<b$, then $a=a'',c=a'$ would contradict our assumption. Therefore, $$\exists~ b\in I:~~(\forall~ a,c\in I\setminus \left\{ {b}\right\}:~~ a,c>b)~\lor~(\forall~a,c\in I\setminus \left\{ {b}\right\}:~~ a,c<b).$$ But this would mean that $b$ is the minimum or the maximum of $I$ contradicting the fact that an open interval has neither a minimum nor a maximum. Therefore, our assumption from the beginning must be false and the claim must hold.
However, if you want a shorter proof, you could just take $a=b-\frac{b-m}{2}, c=b+\frac{n-b}{2}$ where $I=(m,n)$ and then prove quite easily that $m<a<b<c<n$.
• Well, in a way, my second proof is your second approach with $\epsilon =n-\frac{b}{2} - \frac{m}{2}$ and $\delta = \frac{b}{2} + \frac{n}{2} - m$ so by choosing $\epsilon , \delta$ carefully, your approach can definitely work. However, I am not sure if you can also finish up your proof like this without choosing a specific value for $\epsilon ,\delta$. – mxian Oct 17 '17 at 17:30