Prove that if $I \subset \mathbb{R}$ is an interval that is open and closed, $I = \emptyset$ or $I = \mathbb{R}$

I began my proof by considering if $I$ was empty; if it is, we are done. However, if $I$ is non-empty, I know I should use the assumption that firstly, a supremum of I exists which would eventually result in a contradiction (because $I$ is also open?) And then do the same with the infimum of I, thus proving that $I = \mathbb{R}$ as $\mathbb{R}$ is the only interval that is unbounded both above and below. I'm not sure how to go about writing the actual proof though. Any help would be greatly appreciated.

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How do you define interval? – Andrés Caicedo Feb 24 '13 at 19:13
@Andres: Has to be order-convex set. – Brian M. Scott Feb 24 '13 at 19:13
More generally, the only sets that are open and closed in $\mathbb{R}$ are $\emptyset$ and $\mathbb{R}$. This is connectedness of $\mathbb{R}$. – 1015 Feb 24 '13 at 19:14
@BrianM.Scott Is that how Mathlete defines it? – Andrés Caicedo Feb 24 '13 at 19:14
@julien I'm aware of that but I'm supposed to prove it using the idea of a supremum and infimum. – Mathlete Feb 24 '13 at 19:15

Your thinking is correct, all that is needed is the proper phrasing of the proof.

Suppose $I \subset \mathbb{R}$ is an interval both open and closed. If it is non-empty, suppose it is bounded above. Then take $t$ to be the supremum of $I$. Since $t$ is a LEAST upper bound, we can find a sequence $x_n \in I$ such that $x_n \rightarrow t$. I is closed so $t \in I$. $I$ is open so there exists $\delta > 0$ such that $B_t(\delta) \subset I$, but then $t + \frac{\delta}{2} \in I$ contradicting the fact tat $t$ is an upper bound for $I$. Hence $I$ is not bounded above. Similarly, $I$ is not bounded below. To see that $I = \mathbb{R}$, take $x \in \mathbb{R}$. $I$ is not bounded, so there exists $x_1, x_2 \in I$ such that $x_1 < x < x_2$. But then since $I$ is an interval, $x \in I$. Hence $\mathbb{R} \subset I$ and since $I \subset \mathbb{R}$, $I = \mathbb{R}$.

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Thank you, that's very clear. – Mathlete Feb 24 '13 at 19:46
@Mathlete no problem - I'm glad to help! – Tom Oldfield Feb 24 '13 at 19:47

HINT: Suppose that $\varnothing\ne I\ne\Bbb R$, and let $x\in\Bbb R\setminus I$. Show that either $[x,\to)\cap I=\varnothing$, in which case $x$ is an upper bound for $I$, or $(\leftarrow,x]\cap I=\varnothing$, in which case $x$ is a lower bound for $I$. The cases are similar, so let’s look at the first case, in which $I$ is bounded above. Let $u=\sup I$. Either $u\in I$, or $u\notin I$. In one case $I$ is not open, and in one case it’s not closed.

(In case you’ve not seen the arrow notation for rays, $[x,\to)$ is also written $[x,\infty)$.)

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If both $\sup I$ and $\inf I$ are not real numbers then $I\in\{\emptyset,\mathbb{R}\}$.

So you can assume at least one of them is a real number. Say $\alpha=\sup I\in\mathbb{R}$. If $\alpha\in I$ then $I$ cannot be open, as any open set containing $\alpha$ would intersect the complement of $I$. Finally, if $\alpha\not\in\ I$ then the limit of the sequence $\{\alpha-\frac{1}{n}\}_{n\geq n_0}$ is not in $I$ and therefore $I$ is not closed.

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