## Exercise statement

Prove Lemma 11.1.4.

**Lemma 11.1.4.** Let be a subset of the real line. Then the following two statements are logically equivalent:

(a) is bounded and connected.

(b) is a bounded interval.

## Hints

- In order to show that (a) implies (b) in the case when is non-empty, consider the supremum and infimum of .

## How to think about the exercise

This exercise is pretty straightforward, but one must be careful to stick to the definitions instead of using one’s intuitive notion of intervals; the whole point of the exercise is to make sure our formalization of ideas like “bounded” and “connected” and “interval” is correct. An example of what I mean: it seems like Tao defines “bounded interval” and then later on defines “bounded set”. Even though “bounded interval” contains the word “bounded”, the book doesn’t seem to show that bounded intervals are indeed bounded in the sense of bounded set. So part of the point of this exercise is to demonstrate this, and to see that nothing unexpected has happened.

## Model solution

We first show that (b) implies (a). Suppose that is a bounded interval (see Examples 9.1.3 for definition). Thus is of the form , , , or for real numbers . We will just do one of the cases, namely the case , as the proof in the other cases is very similar. We first show that is bounded. We claim that is a bound (the is a technicality, since Definition 9.1.22 requires for some odd reason). Indeed, if then we have by definition of the interval notation. Thus we have

Thus we have . Since was arbitrary, this shows that as required, so is bounded.

Next we show that is connected. Let be elements of such that . We must show that is a subset of . Let be a number such that . We must show that . But since we have , and since we have . Thus by transitivity of inequalities we have , so . This proves that is connected.

Now we show that (a) implies (b). Suppose is bounded and connected. We have two cases. Suppose first that is empty. Then is equal to an empty bounded interval such as .

Now suppose is non-empty. Then define and . Since is a bounded and non-empty set, by Theorem 5.5.9 both and are real numbers. We have four cases depending on whether and . Since the proof is similar in each case, we will just show the case when and . In this case, we will show that equals the bounded interval . To do this, we will show that and . We first show that . Let . Since and is the least upper bound for , there exists some such that (otherwise would be a smaller upper bound for ). Now we have such that ; since is connected, this means that as desired.

Now we show that . Let . Since is the infimum of and is the supremum of , we know that . Thus to show that we just need to show that . But this is easy, since we assumed that .