November | 2020 | Tao Analysis Solutions

Exercise statement

Prove Lemma 11.1.4.

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

(a) $X$ is bounded and connected.
(b) $X$ is a bounded interval.

Hints

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

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 $X$ is a bounded interval (see Examples 9.1.3 for definition). Thus $X$ is of the form $(a,b)$ , $[a,b]$ , $(a,b]$ , or $[a,b)$ for real numbers $a,b \in \mathbf R$ . We will just do one of the cases, namely the case $X=[a,b)$ , as the proof in the other cases is very similar. We first show that $X$ is bounded. We claim that $M := \max(|a|, |b|) + 1$ is a bound (the $+1$ is a technicality, since Definition 9.1.22 requires $M > 0$ for some odd reason). Indeed, if $x \in X$ then we have $a \leq x < b$ by definition of the interval notation. Thus we have

$-M < -\max(|a|, |b|) \leq -|a| \leq a \leq x < b \leq |b| \leq \max(|a|, |b|) < M$

Thus we have $x \in [-M,M]$ . Since $x \in X$ was arbitrary, this shows that $X \subseteq [-M, M]$ as required, so $X$ is bounded.

Next we show that $X$ is connected. Let $x,y$ be elements of $X$ such that $x < y$ . We must show that $[x,y]$ is a subset of $X$ . Let $z$ be a number such that $x \leq z \leq y$ . We must show that $z \in X$ . But since $x \in X$ we have $a \leq x$ , and since $y \in X$ we have $y < b$ . Thus by transitivity of inequalities we have $a \leq x \leq z \leq y < b$ , so $z \in [a, b) = X$ . This proves that $X$ is connected.

Now we show that (a) implies (b). Suppose $X$ is bounded and connected. We have two cases. Suppose first that $X$ is empty. Then $X$ is equal to an empty bounded interval such as $(1,1)$ .

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

Now we show that $X \subseteq [a, b)$ . Let $x \in X$ . Since $a$ is the infimum of $X$ and $b$ is the supremum of $X$ , we know that $a \leq x \leq b$ . Thus to show that $x \in [a,b)$ we just need to show that $x \ne b$ . But this is easy, since we assumed that $b \notin X$ .

How to think about the exercise

This is a straightforward exercise, but it requires some care in two respects:

It can be a little notationally confusing dealing with all of the absolute value signs so it is easy to think one has proved something when there is in fact a slight flaw in the proof.

If one wants to write up the proof in the shortest way possible, thinking of which cases to use can be tricky.

Model solution

First suppose $0 < |x| < 1$ . Since we have $-|x|^n = -|x^n| \leq x^n \leq |x^n| = |x|^n$ for each $n$ , we can use the squeeze test: by Proposition 6.3.10 we know that $\lim_{n\to\infty} |x|^n = 0$ , so by the limit laws we have $\lim_{n\to\infty} -|x|^n = 0$ as well. This means that the sequence $(x^n)_{n=1}^\infty$ must also converge to zero.

Next suppose $x=1$ . Then we have $x^n = 1^n = 1$ for all $n$ , so $(x^n)_{n=1}^\infty$ is the constant sequence, which converges to $1$ . A similar proof can be given when $x=0$ ; in this case $(x^n)_{n=1}^\infty$ is constantly zero, so converges to zero.

If $x=-1$ , the limit $\lim_{n\to\infty} (-1)^n$ does not exist as the sequence $((-1)^n)_{n=1}^\infty = (-1,1,-1,1,-1,\ldots)$ oscillates between $-1$ and $1$ . To prove that $\lim_{n\to\infty} (-1)^n$ does not exist, let $\varepsilon := 1$ . Then no matter how big $N \geq 1$ is, we can pick $j := N$ and $k:=N+1$ . We have $|(-1)^j - (-1)^k| = 2 > \varepsilon$ . Thus the sequence is not Cauchy, so by Theorem 6.4.18 it is not convergent.

Finally suppose $|x| > 1$ . Suppose for sake of contradiction that $(x^n)_{n=1}^\infty$ converges to some limit $L \in \mathbf R$ . Then by the limit laws, $\lim_{n\to\infty} |x^n| = \lim_{n\to\infty} \max(x^n, -x^n) = \max(L,-L) = |L|$ . But $|x^n| = |x|^n$ , so this means that $(|x|^n)_{n=1}^\infty$ converges to $|L|$ . This contradicts Exercise 6.3.4.