Exercise 6.4.3, part (c)

This post was written by Berke Özgür Arslan, who has generously offered to help Issa complete the blog. It has been edited and converted into the WordPress blog post format by Issa.

Exercise statement

Prove part (c) of Proposition 6.4.12. (Since this is a long exercise, Issa has decided to split it up across different blog posts.)

Proposition 6.4.12. Let (a_{n})_{n=m}^{\infty} be a sequence of real numbers, let L^+ be the limit superior of this sequence, and let L^- be the limit inferior of this sequence (thus both L^+ and L^- are extended real numbers).

(a) For every x > L^+, there exists an N \geq m such that a_n < x for all n \geq N. (In other words, for every x > L^+, the elements of the sequence (a_{n})_{n=m}^{\infty} are eventually less than x.) Similarly, for every y < L^- there exists an N \geq m such that a_n > y for all n \geq N.

(b) For every x < L^+, and every N \geq m, there exists an n \geq N such that a_n > x. (In other words, for every x < L^+, the elements of the sequence (a_{n})_{n=m}^{\infty} exceed x infinitely often.) Similarly, for every y > L^- and every N \geq m, there exists an n \geq N such that a_n < y.

(c) We have \inf(a_{n})_{n=m}^{\infty} \leq L^- \leq L^+ \leq \sup(a_{n})_{n=m}^{\infty}.

(d) If c is any limit point of (a_{n})_{n=m}^{\infty}, then we have L^- \leq c \leq L^+.

(e) If L^+ is finite, then it is a limit point of (a_{n})_{n=m}^{\infty}. Similarly, if L^- is finite, then it is a limit point of (a_{n})_{n=m}^{\infty}.

(f) Let c be a real number. If (a_{n})_{n=m}^{\infty} converges to c, then we must have L^+ = L^- = c. Conversely, if L^+ = L^- = c, then (a_{n})_{n=m}^{\infty} converges to c.

Hints

This is a tricky exercise and there are a few ways to approach it. One way is to split by cases based on whether L^+ and L^- are finite, +\infty, or -\infty (it should also be possible to split by cases depending on whether the original sequence (a_n)_{n=m}^\infty is bounded or not). For this approach, you may want to decide whether the sequences (a_N^+)_{N=m}^{\infty} and (a_N^-)_{N=m}^{\infty} are increasing or decreasing, then to compare their terms, and then to use Proposition 6.3.8 followed by a version of Corollary 5.4.10 for sequences of real numbers. For the cases in which L^+ and L^- may not be finite, use part (a) of the proposition.

How to think about the exercise

The concepts of limit superior and limit inferior may be difficult to grasp when one encounters their definitions for the first time. A very helpful picture to keep in mind is the piston analogy that Tao gives before the statement of Proposition 6.4.12. As we keep removing elements a_1, then a_2, and so on, the supremum piston can only slip leftward. This suggests that the sequence a_1^+, a_2^+, a_3^+, \ldots is a decreasing sequence. One of the few results we have so far about decreasing sequences is Proposition 6.3.8 , so it is a good bet that we will want to make use of that result.

So the first thing to try in order to tackle this proof is to treat the sequences (a_{N}^+)_{N=m}^{\infty} and (a_{N}^-)_{N=m}^{\infty} like we would treat any regular sequence. We are dealing here with the infimum and the supremum of these sequences respectively; so it would be nice to use Proposition 6.3.8 and convert the infimum and supremum of these sequences into their limits. But this is only possible when (a_{N}^+)_{N=m}^{\infty} is decreasing and bounded below, and (a_{N}^-)_{N=m}^{\infty} is increasing and bounded above. For the boundedness, luckily, the cases in which these sequences are not bounded correspond to the cases in which L^+ and L^- are either +\infty or -\infty, so we can treat these cases separately, and it is seen easily that these cases do not violate the proposition.

In order to see whether (a_{N}^-)_{N=m}^{\infty} is increasing or not, we need to refer to its definition. For some number M, we know that a_M^- is the infimum of the terms a_M, a_{M+1}, a_{M+2}, \ldots starting with the index M. Therefore, informally, we can say that the infimum of a sequence starting from a smaller index, say, 5, cannot be greater than of a sequence starting from a greater index, say, 10. This is because the sequence starting from index 5 contains the sequence starting from index 10, and it is “easier” to find an even smaller infimum when we have more elements to choose from. (If that was difficult to follow, you can also consider a piston moving to the right and slipping rightward.) Then we must conclude that the sequence of infima is an increasing sequence. You should figure out an analogous argument for the sequence of suprema. Then we just use Proposition 6.3.8 to conclude that limsup and liminf are just the limits of these sequences. (This also justifies the notation \limsup and \liminf: \limsup is the limit of the suprema, i.e. \displaystyle \limsup_{n\to\infty} a_n = \lim_{n\to \infty} \sup(a_N)_{N=n}^\infty, and similarly for \liminf.)

After this, we need to make another important observation. If we pick yet again any number M, then a_M^- is the infimum of the sequence (a_n)_{n=M}^{\infty}, and a_M^+ is its supremum. Clearly, we have a_M^- \leq a_M^+, and by the analogue of Corollary 5.4.10 for real sequences this means that their limits will behave accordingly; and we get the desired result.

While considering the cases where L^+ and L^- may not be finite, the following table might help you to keep track of what you need to show. Check marks indicate that their corresponding cases already satisfy the proposition because +\infty is always larger than any extended real number, and -\infty is smaller than any extended real number, so there is nothing to show. In the proof, we show that the cases indicated with a question mark cannot exist. Try to see why the assertions (1) “If L^+ \in \mathbf{R}, then L^- \neq +\infty”, and (2) “If L^+ = -\infty, then L^- = -\infty as well” exclude these cases.

L^-
L^+ \mathbf R +\infty -\infty
\mathbf R shown ?
+\infty
-\infty ? ?

Model solution 1

Consider first the inequality L^+ \leq \sup(a_{n})_{n=m}^{\infty}. Now, L^+ is defined to be the infimum of the sequence (a_{N}^+)_{N=m}^{\infty} where a_{N}^+ := \sup(a_{n})_{n=N}^{\infty}, so it should be smaller than any particular element of this sequence. Therefore L^+ \leq a_{N}^+ for all N \geq m, and in particular for N=m, we have L^+ \leq a_{m}^+ = \sup(a_{n})_{n=m}^{\infty}. A similar argument gives the inequality \inf(a_{n})_{n=m}^{\infty} \leq L^-. Thus we only need to show that L^- \leq L^+ to finish part (c).

We first assume that both L^+ and L^- are real numbers, i.e. neither of them are + \infty or - \infty. In this case, we first want to show that the sequences (a_N^-)_{N=m}^{\infty} and (a_N^+)_{N=m}^{\infty} are sequences of real numbers, i.e. that none of the elements are +\infty or -\infty. Consider first the sequence (a_N^+)_{N=m}^{\infty}. We show each of the possibilities that an element is -\infty or +\infty leads to a contradiction:

  • If some element a_M^+ = -\infty, then since L^+ is the infimum of this sequence we must have L^+ \leq a_M^+ = -\infty, a contradiction of Definition 6.2.3 (a real number cannot be less than or equal to -\infty).
  • Suppose some element a_M^+ = +\infty. We first show that the original sequence (a_n)_{n=m}^\infty is not bounded above. If there were some upper bound K \in \mathbf R for the whole sequence, then this K would also be an upper bound for the sequence (a_n)_{n=M}^\infty starting at n=M. Since a^+_M is the least upper bound of this sequence and K is merely an upper bound, we have +\infty = a^+_M \leq K, a contradiction of the fact that K is a real number. Thus the original sequence (a_n)_{n=m}^\infty is not bounded above. But this means that (a^+_N)_{N=m}^\infty = (+\infty, +\infty, +\infty, \ldots), so the infimum of this sequence is L^+ = +\infty, a contradiction.

A similar argument shows that (a_N^-)_{N=m}^{\infty} is a sequence of real numbers.

Now let us be reminded that if A and B are sets of real numbers such that A \subseteq B, then \sup (A) \leq \sup (B), and \inf (B) \leq \inf (A). Now, for some N \geq m, by definition, we have a_N^- = \inf (a_n)_{n=N}^{\infty} := \inf \{ a_n : n \geq N \} (and a similar definition for the supremum). Thus if M \geq N, then \{ a_n : n \geq M \} \subseteq \{ a_n : n \geq N \}, and we must have a_M^- \geq a_N^-. Thus the sequence (a_N^-)_{N=m}^{\infty} is an increasing sequence. By a similar argument, we see that (a_N^+)_{N=m}^{\infty} is a decreasing sequence.

Since L^- = \sup (a_N^-)_{N=m}^{\infty} is a real number, the increasing sequence (a_N^-)_{N=m}^{\infty} is bounded above, and since every increasing sequence which is bounded above is convergent by Proposition 6.3.8, we have \lim \limits_{N \to \infty} a_N^- = \sup (a_N^-)_{N=m}^{\infty} = L^-. With a similar argument for decreasing sequences which are bounded below we also have \lim \limits_{N \to \infty} a_N^+ = L^+. (Note: it is the application of Proposition 6.3.8 that required the sequences (a_N^-)_{N=m}^{\infty} and (a_N^+)_{N=m}^{\infty} to consist solely of real numbers.)

Now let N \geq m be any number. The infimum of a_N, a_{N+1}, a_{N+2}, \ldots is smaller than a_N, and the supremum of the same sequence is larger than a_N, so we have:

a_N^- = \inf (a_{n})_{n=N}^{\infty} \leq a_N \leq \sup (a_{n})_{n=N}^{\infty} = a_N^+

In other words, we have the term-wise comparison a_N^- \leq a_N^+. Since this is true for every N \geq m, we have by the analogue of Corollary 5.4.10 for real sequences that \lim_{N\to\infty} a_N^- \leq \lim_{N\to\infty} a_N^+. To summarize, we have:

L^- = \liminf \limits_{n \to \infty} a_n = \lim \limits_{N \to \infty} a_N^- \leq \lim \limits_{N \to \infty} a_N^+ = \limsup \limits_{n \to \infty} a_n = L^+

as desired.

Now, we consider the cases in which L^+ and L^- are possibly +\infty or -\infty. If L^+ = +\infty or if L^- = -\infty, there is nothing to show. Thus we only need to show two things: (1) if L^+ \in \mathbf{R}, then L^- \neq +\infty, and (2) if L^+ = -\infty, then L^- = -\infty as well. For (1), suppose for the sake of contradiction that L^+ \in \mathbf{R}, and L^- = +\infty. Since L^- = +\infty, then by part (a) of the proposition, for every number y \in \mathbf{R}, we have that y < +\infty = L^-, so there exists an M \geq m such that a_n > y for all n \geq M. This means that the sequence (a_n)_{n=m}^{\infty} doesn’t have an upper bound. Then for each N \geq m, we must have a_N^+ = \sup (a_n)_{n=N}^{\infty} = +\infty. But then we have \inf(a_N^+)_{N=m}^{\infty} = L^+ = +\infty, a contradiction. Similarly, for (2), if L^+ = -\infty, then again by (a), for every number x \in \mathbf{R}, since x > -\infty = L^+, there exists an M \geq m such that a_n < x for all n \geq M, meaning that the sequence the sequence (a_n)_{n=m}^{\infty} doesn’t have an lower bound, i.e. for each N \geq m, we have a_N^- = \inf (a_n)_{n=N}^{\infty} = -\infty. And then \sup(a_N^-)_{N=m}^{\infty} = L^- = -\infty, as desired.

Model solution 2

(This section was written by Issa.)

There is another, shorter proof of part (c) that does not require breaking up the proof into cases. This proof is very neat, and I encourage you to check it out, but I also believe it is more difficult for beginners to discover. I may write it up here in detail later, but for now please check out this writeup by Sangchul Lee; it is Exercise 8(iii) on page 5.

Analogue of Corollary 5.4.10 for real sequences

In this post, I will state an analogue of Corollary 5.4.10 for sequences of real numbers. This result easily follows in the book using Lemma 6.4.13 together with Proposition 6.4.12(f), but having this result is handy for proving parts of Proposition 6.4.12, so in order to avoid circularity/anachronisms, we will prove it here using our bare hands.

Exercise statement

Lemma. Let (a_n)_{n=m}^\infty and (b_n)_{n=m}^\infty be convergent sequences of real numbers such that a_n \geq b_n for all n \geq m. Then \lim_{n\to\infty} a_n \geq \lim_{n\to\infty} b_n.

Model solution 1

Let us write x:=\lim_{n\to\infty} a_n and y:=\lim_{n\to\infty} b_n. We are trying to show that x \geq y. Suppose for the sake of contradiction that x < y. Then y-x > 0 so for \varepsilon := (y-x)/3 > 0 we can find some N_1 \geq m such that |a_n - x| \leq \varepsilon for all n \geq N_1, and some N_2 \geq m such that |b_n - y| \leq \varepsilon for all n \geq N_2. Thus if we pick N := \max(N_1, N_2) then since N \geq N_1 and N \geq N_2 we have both |a_N - x| \leq \varepsilon and |b_N - y| \leq \varepsilon. These two inequalities say in particular that -\varepsilon \leq x-a_N and -\varepsilon \leq b_N - y. Thus we have

\begin{aligned} b_N - a_N &= (b_N - y) + (y - x) + (x - a_N) \\ &\geq -\varepsilon + 3\varepsilon - \varepsilon \\ & = \varepsilon \\ &> 0\end{aligned}

This means that b_N > a_N, which contradicts the fact that a_n \geq b_n for all n \geq m. Thus we conclude that x \geq y after all.

Model solution 2

The following solution is due to Berke Özgür Arslan.

In this proof, we mimic the proof of Corollary 5.4.10 by using the limit laws, in particular Theorem 6.1.19(d). We want to show that \lim_{n\to\infty} a_n \geq \lim_{n\to\infty} b_n, but this is the same thing as showing \lim_{n\to\infty} a_n - \lim_{n\to\infty} b_n \geq 0. By the limit laws, \lim_{n\to\infty} a_n - \lim_{n\to\infty} b_n = \lim_{n\to\infty} (a_n - b_n), so it suffices to show that \lim_{n\to\infty} (a_n - b_n) \geq 0.

Thus our goal now is to consider the sequence (a_n - b_n)_{n=m}^\infty, and show that its limit is non-negative. To simplify the exposition, let us define c_n := a_n - b_n. Then since a_n \geq b_n for each n \geq m, we have that c_n = a_n - b_n \geq 0 for each n \geq m.

Using our new notation of c_n, our goal is the following: we want to show that if (c_n)_{n=m}^\infty is a sequence such that c_n \geq 0 for all n\geq m, then \lim_{n\to\infty} c_n \geq 0.

To show this we prove by contradiction. We suppose L := \lim_{n\to\infty} c_n < 0. Then by the definition of sequence convergence, for any \varepsilon > 0, there exists N \geq m such that  L - \varepsilon \leq c_n \leq L + \varepsilon for each n \geq N. But if we let \varepsilon := - L / 2 > 0, then we could find some n such that 3L / 2 \leq c_n \leq L / 2 < 0. But since c_n \geq 0 for all n \geq m, this is a contradiction. Thus we have \lim_{n\to\infty} c_n \geq 0.

Announcing Tao Analysis Flashcards

Hello everyone, today I am announcing a sister-project to this blog that I have been working on for a while:

Tao Analysis Flashcards is a website which hosts question-and-answer flashcards on the content in Analysis I. Each set of flashcards is associated with a section from the book, and is meant to be completed after you finish reading the section. The idea is that instead of a “read the section” → “do the exercises” loop, we make it a “read the section” → “check understanding of reading by doing flashcards” → “do the exercises” loop. I’m hoping the extra step will do some combination of making you pay more attention to the reading, clarifying some points not raised in the reading, and checking your understanding so you feel more confident you understood each section.

Learning mathematics is hard. Even harder is to learn mathematics on your own, which I believe most readers of my blog are doing when they work through Analysis I. But even harder still is to learn mathematics and actually retain it over the long term. So many times I’ve thought I understood a topic, only to realize several months later that I couldn’t recall most of what I had learned! And that brings us to another aspect of these flashcards: they are spaced repetition flashcards, which means you will get email reminders to review the cards over time, allowing you to efficiently retain the knowledge even as months or years pass.

Please let me know in comments your thoughts, especially if a card seems confusing or too difficult.

Exercise 3.1.11

Exercise statement

Show that the axiom of replacement implies the axiom of specification.

Hints

None.

How to think about the exercise

What does it mean for one axiom to imply another? In this case, both axioms are about constructing sets, so we just want to show that any set that we can construct using the axiom of specification can be constructed using the axiom of replacement instead. The axiom of specification allows us to construct sets of the form \{x \in A : P(x) \text{ is true}\}. So our goal is, given some set A and some property (predicate) P, to construct the set \{x \in A : P(x) \text{ is true}\} using the axiom of replacement.

To use the axiom of replacement, we must feed it a two-place property and a set. The set can just be A: we want to produce some subset of A, so it makes sense that we start with A and do some “replacement” on its elements somehow. So that leaves us to determine what the two-place property will be. Let’s call that property Q, since the variable P is already taken. Once we define what Q is, the axiom of replacement allows us to construct the set \{y : Q(x,y) \text{ is true for some }x \in A\}.

How do we decide what Q to use? Well, we want to keep only those elements x of A such that P(x) is true. So it makes sense to require P(x) as part of Q(x,y). And the elements of this new set should be the same elements from A, so we don’t want to apply any transformation to get the ys. So we can just let y = x. In other words, we can define Q(x,y) to be “y=x and P(x)” (it’s fine to say “y=x and P(y)” as well). Thus we have the set \{y : y=x\text{ and }P(x)\text{ for some }x \in A\}.

There is another way to look at this exercise, assuming you are comfortable with functions. The axiom of replacement basically says that if A is a set and f is an operation on elements of A, then \{f(x) : x \in A\} is a set. Here the operation f may return an undefined result (because for each x, the statement P(x,y) is true for at most one y rather than exactly one y). So to construct the set \{x \in A : P(x) \text{ is true}\}, we can define f(x) to be x if P(x) is true, and leave f(x) undefined if P(x) is false.

If you feel uncomfortable about f having undefined values, see Exercise 3.4.7, and also note that we can do everything with functions also: we split into cases depending on whether the resulting set will be empty. If A is empty or P(x) is false for all x\in A then the resulting set will be empty so we don’t need to do anything to construct it. Otherwise, since A is non-empty and P(x) is true for at least one x \in A, we can pick some element x_0 \in A for which P(x_0) is true and let this element be the default output of f for inputs we want to exclude. In other words, f : A \to A is the function defined by:

\displaystyle f(x) := \begin{cases}x & \text{if }P(x)\text{ is true} \\ x_0 & \text{otherwise}\end{cases}

But the discussion involving f is a bit informal, as we have not introduced functions at this point in the book. But nevertheless this way of looking at this problem provides good intuition.

Model solution

Let A be a set, and let P(x) be a statement pertaining to objects x \in A. To show that the axiom of replacement implies the axiom of specification, we will construct the set \{x \in A : P(x) \text{ is true}\} using just the axiom of replacement. Let Q(x,y) be the statement “y=x and P(x)”. To use the axiom of replacement, we must verify that for each x \in A, the statement Q(x,y) is true for at most one y. But for each x\in A, there is exactly one y such that y=x, so it follows that for the full statement of Q(x,y) there is at most one y satisfying the statement. The axiom of replacement thus allows us to construct the set \{y : y=x\text{ and }P(x)\text{ for some }x \in A\}. We claim that this is the same set as \{x \in A : P(x) \text{ is true}\}. We show the inclusion both ways:

  • Suppose z \in \{x \in A : P(x) \text{ is true}\}. Thus z \in A and P(z) is true. But this means that z=x and P(x) for some x \in A; specifically, there is z \in A such that z=z and P(z). Thus z \in \{y : y=x\text{ and }P(x)\text{ for some }x \in A\}.
  • Suppose z \in \{y : y=x\text{ and }P(x)\text{ for some }x \in A\}. This means that for some x \in A, we have z=x and P(x). Since z=x this means we have P(z) and z\in A as well. Thus z \in \{x \in A : P(x) \text{ is true}\}.

Since both directions of the inclusion hold, this means that the two sets are equal.

Exercise 8.3.5

Exercise statement

Show that no power set (i.e., a set of the form 2^X for some set X) can be countably infinite.

Hints

  1. You do not need the axiom of choice for this exercise (though of course if you have read Section 8.4 you may use it).

How to think about the exercise

I think the proof below is pretty straightforward, and I don’t have too much to say beyond that.

Model solution

We have three cases depending on the cardinality of X. If X is finite, then by Exercise 8.3.1 the power set 2^X has finite cardinality, so we are done. If X is countably infinite, then by Cantor’s theorem (Theorem 8.3.1) we see that 2^X cannot be countably infinite, so we are done. Thus it remains to show the result when X is uncountable, so let X be an uncountable set.

Let us introduce some notation. We write |A| \leq |B| to mean that there is an injection from A to B. The intuition for this notation comes from Exercise 8.3.3. The relation \leq has the transitivity property that if |A| \leq |B| and |B| \leq |C|, then |A| \leq |C|: if f : A \to B is an injection and g : B \to C is an injection, then g \circ f : A \to C is an injection by Exercise 3.3.2. (The relation \leq is anti-symmetric thanks to Exercise 8.3.3, and it is also reflexive because the identity map \mathrm{id} : A \to A is injective, so it is a reflexive, anti-symmetric, and transitive relation, i.e. a partial order. Thus we are justified in using the suggestive notation \leq.)

We have |X| \leq |2^X| because f : X \to 2^X defined by f(x) := \{x\} is an injection.

Now suppose for sake of contradiction that 2^X is countably infinite. Then there is a bijection between 2^X and \mathbf N, so in particular we have |2^X| \leq |\mathbf N|. We also showed above that |X| \leq |2^X|. So by transitivity of \leq, we have |X| \leq |\mathbf N|. Thus there is some injection g : X \to \mathbf N. Now consider the image g(X), which is a subset of \mathbf N. By Corollary 8.1.6, g(X) is at most countable. Define the function h : X \to g(X) by h(x) := g(x). This is a bijection: it is injective because g is injective, and it is surjective because g(X) = \{g(x) : x \in X\} is exactly the set of elements that g maps to. The bijection h shows that X has equal cardinality to g(X), which is at most countable. So X is at most countable, which contradicts the assumption that X is uncountable. This contradiction shows that 2^X cannot be countably infinite, which completes the proof.

Exercise 10.1.1

Exercise statement

Suppose that X is a subset of \mathbf R, x_0 is a limit point of X, and f : X \to \mathbf R is a function which is differentiable at x_0. Let Y \subseteq X be such that x_0 \in Y, and x_0 is also a limit point of Y. Prove that the restricted function f|_Y : Y \to \mathbf R is also differentiable at x_0, and has the same derivative as f at x_0. Explain why this does not contradict the discussion in Remark 10.1.2.

Hints

  1. Use Definition 9.3.6.

How to think about the exercise

This is a simple exercise and is a matter of putting together the right definitions.

Model solution

We want to show that

\displaystyle \lim_{y \to x_0; y \in Y\setminus\{x_0\}} \frac{f|_Y(y) - f|_Y(x_0)}{y-x_0} = f'(x_0)

To do this, we will return to the definition of a limit, Definition 9.3.6. To show the limit exists, we must first verify that x_0 is an adherent point of Y\setminus\{x_0\}. But this is the case since x_0 is a limit point of Y. Now  let \varepsilon > 0. We want to find some \delta > 0 such that for all y \in Y\setminus\{x_0\}, if |y-x_0| < \delta then \left|\frac{f|_Y(y) - f|_Y(x_0)}{y-x_0} - f'(x_0)\right| \leq \varepsilon.

How can we find such a \delta? The only information we are given from which we could find a \delta is the fact that f is differentiable at x_0. Since f is differentiable at x_0, we know that for our \varepsilon > 0 in particular, there exists some \delta > 0 such that for all x \in X \setminus \{x_0\}, if |x-x_0| < \delta then \left|\frac{f(x) - f(x_0)}{x-x_0} - f'(x_0)\right| \leq \varepsilon.

So let’s use this \delta > 0. Now that we have a \delta, we must show that all y \in Y\setminus\{x_0\}, if |y-x_0| < \delta then \left|\frac{f|_Y(y) - f|_Y(x_0)}{y-x_0} - f'(x_0)\right| \leq \varepsilon. So let y \in Y\setminus\{x_0\}, and suppose |y-x_0| < \delta. We must show that \left|\frac{f|_Y(y) - f|_Y(x_0)}{y-x_0} - f'(x_0)\right| \leq \varepsilon.

Now Y \subseteq X, so this means Y \setminus \{x_0\} \subseteq X \setminus \{x_0\}. Thus we have y \in X\setminus \{x_0\}.

We know from the differentiability condition for f at x_0 that for all x \in X \setminus \{x_0\}, if |x-x_0| < \delta then \left|\frac{f(x) - f(x_0)}{x-x_0} - f'(x_0)\right| \leq \varepsilon. Since our y satisfies these conditions, we have \left|\frac{f(y) - f(x_0)}{y-x_0} - f'(x_0)\right| \leq \varepsilon.

To complete the proof, recall how f|_Y is defined. If y' \in Y, then f|_Y(y') := f(y'). Since y,x_0 \in Y, we have f|_Y(y) = f(y) and f|_Y(x_0) = f(x_0). Thus we have \left|\frac{f|_Y(y) - f|_Y(x_0)}{y-x_0} - f'(x_0)\right|=\left|\frac{f(y) - f(x_0)}{y-x_0} - f'(x_0)\right| \leq \varepsilon.

This does not contradict the discussion in Remark 10.1.2 because in the example there, 3 was not an adherent point of [1,2], so the limit was undefined. In contrast for this exercise we assumed that x_0 was an adherent point of Y \setminus \{x_0\} (i.e. a limit point of Y). And so if we used the example in Remark 10.1.2, the proof above would fail at the point where we tried to verify that x_0 is an adherent point of Y\setminus\{x_0\}.

The important point is that if x_0 were not a limit point of Y, then we could always pick \delta > 0 small enough so that the expression \frac{f|_Y(y) - f|_Y(x_0)}{y-x_0} is undefined when y is further restricted to the set \{y \in Y \setminus \{x_0\} : |y - x_0| < \delta\}. This would mean that the limit is always defined no matter what L we use (since the implication in the definition of limit would be vacuous), so that in effect the derivative would equal any number possible. This would make the notation f|_Y'(x_0) ambiguous, and this concept is useless anyway, which is why we choose to not define the limit in this case.

Exercise 11.1.1

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

  1. 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.

Exercise 6.5.2

Exercise statement

Prove Lemma 6.5.2.

Lemma 6.5.2. Let x be a real number. Then the limit \lim_{n\to\infty} x^n exists and is equal to zero when |x| < 1, exists and is equal to 1 when x=1, and diverges when x=-1 or when |x| > 1.

Hints

  1. Use Proposition 6.3.10, Exercise 6.3.4, and the squeeze test.

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.

Exercise 7.2.1

Exercise statement

Is the series \sum_{n=1}^\infty (-1)^n convergent or divergent? Justify your answer. Can you now resolve the difficulty in Example 1.2.2?

Hints

None.

How to think about the exercise

This is a straightforward exercise.

Model solution 1

Let (a_n)_{n=1}^\infty be the sequence defined by a_n := 1 for all n \geq 1. This sequence is non-negative and decreasing, since a_n = 1 \geq 0 and a_n = 1 \geq 1 = a_{n+1} for every n \geq 1. We also have \lim_{n\to\infty} a_n = 1 \ne 0. Thus we can apply the alternating series test (Proposition 7.2.12) to conclude that the series \sum_{n=1}^\infty (-1)^n a_n = \sum_{n=1}^\infty (-1)^n diverges.

The reasoning in Example 1.2.2 is not valid because the series does not converge, so variable S is not a real number and thus cannot be manipulated in the usual way using the laws of algebra.

(Thanks to William for this solution.)

Model solution 2

The series \sum_{n=1}^\infty (-1)^n diverges by the zero test, Corollary 7.2.6. Indeed, 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. (Alternatively, we could have used Lemma 6.5.2.)

The reasoning in Example 1.2.2 is not valid because the series does not converge, so variable S is not a real number and thus cannot be manipulated in the usual way using the laws of algebra.

Exercise 9.3.4

Exercise statement

Propose a definition for limit superior \limsup_{x\to x_0; x\in E} f(x) and limit inferior \liminf_{x\to x_0; x\in E} f(x), and then propose an analogue of Proposition 9.3.9 for your definition. (For an additional challenge: prove that analogue.)

Hints

None.

How to think about the exercise

Let’s recall how the limit superior was defined for sequences of real numbers. Given a sequence (a_n)_{n=0}^\infty, we first defined the auxiliary sequence (a^+_N)_{N=0}^\infty by a^+_N := \sup(a_n)_{n=N}^\infty, then we took the infimum of the sequence (which is equivalent to taking the limit, since the sequence is monotonic so always converges): \limsup_{n\to\infty} a_n := \inf(a^+_N)_{N=0}^\infty = \lim_{N \to \infty} a^+_N.

So to define a limit superior, we need to take the supremum over something (and this something must be indexed somehow), then take the limit using the index. To help us see the similarities more with the real-valued functions case, let’s rewrite the limit superior for sequences as \limsup_{n\to\infty} a_n = \lim_{N \to \infty} \sup \{a_n : n \geq N\}.

So our goal now is to figure out the set we are taking the supremum over. Which set? This set must get smaller over time, so that the limit is defined (otherwise we wouldn’t have a guarantee that the sequence is monotonic). In other words, we want to “go further” in some relevant sense. In the case of sequences, we “go further” in the limit by taking larger values of n, but in the case of functions we “go further” by taking x values closer to x_0. So the trick is to take a set that shrinks around the point x_0. In particular, we can take \{f(x) : x \in E \text{ and } |x - x_0| < \delta\}. The smaller the value of \delta, the smaller the set. So we can define \limsup_{x\to x_0; x\in E} f(x) := \lim_{\delta \to 0} \sup\{f(x) : x \in E \text{ and } |x - x_0| < \delta\}. Here it is understood that when we take the limit as \delta \to 0 only positive values of \delta are considered (we could be clearer about this by writing something like \lim_{\delta \to 0; \delta > 0}). (TODO: Tao’s book doesn’t actually define divergent limits, so this part does fail in some edge cases, and so we either need to define such limits or use inf/sup. Thanks to crabman for the catch.)

(THIS IS WRONG. I WILL FIX LATER. Thanks to Karim Taha for the catch.) Now how do we state the analogue of Proposition 9.3.9? For part (a), instead of \lim_{x \to x_0; x\in E} f(x) = L we want to say \limsup_{x\to x_0; x\in E} f(x). For part (b), we can take a sequence (a_n)_{n=0}^\infty converging to x_0, but instead of saying \lim_{n\to \infty} f(a_n) = L, we can use the limit superior for sequences to say \limsup_{n\to \infty} f(a_n) = L.

If you want to think about this topic more, see the first edition of Pugh’s Real Mathematical Analysis, Chapter 3 Exercise 26, which makes the “go further” idea above more rigorous.

Model solution

We define the limit superior \limsup_{x\to x_0; x\in E} f(x) as \lim_{\delta \to 0; \delta > 0} \sup\{f(x) : x \in E \text{ and } |x - x_0| < \delta\}. Alternatively, using infimum instead of a limit we could define it as \inf\{\sup\{f(x) : x \in E \text{ and } |x - x_0| < \delta\} : \delta > 0\}.

(THIS IS WRONG. I WILL FIX LATER.) The analogue of Proposition 9.3.9 is as follows: Let X be a subset of \mathbf R, let f : X \to \mathbf R be a function, let E be a subset of X, let x_0 be an adherent point of E, and let L be a real number. Then the following two statements are logically equivalent:

(a) \limsup_{x\to x_0; x\in E} f(x)=L
(b) For every sequence (a_n)_{n=0}^\infty which consists entirely of elements of E, which converges to x_0, we have \limsup_{n\to\infty} f(a_n)=L.

The case for limit inferior is very similar: just interchange “superior” and “inferior” everywhere.

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