This is just a short announcement that I’ve put up a new page about an independent review/grading of the solutions on this blog.

# 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 be a sequence of real numbers, let be the limit superior of this sequence, and let be the limit inferior of this sequence (thus both and are extended real numbers).

(a) For every , there exists an such that for all . (In other words, for every , the elements of the sequence are eventually less than .) Similarly, for every there exists an such that for all .

(b) For every , and every , there exists an such that . (In other words, for every , the elements of the sequence exceed infinitely often.) Similarly, for every and every , there exists an such that .

(c) We have .

(d) If is any limit point of , then we have .

(e) If is finite, then it is a limit point of . Similarly, if is finite, then it is a limit point of .

(f) Let be a real number. If converges to , then we must have . Conversely, if , then converges to .

## 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 and are finite, , or (it should also be possible to split by cases depending on whether the original sequence is bounded or not). For this approach, you may want to decide whether the sequences and 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 and 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 , then , and so on, the supremum piston *can only slip leftward*. This suggests that the sequence 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 and 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 is decreasing and bounded below, and is increasing and bounded above. For the boundedness, luckily, the cases in which these sequences are not bounded correspond to the cases in which and are either or , 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 is increasing or not, we need to refer to its definition. For some number , we know that is the infimum of the terms starting with the index . 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 and : is the limit of the suprema, i.e. , and similarly for .)

After this, we need to make another important observation. If we pick yet again any number , then is the infimum of the sequence , and is its supremum. Clearly, we have , 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 and 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 is always larger than any extended real number, and 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 , then ”, and (2) “If , then as well” exclude these cases.

shown | ? | ✓ | ||

✓ | ✓ | ✓ | ||

? | ? | ✓ |

## Model solution 1

Consider first the inequality . Now, is defined to be the infimum of the sequence where , so it should be smaller than any particular element of this sequence. Therefore for all , and in particular for , we have . A similar argument gives the inequality . Thus we only need to show that to finish part (c).

We first assume that both and are real numbers, i.e. neither of them are or . In this case, we first want to show that the sequences and are sequences of real numbers, i.e. that none of the elements are or . Consider first the sequence . We show each of the possibilities that an element is or leads to a contradiction:

- If some element , then since is the infimum of this sequence we must have , a contradiction of Definition 6.2.3 (a real number cannot be less than or equal to ).
- Suppose some element . We first show that the original sequence is not bounded above. If there were some upper bound for the whole sequence, then this would also be an upper bound for the sequence starting at . Since is the
*least*upper bound of this sequence and is merely*an*upper bound, we have , a contradiction of the fact that is a real number. Thus the original sequence is not bounded above. But this means that , so the infimum of this sequence is , a contradiction.

A similar argument shows that is a sequence of real numbers.

Now let us be reminded that if and are sets of real numbers such that , then , and . Now, for some , by definition, we have (and a similar definition for the supremum). Thus if , then , and we must have . Thus the sequence is an increasing sequence. By a similar argument, we see that is a decreasing sequence.

Since is a real number, the increasing sequence is bounded above, and since every increasing sequence which is bounded above is convergent by Proposition 6.3.8, we have . With a similar argument for decreasing sequences which are bounded below we also have . (Note: it is the application of Proposition 6.3.8 that required the sequences and to consist solely of real numbers.)

Now let be any number. The infimum of is smaller than , and the supremum of the same sequence is larger than , so we have:

In other words, we have the term-wise comparison . Since this is true for every , we have by the analogue of Corollary 5.4.10 for real sequences that . To summarize, we have:

as desired.

Now, we consider the cases in which and are possibly or . If or if , there is nothing to show. Thus we only need to show two things: (1) if , then , and (2) if , then as well. For (1), suppose for the sake of contradiction that , and . Since , then by part (a) of the proposition, for every number , we have that , so there exists an such that for all . This means that the sequence doesn’t have an upper bound. Then for each , we must have . But then we have , a contradiction. Similarly, for (2), if , then again by (a), for every number , since , there exists an such that for all , meaning that the sequence the sequence doesn’t have an lower bound, i.e. for each , we have . And then , 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 and be convergent sequences of real numbers such that for all . Then .

## Model solution 1

Let us write and . We are trying to show that . Suppose for the sake of contradiction that . Then so for we can find some such that for all , and some such that for all . Thus if we pick then since and we have both and . These two inequalities say in particular that and . Thus we have

This means that , which contradicts the fact that for all . Thus we conclude that 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 , but this is the same thing as showing . By the limit laws, , so it suffices to show that .

Thus our goal now is to consider the sequence , and show that its limit is non-negative. To simplify the exposition, let us define . Then since for each , we have that for each .

Using our new notation of , our goal is the following: we want to show that if is a sequence such that for all , then .

To show this we prove by contradiction. We suppose . Then by the definition of sequence convergence, for any , there exists such that for each . But if we let , then we could find some such that . But since for all , this is a contradiction. Thus we have .

# 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 . So our goal is, given some set and some property (predicate) , to construct the set 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 : we want to produce some subset of , so it makes sense that we start with 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 , since the variable is already taken. Once we define what is, the axiom of replacement allows us to construct the set .

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

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

If you feel uncomfortable about 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 is empty or is false for all then the resulting set will be empty so we don’t need to do anything to construct it. Otherwise, since is non-empty and is true for at least one , we can pick some element for which is true and let this element be the default output of for inputs we want to exclude. In other words, is the function defined by:

But the discussion involving 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 be a set, and let be a statement pertaining to objects . To show that the axiom of replacement implies the axiom of specification, we will construct the set using just the axiom of replacement. Let be the statement “ and ”. To use the axiom of replacement, we must verify that for each , the statement is true for at most one . But for each , there is exactly one such that , so it follows that for the full statement of there is at most one satisfying the statement. The axiom of replacement thus allows us to construct the set . We claim that this is the same set as . We show the inclusion both ways:

- Suppose . Thus and is true. But this means that and for some ; specifically, there is such that and . Thus .
- Suppose . This means that for some , we have and . Since this means we have and as well. Thus .

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 for some set ) can be countably infinite.

## Hints

- 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 . If is finite, then by Exercise 8.3.1 the power set has finite cardinality, so we are done. If is countably infinite, then by Cantor’s theorem (Theorem 8.3.1) we see that cannot be countably infinite, so we are done. Thus it remains to show the result when is uncountable, so let be an uncountable set.

Let us introduce some notation. We write to mean that there is an injection from to . The intuition for this notation comes from Exercise 8.3.3. The relation has the transitivity property that if and , then : if is an injection and is an injection, then is an injection by Exercise 3.3.2. (The relation is anti-symmetric thanks to Exercise 8.3.3, and it is also reflexive because the identity map 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 .)

We have because defined by is an injection.

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

# Exercise 10.1.1

## Exercise statement

Suppose that is a subset of , is a limit point of , and is a function which is differentiable at . Let be such that , and is also a limit point of . Prove that the restricted function is also differentiable at , and has the same derivative as at . Explain why this does not contradict the discussion in Remark 10.1.2.

## Hints

- 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

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 is an adherent point of . But this is the case since is a limit point of . Now let . We want to find some such that for all , if then .

How can we find such a ? The only information we are given from which we could find a is the fact that is differentiable at . Since is differentiable at , we know that for our in particular, there exists some such that for all , if then .

So let’s use this . Now that we have a , we must show that all , if then . So let , and suppose . We must show that .

Now , so this means . Thus we have .

We know from the differentiability condition for at that for all , if then . Since our satisfies these conditions, we have .

To complete the proof, recall how is defined. If , then . Since , we have and . Thus we have .

This does not contradict the discussion in Remark 10.1.2 because in the example there, was not an adherent point of , so the limit was undefined. In contrast for this exercise we assumed that was an adherent point of (i.e. a limit point of ). 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 is an adherent point of .

The important point is that if were not a limit point of , then we could always pick small enough so that the expression is undefined when is further restricted to the set . This would mean that the limit is *always* defined no matter what 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 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 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 .

# Exercise 6.5.2

## Exercise statement

Prove Lemma 6.5.2.

**Lemma 6.5.2.** Let be a real number. Then the limit exists and is equal to zero when , exists and is equal to when , and diverges when or when .

## Hints

- 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 . Since we have for each , we can use the squeeze test: by Proposition 6.3.10 we know that , so by the limit laws we have as well. This means that the sequence must also converge to zero.

Next suppose . Then we have for all , so is the constant sequence, which converges to . A similar proof can be given when ; in this case is constantly zero, so converges to zero.

If , the limit does not exist as the sequence oscillates between and . To prove that does not exist, let . Then no matter how big is, we can pick and . We have . Thus the sequence is not Cauchy, so by Theorem 6.4.18 it is not convergent.

Finally suppose . Suppose for sake of contradiction that converges to some limit . Then by the limit laws, . But , so this means that converges to . This contradicts Exercise 6.3.4.

# Exercise 7.2.1

## Exercise statement

Is the series 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 be the sequence defined by for all . This sequence is non-negative and decreasing, since and for every . We also have . Thus we can apply the alternating series test (Proposition 7.2.12) to conclude that the series diverges.

The reasoning in Example 1.2.2 is not valid because the series does not converge, so variable 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 diverges by the zero test, Corollary 7.2.6. Indeed, the limit does not exist as the sequence oscillates between and . To prove that does not exist, let . Then no matter how big is, we can pick and . We have . 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 is not a real number and thus cannot be manipulated in the usual way using the laws of algebra.