# Exercise 3.1.6

## Exercise statement

Prove Proposition 3.1.28.

Proposition 3.1.28 (Sets form a boolean algebra). Let $A,B,C$ be sets, and let $X$ be a set containing $A,B,C$ as subsets.

(a) (Minimal element) We have $A\cup \emptyset = A$ and $A \cap \emptyset = \emptyset$.
(b) (Maximal element) We have $A \cup X = X$ and $A \cap X = A$.
(c) (Identity) We have $A \cap A = A$ and $A \cup A = A$.
(d) (Commutativity) We have $A \cup B = B \cup A$ and $A \cap B = B \cap A$.
(e) (Associativity) We have $(A\cup B) \cup C = A \cup (B \cup C)$ and $(A \cap B) \cap C = A \cap (B \cap C)$.
(f) (Distributivity) We have $A \cap (B\cup C) = (A \cap B) \cup (A\cap C)$ and $A\cup (B\cap C) = (A\cup B) \cap (A \cup C)$.
(g) (Partition) We have $A \cup (X \setminus A) = X$ and $A \cap (X \setminus A) = \emptyset$.
(h) (De Morgan laws) We have $X \setminus (A \cup B) = (X \setminus A) \cap (X \setminus B)$ and $X\setminus (A \cap B) = (X \setminus A) \cup (X \setminus B)$.

## Hints

1. One can use some of these claims to prove others.
2. Some of the claims have also appeared previously in Lemma 3.1.13.

## How to think about the exercise

The strategy for most of the parts of this exercise is to show that both $A \subseteq B$ and $B \subseteq A$, from which we can conclude $A=B$. (Just to be clear, the $A,B$ here are not the same as the $A,B$ in the exercise itself.) To show that $A \subseteq B$, we suppose that $x \in A$, and then show that $x \in B$. You might be nervous about this maneuver, since we didn’t check that $A$ is non-empty. By single choice (Lemma 3.1.6) we can only say that $x \in A$ when $A$ is non-empty, right? What is going on, and does this make all of our proofs invalid? Might we need to redo all of our proofs to split into cases depending on whether each set is empty or non-empty? It turns out that all of the proofs we do here are correct as stated, without having to consider empty sets separately. But we do need to be careful; indeed, I believe Exercise 3.5.6 is designed to demonstrate why we must be careful.

So why are all of the proofs here fine as stated? To show that $A \subseteq B$, we must show that for all objects $x$, if $x \in A$ then $x \in B$. So we let $x$ be an arbitrary object, and we suppose that $x \in A$, in order to show $x \in B$. We aren’t claiming that an arbitrary object $x$ is contained in $A$, and we aren’t even claiming that $A$ is non-empty. We are just saying to assume this as a hypothesis, in order to demonstrate an implication. This is the same as trying to show an implication $P \implies Q$ and supposing that $P$ is true. If $A$ is empty, the proof is still valid since the implication is vacuous in this case.

If you are suspicious of some of the proofs here (e.g. the De Morgan laws) because they seem to rely on verbal trickery, you are right. We can’t do any better without going into the details of first-order logic, and even then it won’t be completely satisfying because a mind must already contain certain rules in order to do any mathematics at all; see here for a longer explanation.

I’m not entirely sure what Tao meant by “One can use some of these claims to prove others”; I can’t see a way to apply any of the parts to some other part in a way that saves a lot of time.

This exercise is pretty long and tedious, and it can become pretty tempting to quickly rush through it. If you don’t want to do the whole thing, and instead want to pick a couple of parts and do those carefully, then I would recommend doing parts (f) and/or (h).

The distributivity law, part (f), is probably the most interesting one to prove. My recommendation is to not appeal to the fact that logical “or” and logical “and” distribute over each other, as that fact has not been covered in the text. If you prefer, you could first show that logical “or” and logical “and” distribute over each other, after which proving distributivity for set operations becomes easy. The important point is that you show distributivity in one way or another “from first principles” rather than just assuming it.

## Model solution

(a) We have $A\cup \emptyset = A$ by Lemma 3.1.13. Thus we only have to show $A \cap \emptyset = \emptyset$. There cannot be any object $x \in A \cap \emptyset$, because if there were such an object, we would have $x \in A$ and $x \in \emptyset$, which contradicts the fact that $\emptyset$ contains no elements. Thus $A\cap \emptyset$ is empty, which means $A \cap \emptyset = \emptyset$ by the uniqueness of the empty set. (See the text following Axiom 3.2 for the uniqueness of the empty set.)

(One can also use Exercise 3.1.5 and the commutativity of the intersection operation for this part; part (d) can be proved without using part (a), so there is no circularity.)

(b) We first show $A \cup X = X$. If $x \in A \cup X$, then $x \in A$ or $x \in X$. If $x \in A$, then since $A$ is subset of $X$ we have $x \in X$ as required. If $x \in X$, then we are already done. In either case we have $x \in X$, so this shows that $A \cup X \subseteq X$. Now suppose conversely that $x \in X$. We must show that $x \in A$ or $x \in X$; but the latter statement is true, so $x \in A \cup X$, which means that $X \subseteq A\cup X$.

Next we show that $A \cap X = A$. Suppose first that $x \in A \cap X$. Then $x \in A$ and $x \in X$, so in particular $x \in A$. Conversely, suppose $x \in A$. Then since $A$ is a subset of $X$, we have $x \in X$. Thus we have both $x \in A$ and $x \in X$, which means $x \in A \cap X$ as required.

(One can also use Exercise 3.1.5 for this part, but I am doing the exercises out of order and I still haven’t done that one.)

(c) We have $A \cup A = A$ by Lemma 3.1.13. Thus we only have to show that $A \cap A = A$. Suppose $x \in A \cap A$. Then $x \in A$ and $x \in A$, so in particular $x \in A$. This shows that $A \cap A \subseteq A$. Now suppose $x \in A$. Then $x \in A$ and $x \in A$, so $x \in A \cap A$. This shows that $A \subseteq A \cap A$. Combining these two parts, we have $A\cup A = A$ as required.

(One can also use Exercise 3.1.5 for this part, but I am doing the exercises out of order and I still haven’t done that one.)

(Since $A \subseteq A$, we can also take $X=A$ and apply part (b).)

(d) We showed that the union operation is commutative in Lemma 3.1.13, so we only have to show that $A \cap B = B \cap A$. Suppose $x \in A \cap B$. Then $x \in A$ and $x \in B$. We can reorder logical “and” statements, so we have $x \in B$ and $x \in A$, which means $x \in B \cap A$ as required. The converse is proved in the same way, just switching the roles of $A$ and $B$. Thus $A \cap B = B \cap A$ as required.

(e) The union operation is associative by Lemma 3.1.13, so we only have to show that $(A \cap B) \cap C = A \cap (B \cap C)$. Suppose $x \in (A \cap B) \cap C$. This means that $x \in A \cap B$ and $x \in C$. To say that $x \in A \cap B$ means to say that $x \in A$ and $x \in B$. Since $x \in B$ and $x \in C$, this means $x \in B \cap C$. Since we also know that $x \in A$, we thus have $x \in A \cap (B \cap C)$. The converse is proved in the same way. Thus $(A \cap B) \cap C = A \cap (B \cap C)$ as required.

(f) We first show that $A \cap (B\cup C) = (A \cap B) \cup (A\cap C)$. Suppose $x \in A \cap (B \cup C)$. This means that $x \in A$ and $x \in B \cup C$. Since $x \in B \cup C$, this means that $x \in B$ or $x \in C$, so now we have two cases. If $x \in B$, then since we already know that $x \in A$, we have $x \in A \cap B$. Thus we have $x \in A \cap B$ or $x \in A \cap C$ (since the first statement is true), so $x \in (A \cap B)\cup(A \cap C)$. Alternatively if $x \in C$, then we already know $x \in A$ so $x \in A \cap C$. Thus we have $x \in A \cap B$ or $x \in A \cap C$ (since the second statement is true), which means that $x \in (A \cap B)\cup (A\cap C)$. In either case we have shown that $x \in (A \cap B)\cup (A\cap C)$, so this proves that $A \cap (B\cup C) \subseteq (A \cap B) \cup (A\cap C)$. Conversely suppose that $x \in (A \cap B)\cup(A\cap C)$. This means that either $x \in A \cap B$ or $x \in A \cap C$. In the former case, we have $x \in A$ and $x \in B$. Since $x \in B$, we have $x \in B$ or $x \in C$, which means $x \in B \cup C$. Thus we have both $x \in A$ and $x \in B \cup C$, which means $x \in A\cap (B\cup C)$. Similarly in the latter case where $x \in A \cap C$, we can show that $x \in A \cap (B\cup C)$. In either case we have $x \in A\cap (B\cup C)$, so $(A \cap B)\cup(A\cap C) \subseteq A \cap (B\cup C)$ as required.

Next we show that $A\cup (B\cap C) = (A\cup B) \cap (A \cup C)$. Suppose that $x \in A\cup (B\cap C)$. Thus we have $x \in A$ or $x \in B \cap C$. If $x \in A$, then $x \in A$ or $x \in B$ (since the first statement is true), so $x \in A \cup B$. We also have $x \in A$ or $x \in C$ (since again the first statement is true), so $x \in A\cup C$. Thus we have $x \in A\cup B$ and $x \in A \cup C$, so $x \in (A\cup B)\cap (A\cap C)$. On the other hand, if $x \in B\cap C$, then we have both $x \in B$ and $x \in C$. Thus we have $x \in A$ or $x \in B$ (since the second statement is true), so $x \in A \cup B$. Similarly, we have $x \in A$ or $x \in C$ (since again the second statement is true), so $x \in A \cup C$. We thus have both $x \in A \cup B$ and $x \in A \cup C$, so $x \in (A\cup B) \cap (A \cup C)$ as required. In both cases, we have $x \in (A\cup B) \cap (A \cup C)$, so this shows that $A\cup (B\cap C) \subseteq (A\cup B) \cap (A \cup C)$. Conversely suppose $x \in (A\cup B) \cap (A \cup C)$. Then we have both $x \in A\cup B$ and $x \in A\cup C$. Now we have two cases, $x \in A$ or $x \notin A$. If $x \in A$, then $x \in A$ or $x \in B\cap C$ (since the first statement is true), so $x \in A \cup (B\cap C)$. On the other hand, if $x \notin A$, then since $x \in A\cup B$ we have $x \in A$ or $x \in B$, and since the first option is impossible we conclude that $x \in B$. Similarly we have $x \in C$ since $x \in A \cup C$ and $x\notin A$. Thus in this case $x \in B\cap C$, so we have $x \in A$ or $x \in B\cap C$ (since the second statement is true), which means $x \in A \cup (B\cap C)$. In either case we have shown that $x \in A \cup (B\cap C)$, so this shows that $(A\cup B) \cap (A \cup C) \subseteq A \cup (B\cap C)$.

(g) We first show that $A \cup (X \setminus A) = X$. Suppose $x \in A \cup (X \setminus A)$. Then $x \in A$ or $x \in X \setminus A$, so we have two cases. If $x \in A$, then $x \in X$ since $A$ is a subset of $X$. On the other hand, if $x \in X \setminus A$, then we have $x \in X$ and $x \notin A$, so in particular $x \in X$. In either case we have $x \in X$, so $A \cup (X \setminus A) \subseteq X$. Conversely suppose that $x \in X$. We have two cases, either $x \in A$ or $x \notin A$. If $x \in A$ then $x \in A$ or $x \in X \setminus A$ (since the first statement is true), so $x \in A \cup (X \setminus A)$ as required. If $x \notin A$, then since $x \in X$ this means $x \in X \setminus A$. Thus $x \in A$ or $x \in X \setminus A$ (since the second statement is true), so $x \in A \cup (X \setminus A)$. In either case we have $x \in A \cup (X \setminus A)$, so $X \subseteq A \cup (X \setminus A)$.

Next we show that $A \cap (X \setminus A) = \emptyset$. To do this, we shall show that $A \cap (X \setminus A)$ is empty; since the empty set is unique this will be sufficient. We want to show that there is no object $x$ such that $x \in A \cap (X \setminus A)$. Suppose for sake of contradiction that such an object $x$ exists. Then we have $x \in A$ and $x \in X \setminus A$. The latter statement means $x \in X$ and $x \notin A$. Thus we have both $x \in A$ and $x \notin A$, a contradiction.

(h) First we show that $X \setminus (A \cup B) = (X \setminus A) \cap (X \setminus B)$. Suppose $x \in X \setminus (A \cup B)$. This means $x \in X$ and $x \notin A \cup B$. The latter statement means that it is not the case that $x\in A$ or $x \in B$; thus $x \notin A$ and $x \notin B$. But now we have $x \in X$ and $x \notin A$, which means $x \in X \setminus A$, and we also have both $x \in X$ and $x \notin B$, so $x \in X \setminus B$. Thus we have $x \in (X \setminus A) \cap (X\setminus B)$ as required. Conversely, suppose $x \in (X \setminus A) \cap (X \setminus B)$. Thus we have both $x \in X \setminus A$ and $x \in X\setminus B$. The first of these decomposes to $x \in X$ and $x \notin A$, and the second decomposes to $x \in X$ and $x \notin B$. To say that $x \notin A$ and $x \notin B$ is to say that $x$ is not in either of these sets, so $x \notin A\cup B$. Since we also know $x \in X$, this means $x \in X \setminus (A \cup B)$.

Next we show that $X\setminus (A \cap B) = (X \setminus A) \cup (X \setminus B)$. Suppose first that $x \in X\setminus (A \cap B)$. This means $x \in X$ and $x \notin A \cap B$. The latter statement means that $x$ is not in both of the sets $A$ and $B$, so it is either not in $A$ or not in $B$, i.e. $x \notin A$ or $x \notin B$. We now have two cases. If $x \notin A$ then since we already know $x \in X$, we have $x \in X \setminus A$. Thus $x \in X \setminus A$ or $x \in X \setminus B$ (since the first statement is true), which means $x \in (X \setminus A)\cup(X \setminus B)$. On the other hand, if $x \notin B$ then we similarly have $x \in (X \setminus A)\cup(X \setminus B)$. Thus $X\setminus (A \cap B) \subseteq (X \setminus A) \cup (X \setminus B)$. Conversely suppose $x \in (X \setminus A) \cup (X \setminus B)$. Then we have either $x \in X \setminus A$ or $x \in X \setminus B$. In the first case, we have both $x \in X$ and $x \notin A$. If $x \in A\cap B$ then we would have $x \in A$, a contradiction, so we must have $x \notin A\cap B$. Thus we have $x \in X$ and $x \notin A\cap B$, which means $x \in X \setminus (A \cap B)$. In the second case, we can similarly show that $x \in X \setminus (A \cap B)$. In either case we have $x \in X \setminus (A \cap B)$ so $(X \setminus A) \cup (X \setminus B) \subseteq X \setminus (A \cap B)$.

# Exercise 6.1.7

## Exercise statement

Show that Definition 6.1.16 is consistent with Definition 5.1.12 (i.e., prove an analogue of Proposition 6.1.4 for bounded sequences instead of Cauchy sequences).

Definition 6.1.16 (Bounded sequences). A sequence $(a_n)_{n=m}^\infty$ of real numbers is bounded by a real number $M$ iff we have $|a_n| \leq M$ for all $n \geq m$. We say that $(a_n)_{n=m}^\infty$ is bounded iff it is bounded by $M$ for some real number $M > 0$.

Definition 5.1.12 (Bounded sequences). Let $M \geq 0$ be rational. A finite sequence $a_1, a_2, \ldots, a_n$ is bounded by $M$ iff $|a_i| \leq M$ for all $1 \leq i \leq n$. An infinite sequence $(a_n)_{n=1}^\infty$ is bounded by $M$ iff $|a_i| \leq M$ for all $i \geq 1$. A sequence is said to be bounded iff it is bounded by $M$ for some rational $M \geq 0$.

None.

## How to think about the exercise

The finite sequence part of Definition 5.1.12 isn’t relevant to Definition 6.1.16, so you can just ignore it.

The “bounded by” part is silly, and I don’t think you really need to do it. If we filter to the case when both definitions make sense (i.e. rational sequence, positive rational $M$), then both definitions are exactly the same, so there is nothing to prove. I’ve included this part in the solution anyway.

If you find this exercise confusing, really read and understand the proof of Proposition 6.1.4 given in the book.

## Model solution

Let $(a_n)_{n=m}^\infty$ be a sequence of rational numbers, and let $M > 0$ be rational. If $(a_n)_{n=m}^\infty$ is bounded by $M$ in the sense of Definition 6.1.16, then $|a_n| \leq M$ for all $n \geq m$. But this is exactly what it means for $(a_n)_{n=m}^\infty$ to be bounded by $M$ in the sense of Definition 5.1.12. The converse can be proved in the same way. So the definition of “bounded by” is consistent between the two definitions.

Now suppose $(a_n)_{n=m}^\infty$ is bounded in the sense of Definition 6.1.16. Thus there is some real number $M > 0$ such that $|a_n| \leq M$ for all $n \geq m$. To show that $(a_n)_{n=m}^\infty$ is bounded in the sense of Definition 5.1.12, we must find a rational number $M' \geq 0$ such that $|a_n| \leq M'$ for all $n \geq m$. By the Archimedean property, we can find a positive integer $M'$ such that $M' > M$. Every integer is also a rational number, so $M'$ is rational. Now we have $|a_n| \leq M < M'$ for all $n \geq m$ as required.

Conversely, suppose $(a_n)_{n=m}^\infty$ is bounded in the sense of Definition 5.1.12. Then we have a rational number $M \geq 0$ such that $|a_n| \leq M$ for all $n \geq m$. To show that the sequence is bounded in the sense of Definition 6.1.16, we must find a positive real number $M' > 0$ such that $|a_n| \leq M'$. Take $M' := M+1$, which is a real number. Since $M\geq 0$, we have $M' = M+1 \geq 1 > 0$ so $M'$ is positive. We also have $|a_n| \leq M < M+1 = M'$ for all $n \geq m$. Thus $(a_n)_{n=m}^\infty$ is bounded in the sense of Definition 6.1.16.

# Exercise 6.1.6

## Exercise statement

Prove Proposition 6.1.15, using the following outline. Let $(a_n)_{n=m}^\infty$ be a Cauchy sequence of rationals, and write $L := \mathrm{LIM}_{n\to\infty} a_n$. We have to show that $(a_n)_{n=m}^\infty$ converges to $L$. Let $\varepsilon > 0$. Assume for sake of contradiction that the sequence $(a_n)_{n=m}^\infty$ is not eventually $\varepsilon$-close to $L$. Use this, and the fact that $(a_n)_{n=m}^\infty$ is Cauchy, to show that there is an $N \geq m$ such that either $a_n > L + \varepsilon/2$ for all $n\geq N$, or $a_n < L -\varepsilon/2$ for all $n \geq N$. Then use Exercise 5.4.8.

Proposition 6.1.15 (Formal limits are genuine limits). Suppose that $(a_n)_{n=1}^\infty$ is a Cauchy sequence of rational numbers. Then $(a_n)_{n=1}^\infty$ converges to $\mathrm{LIM}_{n\to\infty} a_n$, i.e.

$\displaystyle \mathrm{LIM}_{n\to\infty} a_n = \lim_{n\to\infty} a_n$.

None.

## How to think about the exercise

Minor note: the starting index of the sequence is different between the proposition statement and the exercise. This doesn’t really matter, as we already showed in Exercise 6.1.3.

## Model solution

Let $(a_n)_{n=m}^\infty$ be a Cauchy sequence of rationals, and write $L := \mathrm{LIM}_{n\to\infty} a_n$. We have to show that $(a_n)_{n=m}^\infty$ converges to $L$. Assume for sake of contradiction that $(a_n)_{n=m}^\infty$ does not converge to $L$. This means that there is some $\varepsilon > 0$ such that $(a_n)_{n=m}^\infty$ is not eventually $\varepsilon$-close to $L$. In other words, there is some $\varepsilon > 0$ such that for every $N \geq m$, there is some $n \geq N$ such that $|a_n - L| > \varepsilon$. So choose this $\varepsilon > 0$. Since $(a_n)_{n=m}^\infty$ is Cauchy, there is some $N \geq m$ such that $|a_n - a_k| \leq \varepsilon/2$ for all $n,k \geq N$; fix this $N$. Since $(a_n)_{n=m}^\infty$ is not eventually $\varepsilon$-close to $L$, for this $N$ in particular, there is some $n_0 \geq N$ such that $|a_{n_0} - L| > \varepsilon$; fix this $n_0$. Since $n_0\geq N$, we can fix $k$ above to be $n_0$, so that $|a_n - a_{n_0}| \leq \varepsilon/2$ for all $n \geq N$. We can rewrite this using Exercise 5.4.6 so that $a_{n_0}-\varepsilon/2 \leq a_n \leq a_{n_0}+\varepsilon/2$ for all $n \geq N$. Now we have two cases, depending on the sign of $a_{n_0} - L$.

If $a_{n_0} - L$ is non-negative, then $a_{n_0} - L = |a_{n_0} - L| > \varepsilon$. We also have $a_{n_0} -\varepsilon/2 \leq a_n \leq a_{n_0}+\varepsilon/2$ for all $n \geq N$, so in particular $a_{n_0}-\varepsilon/2 \leq a_n$ for all $n \geq N$. Adding this inequality to the inequality $\varepsilon < a_{n_0} - L$ we obtain $a_{n_0} + \varepsilon/2 < a_{n_0} + a_n - L$. This simplifies to $a_n > L+\varepsilon/2$, and holds for all $n \geq N$. By Exercise 5.4.8 (formally, we replace $(a_n)_{n=m}^\infty$ by an equivalent sequence $(b_n)_{n=m}^\infty$ by changing the finite start of the sequence so that the inequality $b_n > L+\varepsilon/2$ holds for all $n \geq m$), this implies $L \geq L + \varepsilon/2$, a contradiction since $\varepsilon/2$ is positive.

If $a_{n_0} - L$ is negative, then $-(a_{n_0} - L) = |a_{n_0} - L| > \varepsilon$, i.e. $\varepsilon < -a_{n_0} + L$. We also have $a_{n_0}-\varepsilon/2 \leq a_n \leq a_{n_0}+\varepsilon/2$ for all $n \geq N$, so in particular $a_n \leq a_{n_0}+\varepsilon/2$ for all $n \geq N$. Adding these two inequalities, we have $\varepsilon + a_n < L + \varepsilon/2$, i.e. $a_n < L - \varepsilon/2$, which holds for all $n \geq N$. Using Exercise 5.4.8 we obtain $L < L - \varepsilon/2$, a contradiction.

Both possibilities lead to a contradiction, so $(a_n)_{n=m}^\infty$ must converge to $L$.

# Exercise 6.1.5

## Exercise statement

Prove Proposition 6.1.12.

Proposition 6.1.12 (Convergent sequences are Cauchy). Suppose that $(a_n)_{n=m}^\infty$ is a convergent sequence of real numbers. Then $(a_n)_{n=m}^\infty$ is also a Cauchy sequence.

## Hints

1. Use the triangle inequality, or Proposition 4.3.7.

## How to think about the exercise

This is a straightforward exercise, so I don’t have anything to say.

## Model solution

Let $(a_n)_{n=m}^\infty$ be a convergent sequence of real numbers. We want to show that $(a_n)_{n=m}^\infty$ is Cauchy, so let $\varepsilon > 0$ be a real number. We must find an $N \geq m$ such that $|a_j - a_k| \leq \varepsilon$ for all $j,k \geq N$. Since $(a_n)_{n=m}^\infty$ is convergent, it converges to some real number $L$, and since $\varepsilon/2$ is positive, we have some $N \geq m$ such that $|a_n - L| \leq \varepsilon/2$ for all $n \geq N$. Choose this $N$. Then if $j,k \geq N$, we have both $|a_j - L| \leq \varepsilon/2$ and $|a_k - L| \leq \varepsilon/2$. By the triangle inequality, we have $d(a_j,a_k) \leq d(a_j,L) + d(L,a_k)$. Thus we have $|a_j - a_k| \leq |a_j-L| + |a_k -L| \leq \varepsilon/2 + \varepsilon/2 = \varepsilon$. In other words, for all $j,k \geq N$ we have $|a_j - a_k|\leq \varepsilon$ as required.

# Exercise 6.1.4

## Exercise statement

Let $(a_n)_{n=m}^\infty$ be a sequence of real numbers, let $c$ be a real number, and let $k \geq 0$ be a non-negative integer. Show that $(a_n)_{n=m}^\infty$ converges to $c$ if and only if $(a_{n+k})_{n=m}^\infty$ converges to $c$.

None.

## How to think about the exercise

If this exercise seems confusing to you, I would encourage you to define $b_n := a_{n+k}$ and then show that $(a_n)_{n=m}^\infty$ converges to $c$ if and only if $(b_n)_{n=m}^\infty$ converges to $c$.

## Model solution

Suppose first that $(a_n)_{n=m}^\infty$ converges to $c$. We want to show that $(a_{n+k})_{n=m}^\infty$ converges to $c$, so let $\varepsilon > 0$ be a real number. Since $(a_n)_{n=m}^\infty$ converges to $c$, there is some $N \geq m$ such that $|a_n - c| \leq \varepsilon$ for all $n \geq N$. We want to find an $N' \geq m$ such that $|a_{n+k} - c|\leq\varepsilon$ for all $n \geq N'$. Choose $N' := N$. If $n \geq N' = N$, then $n+k \geq n \geq N$, so we have $|a_{n+k} - c| \leq \varepsilon$. Thus $(a_{n+k})_{n=m}^\infty$ converges to $c$.

Conversely suppose that $(a_{n+k})_{n=m}^\infty$ converges to $c$. We want to show $(a_n)_{n=m}^\infty$ converges to $c$, so let $\varepsilon > 0$ be a real number. Since $(a_{n+k})_{n=m}^\infty$ converges to $c$, we have an $N \geq m$ such that $|a_{n+k} - c|\leq \varepsilon$ for all $n \geq N$. We want to find an $N' \geq m$ such that $|a_n - c|\leq \varepsilon$ for all $n \geq N'$. Pick $N' := N+k$. If $n \geq N' = N+k$, then $n-k \geq N$ so we have $|a_{(n-k)+k} - c| \leq \varepsilon$. But $a_{(n-k)+k} = a_n$ so this means $|a_n - c| \leq \varepsilon$. Thus $(a_n)_{n=m}^\infty$ converges to $c$ as required.

# Exercise 6.1.3

## Exercise statement

Let $(a_n)_{n=m}^\infty$ be a sequence of real numbers, let $c$ be a real number, and let $m' \geq m$ be an integer. Show that $(a_n)_{n=m}^\infty$ converges to $c$ if and only if $(a_n)_{n=m'}^\infty$ converges to $c$.

None.

## How to think about the exercise

This is a straightforward exercise, so I don’t have anything to say.

## Model solution

Suppose first that $(a_n)_{n=m}^\infty$ converges to $c$. We want to show that $(a_n)_{n=m'}^\infty$ converges to $c$, so let $\varepsilon > 0$ be a real number. Since $(a_n)_{n=m}^\infty$ converges to $c$, we know that there is some $N \geq m$ such that $(a_n)_{n=N}^\infty$ is $\varepsilon$-close to $c$. Define $N' := N+m'$. Then $N' \geq m'$, and if $n\geq N'$ then $n \geq N$ so $|a_n - c|\leq \varepsilon$. Thus $(a_n)_{n=N'}^\infty$ is $\varepsilon$-close to $c$, which means that $(a_n)_{n=m'}^\infty$ is eventually $\varepsilon$-close to $c$. Since $\varepsilon > 0$ was arbitrary, this shows that $(a_n)_{n=m'}^\infty$ converges to $c$, as desired.

Conversely suppose that $(a_n)_{n=m'}^\infty$ converges to $c$. We want to show that $(a_n)_{n=m}^\infty$ converges to $c$, so let $\varepsilon > 0$ be a real number. We know that $(a_n)_{n=m'}^\infty$ is eventually $\varepsilon$-close to $c$, so there is an $N\geq m'$ such that $(a_n)_{n=N}^\infty$ is $\varepsilon$-close to $c$. We have $N\geq m' \geq m$, so this same $N$ will work: $(a_n)_{n=m}^\infty$ converges to $c$ as desired.

# Exercise 6.1.2

## Exercise statement

Let $(a_n)_{n=m}^\infty$ be a sequence of real numbers, and let $L$ be a real number. Show that $(a_n)_{n=m}^\infty$ converges to $L$ if and only if, given any real $\varepsilon > 0$, one can find an $N \geq m$ such that $|a_n - L| \leq \varepsilon$ for all $n \geq N$.

None.

## How to think about the exercise

In most real analysis texts, the result of this exercise is the definition of convergence. This means that this exercise is only an exercise because Tao’s definition of convergence “wraps up” all the concepts behind terminology like eventual $\varepsilon$-closeness.

I think this can be a pretty tricky exercise, even though it’s very simple. It can seem like we’re just shuffling symbols in a meaningless way. But there is meaning.

## Model solution

Suppose that $(a_n)_{n=m}^\infty$ converges to $L$. Let $\varepsilon > 0$ be a real number. Since $(a_n)_{n=m}^\infty$ converges to $L$, we see that $(a_n)_{n=m}^\infty$ is eventually $\varepsilon$-close to $L$. This means that there exists an $N \geq m$ such that $(a_n)_{n=N}^\infty$ is $\varepsilon$-close to $L$. But saying $(a_n)_{n=N}^\infty$ is $\varepsilon$-close to $L$ is the same thing as saying that $|a_n - L| \leq \varepsilon$ for every $n \geq N$. In other words, we have found an $N \geq m$ such that $|a_n - L| \leq \varepsilon$ for all $n \geq N$, as desired.

Conversely suppose that given any real $\varepsilon > 0$, one can find an $N \geq m$ such that $|a_n - L| \leq \varepsilon$ for all $n \geq N$. We want to show that $(a_n)_{n=m}^\infty$ converges to $L$. Let $\varepsilon > 0$ be a real number. We must show that $(a_n)_{n=m}^\infty$ is eventually $\varepsilon$-close to $L$, i.e. we must find some $N \geq m$ such that $(a_n)_{n=N}^\infty$ is $\varepsilon$-close to $L$. We know that we can find an $N \geq m$ such that $|a_n - L| \leq \varepsilon$ for all $n \geq N$, so pick this $N$. We will show that $(a_n)_{n=N}^\infty$ is $\varepsilon$-close to $L$. Indeed, if $n \geq N$, we have $|a_n - L| \leq \varepsilon$, as desired.

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