Solutions

Exercise 2.2.4

March 31, 2020 Issa Rice 10 Comments

Exercise statement

Justify the three statements marked (why?) in the proof of Proposition 2.2.13.

Hints

You might want to use Proposition 2.2.12.

How to think about the exercise

This is a simple exercise, so I don’t have any comments.

Model solution

When $a=0$ we have $0\leq b$ for all $b$ : We want to show that $b = 0+n$ for some $n$ . Take $n:=b$ . Then $0+n = 0+b = b$ by Definition 2.2.1.

If $a > b$ , then $a{{+}\!{+}} > b$ : To show $a{{+}\!{+}} > b$ , we can instead show $a{{+}\!{+}} \geq b{{+}\!{+}}$ by Proposition 2.2.12(e). Since $a > b$ we have $a \geq b$ , so by Proposition 2.2.12(d) we have $a{{+}\!{+}} = a+1 \geq b+1 = b{{+}\!{+}}$ .

If $a=b$ , then $a{{+}\!{+}} > b$ : Since $a=b$ , we want to show that $a{{+}\!{+}} > a$ . By Proposition 2.2.12(e) it suffices to show $a{{+}\!{+}} \geq a{{+}\!{+}}$ , but this is true since $a{{+}\!{+}} = (a{{+}\!{+}}) + 0$ .

Solutions

Exercise 2.2.3

March 30, 2020 Issa Rice 19 Comments

Exercise statement

Prove Proposition 2.2.12.

Proposition 2.2.12 (Basic properties of order for natural numbers). Let $a,b,c$ be natural numbers. Then

(a) (Order if reflexive) $a \geq a$ .
(b) (Order is transitive) If $a \geq b$ and $b \geq c$ , then $a\geq c$ .
(c) (Order is anti-symmetric) If $a\geq b$ and $b\geq a$ , then $a=b$ .
(d) (Addition preserves order) $a \geq b$ if and only if $a+c \geq b+c$ .
(e) $a<b$ if and only if $a{{+}\!{+}} \leq b$ .
(f) $a<b$ if and only if $b=a+d$ for some positive number $d$ .

Hints

You will need many of the preceding propositions, corollaries, and lemmas.

How to think about the exercise

Each part of the exercise is pretty straightforward. It is tiring to write down everything, but at this stage it is good for your soul.

Model solution

(a) To show that $a\leq a$ we must show that $a = a + m$ for some natural number $m$ . Pick $m:=0$ . Then by Lemma 2.2.2 we see that $a+m = a+0 = a$ as required.

(b) Suppose $a \geq b$ and $b \geq c$ , thus there are natural numbers $n,m$ such that $a = b+n$ and $b=c+m$ . We want to show that $a = c + r$ for some natural number $r$ . Pick $r := m+n$ . Then $c+r = c + (m+n) = (c+m) + n = b+n =a$ by the associativity of addition (Proposition 2.2.5).

(c) Suppose $a\geq b$ and $b\geq a$ , thus there are natural numbers $n,m$ such that $a = b+n$ and $b = a+m$ . Thus we see that $a = (a+m)+n = a + (m+n)$ by associativity of addition (Proposition 2.2.5). By the cancellation law (Proposition 2.2.6) we see that $0 = m+n$ and so by Corollary 2.2.9 we conclude that $m=0$ and $n=0$ . Thus $a = b+n = b + 0 = b$ by Lemma 2.2.2.

(d) Suppose $a \geq b$ , so that we have a natural number $n$ such that $a = b+n$ . By adding $c$ to both sides, we obtain $a+c = (b+n)+c$ . By associativity and commutativity of addition, we have $(b+n)+c = b + (n+c) = b+(c+n) = (b+c)+n$ . Thus $a+c = (b+c) + n$ , which means $a+c \geq b+c$ .

Now suppose $a+c \geq b+c$ , thus $a+c = (b+c) + n$ for some natural number $n$ . We have $(b+c) + n = b+(c+n) = b+(n+c) = (b+n)+c$ by associativity and commutativity of addition. Thus $a+c = (b+n)+c$ , so by cancellation we have $a = b+n$ , which means $a \geq b$ .

(e) Suppose $a<b$ . Thus $a \leq b$ , which means $b = a+n$ for some natural number $n$ , and we also know that $a\ne b$ . If $n=0$ , then $b=a+0=a$ which would contradict the fact that $a\ne b$ . Thus $n$ is positive. By Lemma 2.2.10 this means that there is a natural number $m$ such that $m{{+}\!{+}} = n$ . But now we have

$\begin{aligned}b &= a + (m{{+}\!{+}}) \\ &= (a+m){{+}\!{+}} \\ &= (m+a){{+}\!{+}} \\ &= m+(a{{+}\!{+}}) \\& = (a{{+}\!{+}}) + m\end{aligned}$

by repeated applications of Lemma 2.2.3 and Proposition 2.2.4. Thus we see that $b \geq a{{+}\!{+}}$ .

Now suppose $a{{+}\!{+}} \leq b$ . Thus $b = (a{{+}\!{+}}) + n$ for some natural number $n$ . We have $(a{{+}\!{+}}) + n = (a+n){{+}\!{+}} = a+ (n{{+}\!{+}})$ by Definition 2.2.1 and Lemma 2.2.3. Thus $b = a+ (n{{+}\!{+}})$ , which means that $b \geq a$ . If $b = a$ , then by cancellation we would have $0 = n{{+}\!{+}}$ , which contradicts Axiom 2.3. Thus $b \ne a$ , which combined with $b\geq a$ means that $b > a$ as required.

(f) Suppose $a<b$ . In the first part of the proof of part (e), we already showed that $b = a+n$ for positive $n$ .

Now suppose $b=a+d$ for some positive number $d$ . Since $d$ is a natural number, we have $b \geq a$ . It remains to show that $a\ne b$ . Suppose for contradiction that $a=b$ . Then by cancellation we have $0=d$ , which means $d$ is not positive, a contradiction. Thus $a\ne b$ as required.

Solutions

Exercise 5.4.1

March 29, 2020 Issa Rice 4 Comments

Exercise statement

Prove Proposition 5.4.4.

Proposition 5.4.4 (Basic properties of positive reals). For every real number $x$ , exactly one of the following three statements is true: (a) $x$ is zero; (b) $x$ is positive; (c) $x$ is negative. A real number $x$ is negative if and only if $-x$ is positive. If $x$ and $y$ are positive, then so are $x+y$ and $xy$ .

Hints

If $x$ is not zero, and $x$ is the formal limit of some sequence $(a_n)_{n=1}^\infty$ , then this sequence cannot be eventually $\varepsilon$ -close to the zero sequence $(0)_{n=1}^\infty$ for every single $\varepsilon > 0$ . Use this to show that the sequence $(a_n)_{n=1}^\infty$ is eventually either positively bounded away from zero or negatively bounded away from zero.

How to think about the exercise

This is a pretty involved exercise, and just writing up the proof is pretty tiring, so I’m struggling to come up with things to say. I suggest drawing a picture.

Bonus question: which part of the exercise is the hint talking about?

Model solution

We first show that at most one of (a), (b), (c) is true.

$x$ cannot be both zero and positive: Suppose $x$ is positive, so that $x = \mathrm{LIM}_{n\to\infty} a_n$ for some Cauchy sequence $(a_n)_{n=1}^\infty$ which is positively bounded away from zero. Thus there is some $c >0$ such that $a_n \geq c$ for all $n \geq 1$ . This means that $|a_n - 0| > c/2$ , so $(a_n)_{n=1}^\infty$ and $(0)_{n=1}^\infty$ are not eventually $c/2$ -close, so these two sequences are not equivalent. Thus $x \ne 0$ .
$x$ cannot be both zero and negative: This is similar to the above case. If $x$ is negative, then $x = \mathrm{LIM}_{n\to\infty} a_n$ for some Cauchy sequence $(a_n)_{n=1}^\infty$ which is negatively bounded away from zero. Thus $a_n \leq -c$ for some $-c < 0$ . This means $|a_n - 0| = -a_n > c/2$ so $(a_n)_{n=1}^\infty$ is not equivalent to the zero sequence, and so $x \ne 0$ .
$x$ cannot be both positive and negative: Suppose for sake of contradiction that $x$ is both positive and negative. Thus $x = \mathrm{LIM}_{n\to\infty} a_n$ for some Cauchy sequence $(a_n)_{n=1}^\infty$ which is positively bounded away from zero, and $x = \mathrm{LIM}_{n\to\infty} b_n$ for some Cauchy sequence $(b_n)_{n=1}^\infty$ which is negatively bounded away from zero. Thus we have positive rationals $c,d > 0$ such that $a_n \geq c > 0$ and $b_n \leq -d < 0$ for all $n\geq 1$ . Since $a_n$ is positive and $b_n$ is negative, we see that $a_n - b_n$ is positive, thus $|a_n - b_n| = a_n - b_n$ . Adding the inequalities above we get $|a_n-b_n| = a_n - b_n \geq c+d > (c+d)/2$ . Thus the sequences $(a_n)_{n=1}^\infty$ and $(b_n)_{n=1}^\infty$ are not eventually $(c+d)/2$ -close, which contradicts the fact that they are equivalent sequences.

Next we show that at least one of (a), (b), (c) is true. To do this we will show that if $x \ne 0$ then $x$ is either positive or negative. Suppose $x$ is not zero. By Lemma 5.3.14 we know that $x$ can be written as $x = \mathrm{LIM}_{n\to\infty} a_n$ for some Cauchy sequence $(a_n)_{n=1}^\infty$ which is bounded away from zero, i.e. we have some rational $c > 0$ such that $|a_n| \geq c$ for all $n \geq 1$ . Since $(a_n)_{n=1}^\infty$ is Cauchy, eventually the sequence is $c/2$ -steady. Call this point $N$ so that $|a_j - a_N| \leq c/2$ for all $j\geq N$ . Since the sequence $(a_n)_{n=1}^\infty$ is bounded away from zero, $a_N$ cannot be zero, so it is either positive or negative.

If $a_N > 0$ , then $a_N = |a_N| \geq c$ . From $|a_j - a_N| \leq c/2$ we have in particular $a_N - a_j \leq c/2$ (by Proposition 4.3.3(c)) which means $a_N - c/2 \leq a_j$ ; now using $c \leq a_N$ we have $0 < c/2 \leq a_N - c/2 \leq a_j$ , which holds for all $j \geq N$ . This shows that $(a_n)_{n=N}^\infty$ is positively bounded away from zero. This is almost what we need, except that as in the proof of Lemma 5.3.14, we cannot guarantee that the start of the sequence is above our $c/2$ threshold. To fix this, define a new sequence $b_n := c/2$ if $n < N$ and $b_n := a_n$ if $n\geq N$ . This new sequence is positively bounded away from zero, and is equivalent to $(a_n)_{n=1}^\infty$ . Thus $x$ is positive.
If $a_N < 0$ then $-a_N = |a_N| \geq c$ so $a_N \leq -c$ . By a reasoning similar to the previous case, we can show that $a_j \leq -c/2$ for all $j \geq N$ , and thus show that $x$ is negative.

Next we show that $x$ is negative if and only if $-x$ is positive. Suppose $x$ is negative. Thus $x = \mathrm{LIM}_{n\to\infty} a_n$ for some sequence $(a_n)_{n=1}^\infty$ which is negatively bounded away from zero, which means that we have a rational $-c < 0$ such that $a_n \leq -c$ for all $n \geq 1$ . By Proposition 4.2.9(d) we can add $-a_n+c$ to both sides of the inequality to get $c \leq -a_n$ , which holds for all $n\geq 1$ . Since $c$ is positive, this means that the sequence $(-a_n)_{n=1}^\infty$ is positively bounded away from zero, so $-x = \mathrm{LIM}_{n\to\infty} (-a_n)$ is positive. The converse can be proven in the same way.

Finally we show that if $x$ and $y$ are positive, then so are $x+y$ and $xy$ . Suppose $x$ and $y$ are positive, so that we can write them as $x = \mathrm{LIM}_{n\to\infty} a_n$ and $y = \mathrm{LIM}_{n\to\infty} b_n$ for some Cauchy sequences $(a_n)_{n=1}^\infty$ and $(b_n)_{n=1}^\infty$ which are positively bounded away from zero. This means that there are positive rational numbers $c,d > 0$ such that $a_n \geq c$ and $b_n \geq d$ for all $n \geq 1$ . By Proposition 4.2.9 parts (d) and (e) we see that $a_n + b_n \geq a_n + d \geq c+d$ and $a_n b_n \geq a_n d \geq cd$ for all $n \geq 1$ . Since $c+d$ and $cd$ are positive, this means the sequences $(a_n + b_n)_{n=1}^\infty$ and $(a_n b_n)_{n=1}^\infty$ are positively bounded away from zero. Thus $x+y$ and $xy$ are positive.

Solutions

Exercise 5.3.5

March 28, 2020 Issa Rice Leave a comment

Exercise statement

Show that $\mathrm{LIM}_{n\to\infty} 1/n = 0$ .

Hints

You might find Proposition 4.4.1 and Proposition 4.3.12 helpful.

How to think about the exercise

This is a simple exercise, so I don’t have any comments.

Model solution

We want to show that $(1/n)_{n=1}^\infty$ is equivalent to $(0)_{n=1}^\infty$ , i.e. we want to show that for every rational $\varepsilon > 0$ there exists $N\geq 1$ such that for all $n\geq N$ we have $|1/n - 0| \leq \varepsilon$ .

Let a rational $\varepsilon > 0$ be given. By Proposition 4.4.1 there exists some natural number $N$ such that $N > 1/\varepsilon$ . (Since $1/\varepsilon > 0$ , we see that $N$ cannot be zero, so its reciprocal is defined.) Now let $n \geq N$ . Then we have $n\geq 1/\varepsilon > 0$ so by Proposition 4.3.12(b) we have $0 < 1/n \leq \varepsilon$ . Since $|1/n - 0| = 1/n$ , this completes the proof.

Solutions

Exercise 5.3.4

March 27, 2020 Issa Rice 3 Comments

Exercise statement

Let $(a_n)_{n=1}^\infty$ be a sequence of rational numbers which is bounded. Let $(b_n)_{n=1}^\infty$ be another sequence of rational numbers which is equivalent to $(a_n)_{n=1}^\infty$ . Show that $(b_n)_{n=1}^\infty$ is also bounded.

Hints

Use Exercise 5.2.2.

How to think about the exercise

This is a simple exercise, so I don’t have any comments.

Model solution

Since $(a_n)_{n=1}^\infty$ and $(b_n)_{n=1}^\infty$ are equivalent, they are eventually $\varepsilon$ -close for every rational $\varepsilon > 0$ . Thus for $\varepsilon:=1$ in particular, $(a_n)_{n=1}^\infty$ and $(b_n)_{n=1}^\infty$ are eventually $1$ -close. By Exercise 5.2.2, we thus see that $(a_n)_{n=1}^\infty$ is bounded if and only if $(b_n)_{n=1}^\infty$ is bounded. Since $(a_n)_{n=1}^\infty$ is bounded by hypothesis, we see that $(b_n)_{n=1}^\infty$ is bounded as required.

Solutions

Exercise 5.3.3

March 27, 2020 Issa Rice 1 Comment

Exercise statement

Let $a,b$ be rational numbers. Show that $a=b$ if and only if $\mathrm{LIM}_{n\to\infty} a = \mathrm{LIM}_{n\to\infty} b$ (i.e., the Cauchy sequences $a,a,a,a,\ldots$ and $b,b,b,b, \ldots$ are equivalent if and only if $a=b$ ). This allows us to embed the rational numbers inside the real numbers in a well-defined manner.

Hints

For the “if” direction, prove by contradiction or contrapositive.

How to think about the exercise

I don’t have much to say; this is a pretty simple exercise. One comment I’ll make is that since we are working with the “equivalent sequences” definition, we have to deal with the three nested layers of quantifiers ( $\forall \varepsilon > 0\ \exists N\geq 1\ \forall n\geq N$ ). But here, our sequences don’t actually depend on $n$ . This makes it tempting to skip all the work of dealing with the quantifiers, but I think that’s wrong; we still have to unroll all the quantifiers.

Model solution

Suppose $a=b$ . Then given $\varepsilon > 0$ , pick $N:=1$ . So if $n \geq N$ we have $|a-b| = 0 < \varepsilon$ , which shows that $(a)_{n=1}^\infty$ and $(b)_{n=1}^\infty$ are equivalent.

Now suppose $a\ne b$ . We want to show that $(a)_{n=1}^\infty$ and $(b)_{n=1}^\infty$ are not equivalent. This means that we must find an $\varepsilon > 0$ such that for all $N \geq 1$ there exists $n \geq N$ such that $|a-b| > \varepsilon$ . Choose $\varepsilon := |a-b|/2$ , which is positive since $a\ne b$ . Let $N\geq 1$ . Then for $n = N$ we have $|a-b| > |a-b|/2 = \varepsilon$ .

Solutions

Exercise 5.3.2

March 26, 2020 Issa Rice 9 Comments

Exercise statement

Prove Proposition 5.3.10.

Proposition 5.3.10 (Multiplication is well defined). Let $x = \mathrm{LIM}_{n\to\infty} a_n$ , $y = \mathrm{LIM}_{n\to\infty} b_n$ , and $x' = \mathrm{LIM}_{n\to\infty} a'_n$ be real numbers. Then $xy$ is also a real number. Furthermore, if $x=x'$ , then $xy=x'y$ .

Hints

Proposition 4.3.7 may be useful.

How to think about the exercise

I think the tricky part about this exercise is choosing the right $\varepsilon$ (called $\varepsilon'$ in model solution 1) and $\delta$ for Proposition 4.3.7. Two general principles I have are:

Work backwards from the final expression, and don’t worry about all of the nested quantifiers until you are ready to write up the formal proof.
Use the $\min$ operator to require multiple constraints, so that you can avoid dealing with messy algebra.

How does this work on this exercise? Well, eventually I get to something like $\varepsilon'M_2 + \delta M_1 + \varepsilon'\delta$ and I want this to be at most $\varepsilon$ . How would I pick the right $\varepsilon', \delta$ ? Well, there are three terms, so it would be very convenient if I could get each term to come out to $\varepsilon/3$ . So looking at the first term, $\varepsilon'M_2 \leq \varepsilon/3$ which means $\varepsilon' \leq \frac{\varepsilon}{3M_2}$ . We might want to tack on another constraint, but at least for now we can put $\varepsilon' := \frac{\varepsilon}{3M_2}$ . Similarly, $\delta$ must be smaller than $\frac{\varepsilon}{3M_1}$ . Finally, we want $\varepsilon'\delta \leq \varepsilon/3$ . Here we have control over both factors, so there’s many ways to do this. The one that I chose was to simply require one of them to be smaller than $\varepsilon/3$ , and require the other one to be smaller than one.

Model solution 1

To show that $xy$ is a real number, we need to show that $(a_n b_n)_{n=1}^\infty$ is a Cauchy sequence. Let $\varepsilon > 0$ be given. Since $(a_n)_{n=1}^\infty$ is a Cauchy sequence, by Lemma 5.1.15 it is bounded, so there is a number $M_1 > 0$ which bounds it. Similarly, there is a number $M_2 > 0$ which bounds $(b_n)_{n=1}^\infty$ . Since $(a_n)_{n=1}^\infty$ is Cauchy, for $\varepsilon' := \min\left(\frac{\varepsilon}{3M_2}, \frac\varepsilon 3\right)$ we see that this sequence is eventually $\varepsilon'$ -steady. Similarly, since $(b_n)_{n=1}^\infty$ is Cauchy, for $\delta := \min\left(\frac{\varepsilon}{3M_1}, 1\right)$ the sequence is eventually $\delta$ -steady. Thus there exists $N\geq 1$ such that for all $j,k \geq N$ we have both of the following:

$a_j$ and $a_k$ are $\varepsilon'$ -close
$b_j$ and $b_k$ are $\delta$ -close

Thus, by Proposition 4.3.7(h) we see that $a_jb_j$ and $a_kb_k$ are $(\varepsilon'|b_j| + \delta|a_j| + \varepsilon'\delta)$ -close. If we can show that $\varepsilon'|b_j| + \delta|a_j| + \varepsilon'\delta \leq \varepsilon$ , this will prove that $(a_n b_n)_{n=1}^\infty$ is Cauchy. Term by term, we see that:

$\varepsilon'|b_j| \leq \varepsilon' M_2 \leq \frac{\varepsilon}{3M_2} \cdot M_2 = \frac\varepsilon3$
$\delta|a_j| \leq \delta M_1 \leq \frac{\varepsilon}{3M_1} \cdot M_1 = \frac\varepsilon3$
$\varepsilon'\delta \leq \frac\varepsilon 3 \cdot 1 = \frac\varepsilon3$

Thus adding the inequalities we see that $\varepsilon'|b_j| + \delta|a_j| + \varepsilon'\delta \leq \varepsilon$ as required.

Now we show that if $x=x'$ then $xy=x'y$ . Suppose $x=x'$ , so that $(a_n)_{n=1}^\infty$ and $(a'_n)_{n=1}^\infty$ are equivalent sequences. Since $(b_n)_{n=1}^\infty$ is a Cauchy sequence, there is some number $M > 0$ which bounds it. Now let $\varepsilon > 0$ . Then since $(a_n)_{n=1}^\infty$ and $(a'_n)_{n=1}^\infty$ are equivalent sequences, they are eventually $\varepsilon/M$ -close. Thus there exists $N\geq 1$ such that $a_n$ and $a'_n$ are $\varepsilon/M$ -close for each $n\geq N$ . Let $n\geq N$ . By Proposition 4.3.7(g), $a_nb_n$ and $a'_nb_n$ are $(\varepsilon/M)|b_n|$ -close. Thus if we can show that $(\varepsilon/M)|b_n| \leq \varepsilon$ , we will be done. But this follows from the fact that $|b_n| \leq M$ .

Model solution 2

In my opinion, there is a simpler way to do the first part where we show $xy$ is a real number. It doesn’t make use of Proposition 4.3.7 though.

The trick is to add and subtract the same term, so we have

$\begin{aligned}|a_jb_j - a_kb_k| &= |a_jb_j - a_jb_k + a_jb_k - a_kb_k| \\ &\leq |a_j||b_j-b_k| + |b_k||a_j - a_k|\end{aligned}$

We know we can get $|b_j-b_k|$ and $|a_j - a_k|$ small (since these sequences are Cauchy), and we can also bound $|a_j|$ and $|b_k|$ (since Cauchy sequences are bounded). Thus given bounds $M_1, M_2 > 0$ on $(a_n)_{n=1}^\infty$ and $(b_n)_{n=1}^\infty$ , respectively, and given an $\varepsilon > 0$ , choose $N$ large enough so that $|a_j - a_k| \leq \frac\varepsilon{2M_2}$ and $|b_j-b_k| \leq \frac\varepsilon{2M_1}$ for all $j,k \geq N$ . Now we have $|a_jb_j - a_kb_k| \leq \varepsilon$ for $j,k\geq N$ as required.

Solutions

Exercise 5.3.1

March 26, 2020 Issa Rice Leave a comment

Exercise statement

Prove Proposition 5.3.3.

Proposition 5.3.3 (Formal limits are well-defined). Let $x = \mathrm{LIM}_{n \to \infty} a_n$ , $y = \mathrm{LIM}_{n \to \infty} b_n$ , and $z = \mathrm{LIM}_{n \to \infty} c_n$ be real numbers. Then, with the above definition of equality for real numbers, we have $x=x$ . Also, if $x=y$ , then $y=x$ . Finally, if $x=y$ and $y=z$ , then $x=z$ .

Hints

You may find Proposition 4.3.7 to be useful.

How to think about the exercise

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

Model solution

To show that $x=x$ we need to show that $(a_n)_{n=1}^\infty$ is equivalent to $(a_n)_{n=1}^\infty$ . So let $\varepsilon > 0$ . Then for $N:=1$ we have $|a_n - a_n| = 0 < \varepsilon$ for every $n\geq N$ , so $x=x$ as desired.

To show that $x=y$ implies $y=x$ , suppose $(a_n)_{n=1}^\infty$ and $(b_n)_{n=1}^\infty$ are equivalent. This means that for every $\varepsilon > 0$ there is some $N\geq1$ such that for all $n\geq N$ we have $|a_n-b_n| \leq \varepsilon$ . By Proposition 4.3.3(d) we know that $|a_n-b_n| = |b_n-a_n|$ , so the above is actually also saying that $(b_n)_{n=1}^\infty$ and $(a_n)_{n=1}^\infty$ are equivalent. More formally, let $\varepsilon > 0$ . Then by the equivalence of $(a_n)_{n=1}^\infty$ and $(b_n)_{n=1}^\infty$ we are given some $N\geq 1$ . Then for $n\geq N$ we have $|b_n-a_n| = |a_n-b_n| \leq \varepsilon$ .

Finally suppose $x=y$ and $y=z$ . This means that $(a_n)_{n=1}^\infty$ and $(b_n)_{n=1}^\infty$ are equivalent, and that $(b_n)_{n=1}^\infty$ and $(c_n)_{n=1}^\infty$ are equivalent. Let $\varepsilon >0$ be given. Then there exists $N_1 \geq 1$ such that $a_n$ is $\varepsilon/2$ -close to $b_n$ for all $n\geq N_1$ , and there exists $N_2\geq 1$ such that $b_n$ is $\varepsilon/2$ -close to $c_n$ for all $n\geq N_2$ . So if we pick $N:=\max(N_1,N_2)$ then for $n\geq N$ we have both $n\geq N_1$ and $n\geq N_2$ . Thus by Proposition 4.3.7(c) we see that $a_n$ and $c_n$ are $\varepsilon$ -close. Thus $x=z$ as required.

Solutions

Exercise 5.2.2

March 25, 2020 Issa Rice 13 Comments

Exercise statement

Let $\varepsilon > 0$ . Show that if $(a_n)_{n=1}^\infty$ and $(b_n)_{n=1}^\infty$ are eventually $\varepsilon$ -close, then $(a_n)_{n=1}^\infty$ is bounded if and only if $(b_n)_{n=1}^\infty$ is bounded.

Hints

Use the triangle inequality.

How to think about the exercise

This exercise is similar to both Exercise 5.1.1 and Exercise 5.2.1. I don’t have anything to add beyond what I said in those posts.

Model solution

Suppose $(a_n)_{n=1}^\infty$ and $(b_n)_{n=1}^\infty$ are eventually $\varepsilon$ -close, and suppose $(a_n)_{n=1}^\infty$ is bounded. Since $(a_n)_{n=1}^\infty$ is bounded, there exists a number $M_1 \geq 0$ such that $|a_n| \leq M_1$ for all $n \geq 1$ . Since $(a_n)_{n=1}^\infty$ and $(b_n)_{n=1}^\infty$ are eventually $\varepsilon$ -close, there exists $N \geq 1$ such that $|a_n - b_n| \leq \varepsilon$ for all $n \geq N$ . By the triangle inequality we have $|b_n| - |a_n| \leq |a_n - b_n|$ (to get this, you can start with the ordinary triangle inequality $|x+y| \leq |x| + |y|$ and then substitute $x:= a_n$ and $y := b_n - a_n$ ), so $|b_n| \leq |a_n| + \varepsilon \leq M_1 + \varepsilon$ for all $n \geq N$ . This bounds the sequence $(b_n)_{n=N}^\infty$ (i.e. the infinite tail of $(b_n)_{n=1}^\infty$ ). By Lemma 5.1.14, the finite sequence $b_1, \ldots, b_{N-1}$ is also bounded, say by the number $M_2 \geq 0$ . Thus we can take $M := M_1 + M_2 + \varepsilon$ as a bound on $(b_n)_{n=1}^\infty$ .

For the other direction, just interchange $(a_n)_{n=1}^\infty$ and $(b_n)_{n=1}^\infty$ in the above.

Solutions

Exercise 5.2.1

March 24, 2020 Issa Rice 5 Comments

Exercise statement

Show that if $(a_n)_{n=1}^\infty$ and $(b_n)_{n=1}^\infty$ are equivalent sequences of rationals, then $(a_n)_{n=1}^\infty$ is a Cauchy sequence if and only if $(b_n)_{n=1}^\infty$ is a Cauchy sequence.

Hints

Try to use the triangle inequality.

How to think about the exercise

The intuition for this exercise is simple: eventually all of the $a$ s are close to each other, and the $b$ s are close to the $a$ s, so all of the $b$ s must be close to each other as well.

For this kind of exercise I immediately know that I’m going to be juggling a bunch of $\varepsilon$ s and $N$ s and stringing everything together using the triangle inequality somehow. Why do I know this? Mainly because I’ve worked enough such exercises before.

We’re going to be assuming that $(a_n)_{n=1}^\infty$ is Cauchy and showing that $(b_n)_{n=1}^\infty$ is Cauchy, so the end result is that we want to show for every $\varepsilon > 0$ there is some $N\geq 1$ such that $|b_j - b_k| \leq \varepsilon$ for every $j,k \geq N$ . This means that we will have to find some $N$ that will work, and this $N$ will somehow have to come from the fact that $(a_n)_{n=1}^\infty$ is Cauchy and/or the fact that $(a_n)_{n=1}^\infty$ and $(b_n)_{n=1}^\infty$ are equivalent sequences. Since at the moment it’s not clear what $N$ could possibly work, I find it easiest with this kind of exercise to just forget about all of the quantifiers and see what we would need at the very end.

Well, we want to show $|b_j - b_k| \leq \varepsilon$ . And by the end of the proof, we will have something like $|a_j - a_k| \leq \varepsilon$ to work with (because $(a_n)_{n=1}^\infty$ is Cauchy), and also something like $|a_n - b_n| \leq \varepsilon$ to work with (because $(a_n)_{n=1}^\infty$ and $(b_n)_{n=1}^\infty$ are equivalent). So we need to bridge the gap between the inequalities we have and the inequality we want, and that is where the triangle inequality will come in. We want to write $|b_j - b_k|$ as $|b_j - a_j + a_j - a_k + a_k - b_k|$ by adding and subtracting the same terms. How did I figure this out? Again, it’s mostly from doing similar exercises before. I needed to combine various inequalities, so I wanted to add and subtract some term, but I just didn’t know which. To figure out which term, I knew that either the index had to match (to make use of the sequence equivalence property) or I need to have two $a$ s with different indexes (to make use of the Cauchy property). So a moment-by-moment thought process might look something like this:

write $|b_j$
there is a $b_j$ so I need to match the index with $-a_j$
since I wrote $-a_j$ now I need to cancel this out with $+a_j$ , so that I am effectively adding $0$
that still leaves $-b_k$ hanging there
to balance the $-b_k$ I need to match the index with $+a_k$
since I wrote $+a_k$ now I need to cancel this out with $-a_k$
that leaves $+a_j -a_k$ in the middle—how convenient!

Model solution

Let $(a_n)_{n=1}^\infty$ and $(b_n)_{n=1}^\infty$ be equivalent sequences of rational numbers. Suppose $(a_n)_{n=1}^\infty$ is Cauchy. We want to show that $(b_n)_{n=1}^\infty$ is Cauchy. So let $\varepsilon > 0$ be given. Since $(a_n)_{n=1}^\infty$ is Cauchy, we have some $N_1 \geq 1$ such that $|a_j - a_k| \leq \varepsilon/3$ for all $j,k \geq N_1$ . Also, since $(a_n)_{n=1}^\infty$ and $(b_n)_{n=1}^\infty$ are equivalent, there is some $N_2 \geq 1$ such that $|a_n - b_n| \leq \varepsilon/3$ for all $n \geq N_2$ . Pick $N := \max(N_1,N_2)$ , and let $j,k \geq N$ . Since $j,k \geq N_1$ we have $|a_j - a_k| \leq \varepsilon/3$ . Also since $j,k \geq N_2$ we have $|b_j - a_j| \leq \varepsilon/3$ and $|a_k - b_k| \leq \varepsilon/3$ . By the triangle inequality we have $|b_j - b_k| \leq |b_j - a_j| + |a_j - a_k| + |a_k - b_k| \leq \varepsilon$ as required.

For the other direction of the implication, we can just repeat the proof above by switching the places of $(a_n)_{n=1}^\infty$ and $(b_n)_{n=1}^\infty$ .