A.1 Chapter 1

Let $n\in\mathbb{N}$. Observe that $0\leq n+5\leq n+5n=6n$ whilst $8n^{2}-7\geq 8n^{2}-7n^{2}=n^{2}>0$. Combining these observations,

and so $6$ is an upper bound for $A$. Since an upper bound exists, $A$ is bounded above.

Unless every student is at least $1.7$ meters tall (which would be a major statistical anomaly!), (I) should be false. On the other hand, unless every student is under $1.7$ meters tall (which would again be a major statistical anomaly!), (II) should be true.

The negations of (I) and (II) are, respectively:

1. ($\neg$ I)
   ​

   There exists a student $S$ such that the height of student $S$ is less than $1.7$ meters.

2. ($\neg$ II)
   ​

   For all students $S$, the height of student $S$ is less than $1.7$ meters.

It is very likely that ($\neg$ I) holds and ($\neg$ II) does not hold.

If $x\in\mathbb{R}$ is an upper bound for $A\subseteq\mathbb{R}$, then $y:=x+1$ is also an upper bound for $A$ and $y\neq x$.

(i) Write $A:=\{a\in\mathbb{R}:3\leq a\leq 4\}$. We claim that every $x\geq 4$ is an upper bound for $A$. Indeed, if $x\geq 4$, then $a\leq 4\leq x$ so $a\leq x$ for all $a\in A$.

On the other hand, we claim that every $x<4$ is not an upper bound for $A$. Indeed, if $x<4$, then $4\in A$ and $x\not\geq 4$, so that $x$ is not an upper bound for $A$.

The above shows that $\{x\in\mathbb{R}:x\geq 4\}$ is precisely the set of upper bounds for $A$.

(ii) Write $B:=B_{1}\cup B_{2}$ where $B_{1}:=\{a\in\mathbb{R}:0\leq a\leq 1\}$ and $B_{2}:=\{a\in\mathbb{R}:3\leq a\leq 4\}$. We claim that every $x\geq 4$ is an upper bound for $B$. Indeed, if $x\geq 4$, then $a\leq 4\leq x$ so $a\leq x$ for all $a\in B_{1}$. Similarly, $a\leq 1<4\leq x$ so $a\leq x$ for all $a\in B_{2}$. Thus, $a\leq x$ for all $a\in B$, so $x$ is an upper bound for $B$.

On the other hand, we claim that every $x<4$ is not an upper bound for $B$. Indeed, if $x<4$, then $4\in B$ and $x\not\geq 4$, so that $x$ is not an upper bound for $B$.

The above shows that $\{x\in\mathbb{R}:x\geq 4\}$ is precisely the set of upper bounds for $B$.

Suppose $s_{1}$, $s_{2}\subseteq\mathbb{R}$ are both least upper bounds for $A\subseteq\mathbb{R}$. Since $s_{1}$ is an upper bound for $A$ and $s_{2}$ is a least upper bound for $A$, we must have $s_{2}\leq s_{1}$. By a symmetric argument, we also have $s_{1}\leq s_{2}$. Hence $s_{1}=s_{2}$.

(i) We know from Exercise 1.9 (i) that the set of upper bounds for $A:=\{a\in\mathbb{R}:3\leq a\leq 4\}$ is $\{x\in\mathbb{R}:x\geq 4\}$. Thus, the least upper bound for $A$ is $4$; that is, $\sup A=4$.

We know from Exercise 1.9 (i) that the set of upper bounds for $B:=\{a\in\mathbb{R}:0\leq a\leq 1\}\cup\{a\in\mathbb{R}:3\leq a\leq 4\}$ is $\{x\in\mathbb{R}:x\geq 4\}$. Thus, the least upper bound for $B$ is $4$; that is, $\sup B=4$.

Let $a,b\in\mathbb{R}$ with $a\leq b$.

We claim that $\sup\,[a,b]=b$.

1. 1.
   ​

   Since, by definition, $t\leq b$ for all $t\in[a,b]$, we see that $b$ is an upper bound for $[a,b]$.

2. 2.
   ​

   Suppose $x\in\mathbb{R}$ and $x<b$. Then $b\in[a,b]$ and $x\not\geq b$ so that $x$ is not an upper bound for $[a,b]$. Thus, any upper bound $x$ for $[a,b]$ must satisfy $x\geq b$.

Hence $b$ is the least upper bound for $[a,b]$; in other words, $\sup[a,b]=b$.

We claim that $\sup\,(-\infty,b]=b$. The argument is essentially the same as that used above.

1. 1.
   ​

   Since, by definition, $t\leq b$ for all $t\in(-\infty,b]$, we see that $b$ is an upper bound for $(-\infty,b]$.

2. 2.
   ​

   Suppose $x\in\mathbb{R}$ and $x<b$. Then $b\in(-\infty,b]$ and $x\not\geq b$ so that $x$ is not an upper bound for $(-\infty,b]$. Thus, any upper bound $x$ for $(-\infty,b]$ must satisfy $x\geq b$.

Hence $b$ is the least upper bound for $(-\infty,b]$; in other words, $\sup(-\infty,b]=b$.

The sets $[a,\infty)$ and $\mathbb{R}$ are not bounded above, so each of these sets cannot have a supremum. On the other hand, we know from Remark 1.16 that the empty set does not have a supremum.

The following sketch illustrates the proof that $\sup(0,1)=1$: for every $x\in(0,1)$, the element $t:=\frac{1+x}{2}\in(0,1)$ lies to the right of $x$ on the number line.

Let $a,b\in\mathbb{R}$ with $a<b$.

We claim that $\sup\,(a,b)=\sup\,(a,b]=\sup\,[a,b)=\sup\,(-\infty,b)=b$. To treat all these cases simultaneously, let $I$ denote one of the sets $(a,b)$, $(a,b]$ or $[a,b)$.

1. 1.
   ​

   Since, by definition, $t\leq b$ for all $t\in I$, we see that $b$ is an upper bound for $I$.

2. 2.
   ​

   Suppose $x\in\mathbb{R}$ and $x<b$. If $x\leq a$, then $(a+b)/2\in I$ and $x\not\geq(a+b)/2$, so that $x$ is not an upper bound for $I$. This argument continues to make sense when $I=(-\infty,b)$, provided we just choose $a\in\mathbb{R}$ to be some number satisfying $a<b$. On the other hand, if $a<x<b$, then the midpoint $t:=\frac{b+x}{2}$ satisfies $x<t<b$. Moreover, $t\in I$ and $t>x$, so that $x$ is not an upper bound for $I$. This shows that if $x\in\mathbb{R}$ is an upper bound for $I$, then $x\geq b$.

Hence $b$ is the least upper bound for $I$; in other words, $\sup I=b$. A minor modification of this argument also shows $\sup(-\infty,b)=b$.

The sets $(a,\infty)$ and $\mathbb{R}$ are not bounded above, so each of these sets cannot have a supremum. On the other hand, we know from Remark 1.16 that the empty set does not have a supremum.

Suppose $M_{1}$ and $M_{2}$ are both maxima for $A$. Then $M_{1}\in A$ and so, since $M_{2}$ is a maximum (and therefore an upper bound) for $A$, we have $M_{1}\leq M_{2}$. By a symmetric argument, $M_{2}\leq M_{1}$ and so we must have $M_{1}=M_{2}$.

(i) Let $A:=\{1,2,3\}$. Since $3\in A$ and $1\leq 3$, $2\leq 3$ and $3\leq 3$, it follows that $3$ is a maximum for $A$; that is, $\max A=3$.

On the other hand, we claim $\sup A=3$.

1. 1.
   ​

   The above observations show $3$ is an upper bound for $A$.

2. 2.
   ​

   If $x<3$, then $3\in A$ and $x\not\geq 3$ so that $x$ is not an upper bound for $A$. Thus, any upper bound $x\in\mathbb{R}$ for $A$ must satisfy $x\geq 3$.

This shows $3$ is the least upper bound for $A$; that is, $\sup A=3$.

(ii) Let $B:=\{2/n:n\in\mathbb{N}\}$. Since $2=2/1\in B$ and $2/n\leq 2/1=2$ for all $n\in\mathbb{N}$ (and so $a\leq 2$ for all $a\in B$), it follows that $2$ is a maximum for $B$; that is, $\max B=2$.

On the other hand, we claim $\sup B=2$.

1. 1.
   ​

   The above observations show $2$ is an upper bound for $B$.

2. 2.
   ​

   If $x<2$, then $2\in B$ and $x\not\geq 2$ so that $x$ is not an upper bound for $B$. Thus, any upper bound $x\in\mathbb{R}$ for $B$ must satisfy $x\geq 2$.

This shows $2$ is the least upper bound for $B$; that is, $\sup B=2$.

(i) Suppose a maximum $M$ for $(0,1)$ exists. Then $a\leq M$ for all $a\in(0,1)$, so that $M$ is an upper bound for $(0,1)$. Since $\sup(0,1)=1$ is the least upper bound for $(0,1)$, it follows that $1\leq M$.

(ii) Since $M\geq 1$, it follows that $M\not<1$ and so $M\notin(0,1)$. But this contradicts the fact that $M$ is a maximum for $(0,1)$ and therefore satisfies $M\in(0,1)$.

Let $A:=\{2-1/n:n\in\mathbb{N}\}$. We know that $\sup A=2$ and that $2\not\in A$ (since $2-1/n<2$ for all $n\in\mathbb{N}$). Hence, by a result of Worksheet 1, we know that $A$ does not have a maximum.

Alternatively, if $a\in A$, then $a=2-1/n$ for some $n\in\mathbb{N}$. However, $b:=2-1/(n+1)\in A$ and $b>a$. Thus, $a$ cannot be a maximum. Since $a\in A$ was chosen arbitrarily, this shows no maximum exists.

(i) For $B:=\Big{\{}\frac{3n^{2}-1}{n^{2}}:n\in\mathbb{N}\Big{\}}$, we claim that $\sup B=3$

1. 1.
   ​

   We may write $\frac{3n^{2}-1}{n^{2}}=3-\frac{1}{n^{2}}<3$ for all $n\in\mathbb{N}$. Hence $3$ is an upper bound for $B$.

2. 2.
   ​

   Suppose $x\in\mathbb{R}$ with $x<3$, so that $r:=3-x>0$. Then there exists some $N\in\mathbb{N}$ such that $N>1/r$ and therefore $1/N<r$. Moreover, $3-1/N^{2}\in B$ and $3-1/N^{2}\geq 3-1/N>3-r=x$, so that $x$ is not an upper bound for $B$. Here we have used the fact $N\geq 1$, so that $N^{2}\geq N$. This shows that if $x\in\mathbb{R}$ is an upper bound for $B$, then $x\geq 3$.

Thus, $3$ is the least upper bound for $B$; that is, $\sup B=3$.

We observed above that $\frac{3n^{2}-1}{n^{2}}<3$ for all $n\in\mathbb{N}$, and so $a<3$ for all $a\in B$. In particular, $\sup B=3\notin B$. Since $B$ does not contain its supremum, it does not have a maximum by a result of Worksheet 1.

Alternatively, if $a\in B$, then $a=3-1/n^{2}$ for some $n\in\mathbb{N}$. However, $b:=3-1/(n+1)^{2}\in B$ and $b>a$. Thus, $a$ cannot be a maximum. Since $a\in B$ was chosen arbitrarily, this shows no maximum exists.

(ii) For $C:=\mathbb{Q}\cap(0,1)$, we claim $\sup C=1$.

1. 1.
   ​

   Since, by definition, $a<1$ for all $a\in(0,1)$, we have $a<1$ for all $a\in C$ and so $1$ is an upper bound for $C$.

2. 2.
   ​

   Let $x<1$. If $x\leq 0$, then $1/2\in C$ and $x\not\geq 1/2$, so $x$ is not an upper bound for $C$. On the other hand, suppose $0<x<1$, so that $r:=1-x>0$. There exists some $N\in\mathbb{N}$ such that $N>1/r$ and therefore $1/N<r$. Moreover, $1-1/N>1-r=1-(1-x)=x$. Since $1-1/N\in C$ and $x\not\geq 1-1/N$, it follows that $x$ is not an upper bound for $C$. This shows that if $x\in\mathbb{R}$ is an upper bound for $C$, then $x\geq 1$.

Thus, $1$ is the least upper bound for $C$; that is, $\sup C=1$.

We observed above that $a<1$ for all $a\in C$. In particular, $\sup C=1\notin C$. Since $C$ does not contain its supremum, it does not have a maximum by a result of Worksheet 1.

Alternatively, if $a\in C$, then $a\in\mathbb{Q}$ and $0<a<1$. If we define $b:=(1+a)/2$, then it follows that $b\in\mathbb{Q}$ and $0<b<(1+1)/2=1$. Thus, $b\in C$ and $b>a$ and so $a$ cannot be a maximum. Since $a\in C$ was chosen arbitrarily, this shows no maximum exists.

We claim that for $A:=\left\{2-1/n:n\in\mathbb{N}\right\}$, $\sup A=2$.

1. 1.
   ​

   Since $2-1/n\leq 2$ for all $n\in\mathbb{N}$, it follows that $2$ is an upper bound for $A$.

2. 2.
   ​

   Let $\varepsilon>0$ be given. There exists some $N\in\mathbb{N}$ such that $N>1/\varepsilon$ and therefore $0<1/N<\varepsilon$. If we define $a:=2-1/N$, then $a\in A$ and

   $$a=2-\frac{1}{N}>2-\varepsilon.$$

   Thus, for any given $\varepsilon>0$ we can find $a\in A$ such that $2-\varepsilon<a\leq 2$, which verifies the approximation property.

By Lemma 1.31, we conclude that $\sup A=2$, as claimed.

1 $\Rightarrow$ 2. Suppose $\sup A$ exists and $s=\sup A$. Let $\varepsilon>0$ be given, so that $s-\varepsilon<s$. It follows that $s-\varepsilon$ cannot be an upper bound for $A$, since $s$ is the least upper bound for $A$ (by property 2 of Definition 1.10). This mean that there must exist some $a\in A$ such that $s-\varepsilon<a$. On the other hand, since $s$ is an upper bound for $A$ (by property 1 of Definition 1.10), we also have $a\leq s$. Combining these observations, $s-\varepsilon<a\leq s$, as required.

Consider the set $A:=\big{\{}\frac{12n+5}{3n+2}:n\in\mathbb{N}\big{\}}$. We claim that $\sup A=4$.

1. 1.
   ​

   Writing $12n+5=4(3n+2)-3$, it follows that $\frac{12n+5}{3n+2}=4-\frac{3}{3n+2}\leq 4$ for all $n\in\mathbb{N}$. Thus, $4$ is an upper bound for $A$.

2. 2.
   ​

   We now show $4$ satisfies the approximation property from Lemma 1.31.

   Let $\varepsilon>0$ be given. There exists some $N\in\mathbb{N}$ such that $N>3/\varepsilon$ and therefore $0<1/N<\varepsilon/3$. If we define $a:=\frac{12N+5}{3N+2}$, then $a\in A$ and

   $$a=4-\frac{3}{3N+2}>4-\frac{3}{N}>4-\varepsilon,$$

   where we used the fact that $\frac{3}{3N+2}<\frac{3}{N}$ for $N\in\mathbb{N}$. Thus, $4-\varepsilon<a\leq 4$, which verifies the approximation property.

By Lemma 1.31, we conclude that $\sup A=4$, as claimed.

Note: In this proof, we chose $N$ such that $N>\frac{3}{\varepsilon}$, but here (and often elsewhere) other choices are possible. For instance, we could choose $M$ such that $M>\frac{1}{\varepsilon}$ and note that for $a=\frac{12M+5}{3M+2}$ we have

using the fact that $3M+2>3M$ to see that $\frac{3}{3M+2}<\frac{3}{3M}=\frac{1}{M}$.

The additional statements mean the following:

1. (III)
   ​

   Every pencil is used by at least one student; there are no leftover pencils which go unused.

2. (IV)
   ​

   There is at least one student who writes with every single pencil.

Note that (I), (II), (III) and (IV) all have different meanings. However, (II) $\Rightarrow$ (I), as we already observed, and (IV) $\Rightarrow$ (III) (indeed, if one student uses every single pencil, then every single pencil is used by at least one student).

For the negations:

1. ($\neg$ I)
   ​

   There exists a student $S$ such that for all pencils $P$ student $S$ does not write with pencil $P$.

2. ($\neg$ II)
   ​

   For all pencils $P$, there exists a student $S$ such that student $S$ does not write with pencil $P$.

3. ($\neg$ III)
   ​

   There exists a pencil $P$ such that for all students $S$, student $S$ does not write with pencil $P$.

4. ($\neg$ IV)
   ​

   For all students $S$ there exists a pencil $P$ such that student $S$ does not write with pencil $P$.

These statements can be expressed in more everyday language as follows:

1. ($\neg$ I)
   ​

   There is a student who does not write with any pencil.

2. ($\neg$ II)
   ​

   No pencil is used by all the students.

3. ($\neg$ III)
   ​

   There is a pencil no students writes with.

4. ($\neg$ IV)
   ​

   No student writes with all the pencils.

The additional statements mean the following:

1. (V)
   ​

   Every pencil is used by every student.

2. (VI)
   ​

   There is at least one student who writes with at least one pencil.

Note that (I) – (VI) all have different meanings. However, (V) $\Rightarrow$ (II) $\Rightarrow$ (I) $\Rightarrow$ (VI) and (V) $\Rightarrow$ (IV) $\Rightarrow$ (III) $\Rightarrow$ (VI).

For the negations:

1. ($\neg$ V)
   ​

   There exists a pencil $P$ and there exists a student $S$ such that student $S$ does not write with pencil $P$.

2. ($\neg$ VI)
   ​

   For all pencil $P$ and for all students $S$, student $S$ does not write with pencil $P$.

These statements can be expressed in more everyday language as follows:

1. ($\neg$ V)
   ​

   Not every student writes with every pencil.

2. ($\neg$ VI)
   ​

   No student writes with any pencil.

(i) If $m\in\mathbb{R}$ is a minimum for $A$, then $m\in A$ so $A$ is nonempty. Furthermore, $a\geq m$ for all $a\in A$ and so $m$ is a lower bound for $A$. Hence $A$ is nonempty and bounded below, so $\inf A$ exists. Moreover, since $\inf A$ is the greatest lower bound for $A$, it follows that $\inf A\geq m$. However, we also have $m\in A$ and since $\inf A$ is a lower bound for $A$, it follows that $m\geq\inf A$. Hence, $m=\inf A$ as required.

(ii) Simply take $(0,1]$. Then given any $m\in(0,1)$ we can form $a:=m/2$. It follows that $a\in(0,1]$ and $a<m$, so that $m$ cannot be a minimum for $(0,1]$. Since $m\in A$ was chosen arbitrarily, this shows no minimum exists.

We have $(s-1/n)^{2}=s^{2}-2s/n+1/n^{2}\geq s^{2}-2s/n$. To bound this below by 2, we need to choose $n$ so that

or, equivalently, $1/n<(s^{2}-2)/2s$.

(i) We claim that the set $A:=\{a\in\mathbb{R}:a^{2}<x\}$ is nonempty and bounded above. To see this, we consider the cases $0<x<1$ and $x\geq 1$ separately. If $0<x<1$, then $x^{2}<x$ and so $x\in A$ and the set is nonempty. Furthermore, we must have $a<1$ for all $a\in A$ (since if $a\geq 1$, then $a^{2}\geq 1$ and so $a\notin A$). Hence, $1$ is an upper bound for $A$. On the other hand, if $x\geq 1$, then $1/4<1\leq x$ so $1/2\in A$ and the set is nonempty. Furthermore, $x^{2}\geq x$ and so $x$ must be an upper bound for $A$ (since if $a\geq x$, then $a^{2}\geq x^{2}\geq x$ and so $a\notin A$). In either case, we see that $A$ is nonempty and bounded above.

Since $A$ is nonempty and bounded above, by the completeness axiom $s:=\sup A$ exists. In the case $0<x<1$, we know $x\in A$ and so, since $s$ is an upper bound for $A$, we must have $s\geq x>0$. On the other hand, if $x\geq 1$, then a similar line of reasoning shows $s\geq 1/2>0$. In either case, $s>0$.

(ii) We claim that $s^{2}=x$: in other words, $s=\sqrt{x}$.

Suppose $s^{2}>x$, so that $s^{2}-x>0$ and we can therefore find some $n\in\mathbb{N}$ with $0<1/n<(s^{2}-x)/2s$. Thus,

where the second step is due to the fact that $1/n^{2}\geq 0$ for $n\geq 1$ and the choice of $n$. From (A.1) and the definition of $A$, it follows that $s>s-1/n>a$ for all $a\in A$. However, this contradicts the fact $s$ is the least upper bound for $A$, and so $s^{2}\not>x$.

Suppose $s^{2}<x$, so that $x-s^{2}>0$ and we can therefore find some $n\in\mathbb{N}$ with $0<1/n<(x-s^{2})/(1+2s)$. Thus,

where the second step is similar to that in (A.1), this time using $1/n^{2}\leq 1/n$ for $n\geq 1$. From (A.2) and the definition of $A$, it follows that $s+1/n\in A$ and clearly $s+1/n>s$. However, this contradicts the fact $s$ is an upper bound for $A$, and so $s^{2}\not<x$.

We have shown $s^{2}\not>x$ and $s^{2}\not<x$. It therefore follows that $s^{2}=x$, as claimed.

Clearly $1\in S$ and $a\leq 2$ for all $a\in S$, so $S$ is nonempty and bounded above. Repeating the argument used in Example 1.44 shows that any least upper bound $s$ for $S$ must satisfy $s^{2}=2$. However, there is no rational number with this property. Thus, $S$ does not have a least upper bound in $\mathbb{Q}$.

(i) Let $x\in\mathbb{R}$ and $\varepsilon>0$. Choosing $y:=x\in\mathbb{R}$, we have $|x-y|=0<\varepsilon$. Hence $\mathbb{R}$ is dense in $\mathbb{R}$.

(ii) Let $x\in\mathbb{R}$ and $\varepsilon>0$. If $x\neq 0$, then choosing $y:=x\in\mathbb{R}^{\times}$, we have $|x-y|=0<\varepsilon$. On the other hand, if $x=0$, then we can choose $y:=\varepsilon/2$ so that $|x-y|=|0-\varepsilon/2|=\varepsilon/2<\varepsilon$. Hence $\mathbb{R}^{\times}$ is dense in $\mathbb{R}$.

(iii) Let $x\in\mathbb{R}$ and $\varepsilon>0$. If $x\in\mathbb{R}\setminus\mathbb{Z}$, then choosing $y:=x\in\mathbb{R}$, we have $|x-y|=0<\varepsilon$. On the other hand, if $x\in\mathbb{Z}$, then we can choose $y:=x+\min\{1/2,\varepsilon/2\}$. Then $y\in\mathbb{R}\setminus\mathbb{Z}$ and $|x-y|\leq\varepsilon/2<\varepsilon$. Hence $\mathbb{R}\setminus\mathbb{Z}$ is dense in $\mathbb{R}$.

(i) Let $x:=2$ and $\varepsilon:=1/2$. If $|y-x|<\varepsilon$, then $y\geq 2-1/2=3/2$ and so $y\not\in[0,1]$. Thus, there does not exist any element $y\in[0,1]$ satisfying $|y-x|<\varepsilon$, so $[0,1]$ is not dense in $\mathbb{R}$.

(ii) Let $x:=1/2$ and $\varepsilon:=1/4$. If $|y-x|<\varepsilon$, then $1/4<y<3/4$ and so $y\not\in\mathbb{Z}$. Thus, there does not exist any element $y\in\mathbb{Z}$ satisfying $|y-x|<\varepsilon$, so $\mathbb{Z}$ is not dense in $\mathbb{R}$.