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Mathematics LibreTexts

1.4: Real Numbers

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    22639
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    Let \(C\) be the set of all Cauchy sequences of rational numbers. We define a relation on \(C\) as follows: If \(\left\{a_{i}\right\}_{i \in I}\) and \(\left\{b_{j}\right\}_{j \in J}\) are Cauchy sequences in \(\mathbb{Q}\), then \(\left\{a_{i}\right\}_{i \in I} \sim\left\{b_{j}\right\}_{j \in J},\) which we will write more simply as \(a_{i} \sim b_{i},\) if for every rational number \(\epsilon>0,\) there exists an integer \(N\) such that

    \[\left|a_{i}-b_{i}\right|<\epsilon\]

    whenever \(i>N .\) This relation is clearly reflexive and symmetric. To show that it is also transitive, and hence an equivalence relation, suppose \(a_{i} \sim b_{i}\) and \(b_{i} \sim c_{i} .\) Given \(\epsilon \in \mathbb{Q}^{+},\) choose \(N\) so that

    \[\left|a_{i}-b_{i}\right|<\frac{\epsilon}{2}\]

    for all \(i>N\) and \(M\) so that

    \[\left|b_{i}-c_{i}\right|<\frac{\epsilon}{2}\]

    for all \(i>M .\) Let \(L\) be the larger of \(N\) and \(M .\) Then, for all \(i>L\),

    \[\left|a_{i}-c_{i}\right| \leq\left|a_{i}-b_{i}\right|+\left|b_{i}-c_{i}\right|<\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon.\]

    Hence \(a_{i} \sim c_{i}\).

    Definition: set of equivalence classes

    Using the equivalence relation just defined, we call the set of equivalence classes of \(C\) the real numbers, denoted \(\mathbb{R}\).

    Note that if \(a \in \mathbb{Q},\) we may identify \(a\) with the equivalence class of the sequence \(\left\{b_{i}\right\}_{i=1}^{\infty}\) where \(b_{i}=a, i=1,2,3, \ldots,\) and thus consider \(\mathbb{Q}\) to be a subset of \(\mathbb{R} .\)

    Exercise \(\PageIndex{1}\)

    Suppose \(\left\{a_{i}\right\}_{i \in I}\) and \(\left\{b_{i}\right\}_{i \in J}\) are sequences in \(\mathbb{Q}\) with

    \[\lim _{i \rightarrow \infty} a_{i}=\lim _{i \rightarrow \infty} b_{i}.\]

    Show that \(a_{i} \sim b_{i}\).

    1.4.1 Field Properties

    Suppose \(\left\{a_{i}\right\}_{i \in I}\) and \(\left\{b_{j}\right\}_{j \in J}\) are both Cauchy sequences of rational numbers. Let \(K=I \cap J\) and define a new sequence \(\left\{s_{k}\right\}_{k \in K}\) by setting \(s_{k}=a_{k}+b_{k}\). Given any rational \(\epsilon>0,\) choose integers \(N\) and \(M\) such that

    \[\left|a_{i}-a_{j}\right|<\frac{\epsilon}{2}\]

    for all \(i, j>N\) and

    \[\left|b_{i}-b_{j}\right|<\frac{\epsilon}{2}\]

    for all \(i, j>M .\) If \(L\) is the larger of \(N\) and \(M,\) then, for all \(i, j>L\),

    \[\left|s_{i}-s_{j}\right|=\left|\left(a_{i}-a_{j}\right)+\left(b_{i}-b_{j}\right)\right| \leq\left|a_{i}-a_{j}\right|+\left|b_{i}-b_{j}\right|<\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon,\]

    showing that \(\left\{s_{i}\right\}_{k \in K}\) is also a Cauchy sequence. Moreover, suppose \(a_{i} \sim c_{i}\) and \(b_{i} \sim d_{i} .\) Given \(\epsilon \in \mathbb{Q}^{+},\) choose \(N\) so that

    \[\left|a_{i}-c_{i}\right|<\frac{\epsilon}{2}\]

    for all \(i>N\) and choose \(M\) so that

    \[\left|b_{i}-d_{i}\right|<\frac{\epsilon}{2}\]

    for all \(i>M .\) If \(L\) is the larger of \(N\) and \(M,\) then, for all \(i>L\),

    \[\left|\left(a_{i}+b_{i}\right)-\left(c_{i}+d_{i}\right)\right| \leq\left|a_{i}-c_{i}\right|+\left|b_{i}-d_{i}\right|<\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon.\]

    Hence \(a_{i}+b_{i} \sim c_{i}+d_{i} .\) Thus if \(u, v \in \mathbb{R},\) with \(u\) being the equivalence class of \(\left\{a_{i}\right\}_{i \in I}\) and \(v\) being the equivalence class of \(\left\{b_{j}\right\}_{j \in J},\) then we may unambiguously define \(u+v\) to be the equivalence class of \(\left\{a_{i}+b_{i}\right\}_{i \in K},\) where \(K=I \cap J\).

    Suppose \(\left\{a_{i}\right\}_{i \in I}\) and \(\left\{b_{j}\right\}_{j \in J}\) are both Cauchy sequences of rational numbers.

    Let \(K=I \cap J\) and define a new sequence \(\left\{p_{k}\right\}_{k \in K}\) by setting \(p_{k}=a_{k} b_{k} .\) Let \(B>0\) be an upper bound for the set \(\left\{\left|a_{i}\right|: i \in I\right\} \cup\left\{\left|b_{j}\right|: j \in J\right\} .\) Given \(\epsilon>0\) choose integers \(N\) and \(M\) such that

    \[\left|a_{i}-a_{j}\right|<\frac{\epsilon}{2 B}\]

    for all \(i, j>N\) and

    \[\left|b_{i}-b_{j}\right|<\frac{\epsilon}{2 B}\]

    for all \(i, j>M .\) If \(L\) is the larger of \(N\) and \(M,\) then, for all \(i, j>L\),

    \[\begin{aligned}\left|p_{i}-p_{j}\right| &=\left|a_{i} b_{i}-a_{j} b_{j}\right| \\ &=\left|a_{i} b_{i}-a_{j} b_{i}+a_{j} b_{i}-a_{j} b_{j}\right| \\ &=\left|b_{i}\left(a_{i}-a_{j}\right)+a_{j}\left(b_{i}-b_{j} \right)\right|\\ & \leq\left|b_{i}\left(a_{i}-a_{j}\right)\right|+\left|a_{j}\left(b_{i}-b_{j}\right)\right| \\ &=\left|b_{i}\right|\left|a_{i}-a_{j}\right|+\left|a_{j}\right|\left|b_{i}-b_{j}\right| \\ &<B \frac{\epsilon}{2 B}+B \frac{\epsilon}{2 B} \\ &=\epsilon \end{aligned}\]

    Hence \(\left\{p_{k}\right\}_{k \in K}\) is a Cauchy sequence.

    Now suppose \(\left\{c_{i}\right\}_{i \in H}\) and \(\left\{d_{i}\right\}_{i \in G}\) are Cauchy sequences with \(a_{i} \sim c_{i}\) and \(b_{i} \sim d_{i} .\) Let \(B>0\) be an upper bound for the set \(\left\{\left|b_{j}\right|: j \in J\right\} \cup\left\{\left|c_{i}\right|: i \in H\right\}\). Given \(\epsilon>0,\) choose integers \(N\) and \(M\) such that

    \[\left|a_{i}-c_{i}\right|<\frac{\epsilon}{2 B}\]

    for all \(i>N\) and

    \[\left|b_{i}-d_{i}\right|<\frac{\epsilon}{2 B}\]

    for all \(i>M .\) If \(L\) is the larger of \(N\) and \(M,\) then, for all \(i>L\),

    \[\begin{aligned}\left|a_{i} b_{i}-c_{i} d_{i}\right| &=\left|a_{i} b_{i}-b_{i} c_{i}+b_{i} c_{i}-c_{i} d_{i}\right| \\ &=\left|b_{i}\left(a_{i}-c_{i}\right)+c_{i}\left(b_{i}-d_{i} \right)\right|\\ & \leq\left|b_{i}\left(a_{i}-c_{i}\right)\right|+\left|c_{i}\left(b_{i}-d_{i}\right)\right| \\ &=\left|b_{i}\right|\left|a_{i}-c_{i}\right|+\left|c_{i}\right|\left|b_{i}-d_{i}\right| \\ &<B \frac{\epsilon}{2 B}+B \frac{\epsilon}{2 B} \\ &=\epsilon. \end{aligned}\]

    Hence \(a_{i} b_{i} \sim c_{i} d_{i} .\) Thus if \(u, v \in \mathbb{R},\) with \(u\) being the equivalence class of \(\left\{a_{i}\right\}_{i \in I}\) and \(v\) being the equivalence class of \(\left\{b_{j}\right\}_{j \in J},\) then we may unambiguously define \(u v\) to be the equivalence class of \(\left\{a_{i} b_{i}\right\}_{i \in K},\) where \(K=I \cap J .\)

    If \(u \in \mathbb{R},\) we define \(-u=(-1) u .\) Note that if \(\left\{a_{i}\right\}_{i \in I}\) is a Cauchy sequence of rational numbers in the equivalence class of \(u,\) then \(\left\{-a_{i}\right\}_{i \in I}\) is a Cauchy sequence in the equivalence class of \(-u .\)

    We will say that a sequence \(\left\{a_{i}\right\}_{i \in I}\) is bounded away from 0 if there exists a rational number \(\delta>0\) and an integer \(N\) such that \(\left|a_{i}\right|>\delta\) for all \(i>N .\) It should be clear that any sequence which converges to 0 is not bounded away from \(0 .\) Moreover, as a consequence of the next exercise, any Cauchy sequence which does not converge to 0 must be bounded away from \(0 .\)

    Exercise \(\PageIndex{2}\)

    Suppose \(\left\{a_{i}\right\}_{i \in I}\) is a Cauchy sequence which is not bounded away from 0. Show that the sequence converges and \(\lim _{i \rightarrow \infty} a_{i}=0\).

    Exercise \(\PageIndex{3}\)

    Suppose \(\left\{a_{i}\right\}_{i \in I}\) is a Cauchy sequence which is bounded away from 0 and \(a_{i} \sim b_{i} .\) Show that \(\left\{b_{j}\right\}_{j \in J}\) is also bounded away from \(0 .\)

    Now suppose \(\left\{a_{i}\right\}_{i \in I}\) is a Cauchy sequence which is bounded away from 0 and choose \(\delta>0\) and \(N\) so that \(\left|a_{i}\right|>\delta\) for all \(i>N .\) Define a new sequence \(\left\{c_{i}\right\}_{i=N+1}^{\infty}+1\) by setting

    \[c_{i}=\frac{1}{a_{i}}, i=N+1, N+2, \ldots\]

    Given \(\epsilon>0,\) choose \(M\) so that

    \[\left|a_{i}-a_{j}\right|<\epsilon \delta^{2}\]

    for all \(i, j>M .\) Let \(L\) be the larger of \(N\) and \(M .\) Then, for all \(i, j>L,\) we have

    \[\begin{aligned}\left|c_{i}-c_{j}\right| &=\left|\frac{1}{a_{i}}-\frac{1}{a_{j}}\right| \\ &=\left|\frac{a_{j}-a_{i}}{a_{i} a_{j}}\right| \\ &=\frac{\left|a_{j}-a_{i}\right|}{\left|a_{i} a_{j}\right|} \\ &<\frac{\epsilon \delta^{2}}{\delta^{2}} \\ &=\epsilon. \end{aligned}\]

    Hence \(\left\{c_{i}\right\}_{i=N+1}^{\infty}\) is a Cauchy sequence.

    Now suppose \(\left\{b_{j}\right\}_{j \in J}\) is a Cauchy sequence with \(a_{i} \sim b_{i} .\) By Exercise 1.4.3 we know that \(\left\{b_{j}\right\}_{j \in J}\) is also bounded away from \(0,\) so choose \(\gamma>0\) and \(K\) such that \(\left|b_{j}\right|>\gamma\) for all \(j>K .\) Given \(\epsilon>0,\) choose \(P\) so that

    \[\left|a_{i}-b_{i}\right|<\epsilon \delta \gamma .\]

    for all \(i>P .\) Let \(S\) be the larger of \(N, K,\) and \(P .\) Then, for all \(i, j>S,\) we have

    \[\begin{aligned}\left|\frac{1}{a_{i}}-\frac{1}{b_{i}}\right| &=\left|\frac{b_{i}-a_{i}}{a_{i} b_{i}}\right| \\ &=\frac{\left|b_{i}-a_{i}\right|}{\left|a_{i} b_{i}\right|} \\ &<\frac{\epsilon \delta \gamma}{\delta \gamma} \\ &=\epsilon . \end{aligned}\]

    Hence \(\frac{1}{a_{i}} \sim \frac{1}{b_{i}} .\) Thus if \(u \neq 0\) is a real number which is the equivalence class of \(\left\{a_{i}\right\}_{i \in I}(\text { necessarily bounded away from } 0),\) then we may define

    \[a^{-1}=\frac{1}{a}\]

    to be the equivalence class of

    \[\left\{\frac{1}{a_{i}}\right\}_{i=N+1}^{\infty},\]

    where \(N\) has been chosen so that \(\left|a_{i}\right|>\delta\) for all \(i>N\) and some \(\delta>0 .\)

    It follows immediately from these definitions that \(\mathbb{R}\) is a field. That is:

    1. \(a+b=b+a\) for all \(a, b \in \mathbb{R}\);

    2. \((a+b)+c=a+(b+c)\) for all \(a, b, c \in \mathbb{R}\);

    3. \(a b=b a\) for all \(a, b \in \mathbb{R}\);

    4. \((a b) c=a(b c)\) for all \(a, b, c \in \mathbb{R}\);

    5. \(a(b+c)=a b+a c\) for all \(a, b, c \in \mathbb{R}\);

    6. \(a+0=a\) for all \(a \in \mathbb{R}\);

    7. \(a+(-a)=0\) for all \(a \in \mathbb{R}\);

    8. \(1 a=a\) for all \(a \in \mathbb{R}\);

    9. if \(a \in \mathbb{R}, a \neq 0,\) then \(a a^{-1}=1\).

    1.4.2 Order and Metric Properties

    Definition

    Given \(u \in \mathbb{R},\) we say that \(u\) is positive, written \(u>0,\) if \(u\) is the equivalence class of a Cauchy sequence \(\left\{a_{i}\right\}_{i \in I}\) for which there exists a rational number \(\epsilon>0\) and an integer \(N\) such that \(a_{i}>\epsilon\) for every \(i>N .\) A real number \(u \in \mathbb{R}\) is said to be negative if \(-u>0 .\) We let \(\mathbb{R}^{+}\) denote the set of all positive real numbers.

    Exercise \(\PageIndex{4}\)

    Show that if \(u \in \mathbb{R},\) then one and only one of the following is true: \((\mathrm{a}) u>0,(\mathrm{b}) u<0,\) or \((\mathrm{c}) u=0\).

    Exercise \(\PageIndex{5}\)

    Show that if \(a, b \in \mathbb{R}^{+},\) then \(a+b \in \mathbb{R}^{+}\).

    Definition

    Given real numbers \(u\) and \(v,\) we say \(u\) is greater than \(v\), written \(u>v,\) or, equivalently, \(v\) is less than \(u,\) written, \(v<u,\) if \(u-v>0 .\) We write \(u \geq v,\) or, equivalently, \(v \leq u,\) to indicate that \(u\) is either greater than or equal to \(v .\) We say that \(u\) is nonnegative if \(u \geq 0\).

    Exercise \(\PageIndex{6}\)

    Show that \(\mathbb{R}\) is an ordered field, that is, verify the following:

    a. For any \(a, b \in \mathbb{R},\) one and only one of the following must hold: \((i) a<b,\) (ii) \(a=b,(\text { iii) } a>b\).

    b. If \(a, b, c \in \mathbb{R}\) with \(a<b\) and \(b<c,\) then \(a<c\).

    c. If \(a, b, c \in \mathbb{R}\) with \(a<b,\) then \(a+c<b+c\).

    d. If \(a, b \in \mathbb{R}\) with \(a>0\) and \(b>0,\) then \(a b>0\).

    Exercise \(\PageIndex{7}\)

    Show that if \(a, b \in \mathbb{R}\) with \(a>0\) and \(b<0,\) then \(a b<0\).

    Exercise \(\PageIndex{8}\)

    Show that if \(a, b, c \in \mathbb{R}\) with \(a<b,\) then \(a c<b c\) if \(c>0\) and \(a c>b c\) if \(c<0\).

    Exercise \(\PageIndex{9}\)

    Show that if \(a, b \in \mathbb{R}\) with \(a<b,\) then for any real number \(\lambda\) with \(0<\lambda<1, a<\lambda a+(1-\lambda) b<b\).

    Definition

    For any \(a \in \mathbb{R},\) we call

    \[|a|=\left\{\begin{array}{cc}{a,} & {\text { if } a \geq 0,} \\ {-a,} & {\text { if } a<0,}\end{array}\right.\]

    the absolute value of \(a\).

    Exercise \(\PageIndex{10}\)

    Show that for any \(a \in \mathbb{R},-|a| \leq a \leq|a|\).

    Proposition \(\PageIndex{1}\)

    For any \(a, b \in \mathbb{R},|a+b| \leq|a|+|b|\).

    Proof

    If \(a+b \geq 0,\) then

    \[|a|+|b|-|a+b|=|a|+|b|-a-b=(|a|-a)+(|b|-b).\]

    Both of the terms on the right are nonnegative by Exercise \(1.4 .10 .\) Hence the sum is nonnegative and the proposition follows. If \(a+b<0,\) then

    \[|a|+|b|-|a+b|=|a|+|b|+a+b=(|a|+a)+(|b|+b).\]

    Again, both of the terms on the right are nonnegative by Exercise \(1.4 .10 .\) Hence the sum is nonnegative and the proposition follows. \(\quad Q.E.D.\)

    It is now easy to show that the absolute value function satisfies

    1. \(|a-b| \geq 0\) for all \(a, b \in \mathbb{R},\) with \(|a-b|=0\) if and only if \(a=b,\)

    2. \(|a-b|=|b-a|\) for all \(a, b \in \mathbb{R}\),

    3. \(|a-b| \leq|a-c|+|c-b|\) for all \(a, b, c \in \mathbb{R} .\)

    These properties show that the function

    \[d(a, b)=|a-b|\]

    is a metric, and we will call \(|a-b|\) the distance from \(a\) to \(b\).

    Proposition \(\PageIndex{2}\)

    Given \(a \in \mathbb{R}^{+},\) there exist \(r, s \in \mathbb{Q}\) such that \(0<r<a<s\).

    Proof

    Let \(\{u\}_{i \in I}\) be a Cauchy sequence in the equivalence class of \(a .\) Since \(a>0,\) there exists a rational \(\epsilon>0\) and an integer \(N\) such that \(u_{i}>\epsilon\) for all \(i>N .\) Let \(r=\frac{\epsilon}{2} .\) Then \(u_{i}-r>\frac{\epsilon}{2}\) for every \(i>N,\) so \(a-r>0,\) that is, \(0<r<a .\)

    Now choose an integer \(M\) so that \(\left|u_{i}-u_{j}\right|<1\) for all \(i, j>M .\) Let \(s=u_{M+1}+2 .\) Then

    \[s-u_{i}=u_{M+1}+2-u_{i}>1\]

    for all \(i>M .\) Hence \(s>a\). \(\quad\) Q.E.D.

    Proposition \(\PageIndex{3}\)

    \(\mathbb{R}\) is an archimedean ordered field.

    Proof

    Given real numbers \(a\) and \(b\) with \(0<a<b,\) let \(r\) and \(s\) be rational numbers for which \(0<r<a<b<s .\) since \(\mathbb{Q}\) is a an archimedean field, there exists an integer \(n\) such that \(n r>s .\) Hence

    \[n a>n r>s>b.\]

    Q.E.D.

    Proposition \(\PageIndex{4}\)

    Given \(a, b \in \mathbb{R}\) with \(a<b,\) there exists \(r \in \mathbb{Q}\) such that \(a<r<b\).

    Proof

    Let \(\{u\}_{i \in I}\) be a Cauchy sequence in the equivalence class of \(a\) and let \(\{v\}_{j \in J}\) be in the equivalence class of \(b .\) Since \(b-a>0,\) there exists a rational \(\epsilon>0\) and an integer \(N\) such that \(v_{i}-u_{i}>\epsilon\) for all \(i>N .\) Now choose an integer \(M\) so that \(\left|u_{i}-u_{j}\right|<\frac{e}{4}\) for all \(i, j>M .\) Let \(r=u_{M+1}+\frac{\epsilon}{2} .\) Then

    \[\begin{aligned} r-u_{i} &=u_{M+1}+\frac{\epsilon}{2}-u_{i} \\ &=\frac{\epsilon}{2}-\left(u_{i}-u_{M+1}\right) \\ &>\frac{\epsilon}{2}-\frac{\epsilon}{4} \\ &=\frac{\epsilon}{4} \end{aligned}\]

    for all \(i>M\) and

    \[\begin{aligned} v_{i}-r &=v_{i}-u_{M+1}-\frac{\epsilon}{2} \\ &=\left(v_{i}-u_{i}\right)-\left(u_{M+1}-u_{i}\right)-\frac{\epsilon}{2} \\ &>\epsilon-\frac{\epsilon}{4}-\frac{\epsilon}{2} \\ &=\frac{\epsilon}{4} \end{aligned}\]

    for all \(i\) larger than the larger of \(N\) and \(M .\) Hence \(a<r<b . \quad\) Q.E.D.

    1.4.3 Upper and Lower Bounds

    Definition

    Let \(A \subset \mathbb{R}\). If \(s \in \mathbb{R}\) is such that \(s \geq a\) for every \(a \in A,\) then we call \(s\) an upper bound for \(A\). If \(s\) is an upper bound for \(A\) with the property that \(s \leq t\) whenever \(t\) is an upper bound for \(A,\) then we call \(s\) the supremum, or least upper bound, of \(A,\) denoted \(s=\sup A\). Similarly, if \(r \in \mathbb{R}\) is such that \(r \leq a\) for every \(a \in A,\) then we call \(r\) a lower bound for \(A .\) If \(r\) is a lower bound for \(A\) with the property that \(r \geq t\) whenever \(t\) is a lower bound for \(A,\) then we call \(r\) the infimum, or greatest lower bound, of \(A,\) denoted \(r=\inf A .\)

    Theorem \(\PageIndex{5}\)

    Suppose \(A \subset \mathbb{R}, A \neq \emptyset,\) has an upper bound. Then sup \(A\) exists.

    Proof

    Let \(a \in A\) and let \(b\) be an upper bound for \(A .\) Define sequences \(\left\{a_{i}\right\}_{i=1}^{\infty}\) and \(\left\{b_{i}\right\}_{i=1}^{\infty}\) as follows: Let \(a_{1}=a\) and \(b_{1}=b .\) For \(i>1,\) let

    \[c=\frac{a_{i-1}+b_{i-1}}{2}.\]

    If \(c\) is an upper bound for \(A,\) let \(a_{i}=a_{i-1}\) and let \(b_{i}=c_{i}\) otherwise, let \(a_{i}=c\) and \(b_{i}=b_{i-1} .\) Then

    \[\left|b_{i}-a_{i}\right|=\frac{|b-a|}{2^{i-1}}\]

    for \(i=1,2,3, \ldots\) Now, for \(i=1,2,3, \ldots,\) let \(r_{i}\) be a rational number such that \(a_{i}<r_{i}<b_{i} .\) Given any \(\epsilon>0,\) we may choose \(N\) so that

    \[2^{N}>\frac{|b-a|}{\epsilon}.\]

    Then, whenever \(i>N\) and \(j>N\),

    \[\left|r_{i}-r_{j}\right|<\left|b_{N+1}-a_{N+1}\right|=\frac{|b-a|}{2^{N}}<\epsilon.\]

    Hence \(\left\{r_{i}\right\}_{i=1}^{\infty}\) is a Cauchy sequence. Let \(s \in \mathbb{R}\) be the equivalence class of \(\left\{r_{i}\right\}_{i=1}^{\infty} .\) Note that, for \(i=1,2,3, \ldots, a_{i} \leq s \leq b_{i}\).

    Now if \(s\) is not an upper bound for \(A,\) then there exists \(a \in A\) with \(a>s .\) Let \(\delta=a-s\) and choose an integer \(N\) such that

    \[2^{N}>\frac{|b-a|}{\delta}.\]

    Then

    \[b_{N+1} \leq s+\frac{|b-a|}{2^{N}}<s+\delta=a.\]

    But, by construction, \(b_{N+1}\) is an upper bound for \(A .\) Thus s must be an upper bound for \(A .\)

    Now suppose \(t\) is another upper bound for \(A\) and \(t<s .\) Let \(\delta=s-t\) and choose an integer \(N\) such that

    \[2^{N}>\frac{|b-a|}{\delta}.\]

    Then

    \[a_{N+1} \geq s-\frac{|b-a|}{2^{N}}>s-\delta=t,\]

    which implies that \(a_{N+1}\) is an upper bound for \(A .\) But, by construction, \(a_{N+1}\) is not an upper bound for \(A\). Hence \(s\) must be the least upper bound for \(A\), that is, \(s=\sup A .\) \(\quad\) Q.E.D.

    Exercise \(\PageIndex{11}\)

    Show that if \(A \subset \mathbb{R}\) is nonempty and has a lower bound, then inf \(A\) exists. (Hint: You may wish to first show that inf \(A=-\sup (-A),\) where \(-A=\{x:-x \in A\}) .\)

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