Introduction

Hi! These are my attempted solutions to Real Analysis: Foundations by Sergei Ovchinnikov. I wrote this up in August to September in 2024. I do not guarantee any question's veracity or any chapter's completeness. This is just something that I enjoy doing and I decided to show my proofs.

Is an answer wrong? Have another solution? You can submit a pull request and I'll merge it and credit you (as long as I can't find any problems with it).

74/152 problems solved

Chapter 1: Rational Numbers

18/28 problems solved

Exercise 1.1

From Theorem 1.1, \((mp, np) \sim (m, n)\) is true if and only if \(mpn = mnp\) which is easily true.

Exercise 1.2

Suppose \(\phi\) is not one-to-one and there exists \(\phi(m) = \phi(n)\). This implies that \(m/1 = n/1\). This equality is only true if \(m \cdot 1 = n \cdot 1\), so \(m = n\) and \(\phi\) is one-to-one.

\[ \begin{align} \phi(m + n) &= (m + n)/1 \\ &= m/1 + n/1 \\ &= \phi(m) + \phi(n) \\ \end{align} \]

\[ \begin{align} \phi(m \cdot n) &= (m \cdot n)/1 \\ &= m/1 \cdot n/1 \\ &= \phi(m) \cdot \phi(n) \\ \end{align} \]

Exercise 1.3

Proving A1:

When \(a = b\), this is proven by reflexivity. For \(0 + 1 = 1 + 0\), both sides evaluate to \(1\).

Proving A2:

When \(a = 0\), both sides simplify to \(b + c = b + c\) which is true by reflexivity. Looking at the case where \(a = 1\), when \(b = 0\) both sides simplify to \(1 + c = 1 + c\) which is true by reflexivity. Now when \(a = b = 1\), we can look at both cases where \(c = 0\) and \(c = 1\) and see the equality holds and is equal to \(0\) and \(1\) respectively.

Proving A3:

We can see that \(0 + 0 = 0\) and \(1 + 0 = 1\) so \(0\) is the additive identity on \(\mathbb{Z}_2\).

Proving A4:

For \(0\), the additive inverse is \(0\) since \(0 + 0 = 0\). For \(1\), the additive inverse is \(1\) since \(1 + 1 = 0\).

Proving M1:

When \(a = b\), the equality is true by reflexivity. For \(0 \cdot 1 = 1 \cdot 0\), both sides of the equation evaluate to 0.

Proving M2:

When \(a = 0\) or \(b = 0\) or \(c = 0\), the product is \(0\) on both sides. When \(a = b = c = 1\), then both sides evaluate to \(1\).

Proving M3:

We can see that \(0 \cdot 1 = 0\) and \(1 \cdot 1 = 1\) so \(1\) is the multiplicative identity on \(\mathbb{Z}_2\).

Proving M4:

For \(a = 1\), the multiplicative inverse is \(1\) since \(1 \cdot 1 = 1\).

Proving D:

For \(a \cdot (b + c)\), when \(a = 0\) this is true since \(0\) or \(1\) times \(0\) is \(0\). When \(a = 1\), then the equality evaluates to \(b + c = b + c\) which is trivially true by reflexivity.

Exercise 1.4

Suppose the additive inverse is not unique, that there are two elements \(-a\) and \(-a'\) for which \(a + (-a) = 0\) and \(a + (-a') = 0\). From A2, \((-a + a) + -a' = -a + (a + -a')\). Since both \(-a\) and \(-a'\) are additive inverses, this simplifies to \(-a' = -a\). Therefore, the additive inverse must be unique.

Suppose the multiplicative inverse is not unique, that there are two elements \(a^{-1}\) and \(a^{-1}{'}\) for which \(a \cdot a^{-1} = 1\) and \(a \cdot a^{-1}{'} = 1\). From M2, \((a^{-1} \cdot a) \cdot a^{-1}{'} = a^{-1} \cdot (a \cdot a^{-1}{'})\). Since both \(a^{-1}\) and \(a^{-1}{'}\) are multiplicative inverses, this simplifies to \(a^{-1}{'} = a^{-1}\). Therefore, the multiplicative inverse must be unique.

Exercise 1.5

Since \(0 + 0 = 0\), we can use this to find that:

\[ \begin{align} a \cdot 0 &= a \cdot (0 + 0) \\ &= a \cdot 0 + a \cdot 0 \end{align} \]

Therefore, \(a \cdot 0 = 0\).

Exercise 1.6

(a) By M3, \(1 \cdot 1 = 1\). This equation also satisfies M4, showing that \(1^{-1} = 1\).

(b)

\[ \begin{align} (a \cdot b)^{-1} &= a^{-1} \cdot b^{-1} \\ (a \cdot b)^{-1} \cdot (a \cdot b) &= a^{-1} \cdot b^{-1} \cdot (a \cdot b) \\ 1 &= aa^{-1} \cdot bb^{-1} \\ 1 &= 1 \cdot 1 \\ 1 &= 1 \\ \end{align} \]

(c)

\[ c \cdot b = a \iff c \cdot b \cdot b^{-1} = a \cdot b^{-1} \iff c = a/b \]

(d)

\[ \frac{a}{b} \cdot \frac{c}{d} = a \cdot c \cdot \frac{1}{b} \cdot \frac{1}{d} = \frac{a \cdot c}{b \cdot d} \]

(e)

\[ ab^{-1} = ab^{-1}cc^{-1} = (a \cdot c)(b^{-1} \cdot c^{-1}) = (a \cdot c)(b \cdot c)^{-1} \]

(f)

\[ ab^{-1} + cd^{-1} = ab^{-1}dd^{-1} + cd^{-1}bb^{-1} = (ad)(bd)^{-1} + (cb)(bd)^{-1} = (ad + cb)(bd)^{-1} \]

(g)

\[ ab^{-1} = cd^{-1} \iff ab^{-1}bd = cd^{-1}db \iff ad = cb \]

(h)

\[ (ab^{-1})^{-1} = a^{-1}(b^{-1})^{-1} = ba^{-1} \]

(i)

\[ ab^{-1} \div cd^{-1} = ab^{-1} \cdot (cd^{-1})^{-1} = ab^{-1}c^{-1}d = ad(bd)^{-1} \]

Exercise 1.7

Reflexivity: \((f, g) \sim (f, g)\) since \(fg = fg\).

Symmetry: If \(fk = hg\), then \(hg = fk\). Therefore, \((f, g) \sim (h, k)\) implies \((h, k) \sim (f, g)\).

Transitivity: Suppose \((f, g) \sim (h, k)\) and \((h, k) \sim (m, n)\). That is, \(fk = hg\) and \(hn = mk\). Therefore, \((fk)(hn) = (hg)(mk)\) which simplifies to \(fn = gm\). Therefore, \((f, g) \sim (m, n)\).

Exercise 1.8

Proving A1: \[ f/g + h/k = (fk + gh)/gk = (hg + kf)/kg = h/k + f/g \]

Proving A2: \[ \begin{align} (f/g + h/k) + m/n &= (fk + gh)/gk + m/n \\ &= (fkn + ghn + gkm)/gkn \\ &= f/g + (hn + mk)/kn \\ &= f/g + (h/k + m/n) \\ \end{align} \]

Proving A3: The \(0\) rational polynomial is the additive identity since the \(0\) polynomial is the additive identity for integer polynomials.

Proving A4: The additive identity of a rational polynomial \(f/g\) is \(-f/g\) since adding these two gives us \(f/g + (-f/g) = (fg + -gf)/gg = 0/gg = 0\).

Proving M1: \[ f/g \cdot h/k = fh/gk = hf/kg = h/k \cdot f/g \]

Proving M2: \[ (f/g \cdot h/k) \cdot m/n = fh/gk \cdot m/n = fhm/gkn = f/g \cdot hm/kn = f/g \cdot (h/k \cdot m/n) \]

Proving M3: The multiplicative identity is the \(1\) polynomial since \(f/g \cdot 1/1 = f/g\).

Proving M4: The multiplicative inverse of a rational polynomial \(f/g\) is \(g/f\) since multiplying these two gives us \(fg/gf = 1\).

Proving D: \[ \begin{align} f/g \cdot (h/k + m/n) &= f/g \cdot (hn + mk)/kn \\ &= (fhn + fmk)/gkn \\ &= fhn/gkn + fmk/gkn \\ &= fh/gk + fm/gn \\ &= f/g \cdot h/k + f/g \cdot m/n \\ \end{align} \]

Exercise 1.9

In an ordered field, \(0 \neq 1\). If \(1 < 0\), then ... [find contradiction]. Therefore, from the trichotomy property, \(0 < 1\).

Exercise 1.10

We prove this via induction. The case \(n = 0\) is trivial since \(\tau(-0) = \tau(0) = 0 = -0 = -\tau(0)\).

Now, we assume that \(\tau(-n) = -\tau(n)\) and must prove \(\tau(-(n + 1)) = -\tau(n + 1)\). This can be shown be these series of equalities,

\(-\tau(n + 1) = -(\tau(n) + 1) = -\tau(n) - 1 = \tau(-n) - \tau(1)\)

Exercise 1.11

\[ a/b < c/d \iff abd/b < bcd/d \iff ad < bc \]

Exercise 1.12

We can prove this via induction. For the base case, \(n = 3\), we need to show \(a_1 < a_3\). Using the order relation's transitivity property with \(a_1 < a_2\) and \(a_2 < a_3\), we know that \(a_1 < a_3\).

Now, we assume that \(a_1 < a_n\) and must show that \(a_1 < a_{n+1}\). We are given that \(a_n < a_{n+1}\). Using the transitivity property again, we can prove that \(a_1 < a_{n+1}\) for all \(n > 2\).

Exercise 1.13

Exercise 1.14

From Example 1.3, we define \(f/g < h/k\) if \(fk < gh\).

(a) \(f/g < h/k\) iff \(f/g + m/n < h/k + m/n\)

If \(f/g < h/k\), then we know \(fk < gh\) and must prove \((fn + mg)/gn < (hn + mk)/kn\) or equivalently \((fn + mg)kn < (hn + mk)gn\). We can remove the \(n\) on both sides and distribute to get \(fnk + mgk < hng + mkg\). This can be further simplified into \(fk < hg\) which is what we know from \(f/g < h/k\).

If \(f/g + m/n < h/k + m/n\), then we know \((fn + mg)/gn < (hn + mk)/kn\), and must prove \(f/g < h/k\) or equivalently \(fk < gh\). We can multiply \(n\) and add \(mgk\) on both sides to get \(fkn + mgk < ghn + mgk\). This can be rewritten as \((fn + mg)k < (hn + mk)g\). Multiply \(n\) on both sides to get \((fn + mg)kn < (hn + mk)gn\). This is what we know from \((fn + mg)/gn < (hn + mk)/kn\).

(b) \(f/g < h/k\) iff \(f/g \cdot m/n < h/k \cdot m/n\) where \(m/n > 0\)

If \(f/g < h/k\), then we know \(fk < gh\) and must prove \(fm/gn < hm/kn\) or equivalently \(fmkn < hmgn\). We can remove \(mn\) from both sides of the inequality to get \(fk < hg\), which is true from \(f/g < h/k\).

If \(f/g \cdot m/n < h/k \cdot m/n\), then we know \(fmkn < hmgn\) and must prove \(f/g < h/k\) or equivalently \(fk < gh\). We can multiply each side of the inequality by \(mn\) to get \(fkmn < ghmn\), which is true from \(f/g \cdot m/n < h/k \cdot m/n\).

Exercise 1.15

Exercise 1.16

If \(x = y\), then \(|x - y| = |x - x| = |0| = 0\). If \(|x - y| = 0\), then, from the definition of absolute value, \(|x - y| = x - y\). Substituting this into the above equality, we see that \(x - y = 0\). From here, it's simple to see \(x = y\).

If \(x - y\) is positive, \(|y - x| = |-(x - y)| = x - y = |x - y|\). If \(x - y\) is negative, \(|x - y| = |-(y - x)| = y - x = |y - x|\). If \(x - y = 0\), this is trivially true.

We know that \(|x - z| = |x - y + y - z|\).

Exercise 1.17

(a) \(|a| \geq 0\)

If \(a \geq 0\), then \(|a| \geq 0\) by the definition of absolute value. If \(a < 0\), then \(0 < -a\). We can do \(|a| = |a - 0| = |0 - a| = |-a| = -a > 0\) by symmetry to show that \(|a| > 0\).

(b) \(|a| = 0\) iff \(a = 0\)

If \(a = 0\), then \(|a| = 0\). If \(|a| = 0\), then we show through the trichotomy property that \(a = 0\). If \(a > 0\), then \(|a| > 0\). If \(a < 0\), then \(0 < -a\) so \(|a| = |-a| > 0\). Therefore, \(a = 0\).

(c) \(|-a| = |a|\)

\[|a| = |a - 0| = |0 - a| = |-a|\]

(d) \(|a \cdot b| = |a| \cdot |b|\)

If \(a \geq 0\) and \(b \geq 0\), \(|a \cdot b| = a \cdot b = |a| \cdot |b|\).

If \(a < 0\) and \(b < 0\), \(|a \cdot b| = |-a \cdot -b| = -a \cdot -b = a \cdot b = |a| \cdot |b|\).

If \(a \geq 0\) and \(b < 0\), \(|a \cdot b| = |a \cdot -b \cdot -1| = |-(a \cdot -b)| = |a \cdot -b| = a \cdot -b = |a| \cdot |-b| = |a| \cdot |b|\). The same can be done for \(a < 0\) and \(b \geq 0\).

(e) \(-|a| \leq a \leq |a|\)

If \(a \geq 0\), then \(|a| = a\). Therefore, \(a \leq |a|\) and \(-|a| = -a \leq a\) since \(0 \leq a + a\).

If \(a < 0\), then \(|a| = |-a| = -a\). Therefore, \(-|a| \leq a\) and \(a \leq -a = |a|\) since \(a + a \leq 0\).

(f) \(|a| \leq b\) iff \(-b \leq a \leq b\)

If \(a \geq 0\), then \(|a| = a\). Therefore, if \(|a| \leq b\), then it's trivial to see \(a \leq b\). Moreover, \(0 \leq b - a\) which means \(a - b \leq 0\) and so \(-b \leq -a\). Since \(a \geq 0\), \(-b \leq -a \leq a \leq b\). If \(-b \leq a \leq b\), then \(-b \leq |a| \leq b\) and so \(|a| \leq b\).

If \(a < 0\), then \(|a| = -a\). Therefore, if \(|a| \leq b\), then \(-b \leq a\). Moreover, \(0 \leq a + b\) which means \(-a - b \leq 0\) and so \(-a \leq b\). Since \(a < 0\), \(-b \leq a \leq -a \leq b\). If \(-b \leq a \leq b\), then \(-b \leq a\) so \(-a \leq b\) which means \(|a| \leq b\).

(g) \(|a|^2 = a^2\)

If \(a \geq 0\), then \(|a|^2 = a^2\). If \(a < 0\), then \(|a|^2 = |-a|^2 = (-a)^2 = -1 \cdot -1 \cdot a \cdot a = a^2\)

(h) \(|a - b| \geq ||a| - |b||\)

If \(a - b \geq 0\), then \(a \geq b\) and we only need to prove \(a - b \geq ||a| - |b||\). If \(b \geq 0\), then \(a \geq 0\) and so we only need to show \(a - b \geq |a - b| = a - b\) which is trivially true. If \(b < 0\), then we need to show \(a - b \geq ||a| + b|\). In this scenario, if \(a \geq 0\) or \(a < 0\), then this simplifies to \(a - b \geq a + b\) or \(a - b \geq |-(a - b)| = a - b\). The latter is trivially true and the former is true since \(b < 0\).

If \(a - b > 0\), then \(a < b\) and we need to show that \(b - a \geq ||a| - |b||\). If \(a \geq 0\), then \(b \geq 0\) as well and so \(b - a \geq |a - b| = |-(b - a)| = b - a\). If \(a < 0\), then we need to show \(b - a \geq |-a -|b||\). In this scenario, if \(b < 0\), then \(b - a \geq |-a + b| = b - a\) which is trivially true. Moreover, if \(b \geq 0\), then \(b - a \geq |-a - b| = a + b\) which is true since \(a < 0\).

(i) \(|a + b| = |a| + |b|\) iff \(ab \geq 0\)

Assume \(|a + b| = |a| + |b|\). If \(a + b \geq 0\), then \(a + b = |a| + |b|\) which means \(|a| = a\) and \(|b| = b\) so \(ab \geq 0\). If \(a + b < 0\), then \(-(a + b) = -a + -b = |a| + |b|\) which means \(|a| = -a\) and \(|b| = -b\) so \(ab = -a \cdot -b \geq 0\).

If \(ab \geq 0\), then either \(a \geq 0\) and \(b \geq 0\) or \(a \leq 0\) and \(b \leq 0\). If \(a \geq 0\) and \(b \geq 0\), then \(|a + b| = a + b = |a| + |b|\). If \(a \leq 0\) and \(b \leq 0\), then \(|a + b| = |-(a + b)| = -(a + b) = -a + -b = |a| + |b|\).

Exercise 1.18

If \(a \leq b \leq c\), then \(a - b \geq 0\) and \(b - c \geq 0\). Therefore, from Exercise 1.17(i), \(|(a - b) + (b - c)| = |a - b| + |b - c|\). Simplifying this gives us \(|a - b| + |b - c| = |a - c|\).

If \(|a - b| + |b - c| = |a - c|\), then by Exercise 1.17(i), \((a - b)(b - c) \geq 0\). This means \((a - b)\) and \((b - c)\) must both be positive or negative. However, if they are both positive, then \(a - b \geq 0 \implies a \geq b\) and \(b - c \geq 0 \implies b \geq c\) so \(a \geq c\) violating our assumption that \(a \leq c\). When they are both negative, the same reasoning shows that \(a \leq b \leq c\).

Exercise 1.19

From Exercise 1.17(e), to show that \(|x - y| \leq b - a\) we can show that \(-(b - a) \leq x - y \leq b - a\). We also know that since \(a \leq x \leq b\), \(0 \leq x - a \leq b - a\) and \(-(b - a) \leq x - b \leq 0\). Since \(a \leq y \leq b\), this means that \(-(b - a) \leq x - y \leq b - a\).

Exercise 1.20

Exercise 1.21

(a) \(\lim(1/n) = 0\)

To prove this, we must show that \(| 1/n - 0 | < \varepsilon\) for all \(n > N\) or \(1/n < \varepsilon\) since \(n > 0\). First, we can see that \(1/n\) is strictly decreasing since \(1/n > 1/(n+1)\). This is true since \(n+1 > n\) is obviously true. Therefore, if \(N = 1/\varepsilon\), then for all \(n > N\), \(1/n < \varepsilon\) since if \(n = 1/\varepsilon\) then \(1/n = \varepsilon\) and \(1/n\) is strictly decreasing.

(b) \(\lim(1/n^3) = 0\)

We can use the Squeeze Theorem to show this limit is true. We know that \(1/n^3 < 1/n\) for \(n > 0\) since \(n < n^3\). We also know that \(0/1 < 1/n^3\) for \(n > 0\) since \(0 < 1\). Since \(0 < 1/n^3 < 1/n\) and \(\lim(0) = 0\) and \(\lim(1/n) = 0\), by the Squeeze Theorem, \(\lim(1/n^3) = 0\).

(c) \(\lim\frac{2 + n - n^2}{4 + 5n + 3n^2} = -\frac{1}{3}\)

Exercise 1.22

We want to show that \((|a_n|)\) is a Cauchy sequence given that \((a_n)\) is a Cauchy sequence. From the definition of a Cauchy sequence, we know that for every \(\varepsilon > 0\), there is \(N\) such that \(|a_m - a_n| < \varepsilon\) for all \(m, n > N\). From Exercise 1.17(h), we know that \(||a_n| - |a_m|| \leq |a_n - a_m|\) for \(m > n\). Therefore, for every \(\varepsilon > 0\), there is \(N\) such that \(||a_n| - |a_m|| < \varepsilon\) for all \(m, n > N\) so \((|a_n|)\) is a Cauchy sequence.

Exercise 1.23

Let \(b_n\) be a constant sequence that equals \(k\) for all \(n\). This implies \(k a_n = b_n a_n\). Using Theorem 1.20, \(\lim (k a_n) = \lim (b_n a_n) = (\lim b_n)(\lim a_n) = k \lim (a_n)\).

Exercise 1.24

Exercise 1.25

We can prove there is a minimum and a maximum via induction over the size of the finite set of elements of the ordered field, denoted \(n\). For the base case where \(n = 1\), the only element in the set is the minimum and maximum. Now, we assume that there is a minimum and maximum in a finite set of size \(n\) and must prove they exist for a finite set of size \(n + 1\). We can take \(n\) elements of the set with \(n + 1\) and see that it has a minimum and maximum value. Then with the leftover element, we can compare it against the minimum and maximum. If it is less than the minimum, then the element is the minimum of the set. Otherwise the minimum stays the same. If it is more than the maximum, then the element is the maximum of the set. Otherwise the maximum stays the same. Therefore, there is a minimum and maximum of a finite set of elements of an ordered field.

Exercise 1.26

Let \(\varepsilon > 0\). There is \(N \in \mathbb{N}\) such that \(|a_n - a| < \varepsilon\) for all \(n > N\) where \(a_n \to a\). Since \(m, m + 1, \cdots\) is a strictly increasing sequence of natural numbers, there is \(N' \in \mathbb{N}\) such that \(m + n > N\) for all \(n > N'\). It follows that \(|a_{m + n} - a| < \varepsilon\) for all \(n > N'\). Therefore, \(a_{m + n} \to a\).

Exercise 1.27

Exercise 1.28

Chapter 2: Real Numbers

17/31 problems solved

Exercise 2.1

Suppose there are two supremums for a nonempty subset of an ordered field called \(a\) and \(b\). This means they are both upper bounds and they are both less than or equal to all other upper bounds of this subset. As a result, \(a \leq b\) and \(b \leq a\) implying that \(a = b\).

The same argument can be made for infimums as well with the infimums \(a\) and \(b\) having to satisfy \(a \geq b\) and \(b \geq a\). Therefore, \(a = b\) must be true.

Exercise 2.2

Every bounded above subset of a field has an associated subset that is bounded below taking every element in the original subset to its additive inverse. We know it is bounded below since \(x \leq b\) implies \(-b \leq -x\) for all elements in the original subset \(x\) and the original supremum \(b\). Moreover, \(-b\) is the infimum of the new subset since \(b \leq b'\) for any upper bound \(b'\) implies \(-b' \leq -b\) for any lower bound \(-b'\). Therefore, the two definitions are equivalent.

Exercise 2.3

If \(p < r < s < q\), then \(p < r\) and \(s < q\). This means that \(s + p < q + r\). Rearranging, we get \(s - r < q - p\).

Exercise 2.4

Let \(m = n + i\) and we will show via induction that \(a_n \leq a_m\) and \(b_n \geq b_m\) for natural numbers \(n < m\). When \(i = 1\), previous part of the Lemma proves this. Now we assume that for \(m = n + i\), \(a_n \leq a_m\) and \(b_n \geq b_m\) and must prove \(a_n \leq a_{m + 1}\) and \(b_n \geq b_{m + 1}\). From the previous part, we know that \(a_m \leq a_{m + 1}\) and \(b_{m} \geq b_{m + 1}\). From transitivity, this must mean that \(a_n \leq a_{m + 1}\) and \(b_n \geq b_{m + 1}\).

Exercise 2.5

Exercise 2.6

For the sake of contradiction, assume the opposite: that \(\sup A > \inf B\). This means that there exists \(\varepsilon > 0\) such that \(\sup A - \inf B = \varepsilon\). From Theorem 2.1, there exists an element \(x \in A\) where \(x > \sup A - \varepsilon\). Substituting \(\varepsilon\) for \(\sup A - \inf B\) reveals the contradiction, \(x > \inf B\). From our assumptions, \(a \leq b\) for all \(a \in A\) and \(b \in B\) so \(x > \inf B\) violates our premise proving that \(\sup A \leq \inf B\).

Exercise 2.7

Let \(\sup A = a\) and \(\sup B = b\).

First, let's assume \(a < b\). This means that for all \(x \in B\), \(x \leq a\). In other words, \(a\) is an upper bound of \(B\) as well. Moreover, the supremum of \(A \cup B\) must be \(a\) since if there was a smaller upper bound than \(a\) then it would need to also be the supremum of \(A\) as well. Therefore \(\sup(A \cup B) = \sup A\) when \(a < b\).

The same argument can be made when \(b < a\) to show that the supremum of \(A \cup B\) in this scenario would be \(\sup B\). When \(a = b\), then both are supremums of each set so \(\sup(A \cup B) = a = b\). Taking everything in, we get that \(\sup(A \cup B) = \max\{a, b\}\).

Exercise 2.8

Let \(\sup A = a\) and \(\sup B = b\).

We know that \(a + b\) is an upper bound for \(A + B\) since \(x \leq a\) and \(y \leq b\) for all \(x \in A\) and \(y \in B\) so \(x + y \leq a + b\). If there was a smaller lower bound \(c\), then \(0 < \varepsilon = a + b - c\). Since \(a\) and \(b\) are suprema, then there exists \(x \in A\) such that \(\varepsilon/2 > a - x\) and \(y \in B\) such that \(\varepsilon/2 > b - y\). Therefore, \(x + y > a + b - \varepsilon\) so \(\varepsilon > a + b - (x + y)\). Putting everything together,

\[ \begin{align} a + b - (x + y) &< \varepsilon = a + b - c \\ - (x + y) &< - c \\ c &< x + y \\ \end{align} \]

This is an obvious contradiction, the supremum of \(A + B\) cannot be less than an element in \(A + B\). Therefore, \(a + b\) is the supremum of \(A + B\).

Exercise 2.9

Let \(\sup E = f\) and \(\inf E = g\).

We know that, for all \(x \in E\), \(x \leq f\) and \(g \leq x\). Multiplying by the given \(c > 0 \in \mathbb{R}\), we get \(cx \leq cf\) and \(cg \leq cx\). This implies that \(cf\) is an upper bound for \(cE\) and \(cg\) is a lower bound. We can also see that these are the supremum and infimum of \(cE\). This is because we know that \(f \leq f'\) for any upper bound \(f'\) and \(g' \leq g\) for any lower bound \(g'\). Again, multiplying by \(c\) on both sides shows that \(cf\) and \(cg\) are the least upper bound and greatest lower bound respectively when \(c > 0\).

When \(c < 0\), we can obtain similar but different results. Instead, we can multiply by \(-c\) to find \(cf \leq cx\) and \(cx \leq cg\) for \(x \in E\). In other words, \(cf\) is a lower bound and \(cg\) is an upper bound for \(cE\). We can similarly find that \(cf' \leq cf\) and \(cg \leq cg'\) for any upper bound \(f'\) and lower bound \(g'\). Therefore, \(cf\) and \(cg\) are the greatest lower bound and least upper bound respectively when \(c < 0\).

Exercise 2.10

Suppose, for the sake of contradiction, that \(\inf E > \inf E'\). Since it is a lower bound, \(\inf E \leq x\) for any \(x \in E\). Since \(E'\) is a subset of \(E\), this means \(\inf E\) is a lower bound for \(E'\) as well. It is also a greater lower bound than \(\inf E'\) by our assumption resulting in our contradiction. Therefore, \(\inf E \leq \inf E'\).

Showing that \(\inf E' \leq \sup E'\) is trivial and results by definition of upper bound and lower bound.

Suppose, for the sake of contradiction, that \(\sup E' > \sup E\). Since it is an upper bound, \(x \leq \sup E\) for any \(x \in E\). Since \(E'\) is a subset of \(E\), this means \(\sup E\) is an upper bound for \(E'\) as well. It is also a smaller upper bound than \(\sup E'\) by our assumption resulting in our contradiction. Therefore, \(\sup E' \leq \sup E\).

Exercise 2.11

Suppose \(b\) is the supremum. Since \(b\) is the least upper bound and \(b - 1/n < b\), \(b - 1/n\) cannot be an upper bound or it would contradict the definition of supremum. Moreover, since \(b + 1/n > b\) and \(b\) is an upper bound, \(b + 1/n\) must also be an upper bound as well.

Suppose for every natural number \(n\) that \(b - 1/n\) is not an upper bound and \(b + 1/n\) is an upper bound. Since \(\lim(b + 1/n) = b\), this means \(b\) is also an upper bound. Moreover, if there existed a lower upper bound \(b'\) such that \(b' = b - \varepsilon\) where \(\varepsilon > 0\), then for \(n > 1/\varepsilon\) we can see that \(b - \varepsilon\) is not an upper bound so \(b'\) cannot be a supremum. Therefore, \(b\) is the supremum.

Exercise 2.12

For the sake of contradiction, suppose \(\sqrt{2} + \sqrt{3}\) is a rational number. That is, \(\sqrt{2} + \sqrt{3} = p/q\).

Exercise 2.13

Let \(a = p/q\).

For the sake of contradiction, suppose \(a + b = r/s\). Substituting \(a = p/q\), we get \(b = (qr - ps)/qs\) which is a contradiction as \(b\) is irrational.

For the sake of contradiction, suppose \(a - b = r/s\). Substituting \(a = p/q\), we get \(b = (qr + ps)/qs\) which is a contradiction as \(b\) is irrational.

For the sake of contradiction, suppose \(a/b = r/s\). Substituting \(a = p/q\), we get \(b = ps/qr\) which is a contradiction as \(b\) is irrational.

For the sake of contradiction, suppose \(b/a = r/s\). Substituting \(a = p/q\), we get \(b = pr/qs\) which is a contradiction as \(b\) is irrational.

Exercise 2.14

Reflexivity: Since \(\lim(a_n - a_n) = \lim(0) = 0\), the relation is reflexive.

Symmetry: If \((a_n) \sim (b_n)\), then \(\lim(a_n - b_n) = 0\). This means \(\lim(a_n) - \lim(b_n) = 0\) so \(\lim(a_n) = \lim(b_n)\) so \(0 = \lim(b_n) - \lim(a_n)\) so \(0 = \lim(b_n - a_n)\). Therefore, \((b_n) \sim (a_n)\) and the relation is symmetric.

Transitivity: If \((a_n) \sim (b_n)\) and \((b_n) \sim (c_n)\), then \(\lim(a_n - b_n) = 0\) and \(\lim(b_n - c_n) = 0\). This means \(\lim(a_n) = \lim(b_n)\) and \(\lim(b_n) = \lim(c_n)\). Therefore, \(\lim(a_n) = \lim(c_n)\) so \(\lim(c_n - a_n) = 0\) and \((a_n) \sim (c_n)\).

Exercise 2.15

Exercise 2.16

Since \(E\) is a dense subset, this means for any \(a \in \mathbb{F}\) and \(\varepsilon > 0 \in \mathbb{F}\), there is an element \(x \in E\) such that \(|x - a| < \varepsilon\). Equivalently, \(a - \varepsilon < x < a + \varepsilon\). We are given \(x < y\) so \(y = x + \varepsilon\) for some \(\varepsilon > 0\). From the definition of a dense subset, there is \(r \in E\) such that \(x - \varepsilon < r < x + \varepsilon\). Therefore, \(x < r < y\).

Exercise 2.17

Exercise 2.18

(a) \(\varphi\) is a one-to-one mapping

To show \(\varphi\) is a one-to-one mapping, we must show that \(\varphi(x) = \varphi(y) \implies x = y\). The left hand side of the implication shows that \(\varphi(x) - \varphi(y) = 0\) which consequently show \(\varphi(x - y) = 0\). Since \(\varphi(0) = 0\) (this is proven in the next section), this means \(\varphi(x - y) = \varphi(0)\) and so \(x - y = 0\) meaning \(x = y\).

(b) \(\varphi(0) = 0\) and \(\varphi(1) = 1\)

We start by showing that \(\varphi(0) = \varphi(0 + 0) = \varphi(0) + \varphi(0)\). Since \(\varphi(0) = \varphi(0) + \varphi(0)\), this means \(\varphi(0) = 0\).

Similarly, \(\varphi(1) = \varphi(1 \cdot 1) = \varphi(1) \cdot \varphi(1)\). Since \(\varphi(1) = \varphi(1) \cdot \varphi(1)\), this means \(\varphi(1) = 1\).

(c) \(\varphi(\mathbb{F})\) is an ordered subfield of \(\mathbb{G}\)

I'm pretty sure the requirements for an embedding, that \(\varphi(a + b) = \varphi(a) + \varphi(b)\) and \(\varphi(a \cdot b) = \varphi(a) \cdot \varphi(b)\) and \(a < b\) iff \(\varphi(a) < \varphi(b)\), directly prove that \(\varphi(\mathbb{F})\) is an ordered subfield of \(\mathbb{G}\).

Exercise 2.19

Exercise 2.20

Exercise 2.21

Exercise 2.22

Exercise 2.23

Exercise 2.24

Exercise 2.25

Exercise 2.26

Exercise 2.27

For the sake of contradiction, suppose there are two unique cut points \(c, d\) such that \(d - c = \varepsilon > 0\). Since \(c\) is a cut point, \(c \leq b\) for all \(b \in B\). However, since \(d > c\) and from the definition of a cut, there exists \(b \in B\) such that \(b - c < \varepsilon\). In other words, there is \(b \in B\) between \(c\) and \(d\) so \(c < b < d\). Therefore, \(d\) is not a cut point which contradicts our assumption.

If \(c - d = \varepsilon > 0\), a similar argument holds. Since \(c\) is a cut point, \(a \leq c\) for all \(a \in A\). However, since \(d < c\) and the definition of a cut, there exists \(a \in A\) such that \(c - a < \varepsilon\). In other words, there is \(a \in A\) between \(d\) and \(c\) so \(d < a < c\). Therefore, \(d\) is not a cut point which again contradicts our assumption.

From the trichotomy property, this must mean \(c = d\) so there can only be one unique cut point.

Exercise 2.28

Since \(c\) is a cut point, for all \(a \in A\) and \(b \in B\), \(a \leq c\) and \(c \leq b\). This means \(c\) is an upper bound for \(A\) and a lower bound for \(B\).

To show it is the supremum for \(A\), suppose there is a smaller upper bound \(c' < c\). That means \(c - c' = \varepsilon > 0\). From the definition of a cut, there exists \(a \in A\) such that \(c - a < \varepsilon\). In other words, there is \(a \in A\) in between \(c' < a < c\) contradicting our assumption that \(c'\) is an upper bound. Therefore, \(c = \sup A\).

To show it is the infimum for \(B\), suppose there is a greater lower bound \(c' > c\). That means \(c' - c = \varepsilon > 0\). From the definition of a cut, there exists \(b \in B\) such that \(c - b > \varepsilon\). In other words, there is \(b \in B\) in between \(c < b < c'\) contradicting our assumption that \(c'\) is a lower bound. Therefore, \(c = \inf B\).

Exercise 2.29

Exercise 2.30

Exercise 2.31

Chapter 3: Continuous Functions

14/28 problems solved

Exercise 3.1

If for every \(x \in G\), there is an \(\varepsilon\)-neighborhood of \(x\) such that \((x - \varepsilon, x + \varepsilon) \subseteq G\), then \(G\) satisfies the definition of an open subset.

If \(G\) is an open subset, then for all \(x \in G\), there is a neighborhood \((a, b)\) such that \(a < x < b\). Let \(\varepsilon = \min\{x - a, b - x\}\). If \(\varepsilon = x - a\), then \(a < x < b\) implies \(x - \varepsilon < x < b\) and since \(x - a \leq b - x\), this means \(x - \varepsilon < x < x + \varepsilon\). If \(\varepsilon = b - x\), then this also implies \(a < x < x + \varepsilon\) and since \(b - x \leq x - a\), this means \(x - \varepsilon < x < x + \varepsilon\).

Exercise 3.2

If \(f(x) = f(y)\), then \(px + q = py + q \implies px = py \implies x = y\). Therefore, \(f\) is injective. For all \(y \in \mathbb{F}\), there exists \(x = (y - q)/p \in \mathbb{F}\) such that \(f(x) = y\). Therefore, \(f\) is surjective. To show \(f\) preserves order on \(\mathbb{F}\), assume \(x < y\). We can see that \(x < y \implies px < py \implies px + q < py + q \implies f(x) < f(y)\). Putting these together, \(f\) is a bijection preserving order on \(\mathbb{F}\).

Since \(G\) is an open set, any point \(x \in G\) has an \(\varepsilon\)-neighborhood, \((x - \varepsilon, x + \varepsilon) \subseteq G\). This subset corresponds to the inequality \(x - \varepsilon < x < x + \varepsilon\). Since \(f\) is a bijection preserving order on \(\mathbb{F}\), this means \(f(x - \varepsilon) < f(x) < f(x + \varepsilon)\). With some algebra, we can see this is equivalent to \(f(x) - p\varepsilon < f(x) < f(x) + p\varepsilon\). Setting \(\varepsilon' = p\varepsilon\), we can see every element \(f(x) \in f(G)\) has an \(\varepsilon\)-neighborhood so \(f(G)\) is open.

Exercise 3.3

We can assume \(I\) is an open interval \((a, b)\) and our result holds for all other intervals. For any \(x \in I\), we know by definition that \(a < x < b\). Any neighborhood of \(x\), denoted \((a', b')\), would need to satisfy \(a' < x < b'\). If \(a' < a\) or \(b < b'\), then it is trivial to see that \((a', b')\) contains a point in \((a, b)\). If \(a < a'\) and \(b' < b\), then we can see that the element \((x + a') / 2\) is also in both open intervals. Therefore, any point in \(I\) is a limit point.

Exercise 3.4

Exercise 3.5

Suppose \(x - y > 0\). This means \(x - y < \varepsilon\) so \(x - \varepsilon < y\). For any \(a \in (y - \varepsilon, y + \varepsilon)\), by definition \(y - \varepsilon < a < y + \varepsilon\). In terms of \(x\), we can see \(x - 2\varepsilon < a < x\).

Suppose \(-(x - y) > 0\). This means \(y - x < \varepsilon\) so \(y < x + \varepsilon\). For any \(a \in (y - \varepsilon, y + \varepsilon)\), by definition \(y - \varepsilon < a < y + \varepsilon\). In terms of \(x\), we can see \(x < a < x + 2\varepsilon\).

In either case, any element in \((y - \varepsilon, y + \varepsilon)\) will also be in \((x - 2\varepsilon, x + 2\varepsilon)\).

Exercise 3.6

Consider the sequence of points \(a_n = 1/n\) which is in \(\{1/n : n \in \mathbb{N}\}\). We have already shown that the sequence converges to \(0\) previously. Since \(0 \notin \{1/n : n \in \mathbb{N}\}\), the set is not compact.

Exercise 3.7

Exercise 3.8

Since \(\{x_n : n \in \mathbb{N}\}\) is finite, then there must be an element in the set that \((x_n)\) repeats infinitely. If there wasn't, then \((x_n)\) would need to be finite, a contradiction to the size of \(\mathbb{N}\). Let the subsequence \((x'_n)\) be a constant sequence for the element that repeats infinitely. This subsequence is obviously convergent.

Exercise 3.9

Exercise 3.10

Assume \(\lim (x_n) \to x\). For the sake of contradiction, also assume that there exists a neighborhood of \(x\) that does not contain infinitely many terms of \((x_n)\). That is, there are only finitely many \(n\) such that \(x_n \in (x - \varepsilon, x + \varepsilon)\). Since convergent sequences are Cauchy, there exists \(N\) such that for all \(m > N\), \(|x_m - x| < \varepsilon\). Since there are an infinite number of such \(m\), there cannot be finitely many terms of \((x_n)\) in any neighborhood of \(x\).

Assume every neighborhood of \(x\) contains infinitely many terms of \((x_n)\). For the sake of contradiction, also assume that \(\lim (x_n) \nrightarrow x\). However, from our assumption, for any \(\varepsilon > 0\) the neighborhood \((x - \varepsilon, x + \varepsilon)\) will contain infinitely many terms of \((x_n)\). Therefore, for any \(N\) there will always exist \(n > N\) such that \(x - \varepsilon < x_n < x + \varepsilon\). Therefore, \(\lim (x_n) \to x\).

Exercise 3.11

For any sequence of points in the union of two compact sets, there are two subsequences divided by the original sets. At least one of the subsequences must have an infinite number of terms or the original sequence is finite. Since each set is compact, a subsequence of the subsequence must converge to a point in the set. The union of the two compact sets will also contain this point. Therefore, the union is compact.

Exercise 3.12

The intersection of two compact sets must be compact. For the sake of counterargument, assume it is not true. That is, there exists there is a sequence in the intersection of compact sets where every subsequence converges to a point outside of the intersection. Since this sequence was in both compact sets, that means its limit must also be in both compact sets as well. In other words, every subsequence must converge inside the intersection of compact sets.

Any family of compact sets must also be compact in the same way. Assume the intersection of \(k\) compact sets is compact. The same argument can be made to show that the intersection of \(k + 1\) compact sets is also compact.

Exercise 3.13

Since \(E\) and its superset are both compact, the intersection of these two sets must also be compact as shown in Exercise 3.12. The intersection of a set and its subset is the subset so \(E\) is the compact intersection.

Exercise 3.14

Exercise 3.15

Exercise 3.16

(a) \(f + g\)

We want to show that for any \(a \in E\) and \(\varepsilon > 0\), there exists \(\delta > 0\) such that \(|x - a| < \delta \implies |(f + g)(x) - (f + g)(a)| < \varepsilon\). Since we know \(f\) and \(g\) are continuous functions, there exists \(\delta'\) and \(\delta''\) for any \(a \in E\) and \(\varepsilon > 0\) such that \(|x - a| < \delta' \implies |f(x) - f(a)| < \varepsilon/2\) and \(|x - a| < \delta'' \implies |g(x) - g(a)| < \varepsilon/2\). If we let \(\delta = \min\{\delta', \delta''\}\), then we can see that when \(|x - a| < \delta\) then

\[ \begin{align} |(f + g)(x) - (f + g)(a)| &= |f(x) - f(a) + g(x) - g(a)| \\ &\leq |f(x) - f(a)| + |g(x) - g(a)| \\ &< \varepsilon/2 + \varepsilon/2 \\ &= \varepsilon \\ \end{align} \]

Therefore, \(f + g\) is continuous.

(b) \(k \cdot f, k \in \mathbb{F}\)

Since \(f\) is continuous, there exists \(\delta\) such that if \(|x - a| < \delta\) then \(|f(x) - f(a)| < \varepsilon/|k|\) for any \(a\).

\[ \begin{align} |(k \cdot f)(x) - (k \cdot f)(a)| &= |k \cdot (f(x) - f(a))| \\ &= |k| \cdot |f(x) - f(a)| \\ &< |k| \cdot \varepsilon/|k| \\ &= \varepsilon \\ \end{align} \]

Therefore, \(k \cdot f\) is continuous.

(c) \(f \cdot g\)

\[ \begin{align} |(f \cdot g)(x) - (f \cdot g)(a)| &= |f(x) \cdot g(x) - f(a) \cdot f(a)| \\ \end{align} \]

(d) \(f/g\), where \(g(x) \neq 0\) on \(E\)

(e) \(|f|\)

Exercise 3.17

Since \(f\) is continuous, we know that for any \(a \in E\), there exists \(\delta > 0\) such that if \(|x - a| < \delta\), then \(|f(x) - f(a)| < \delta'\). Since \(g\) is also continuous, then for all \(f(a) \in E'\), there exists \(\delta' > 0\) such that if \(|f(x) - f(a)| < \delta'\), then \(|g(f(x)) - g(f(a))| < \varepsilon\). Therefore, \(g \circ f\) is continuous.

Exercise 3.18

Exercise 3.19

Suppose for the sake of counterargument that \([a, b] \nsubseteq E\). This would mean that there exists \(x\) such that \(a \leq x \leq b\) and \(x \notin E\). However, this would mean that \(E = [a, x) \cup (x, b]\). In other words, \(E\) is disconnected, a contradiction. Therefore, \([a, b] \subseteq E\).

Exercise 3.20

Exercise 3.21

Exercise 3.22

Exercise 3.23

First, we show that \(f(x) = x^n, x \in [0, \infty) \subseteq \mathbb{R}\) is a strictly increasing function. To show that it is strictly increasing, we need to show that \(x^n < (x + \varepsilon)^n\) for \(\varepsilon > 0\). We know that \(x^n < x^n + \varepsilon^n < (x + \varepsilon)^n\) for \(x \geq 0\) so \(f(x)\) is strictly increasing. Using Theorem 3.24, we can now see that its inverse function, \(f^{-1}(x) = \sqrt[n]{x}\) is continuous.

Exercise 3.24

Exercise 3.25

Exercise 3.26

Exercise 3.27

Since \(f\) is uniformly continuous, we know that for every \(\varepsilon > 0\) there is a \(\delta > 0\) such that if \(|x - y| < \delta\) then \(|f(x) - f(y)| < \varepsilon\) for any \(x, y \in E\). Therefore, for any point \(a \in E\), then for all \(\varepsilon > 0\) there exists \(\delta > 0\) such that if \(|x - a| < \delta\) then \(|f(x) - f(a)| < \varepsilon\) for any \(x \in E\). This satisfies the definition of continuous so \(f\) is continuous at every point in \(E\).

Exercise 3.28

Chapter 4: Differentiation

11/19 problems solved

Exercise 4.1

Since \(\lim_{x \to c} f(x) = L\), we know that for all \(\varepsilon > 0\), there exists \(\delta > 0\) such that for all \(x \in U_\delta(c)\), \(L - \varepsilon < f(x) < L + \varepsilon\).

Multiplying by \(k > 0\), we get \(kL - k\varepsilon < k \cdot f(x) < kL + k\varepsilon\). For \(\varepsilon' = k\varepsilon > 0\), we know there exists \(\delta > 0\) where all \(x \in U_\delta(c)\) satisfies \(kL - \varepsilon' < k \cdot f(x) < kL + \varepsilon'\). Therefore, \(kf\) converges to \(kL\) as \(x\) approaches \(c\) for \(k > 0\). If \(k < 0\), then changing \(\varepsilon'\) to \(-k\varepsilon\) maintains the same inequality. When \(k = 0\), it is trivial to show the same inequality is true. Therefore, \(kf\) converges to \(kL\) as \(x\) approaches \(c\).

Exercise 4.2

We know that for all \(\varepsilon > 0\), there exists \(\delta > 0\) such that for any \(x \in U_\delta(c)\), \(-\varepsilon < |f(x)| < \varepsilon\). If \(f(x) > 0\), then \(-\varepsilon < f(x) < \varepsilon\). Otherwise, \(-\varepsilon < -f(x) < \varepsilon\) which means \(\varepsilon > f(x) > -\varepsilon\). Therefore, \(-\varepsilon < f(x) < \varepsilon\) so \(\lim_{x \to c} f(x) = 0\).

Exercise 4.3

If \(\lim_{x \to c} f(x) = L\), then for any \(\varepsilon > 0\) there exists \(\delta > 0\) such that for all \(x \in U_\delta(c)\), \(L - \varepsilon < f(x) < L + \varepsilon\). The same holds for \(\lim_{x \to 0} f(x + c) = L\) except for all \(x \in U_\delta(0)\), \(L - \varepsilon < f(x + c) < L + \varepsilon\). We can write any \(x \in U_\delta(c)\) as a \(x' \in U_\delta(0)\) plus \(c\). We can also go the other way for any \(x \in U_\delta(0)\) written as a \(x' \in U_\delta(c)\) minus \(c\). Therefore, the two claims are equivalent.

Exercise 4.4

Exercise 4.5

Exercise 4.6

(a)

\[ \begin{align} (f + g)'(c) &= \lim_{x \to c} \frac{(f + g)(x) - (f + g)(c)}{x - c} \\ &= \lim_{x \to c} \frac{f(x) + g(x) - f(c) - g(c)}{x - c} \\ &= \lim_{x \to c} \frac{f(x) - f(c)}{x - c} + \lim_{x \to c} \frac{g(x) - g(c)}{x - c} \\ &= f'(c) + g'(c) \\ \end{align} \]

(b)

\[ \begin{align} (kf)'(c) &= \lim_{x \to c} \frac{(kf)(x) - (kf)(c)}{x - c} \\ &= \lim_{x \to c} \frac{k \cdot f(x) - k \cdot f(c)}{x - c} \\ &= k \cdot \lim_{x \to c} \frac{f(x) - f(c)}{x - c} \\ &= k f'(c) \\ \end{align} \]

Exercise 4.7

(a)

We can show that \((x^n)' = nx^{n-1}\) for \(n > 0\) using induction. When \(n = 1\), we need to prove that \((x)' = 1\). This is simple since \(\lim_{t \to x} \frac{t - x}{t - x} = 1\). Now we assume \((x^n)' = nx^{n-1}\) and must show that \((x^{n+1})' = (n+1)x^n\). We can do this by following the following equalities \((x^{n+1})' = (x \cdot x^n)' = (x)' \cdot x^n + x \cdot (x^n)' = x^n + x \cdot nx^{n-1} = (n+1)x^n\). Therefore, \((x^n)' = nx^{n-1}\) for \(n > 0\).

(b)

(c)

Exercise 4.8

\[ \begin{align} \frac{d}{dx} \sqrt{x + \sqrt{x + \sqrt{x}}} &= \frac{1}{2}\left(x + \sqrt{x + \sqrt{x}}\right)^{-1/2}\left(\frac{d}{dx}\left(x + \sqrt{x + \sqrt{x}}\right)\right) \\ &= \frac{1}{2}\left(x + \sqrt{x + \sqrt{x}}\right)^{-1/2}\left(1 + \frac{1}{2}(x + \sqrt{x})^{-1/2}\right)\left(\frac{d}{dx}\left(x + \sqrt{x}\right)\right) \\ &= \frac{1}{2}\left(x + \sqrt{x + \sqrt{x}}\right)^{-1/2}\left(1 + \frac{1}{2}(x + \sqrt{x})^{-1/2}\right)\left(1 + \frac{1}{2}x^{-1/2}\right) \\ \end{align} \]

Exercise 4.9

False, consider \(f(x) = |x|\) and \(g(x) = -|x|\). Both functions are not differentiable at \(x = 0\). However, \(f(x) + g(x) = |x| - |x| = 0\) is differentiable at \(x = 0\).

Exercise 4.10

If \(f(x) = g(x) + C\), then the derivative of \(f\) is \(g'(x) + 0 = g'(x)\). Therefore, \(f'(x) = g'(x)\).

If \(f'(x) = g'(x)\), then \(f'(x) - g'(x) = 0\). Since the derivative of \(f(x) - g(x)\) is \(0\), then \(f(x) - g(x) = C\) for some constant \(C\) since the field of real numbers is complete and satisfies the Zero Derivative Property. Therefore, \(f(x) = g(x) + C\).

Exercise 4.11

Exercise 4.12

Exercise 4.13

Exercise 4.14

Since \(x_1 < x_2\), there exists \(c\) such that \(x_2 = x_1 + c\). Moreover, since \(x \in [x_1, x_2]\), there exists \(x = x_1 + d\).

Suppose for the sake of contradiction that there exist \(\lambda, \lambda' \in [0, 1]\) such that \(x = \lambda x_1 + (1 - \lambda) x_2\) and \(x = \lambda' x_1 + (1 - \lambda') x_2\). Rewriting \(x\) and \(x_2\) above and simpifying gives us \(d = x_1 + (1 - \lambda) c\) and \(d = x_1 + (1 - \lambda') c\). These two equations show us that \(\lambda = \lambda'\).

Exercise 4.15

\[ \begin{align} f(x) &\leq \frac{x_2 - x}{x_2 - x_1} f(x_1) + \frac{x - x_1}{x_2 - x_1} f(x_2) \\ (x_2 - x_1) f(x) &\leq (x_2 - x) f(x_1) + (x - x_1) f(x_2) \\ (x_2 - x + x - x_1) f(x) &\leq (x_2 - x) f(x_1) + (x - x_1) f(x_2) \\ (x_2 - x) f(x) + (x - x_1) f(x) &\leq (x_2 - x) f(x_1) + (x - x_1) f(x_2) \\ (x_2 - x) f(x) - (x_2 - x) f(x_1) &\leq (x - x_1) f(x_2) - (x - x_1) f(x) \\ (x_2 - x) (f(x) - f(x_1)) &\leq (x - x_1) (f(x_2) - f(x)) \\ \frac{f(x) - f(x_1)}{x - x_1} &\leq \frac{f(x_2) - f(x)}{x_2 - x} \\ \end{align} \]

Exercise 4.16

First, we show that \(\min\{x_1, \dotsc, x_n\} \leq x\). Since \(x = \lambda_1 x_1 + \dotsb + \lambda_n x_n\), we can rewrite \(x\) so the inequality is \(\min\{x_1, \dotsc, x_n\} \leq \lambda_1 x_1 + \dotsb + \lambda_n x_n\). Let the minimum of \(x_1, \dotsc, x_n\) be \(c\). Since \(c \leq x_k\) for all \(1 \leq k \leq n\), we can write \(d_k = x_k - c > 0\). Substituting \(x_k\) in the original inequality shows us \(c \leq \lambda_1 (c + d_1) + \dotsb + \lambda_n (c + d_n)\) which means \(c \leq (\lambda_1 + \dotsb + \lambda_n) c + (\lambda_1 d_1 + \dotsb + \lambda_n d_n)\). Since \(\lambda_1 + \dotsb + \lambda_n = 1\), \(c \leq c + \lambda_1 d_1 + \dotsb + \lambda_n d_n\) and so this inequality is true.

The argument to show that \(x \leq \max\{x_1, \dotsc, x_n\}\) follows a similar structure. We let \(c\) be the maximum of \(x_1, \dotsc, x_n\) and since \(x_k \leq c\) for all \(1 \leq k \leq n\), we can write \(d_k = c - x_k > 0\). This gives us \(\lambda_1 (c - d_1) + \dotsb + \lambda_n (c - d_n) \leq c\) so \((\lambda_1 + \dotsb + \lambda_n) c - (\lambda_1 d_1 + \dotsb + \lambda_n d_n) \leq c\) and so \(c - (\lambda_1 d_1 + \dotsb + \lambda_n d_n) \leq c\). Therefore, this inequality is true as well.

Exercise 4.17

Exercise 4.18

To show the power function is a convex function over the interval \((a, b)\) where \(a \geq 0\), we can show that its second derivative is always greater than or equal to 0. If \(f(x) = x^p\), then \(f''(x) = p(p - 1) \cdot x^{p - 2}\). Since \(p \geq 1\), \(p(p - 1) \geq 0\). Moreover, \(x^{p - 2} > 0\) for all \(x \in (a, b)\) and the following proves this inequality. Since \(p - 2\) is a rational number, we can write \(p - 2 = a/b\) and so \(\sqrt[b]{x^a}\). Since \(x > 0\), \(x^a > 0\) and taking the \(b\)-th root of a positive number will also be positive so \(x^{p - 2} > 0\). Since \(p(p - 1) \geq 0\) and \(x^{p - 2} > 0\), this means \(p(p - 1) x^{p - 2} \geq 0\) and so the power function is convex.

Exercise 4.19

Chapter 5: Integration

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Exercise 5.1

We can prove that a finite subset of an ordered field containing more than one element can be enumerated by using induction on the size of the subset \(n\). For the base case of \(n = 2\), it is obvious that the two elements can be sorted whether the first is smaller than the second or vice versa. For the inductive case, we can assume that we can order a finite subset of size \(n\) and must show that a finite subset of size \(n + 1\) can be ordered in the same way.

Exercise 5.2

Let \(n\) be the number of elements in the nonempty subset of an ordered field. If \(n = 1\), then the maximum element of the subset is the only element in the subset. For the inductive case, we assume that there is a maximum element for an \(n\) element subset and must show there is a maximum element for a subset with \(n + 1\) elements. We can find the maximum of \(n\) elements in the \(n + 1\) subset and compare this maximum to the leftover element. If the leftover element is greater than the maximum, then it is the maximum of the entire subset due to transitivity. Otherwise, the maximum is still the maximum.

Exercise 5.3

We only need to show that a refinement with one extra element satisfies these since refinements with more elements can be created through successive refinements adding an additional element each time. Let \(q \in Q\) be this element. By definition of refinement, \(P \subset Q\) and \(q \notin P\). We can easily see that all the subintervals of \(Q\) are the same as \(P\) except for the subinterval containing \(q\). For this subinterval in \(Q\), it is split into two at \(q\). Both of these are subintervals of the original subinterval in \(P\). Therefore, the first claim is proven.

Again, we only prove that \(|Q| \leq |P|\) for a refinement with one extra element. If this extra element \(q\) is in the largest subinterval, then this subinterval must shrink in \(Q\) making \(|Q| < |P|\). Otherwise, the largest subinterval does not change so \(|Q| = |P|\). Therefore, \(|Q| \leq |P|\).

Exercise 5.4

(a)

For any \(x \in \mathcal{I}(\mathbb{F})\), we know that for every \(n\) that \(|x| < 1/n\). Since \(1/n \leq 1\), this means \(|x| < 1\) by transitivity. Therefore, \(x\) must also be in \(\mathcal{F}(\mathbb{F})\) so \(\mathcal{I}(\mathbb{F}) \subseteq \mathcal{F}(\mathbb{F})\).

(b)

Suppose for the sake of counterargument that there exists \(x \in \mathbb{F}\) that is not in \(\mathcal{F}(\mathbb{F}) \cup \mathcal{L}(\mathbb{F})\). This \(x\) could not satisfy either requirements of the sets. That is, \(|x|\) must be greater or equal to than any \(n \in \mathbb{N}\). However, if \(|x| = n\) then \(|x| < n + 1\) so \(|x| > n\) for all \(n \in \mathbb{N}\). This is the exact requirement for \(x\) to be in \(\mathcal{L}(\mathbb{F})\).

(c)

If \(x \in \mathcal{L}(\mathbb{F})\), then by definition \(|x| > n\) for any \(n \in \mathbb{N}\). If this same \(x\) was in \(\mathcal{F}(\mathbb{F})\), then there would be a natural number that \(|x|\) was less than. However, this would contradict the requirement to be in \(\mathcal{L}(\mathbb{F})\). Therefore, \(\mathcal{F}(\mathbb{F}) \cap \mathcal{L}(\mathbb{F}) = \emptyset\).

Exercise 5.5

To show that \(\mathcal{I}(\mathbb{F})\) is closed under addition, suppose there exists \(x + y \notin \mathcal{I}(\mathbb{F})\) where \(x, y \in \mathcal{I}(\mathbb{F})\). In other words, there exists \(p \in \mathbb{N}\) where \(|x + y| \geq 1/p\). Since \(x, y \in \mathcal{I}(\mathbb{F})\), \(|x|, |y| < 1/3p\). The following set of equalities and inequalities show that \(x + y\) must be in \(\mathcal{I}(\mathbb{F})\).

\[ |x + y| \leq |x| + |y| < \frac{1}{3p} + \frac{1}{3p} = \frac{2}{3p} < \frac{1}{p} \]

To show that \(\mathcal{L}(\mathbb{F}) \cap \mathbb{F}^+\) is closed under addition, suppose there exists \(x + y \notin \mathcal{L}(\mathbb{F}) \cap \mathbb{F}^+\) where \(x, y \in \mathcal{L}(\mathbb{F}) \cap \mathbb{F}^+\). In other words, there exists \(p \in \mathbb{N}\) where \(|x + y| \leq p\). Since \(x, y \in \mathcal{L}(\mathbb{F}) \cap \mathbb{F}^+\), \(|x| > 3p\) and \(|y| > p\). The following set of equalities and inequalities show that \(x + y\) must be in \(\mathcal{L}(\mathbb{F}) \cap \mathbb{F}^+\).

\[ |x + y| \geq ||x| - |-y|| > |3p - p| = 2p > p \]

To show \(x - y \in \mathcal{L}(\mathbb{F})\), ...

Exercise 5.6

Exercise 5.7

Exercise 5.8

Exercise 5.9

Exercise 5.10

Since \(g(x) \leq f(x)\), then \(0 \leq f(x) - g(x) = (f - g)(x)\). Since \(f - g\) is a non-negative function, that means \(\int_a^b (f - g) \geq 0\). From the linearity property of the Riemann integral, \(\int_a^b f - \int_a^b g \geq 0\), and so \(\int_a^b g \leq \int_a^b f\).

Exercise 5.11

Exercise 5.12

Exercise 5.13

Exercise 5.14

Exercise 5.15

Exercise 5.16

Exercise 5.17

(a)

For any \(x, y \in \mathbb{F}\) where \(x < y\), \(x^+ \leq y^+\) and \(x - x^+ \leq y - y^+\). If both are less than or equal to 0, then \(x^+ = 0 \leq 0 = y^+\) and \(x - x^+ = x - 0 = x \leq y = y - 0 = y - y^+\). If both are greater than 0, then \(x^+ = x \leq y = y^+\) and \(x - x^+ = x - x = 0 \leq 0 = y - y = y - y^+\). If x is less than or equal to 0 and y is greater than 0, then \(x^+ = 0 \leq y = y^+\) and \(x - x^+ = x - 0 = x \leq 0 = y - y = y - y^+\). Therefore, in every case \(x^+\) and \(x - x^+\) are increasing.

(b)

If \(x \leq 0\), then \(x^+ = 0\) and \(x^- = x\) and if \(x > 0\), then \(x^+ = x\) and \(x^- = 0\). Therefore, we can see that \(|x| = x^+ - x^-\) is true since when \(x \leq 0\), the equation turns into \(|x| = 0 - x = -x\) and when \(x > 0\), the equation turns into \(|x| = x - 0 = x\). Therefore, the equation is true for all \(x\).

Exercise 5.18

Chapter 6: Infinite Series

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Exercise 6.1

Exercise 6.2

Exercise 6.3

Exercise 6.4

Exercise 6.5

Exercise 6.6

Exercise 6.7

Exercise 6.8

Exercise 6.9

Exercise 6.10

Exercise 6.11

Exercise 6.12

Exercise 6.13

Exercise 6.14

Appendix A: Natural Numbers and Integers

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Exercise A.1

We can see that \(s\) is one-to-one from P2 of the Peano Axioms.

Every element in \(N \setminus \{1\}\) can be expressed as \(a + s(b)\). From Theorem 1, there is \(s(a + b)\) that equals this element so \(s\) is onto.

Therefore, \(s\) is a bijection.

This can be done over and over with no end and the resulting set can still have a bijection to a smaller one. Therefore, the set \(N\) must be an infinite collection of elements.

Exercise A.2

From P1 of the Peano Axioms, we know that \(s(1) \neq 1\). Given that \(s(a) \neq a\), then \(s(s(a)) \neq s(a)\) from P2 of the Peano Axioms.

Exercise A.3

Given \(a \in N\), define \(M = \{b \in N : a + b \neq b\}\). We can see that \(1 \in M\) since \(a + 1 \neq 1\). Moreover, if \(c \in M\) then \(s(c)\) must also be in \(M\) since \(a + c \neq c\) implies \(s(a + c) \neq s(c)\) and finally \(a + s(c) \neq s(c)\).

From the Axiom of Induction, \(M = N\) so \(a + b \neq b\) for all \(a, b \in N\).

Exercise A.4

(a) Suppose there are two different least elements \(a, b \in N\). By definition, \(a \leq b\) and \(b \leq a\). It is not possible for either \(a < b\) or \(b < a\) without contradicting prior definitions. Therefore, \(a = b\).

(b) If there was an element \(a \in N\) that was not 1 that was the least element of \(N\), then there exists \(n\) such that \(1 = a + n\). However, from P1 of the Peano Axioms, \(s(a) \neq 1\) for all \(a\). This is a contradiction so 1 must be the least element of \(N\).

Suppose there was a greatest element of \(N\) called \(a\). This element is greater than or equal to all other elements in \(N\). However, \(s(a) > a\) which is a contradiction. Therefore, there is no greatest element of \(N\).

Exercise A.5

We know that \((m', n') \sim (m, n)\) and \((p', q') \sim (p, q)\). By definition, this means \(m' + n = n' + m\) and \(p' + q = q' + p\).

For the equality to hold, this requires \((m' + p', n' + q') \sim (m + p, n + q)\). In turn, this means \((m' + p') + (n + q) = (n' + q') + (m + p)\). Using the equalities above, this is true.

Exercise A.6

\[ \begin{align} [m', n'] \cdot [p', q'] &= [m, n] \cdot [p, q] \\ [m' q' + n' p', n' q' + m' p'] &= [m q + n p, n q + m p] \\ \end{align} \]

To prove the above equality, we want to show that \[(m' q' + n' p') + (n q + m p) = (n' q' + m' p') + (m q + n p)\]

Since \((m', n') \sim (m, n)\) and \((p', q') \sim (p, q)\), this means \(m' + n = n' + m\) and \(p' + q = q' + p\).

Exercise A.7

There must be a zero element due to condition R3. To show the zero element is unique, assume there are two zero elements, \(0\) and \(0'\), for the sake of contradiction. By R1, \(0 + 0' = 0' + 0\). Since both are zero elements, this simplifies to \(0 = 0'\). Therefore, the zero element is unique.

Exercise A.8

There must exist a unique additive inverse \(-a\) due to R3. To show this additive inverse is unique, assume there are two additive inverses, \(-a\) and \(-a'\), for the element \(a\). From R2, \(-a + (a + -a') = (-a + a) + -a'\). Since they are additive inverses, ths simplifies to \(-a = -a'\). Therefore, the additive inverse is unique.

Exercise A.9

(a) \(R = \{0\}\)

By exhaustion, we can see that for all elements in \(\{0\}\),

\[ 0 + 0 = 0 + 0 \\ 0 + (0 + 0) = (0 + 0) + 0 \\ \text{There is } 0 \in R \text{ such that } 0 + 0 = 0 \\ 0 \cdot (0 \cdot 0) = (0 \cdot 0) \cdot 0 \\ 0 \cdot (0 + 0) = 0 \cdot 0 + 0 \cdot 0 \text{ and } (0 + 0) \cdot 0 = 0 \cdot 0 + 0 \cdot 0 \]

All of these expressions evaluate to 0.

(b) \(R = \mathbb{Z}_2\)

Proving R1: When \(a = b\), this is proven by reflexivity. For \(0 + 1 = 1 + 0\), both sides evaluate to \(1\).

Proving R2: When \(a = 0\), both sides simplify to \(b + c = b + c\) which is true by reflexivity. Looking at the case where \(a = 1\), when \(b = 0\) both sides simplify to \(1 + c = 1 + c\) which is true by reflexivity. Now when \(a = b = 1\), we can look at both cases where \(c = 0\) and \(c = 1\) and see the equality holds and is equal to \(0\) and \(1\) respectively.

Proving R3: When \(a = 0\), then \(x = 0\) for \(b = 0\) or \(x = 1\) for \(b = 1\). When \(a = 1\), then \(x = 1\) for \(b = 0\) or \(x = 0\) for \(b = 1\).

Proving R4: When \(a = 0\) or \(b = 0\) or \(c = 0\), the product is \(0\) on both sides. When \(a = b = c = 1\), then both sides evaluate to \(1\).

Proving R5: For \(a \cdot (b + c)\), when \(a = 0\) this is true since \(0\) or \(1\) times \(0\) is \(0\). When \(a = 1\), then the equality evaluates to \(b + c = b + c\) which is trivially true by reflexivity.

(c) \(R = 2\mathbb{Z}\)

Since addition and multiplication are taken from \(\mathbb{Z}\), R1, R2, R4, and R5 are all still true. To show R3 is true, we must show \(x \in 2\mathbb{Z}\). Fix \(a = 2m\) and \(b = 2n\). Solving for \(x\), we find \(x = 2n - 2m = 2(n - m)\). Therefore, \(x \in 2\mathbb{Z}\).

(d) \(R = 2^X\)

Proving R1:

\(A \bigtriangleup B = (A \setminus B) \cup (B \setminus A) = (B \setminus A) \cup (A \setminus B) = B \bigtriangleup A\)

Proving R2:

\(A \bigtriangleup (B \bigtriangleup C) = A \bigtriangleup ((B \setminus C) \cup (C \setminus B)) = (A \setminus ((B \setminus C) \cup (C \setminus B))) \cup (((B \setminus C) \cup (C \setminus B)) \setminus A) \)

(e) \(R = \langle F, +, \cdot \rangle\)

Exercise A.10

From Theorem 18, every element has a unique additive inverse. Therefore, if we add the additive inverse of \(c\), \(-c\), to both sides of \(a + c = b + c\), then these cancel to leave us with \(a + b\).

Exercise A.11

For the sake of contradiction, let's assume there are two identity elements of a ring, \(1\) and \(1'\). This means that \(1 \cdot 1' = 1\). However, since this is the identity element, it also means that \(1 \cdot 1' = 1' \cdot 1 = 1'\). Therefore, \(1 = 1'\).

Exercise A.12

Exercise A.13

Exercise A.14