A solutions manual for Topology by James Munkres
4. The Integers and the Real Numbers
1. Prove the following “laws of algebra” for \(\mathbb{R}\), using only axioms (1)–(5):
\(\quad\)(a) If \(x+y=x\), then \(y=0\).
\(\quad\)(b) \(0\cdot x=0\). [Hint: Compute \((x+0)\cdot x\).]
\(\quad\)(c) \(-0=0\).
\(\quad\)(d) \(-(-x) = x\).
\(\quad\)(e) \(x(-y) = -(xy) = (-x)y\).
\(\quad\)(f) \((-1)x=-x\).
\(\quad\)(g) \(x(y-z)=xy-xz\).
\(\quad\)(h) \(-(x+y)=-x-y;-(x-y)=-x+y\).
\(\quad\)(i) If \(x\neq 0\) and \(x\cdot y=x\), then \(y=1\).
\(\quad\)(j) \(x/x=1\) if \(x\neq 0\).
\(\quad\)(k) \(x/1=x\).
\(\quad\)(l) \(x\neq 0\) and \(y \neq 0\Rightarrow xy \neq 0\).
\(\quad\)(m) \((1/y)(1/z) = 1/(yz)\) if \(y, z \neq 0\).
\(\quad\)(n) \((x/y)(w/z) = (xw)/(yz) if y, z \neq 0\).
\(\quad\)(o) \((x/y)+(w/z)=(xz+wy)/(yz)\) if \(y,z\neq 0\).
\(\quad\)(p) \(x \neq 0\Rightarrow 1/x \neq 0\).
\(\quad\)(q) \(1/(w/z)=z/w\) if \(w,z\neq 0\).
\(\quad\)(r) \((x/y)/(w/z) = (xz)/(yw)\) if \(y, w, z \neq 0\).
\(\quad\)(s) \((ax)/y=a(x/y)\) if \(y \neq 0\).
\(\quad\)(t) \((-x)/y = x/(-y) = -(x/y)\) if \(y\neq 0\).
Proof. \(\quad\)(a) \(-x+x+y=-x+x \Rightarrow y=0\).
\(\quad\)(b) \(x\cdot x=(x+0)\cdot x=x\cdot x+0\cdot x\Rightarrow 0\cdot x=0\).
\(\quad\)(c) \(-0=(-0)+0=0\).
\(\quad\)(d) \(-(-x)=-(-x)+(-x)+x=x\).
\(\quad\ldots\)
2. Prove the following “laws of inequalities” for R, using axioms (1)–(6) along with the results of Exercise 1:
\(\quad\)(a) \(x>y\) and \(w>z\Rightarrow x+w>y+z\).
\(\quad\)(b) \(x>0\) and \(y>0\Rightarrow x+y>0\) and \(x\cdot y>0\).
\(\quad\)(c) \(x >0\Leftrightarrow -x <0\).
\(\quad\)(d) \(x>y\Leftrightarrow -x<-y\).
\(\quad\)(e) \(x>y\) and \(z<0\Rightarrow xz<yz\).
\(\quad\)(f) \(x\neq 0\Rightarrow x^2>0\), where \(x^2=x\cdot x\).
\(\quad\)(g) \(-1<0<1\).
\(\quad\)(h) \(x y > 0 \Leftrightarrow x\) and \(y\) are both positive or both negative.
\(\quad\)(i) \(x >0\Rightarrow 1/x >0\).
\(\quad\)(j) \(x>y>0\Rightarrow 1/x<1/y\).
\(\quad\)(k) \(x<y\Rightarrow x<(x+y)/2<y\).
Proof. \(\quad\)(a) \(x+w>x+z>y+z\).
\(\quad\)(b) \(x+y>0+0=0\), \(x\cdot y>0\cdot y=0\).
\(\quad\ldots\)
3. \(\quad\)(a) Show that if \(\mathcal{A}\) is a collection of inductive sets, then the intersection of the elements of \(\mathcal{A}\) is an inductive set.
\(\quad\)(b) Prove the basic properties (1) and (2) of \(\mathbb{Z}_+\).
Proof. \(\quad\)(a) Let \(A\in \mathcal{A}\). \(1\in A\), thus \(1\in \bigcap\mathcal{A}\). If \(x\in\bigcap\mathcal{A}\), then \(x\in A\); thus \(x+1\in A\), and so \(x+1\in\bigcap\mathcal{A}\). Therefore, \(\bigcap\mathcal{A}\) is an inductive set.
\(\quad\)(b) \(\mathbb{Z}_+\) is an inductive set by (a). If \(A\) is an inductive set, then \(\mathbb{Z}_+ \subset A\). Thus if \(A\) is an inductive set of positive integers, then \(A = \mathbb{Z}_+\).\(\quad\square\)
4. \(\quad\)(a) Prove by induction that given \(n \in \mathbb{Z}_+\), every nonempty subset of \(\{1,\ldots, n\}\) has a largest element.
\(\quad\)(b) Explain why you cannot conclude from (a) that every nonempty subset of \(\mathbb{Z}_+\) has a largest element.
Proof. \(\quad\)(a) Every nonempty subset of \(\{1\}\) has the largest element \(1\). Suppose that every nonempty subset of \(\{1,\ldots,n\}\) has a largest element. Let \(S\) be a nonempty subset of \(\{1,\ldots,n+1\}\). If \(n+1\in S\), then \(n+1\) is the largest element of \(S\). Otherwise \(S\) is a nonempty subset of \(\{1,\ldots,n\}\); thus \(S\) has a largest element, and so in either case, \(S\) has a largest element.
\(\quad\)(b) Every nonempty subset of \(\{1,\ldots,n\}\) is finite. But there is an infinite unbounded subset of \(\mathbb{Z}_+\).\(\quad\square\)
5. Prove the following properties of \(\mathbb{Z}\) and \(\mathbb{Z}_+\):
\(\quad\)(a) \(a,b \in \mathbb{Z}_+ \Rightarrow a + b \in \mathbb{Z}_+\). [Hint: Show that given \(a \in \mathbb{Z}_+\), the set \(X =\{x\mid x \in \mathbb{R}\) and \(a+x \in\mathbb{Z}_+\}\) is inductive.]
\(\quad\)(b) \(a,b\in\mathbb{Z}_+\Rightarrow a\cdot b\in\mathbb{Z}_+\).
\(\quad\)(c) Show that \(a\in\mathbb{Z}_+\Rightarrow a-1\in\mathbb{Z}_+ \cup \{0\}\). [Hint: Let \(X=\{x\mid x\in R\) and \(x - 1 \in \mathbb{Z}_+ \cup \{0\}\}\); show that \(X\) is inductive.]
\(\quad\)(d) \(c,d\in\mathbb{Z}\Rightarrow c+d\in\mathbb{Z}\) and \(c-d\in\mathbb{Z}\). [Hint: Prove it first for \(d=1\).]
\(\quad\)(e) \(c,d\in\mathbb{Z}\Rightarrow c\cdot d\in\mathbb{Z}\).
Proof. \(\quad\)(a) \(a+1\in\mathbb{Z}_+\); thus \(1\in X\). If \(x\in X\), then \(a+x\in\mathbb{Z}_+\); thus \(a+x+1=a+(x+1)\in\mathbb{Z}_+\), and so \(x+1\in X\). Hence \(X\) is inductive; thus \(\mathbb{Z}_+\subset X\), and so \(b\in X\). Therefore, \(a+b\in\mathbb{Z}_+\).
\(\quad\)(b) Define \(X =\{x \in \mathbb{R}\) \(\mid\) \(a\cdot x \in\mathbb{Z}_+\}\). Similarly to (a), \(X\) is inductive.
\(\quad\)(c) It is easy to show that \(1\in X\) and if \(x\in X\) then \(x+1\in X\). Thus \(\mathbb{Z}_+\subset X\). Therefore, \(a\in\mathbb{Z}_+\Rightarrow a-1\in\mathbb{Z}_+ \cup \{0\}\).
\(\quad\)(d) Define \(X =\{x \in \mathbb{R}\) \(\mid\) \(c+x\in\mathbb{Z}\) and \(c-x\in\mathbb{Z}\}\). We show that \(X\) is inductive. If \(c\in\mathbb{Z}_+\), then \(c+1\in\mathbb{Z}_+\subset\mathbb{Z}\). If \(c=0\), then \(c+1\in\mathbb{Z}\). If \(c\in\mathbb{Z}_-\), then \(c+1=-(-c-1)\); thus \(-c-1\in\mathbb{Z}_+ \cup \{0\}\) by part (c), and so \(-(-c-1)=c+1\in \mathbb{Z}_- \cup \{0\}\subset \mathbb{Z}\). Thus if \(c\in\mathbb{Z}\), then \(c+1\in\mathbb{Z}\). Similarly to the previous, if \(c\in\mathbb{Z}\), then \(c-1\in\mathbb{Z}\). Hence \(1\in X\). Also similarly to the previous, If \(x\in X\), then \(c+x\in\mathbb{Z}\) and \(c-x\in\mathbb{Z}\); thus \(c+x+1\in\mathbb{Z}\) and \(c-x-1=c-(x+1)\in\mathbb{Z}\), and so \(x+1\in X\). Hence \(X\) is inductive, and so for any \(c\in\mathbb{Z}\) and any \(d\in\mathbb{Z}_+\), \(c+d\in\mathbb{Z}\) and \(c-d\in\mathbb{Z}\). It is obvious that this holds for \(d\in\mathbb{Z}_-\) or \(d=0\). Therefore, this holds for \(d\in\mathbb{Z}\).
\(\quad\)(e) Similarly to the previous.\(\quad\square\)
6. Let \(a \in \mathbb{R}\). Define inductively \[ a^1 = a,\\ a^{n+1} = a^n \cdot a \] for \(n \in \mathbb{Z}_+\). (See §7 for a discussion of the process of inductive definition.) Show that for \(n, m \in \mathbb{Z}_+\) and \(a, b \in \mathbb{R}\), \[ a^na^m =a^{n+m},\\ (a^n)^m =a^{nm},\\ a^mb^m =(ab)^m. \] These are called the laws of exponents. [Hint: For fixed \(n\), prove the formulas by induction on \(m\).]
Proof. \(\quad\)All the laws of exponents hold for \(m=1\). It is easy to show that if they hold for some \(m\) then they also hold for \(m+1\).\(\quad\square\)
7. Let \(a\in \mathbb{R}\) and \(a\neq 0\). Define \(a^0 =1\), and for \(n\in\mathbb{Z}_+\), \(a^{-n} =1/a^n\). Show that the laws of exponents hold for \(a,b \neq 0\) and \(n,m \in \mathbb{Z}\).
Proof. \(\quad\)Trivial.
8. \(\quad\)(a) Show that \(\mathbb{R}\) has the greatest lower bound property.
\(\quad\)(b) Show that \(\text{inf}\{1/n\mid n\in\mathbb{Z}_+\}=0\).
\(\quad\)(c) Show that given \(a\) with \(0<a<1,\text{ inf}\{a^n \mid n\in\mathbb{Z}_+\}=0\). [Hint: Let \(h=(1-a)/a\), and show that \((1+h)^n \ge 1+nh\).]
Proof. \(\quad\)(a) By order property (7) in page 31, \(\mathbb{R}\) has the least upper bound property, and by Exercise 13 of §3, \(\mathbb{R}\) has the greatest lower bound property.
\(\quad\)(b) If \(n>0\), then \(1/n > 0\); thus \(0\) is a lower bound for \(\{1/n\mid n\in\mathbb{Z}_+\}\). If \(x>0\), then there is \(n>1/x\); thus \(1/n<x\), and so every positive number is not a lower bound for the set. Therefore, \(0\) is the greatest lower bound.
\(\quad\)(c) \((1+h)^1=1+1\cdot h\), and if \((1+h)^n\ge 1+nh\), then \((1+h)^{n+1}\ge(1+nh)(1+h)\) \(=\) \(1+(n+1)h+nh^2>1+(n+1)h\). Hence \((1+h)^n=(1/a)^n=1/a^n\ge 1+nh\), and we already have that for any \(x\), there exists \(n\) such that \(1+nh>1/x\) since \(\mathbb{Z}_+\) is unbounded; thus \(a^n<x\), and so there is no positive lower bound. Therefore, \(0\) is the greatest lower bound.\(\quad\square\)
9. \(\quad\)(a) Show that every nonempty subset of \(\mathbb{Z}\) that is bounded above has a largest element.
\(\quad\)(b) If \(x\notin \mathbb{Z}\), show there is exactly one \(n\in\mathbb{Z}\) such that \(n<x<n+1\).
\(\quad\)(c) If \(x-y>1\), show there is at least one \(n\in\mathbb{Z}\) such that \(y<n<x\).
\(\quad\)(d) If \(y<x\), show there is a rational number \(z\) such that \(y<z<x\).
Proof. \(\quad\)(a) Let a nonempty subset of \(\mathbb{Z}\) be \(A\). If \(A\subset \mathbb{Z}_-\), then \(\{-x\mid x\in A\}\) has the smallest element \(a\). \(-a\) is the largest element. If \(0\in A\) and \(A\) has no positive numbers, then \(0\) is the largest element. If \(A\) has positive numbers and \(n\) is a upper bound of \(A\), then by Exercise 4 (a), the largest element of \(A\cap\{1,\cdots,n\}\) is the largest element of \(A\).
\(\quad\)(b) Let \(A=\{a\in\mathbb{Z}\mid a<x\}\). \(A\) is bounded above, thus \(A\) has the largest element \(n\) such that \(n<x\le n+1\). Since \(x\notin\mathbb{Z}\), \(n<x<n+1\).
\(\quad\)(c) \(n<x\le n+1\) for some \(n\in\mathbb{Z}\). Since \(1+y<x\), \(y<n\); otherwise \(n+1<x\). Thus \(y<n<x\).
\(\quad\)(d) \(x-y>0\Rightarrow n>1/(x-y)>0\Rightarrow nx-ny>1\) for some \(n\in\mathbb{Z}\). By (c), there is \(m\in\mathbb{Z}\) such that \(ny<m<nx\); thus \(y<m/n<x\).\(\quad\square\)
10. Show that every positive number \(a\) has exactly one positive square root, as follows:
\(\quad\)(a) Show that if \(x>0\) and \(0\le h<1\), then \[
(x + h)^2 \le x^2 + h(2x + 1),\\
(x - h)^2 \ge x^2 - h(2x).
\] \(\quad\)(b) Let \(x>0\). Show that if \(x^2<a\), then \((x+h)^2<a\) for some \(h>0\); and if \(x^2 >a\), then \((x-h)^2 >a\) for some \(h>0\).
\(\quad\)(c) Given \(a > 0\), let \(B\) be the set of all real numbers \(x\) such that \(x^2 < a\). Show that \(B\) is bounded above and contains at least one positive number. Let \(b=\text{sup }B\); show that \(b^2 =a\).
\(\quad\)(d) Show that if \(b\) and \(c\) are positive and \(b =c\), then \(b=c\).
Proof. \(\quad\)Just follow the instructions.
11. Given \(m\in\mathbb{Z}\), we say that \(m\) is even if \(m/2\in\mathbb{Z}\), and \(m\) is odd otherwise.
\(\quad\)(a) Show that if \(m\) is odd, \(m=2n+1\) for some \(n\in\mathbb{Z}\). [Hint: Choose \(n\) so that \(n < m/2 < n + 1\).]
\(\quad\)(b) Show that if \(p\) and \(q\) are odd, so are \(p\cdot q\) and \(p^n\), for any \(n\in\mathbb{Z}_+\).
\(\quad\)(c) Show that if \(a>0\) is rational, then \(a=m/n\) for some \(m,n\in\mathbb{Z}_+\) where not both \(m\) and \(n\) are even. [Hint: Let \(n\) be the smallest element of the set \(\{x\mid x\in\mathbb{Z}_+\) and \(x\cdot a\in\mathbb{Z}_+\}\).]
\(\quad\)(d) Theorem. \(\sqrt{2}\) is irrational.
Proof. \(\quad\)(a) Since \(m\) is odd, \(m/2\notin\mathbb{Z}\); thus there is \(n\in\mathbb{Z}\) such that \(n<m/2 < n + 1\), and so \(2n<m<2n+2=(2n+1)+1\). Therefore, \(m=2n+1\).
\(\quad\)(b) Let \(p=2m+1\) and \(q=2n+1\) for some \(m,n\in\mathbb{Z}\). \(p\cdot q=2(2mn+m+n)+1\); thus \(p\cdot q\) is odd. \(p^1\) is odd. If \(p^n\) is odd, then \(p^{n+1}=p^n\cdot p\); thus \(p^{n+1}\) is odd, and so \(p^n\) is odd for any \(n\in\mathbb{Z}_+\).
\(\quad\)(c) Let \(m=na\). \(a=m/n\). If both \(m\) and \(n\) are even, then \((n/2)a=m/2\in\mathbb{Z}_+\) and \(n/2<n\), a contradiction.
\(\quad\)(d) Suppose that \(\sqrt{2}\) is rational. \(\sqrt{2}=n/m\) for some \(m,n\in\mathbb{Z}_+\) where not both \(m\) and \(n\) are even. \(2=n^2/m^2\). \(2m^2=n^2\); thus \(n^2\) is even, and so \(n\) is even. Hence \(n=2p\) for some \(p\in\mathbb{Z}_+\); thus \(2m^2=(2p)^2=4p^2\), and so \(m^2\) and \(m\) are even; a contradiction.\(\quad\square\)