Solutions to Chapter 1
Exercise 1.6¶
Suppose \(f:X\to Y\) and \(A\subseteq X\) and \(B\subseteq Y\).
- Prove that \(f(f^{-1}(B))\subseteq B\)
- Give an example where \(f(f^{-1}(B))\neq B\)
- Prove that \(A\subseteq f^{-1}(f(A))\)
- Give an example where \(A\neq f^{-1}(f(A))\)
Answer
Helpful things to know for this exercise :
- \(x\in A \Rightarrow f(x) \in f(A)\) and
- \(y\in f^{-1}(B) \iff f(y)\in B\)
Part 1. Suppose \(x\in f(f^{-1}(B))\). This means there exists some \(y\in f^{-1}(B)\) such that \(f(y)=x\), and since
\(y\in f^{-1}(B) \ \Longrightarrow \ f(y)\in B\), thus \(x\in B\) as needed.
Part 2. Consider \(X=\{x_1\}\) and \(Y=\{y_1,y_2\}\) with a function \(f:X\to Y\) such that \(f(x_1)=y_1\). Set \(B\) to be the subset containing all of \(Y\), then \(f^{-1}(B)=\{x_1\}\) whose image contains just \(y_1\) and is hence unequal to \(B\).
Part 3. Suppose \(x\in A\), then \(f(x)\in f(A)\) then \(x\in f^{-1}(f(A))\) (reverse direction of point 2 in helpful things to know) as needed.
Part 4. Consider \(X=\{x_1, x_2\}\) and \(Y=\{y_1\}\) with a function \(f:X\to Y\) such that \(f(x_1)=f(x_2)=y_1\). Set \(A\) to be the subset containing only \(x_1\), then \(f(A)=\{y_1\}\) and the inverse image of that contains both \(x_1,x_2\) and is thus unequal to \(A\).
Note. In general, equality is achieved in \(f(f^{-1}(B))\neq B\) iff \(f\) is surjective, while equality is achieved in \(A\subseteq f^{-1}(f(A))\) iff \(f\) is injective.
Exercise 1.13¶
Let \(a,b,\epsilon\) be elements of an ordered field.
- Show that if \(a<b+\epsilon\) for every \(\epsilon>0\), then \(a\leq b\).
- Use the first part to show that if \(|a-b|<\epsilon\) for all \(\epsilon>0\), then \(a=b\).
Answer
Part 1. Suppose \(b<a\), thus there must exist some positive \(x\in \mathbb{R}\) such that \(a=b+x\). Since \(a<b+\epsilon\) for every \(\epsilon>0\), setting \(\epsilon := x\) gives \(a<a\) which is a contradiction.
Part 2. This is equivalent to \(-\epsilon < a-b < \epsilon\) for all \(\epsilon\). We get two inequalities out of this :
- \(a<b+\epsilon\) for all \(\epsilon>0\)
- \(b<a+\epsilon\) for all \(\epsilon>0\)
By part 1, we get both \(a\leq b\) and \(b\leq a\) from these inequalities. Thus \(a=b\) as needed.
Exercise 1.14¶
Prove that \(|ab| = |a||b|\) for all real numbers \(a,b\).
Answer
- Define a function \(\text{sign}:\mathbb{R}\to \mathbb{R}\) which takes positive numbers to \(1\), negative numbers to \(-1\) and zero to itself. Then notice that :
- \(|a| = a\cdot \text{sign}(a)\) and
- \(\text{sign}(ab) = \text{sign}(a)\cdot \text{sign}(b)\) for all real \(a,b\).
The proof is straightforward via the properties of the sign function :
Exercise 1.17¶
Suppose \(a,b\in \mathbb{R}\) and \(0<a<b\). Prove that \(a^n<b^n\) for all \(n\in \mathbb{N}\).
Answer
Notice that \(1<b/a\), thus
In particular \(1<(b/a)^n\) for all \(n\). Thus,
for all \(n\) as needed.
Note. The general idea is that for any positive \(x\in \mathbb{R}\) we have \(x<cx\) whenever \(c>1\) and \(cx<x\) whenever \(c<1\) and obviously \(cx=x\) when \(c=1\). We prove this by cases.
Case 1. \(c>1\)
We can write \(c=1+y\) for some positive \(y\). Thus \(cx=(1+y)x=x+yx\). Note that \(yx\) is positive so \(cx\) is strictly greater than \(x\) as required.
Case 2. \(c<1\)
We can write \(1=c+y\) for some positive \(y\). Thus \(cx=(1-y)x=x-yx\). Note that \(yx\) is positive so \(cx\) is strictly less than \(x\) as required.
Exercise 1.21¶
Let \(f:X\to Y\) and suppose \(A_1,A_2\subseteq X\). Show that
\[ f(A_1\cap A_2) \subseteq f(A_1)\cap f(A_2) \]
Answer
Suppose \(x\in f(A_1\cap A_2)\), then there exists a \(y\in A_1\cap A_2\) such that \(f(y) = x\). In particular, \(y\in A_1\) so that \(x=f(y)\in f(A_1)\) and through similar reasoning \(x\in f(A_2)\). Thus \(x\in f(A_1)\cap f(A_2)\) as needed.
Exercise 1.22¶
Give an example of a function \(f\) and a pair of sets \(A,B\) for which
\[f(A\cap B) \neq f(A)\cap f(B)\]
Answer
Let \(f:\mathbb{R}\to \mathbb{R}\) where \(f(x)\mapsto x^2\) and \(A=\{0,2\}\) and \(B=\{0,-2\}\). Then the LHS computes to \(\{0\}\) while the RHS computes to \(\{0,4\}\).
Note. In general, equality is achieved iff the function is injective.
Exercise 1.23¶
Assume that \(A\subseteq B\subseteq \mathbb{R}\) and both are bounded above. Prove that \(\sup(A)\leq \sup(B)\)
Answer
The idea is to show that \(\sup(B)\) is an upper bound of \(A\), hence it should be greater than or equal to the least upper bound (sup) of \(A\). Consider
thus \(\sup(B)\) is an upper bound of \(A\) as required.
Exercise 1.30¶
Let \(A,B\subseteq \mathbb{R}\) be nonempty and bounded above and assume that \(\sup(A)<\sup(B)\)
- Show that there exists an element \(b\in B\) that is an upper bound of \(A\).
- Show that this is not necessarily the case if \(\sup(A)\leq \sup(B)\).
Answer
Part 1. By Theorem 1.24, we can choose an \(\epsilon>0\) and guarantee the existence of a \(b\in B\) such that \(\sup(B)-\epsilon<b\). The idea is to choose an \(\epsilon\) small enough so that \(\sup(A)< \sup(B)-\epsilon < b\). Thus \(b\) is an upper bound of \(A\) as required.
Part 2. Consider \(A=\{1\}\) and \(B=(0,1)\) then \(\sup(A)=\sup(B)=1\) but nothing in \(B\) can be an upper bound of \(A\).
Exercise 1.32¶
Suppose \(A\subseteq \mathbb{R}\) is nonempty and bounded above and \(c\in \mathbb{R}\). Define \(c+A = \{c+a \ \vert \ a\in A\}\) and \(cA = \{ca \ \vert \ a\in A\}\).
- Prove that \(\sup(c+A)=c+\sup(A)\).
- Determine necessary and sufficient conditions on \(c\) and \(A\) for \(\sup(cA)=c\sup(A)\). Give an example of a set \(A\) and a number \(c\) where \(\sup(cA)\neq c\sup(A)\).
Answer
Part 1. First, we show that \(c+\sup(A)\) is an upper bound of \(c+A\). Let \(x\in c+A\), then \(x=c+a\) for some \(a\in A\). Then \(a\leq \sup(A) \ \Rightarrow \ x = c+a\leq c+\sup(A)\). Thus, \(c+\sup(A)\) is an upper bound of \(c+A\).
Next, we show that for any \(\epsilon>0\) there exists an \(x\in c+A\) such that \(c+\sup(A)-\epsilon<x\). Fix an \(\epsilon>0\), then there exists an \(a\in A\) such that \(\sup(A)-\epsilon<a\). Adding \(c\) gives \(c+\sup(A)-\epsilon < a+c\). Since \(a+c\in c+A\) we have achieved our goal.
Part 2. This only holds iff \(c\geq 0\). We shall go case by case.
Case 1. \(c=0\).
No matter what \(A\) is, the set \(cA\) is just \(\{0\}\) so \(\sup(cA)=c\sup(A)=0\).
Case 2. \(c>0\).
We know that \(a\leq \sup(A)\) for any \(a\in A\). Multiplying by \(c\) gives \(ca\leq c\sup(A)\) for all \(a\in A\), thus \(c\sup(A)\) is an upper bound for \(cA\). Next, suppose \(y\) is an upper bound of \(cA\), then \(ca\leq y\) for all \(a\in A\). Dividing by \(c\) gives \(a\leq y/c\) for all \(a\in A\). Thus \(y/c\) is an upper bound of \(A\) so that \(\sup(A)\leq y/c \ Rightarrow \ c\sup(A)\leq y\). Thus \(c\sup(A)\) is the least upper bound for \(cA\).
Case 3. \(c<0\).
Let \(A=\{1,2\}\) and \(c=-1\). Then \(cA = \{-1,-2\}\) so that \(\sup(cA)=-1\) while \(c\sup(A)=-2\). More generally, \(\sup(cA)=cinf(A)\) whenever \(c<0\).
Exercise 1.34¶
For each \(n\in \mathbb{N}\), assume we are given a closed interval \(I_n = [a_n,b_n]\). Also assume that each \(I_{n+1}\) is contained inside \(I_n\). This gives a sequence of increasingly smaller intervals,
\[ I_1\supseteq I_2\supseteq I_3 \supseteq \cdots \]Prove that \(\bigcap_{n=1}^\infty{I_n}\) is nonempty.
Answer
This is obvious, but it is helpful to keep in mind that for any \(n,m\in \mathbb{N}\) such that \(n<m\)
- \(a_n\leq b_n\)
- \(a_n\leq a_m\)
- \(b_m\leq b_n\)
Set \(A=\{a_n \ \vert \ n\in \mathbb{N}\}\). We claim that \(\sup(A)\in \bigcap_{n=1}^\infty{I_n}\). First, we have to show that \(A\) is nonempty and bounded above (so that we know the supremum exists). \(A\) is clearly nonempty. Next, we will show that for any \(n,m\in \mathbb{N}\) we have \(a_n\leq b_m\). Fix an \(n\in \mathbb{N}\). We split by cases. For the case that \(n<m\), consider \(a_n\leq a_m\leq b_m\). For the case \(m\leq n\), consider \(a_n\leq b_n\leq b_m\). Thus every \(b_n\) is an upper bound of \(A\) so \(A\) is bounded above as required.
Next, showing that \(\sup(A)\in \bigcap_{n=1}^\infty{I_n}\) is equivalent to showing that \(a_n\leq \sup(A)\leq b_n\) for all \(n\). The first half of the inequality is obvious. For the second half, recall that each \(b_n\) is an upper bound of \(A\) so that each \(b_n\) has to be greater than or equal to \(\sup(A)\). Thus, we are done.
Exercise 1.35¶
Give an example showing that the conclusion of Exercise 1.34 need not hold if each \(I_n\) is allowed to be an open interval.
Answer
Consider \(I_n := (0, 1/n)\). To show that \(\bigcap_{n=1}^\infty\) is empty amounts to showing that there is no \(x\in \mathbb{R}\) such that \(0<x<1/n\) for all \(n\). Suppose there exists such an \(x\). Recall the Archimedean Principle : For any \(x\in \mathbb{R}\) there exists an \(n\in \mathbb{N}\) such that \(x<n\) (or equivalently \(x\leq n\)). Taking the inverse of this inequality gives \(1/n<x\) for some \(n\) and this contradicts our assumption.