5.1: CHAPTER 1

Exercise \(1.1.2\).

Answer

Applying basic rules of operations on sets yields \[(X \backslash Y) \cap Z=Y^{c} \cap Z=Z \backslash Y .\] and \[Z \backslash(Y \cap Z)=(Z \backslash Y) \cup(Z \backslash Z)=(Z \backslash Y) \cup \emptyset=Z \backslash Y .\] Therefore, \((X \backslash Y) \cap Z=Z \backslash(Y \cap Z)\).

Exercise \(1.2.1\).

Answer

1. For any \(a \in A\), we have \(f(a) \in f(A)\) and, so, \(a \in f^{-1}(f(A))\). This implies \(A \subset f^{-1}(f(A))\). Note that this inclusion does not require the injectivity of \(f\). Now fix any \(a \in f^{-1}(f(A))\). Then \(f(a) \in f(A)\), so there exists \(a^{\prime} \in A\) such that \(f(a)=f\left(a^{\prime}\right)\). Since \(f\) is one-to-one, \(a=a^{\prime} \in A\). Therefore, \(f^{-1}(f(A)) \subset A\) and the equality holds.
2. Fix any \(b \in f\left(f^{-1}(B)\right)\). Then \(b = f(x)\) for some \(x \in f^{-1}(B)\). Thus, \(b = f(x) \in B\) and, hence, \(f\left(f^{-1}(B)\right) \subset B\). This inclusion does not require the surjectivity of \(f\). Now fix \(b \in B\). Since \(f\) is onto, there exists \(x \in X\) such that \(f(x) = b \in B\). Thus, \(x \in f^{-1}(B)\) and, hence, \(b \in f\left(f^{-1}(B)\right)\). We have shown that \(B \subset f\left(f^{-1}(B)\right)\) and the equality holds.

Without the injectivity of \(f\), the equlaity part (a) is no longer valid. Consider \(f(x)=x^{2}\), \(x \in \mathbb{R}\), and \(A=[-1,2]\). Then \(f(A)=[0,4]\) and, hence, \(f^{-1}(f(A))=[-2,2]\), which strictly contains \(A\). It is also not hard to find an example of a function \(f\) and a set \(B\) for which the equality in part (b) does not hold true.

Exercise \(1.3.6\).

Answer

For \(n = 1\), \[\frac{1}{\sqrt{5}}\left[\frac{1+\sqrt{5}}{2}-\frac{1-\sqrt{5}}{2}\right]=\frac{1}{\sqrt{5}} \frac{2 \sqrt{5}}{2}=1 .\] Thus, the conclusion holds for \(n = 1\). It is also easy to verify that the conclusion holds for \(n = 2\).

Suppose that \[a_{k}=\frac{1}{\sqrt{5}}\left[\left(\frac{1+\sqrt{5}}{2}\right)^{k}-\left(\frac{1-\sqrt{5}}{2}\right)^{k}\right]\] for all \(k \leq n\), where \(n \geq 2\). Let us show that \[a_{n+1}=\frac{1}{\sqrt{5}}\left[\left(\frac{1+\sqrt{5}}{2}\right)^{n+1}-\left(\frac{1-\sqrt{5}}{2}\right)^{n+1}\right] .\] By the definition of the sequence and the induction hypothesis, \[\begin{aligned}
a_{n+1} &=a_{n}+a_{n-1} \\
&=\frac{1}{\sqrt{5}}\left[\left(\frac{1+\sqrt{5}}{2}\right)^{n}-\left(\frac{1-\sqrt{5}}{2}\right)^{n}\right]+\frac{1}{\sqrt{5}}\left[\left(\frac{1+\sqrt{5}}{2}\right)^{n-1}-\left(\frac{1-\sqrt{5}}{2}\right)^{n-1}\right] \\
&=\frac{1}{\sqrt{5}}\left[\left(\frac{1+\sqrt{5}}{2}\right)^{n-1}\left(\frac{1+\sqrt{5}}{2}+1\right)-\left(\frac{1-\sqrt{5}}{2}\right)^{n-1}\left(\frac{1-\sqrt{5}}{2}-1\right)\right] \text {.}
\end{aligned}\] Observe that \[\frac{1+\sqrt{5}}{2}+1=\frac{3+\sqrt{5}}{2}=\left(\frac{1+\sqrt{5}}{2}\right)^{2} \text { and } \frac{1-\sqrt{5}}{2}+1=\frac{3-\sqrt{5}}{2}=\left(\frac{1-\sqrt{5}}{2}\right)^{2} .\] Therefore, (5.1) follows easily.

In this exercise, observe that the two numbers \(\frac{1+\sqrt{5}}{2}\) and \(\frac{1-\sqrt{5}}{2}\) are the roots of the quadratic equation \[x^{2}=x+1 .\] A more general result can be formulated as follows. Consider the sequence \(\left\{a_{n}\right\}\) defined by \[\begin{array}{l}
a_{1}=a \text {;}\\
a_{2}=b \text {;}\\
a_{n+2}=\alpha a_{n+1}+\beta a_{n} \text { for } n \in \mathbb{N} \text {.}
\end{array}\] Suppose that the equation \(x^{2}=\alpha x+\beta\) has two solutions \(x_{1}\) and \(x_{2}\). Let \(c_{1}\) and \(c_{2}\) be two constants such that \[\begin{aligned}
c_{1} x_{1}+c_{2} x_{2} &=a \text {;}\\
c_{1}\left(x_{1}\right)^{2}+c_{2}\left(x_{2}\right)^{2} &=b \text {.}
\end{aligned}\] Then we can prove by induction that \[x_{n}=c_{1}\left(x_{1}\right)^{n}+c_{2}\left(x_{2}\right)^{n} \text { for all } n \in \mathbb{N} .\]

This is a very useful method to find a general formula for a sequence defined recursively as above. For example, consider the sequence \[\begin{array}{l}
a_{1}=1 \text {;}\\
a_{2}=1 \text {;}\\
a_{n+2}=a_{n+1}+2 a_{n} \text { for } n \in \mathbb{N} \text {.}
\end{array}\] Solving the equation \(x^{2}=x+2\) yields two solutions \(x_{1}=2\) and \(x_{2}=(-1)\). Thus, \[x_{n}=c_{1} 2^{n}+c_{2}(-1)^{n} ,\] where \(c_{1}\) and \(c_{2}\) are constants such as \[\begin{array}{r}
c_{1}(2)+c_{2}(-1)=1 \text {;}\\
c_{1}(2)^{2}+c_{2}(-1)^{2}=1 \text {.}
\end{array}\] It is not hard to see that \(c_{1}=1 / 3\) and \(c_{2}=-1 / 3\). Therefore, \[a_{n}=\frac{1}{3} 2^{n}-\frac{1}{3}(-1)^{n} \text { for all } n \in \mathbb{N} .\

Exercise \(1.3.8\).

Answer

Hint: Prove first that, for \(k=1,2, \ldots, n\), we have \[\left(\begin{array}{l}
n \\
k
\end{array}\right)+\left(\begin{array}{c}
n \\
k-1
\end{array}\right)=\left(\begin{array}{c}
n+1 \\
k
\end{array}\right)\]

Exercise \(1.4.5\).

Answer

In general, to prove that \(|a| \leq m\), where \(m \geq 0\), we only need to show that \(a \leq m\) and \(-a \leq m\).

For any \(x, y \in \mathbb{R}\), \[|x|=|x-y+y| \leq|x-y|+|y| ,\] This implies \[|x|-|y| \leq|x-y| .\] Similarly, \[|y|=|y-x+x| \leq|x-y|+|x| .\] This implies \[-(|x|-|y|) \leq|x-y| .\] Therefore, \[\| x|-| y|| \leq|x-y| .\

Exercise \(1.5.4\).

Answer

Let us first show that \(A + B\) is bounded above. Since \(A\) and \(B\) are nonempty and bounded above, by the completeness axiom, \(\sup A\) and \(\sup B\) exist and are real numbers. In particular, \(a \leq \sup A\) for all \(a \in A\) and \(b \leq \sup B\) for all \(b \in B\).

For any \(x \in A + B\), there exists \(a \in A\) and \(b \in B\) such that \(x = a + b\). Thus, \(x = a + b \leq \sup A + \sup B\), which shows that \(A + B\) is bounded above.

We will now show that \(\sup A + \sup B\) is the supremum of the set \(A + B\) by showing that \(\sup A + \sup B\) satisfies conditions \((1^{\prime})\) and \((2^{\prime})\) of Proposition 1.5.1.

We have just shows that \(\sup A + \sup B\) is an upper bound of \(A +B\) and, hence, \(\sup A+\sup B\) satisfies condition \((1^{\prime})\).

Now let \(\varepsilon > 0\). Using \(frac{\varepsilon}{2}\) in part \((2^{\prime})\) of Proposition 1.5.1 applied to the sets \(A\) and \(B\), there exits \(a \in A\) and \(b \in B\) such that \[\sup A-\frac{\varepsilon}{2}<a \text { and } \sup B-\frac{\varepsilon}{2}<b .\] It follows that \[\sup A+\sup B-\varepsilon<a+b .\] This proves condition \((2^{\prime})\) of Proposition 1.5.1 applied to the set \(A + B\) that \(\sup A + \sup B = \sup (A + B)\) as desired

Exercise \(1.6.2\).

Answer

Let \(x = \frac{1}{r}\). By Theorem 1.6.2(d), there exists \(m \in \mathbb{R}\) such that \[m-1 \leq \frac{1}{r}<m .\] Since \(1 / r>1\), we get \(m > 1\) and, so, \(m \geq 2\). It follows that \(m-1 \in \mathbb{N}\). Set \(n = m - 1\) and then we get \[\frac{1}{n+1}<r \leq \frac{1}{n} .\]