# Understanding Analysis, Stephen Abbott, Chapter 1.2 Exercises

This is a fairly popular analysis book with many, many questions and no answers. So I will gradually attempt to answer some of them here. I may be wrong sometimes, so feel free to correct me if you can. If you are using this to cheat, then shame on you (this blog entry became the second most viewed after just one week; what's up with that?)!

**Claim**: $\sqrt3$ is irrational.

Let $\sqrt3$ be rational. Then $\sqrt3=p/q$, where $p,q\in\mathbb Z$, $q\neq0$, and $\text{gcd}(p,q)=1$. This last assumption can be made without loss of generality. It follows that

$$3=\frac{p^2}{q^2}\Longrightarrow3q^2=p^2.$$

$p^2$ cannot be even because it would imply that $p$ is even and that $q^2$ is even, which would in turn imply that $q$ is even. However, $p$ and $q$ cannot both be even because we assumed that $\text{gcd}(p,q)=1$. Therefore, $p^2$ must be odd, which implies that $p$ is odd and that $q^2$ is odd, which in turn implies that $q$ is odd. Let $p=2m+1$, and $q=2n+1$, as per the definition of an odd number, where $m,n\in\mathbb Z$. Then

$$\begin{align}3(2n+1)^2&=(2m+1)^2\Longrightarrow\\3(4n^2+4n+1)&=4m^2+4m+1\Longrightarrow\\12n^2+12n+2&=4m^2+4m\Longrightarrow\\6n^2+6n+1&=2m^2+2m\Longrightarrow\\2(3n^2+3n)+1&=2(m^2+m).\end{align}$$

Nevertheless, this last expression is absurd; an odd number cannot be equal to an even number. Therefore, $\sqrt3$ is irrational. This proves the claim.

**Claim**: If for every $n\in\mathbb N$, $A_n\supseteq A_{n+1}$ and $A_n$ contains infinitely-many elements, then $\bigcap_{n\in\mathbb N}A_n$ contains infinitely-many elements.

Let $A_n=\{n,n+1,n+2,\dots\}$ for every $n\in\mathbb N$. Then the hypothesis holds. Moreover, let $x\in\bigcap_{n\in\mathbb N}A_n$. However, $x\notin A_{x+1}$. Therefore, $\bigcap_{n\in\mathbb N}A_n=\emptyset$. This dismisses the claim.

**Claim**: If for every $n\in\mathbb N$, $A_n\supseteq A_{n+1}$ and $A_n$ is a nonempty set containing a finite number of real numbers, then $\bigcap_{n\in\mathbb N}A_n$ is a nonempty set containing a finite number of real numbers.

Let, for every $n\in\mathbb N$, $A_n\supseteq A_{n+1}$, $A_n$ be a nonempty set containing a finite number of real numbers, and $|A_1|=m$. Then there exists an $N\in\mathbb N$ such that $A_n=A_{n+1}$ for every $n\geq N$. Therefore, $\bigcap_{n\in\mathbb N}A_n=A_N$, which is a nonempty set containing a finite number of real numbers. This proves the claim.

**Claim**: $A\cap(B\cup C)=(A\cap B)\cup C$.

Let $A=\{1,2,3\}$, $B=\{2,3,4\}$, and $C=\{3,4,5\}$. Then $A\cap(B\cup C)=\{2,3\}$, and $(A\cap B)\cup C=\{2,3,4,5\}$, which are different. Therefore, the claim is dismissed.

**Claim**: $A\cup(B\cup C)=(A\cup B)\cup C$.

Let $x\in A\cup(B\cup C)$. Then $x\in A$ or $x\in B\cup C$. $x\in B\cup C$ implies that $x\in B$ or $x\in C$. $x\in A$ or $x\in B$ implies that $x\in A\cup B$. $x\in A\cup B$ or $x\in C$ implies that $x\in(A\cup B)\cup C$. Therefore, $A\cup(B\cup C)\subseteq(A\cup B)\cup C$. The converse can be proved in the same manner.

**Claim**: If $A,B\subseteq\mathbb R$ and $g:\mathbb R\to\mathbb R$, then $g(A\cap B)\subseteq g(A)\cap g(B)$.

*Lemma*: $A\subseteq B\implies f(A)\subseteq f(B)$, $A,B\subseteq\mathbb R$, $f:\mathbb R\to\mathbb R$.

$A\cap B\subseteq A\implies f(A\cap B)\subseteq f(A)$.

$A\cap B\subseteq B\implies f(A\cap B)\subseteq f(B)$.

$f(A\cap B)\subseteq f(A)\cap f(B)$.

Given a function $f:D\to\mathbb R$ and a subset $B\subseteq\mathbb R$, let $f^{-1}(B)$ be the set of all points from the domain $D$ that get mapped into $B$; that is, $f^{-1}(B)=\left\{x\in D:f(x)\in B\right\}$. This set is called the

*preimage*of $B$.
Let $f(x)=x^2$. If $A$ is the closed interval $[0,4]$ and $B$ is the closed interval $[-1,1]$, find $f^{-1}(A)$ and $f^{-1}(B)$.

$$\begin{align}f^{-1}(A)&=[-2,2]\text{, and}\\f^{-1}(B)&=[-1,1].\end{align}$$

Does $f^{-1}(A\cap B)=f^{-1}(A)\cap f^{-1}(B)$ in this case?

$$\begin{align}A\cap B&=[0,1],\\f^{-1}(A\cap B)&=[-1,1]\text{, and}\\f^{-1}(A)\cap f^{-1}(B)&=[-1,1].\end{align}$$

Therefore, they are equivalent.

The good behavior of preimages just demonstrated is completely general. Show that for an arbitrary function $g:\mathbb R\to\mathbb R$, it is always true that $g^{-1}(A\cap B)=g^{-1}(A)\cap g^{-1}(B)$ and $g^{-1}(A\cup B)=g^{-1}(A)\cup g^{-1}(B)$ for all sets $A,B\in\mathbb R$.

Let $x\in g^{-1}(A\cap B)$. Then $g(x)\in A\cap B$. This implies that $g(x)\in A$ and $g(x)\in B$. This implies that $x\in g^{-1}(A)$ and $x\in g^{-1}(B)$. This implies that $x\in g^{-1}(A)\cap g^{-1}(B)$.

*To be continued...*

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