Tuesday, May 19, 2015

Understanding Analysis, Stephen Abbott, Chapter 1.2 Exercises

This is a fairly popular analysis book with 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.

Why am I doing this? This is my way of letting off steam, believe it or not. To each his own, right?


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...