Cracking the Code: A Measurability Proof Revealed!

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Proof

: Assume that \(\mathcal{M}\) is measurable. Let \(r_{1},r_{2},r_{3},\dots\) be an enumeration of all rationals in \([-1,1]\), and consider the translates \(\mathcal{M}_{j}=\mathcal{M}+r_{j}\) for \(j=1,2,3,\dots\). First, observe that the sets \(\mathcal{M}_{j}\) are disjoint, because of the fact that \(\mathcal{M}\) contains only one representative of each equivalence class. Moreover each \(\mathcal{M}_{j}\) by construction is contained in \([-1,2]\). Finally, if \(x\in[0,1]\), then \(x\sim x_{\alpha}\) for some \(\alpha\), and therefore \(x-x_{\alpha}=r_{j}\) for some \(j\). Thus we have:

\[[0,1]\subset\bigcup_{j=1}^{\infty}\mathcal{M}_{j}\subset[-1,2].\]

The inclusions above yield:

\[m([0,1])=1\leq\sum_{j=1}^{\infty}m(\mathcal{M}_{j})\leq 3=m([-1,2]).\]

Since \(\mathcal{M}_{j}\) is a translate of \(\mathcal{M}\), we must have

\[m(\mathcal{M})=m(\mathcal{M}_{j})\quad\text{for all}\quad j.\]

Consequently

\[1\leq\sum_{j=1}^{\infty}m(\mathcal{M})\leq 3.\]

This is the desired contradiction, since neither \(m(\mathcal{M})=0\) nor \(m(\mathcal{M})>0\) is possible. Here \(\mathcal{M}\) is non-measurable.

In the following we present two examples concerning Banach-Tarski type decomposition; both examples depend on the Axiom of Choice. Recall that two subsets of the plane are said to be congruent if one can be made to coincide precisely with the other using only translations and rotations in the plane.

In the following examples, the basic idea is divide an object \(X\) into a finite number of disjoint parts and then “rearrange” them into a new object, say \(Y\). In this case we say \(X\) is equivalent by _finite decomposition_ to \(Y\). Note that rearranging a given set means that the set in its initial position is congruent to the set in its final position.

**Example 94**.: We claim that \(\mathbb{N}\) is equivalent by finite decomposition to the natural numbers with a one element “hole” in them. To this end, first create two subsets of \(\mathbb{N}\):

* \(A=\{1,2,3,4,6,7,\dots\}\) (non-multiples of \(5\)).
* \(B=\{5,10,15,\dots\}\) (all multiples of \(5\)).

Notice that

\[A\cap B=\emptyset\quad\text{ and}\quad A\cup B=\mathbb{N}.\]Next, we shift \(B\) toward infinity by 5 units (we remove 5 from \(\mathbb{N}\)), and thus produce the new set

\[B^{*}=\{10,15,20,\dots\}.\]

We still have:

\[A\cap B^{*}=\emptyset\quad\text{ and }\quad A\cup B^{*}=\mathbb{N}\setminus\{5\}.\]

Moreover, \(B\) and \(B^{*}\) are congruent. Thus \(\mathbb{N}\) and \(\mathbb{N}\setminus\{5\}\) are equivalent by finite decomposition.

The Banach-Tarski construction is similar to an older construction due to G. Vitali, which concerns area rather than the volume.

**Example 95** (Doubling Area).: Let \(L_{\theta}\) denote the line segment in \(\mathbb{R}^{2}\) represented as

\[L_{\theta}=\{(r,\theta):\ 0

Observe that union of all such segments is the punctured unit disc \(B^{\prime}\), i.e., \(B^{\prime}\) is the unit disc in \(\mathbb{R}^{2}\) with the origin removed. Next we define an equivalence relation:

\(L_{\theta}\) and \(L_{\phi}\) belong to the same equivalence class if

\[\theta-\phi\qquad\text{ is a rational multiple of }\pi.\]

Consider a set \(S\) that is the union of a set of \(L_{\theta}\) containing exactly one element from each equivalence class (notice how we used the Axiom of Choice). Let

\[S_{n}=\{L_{\theta+2\pi r_{n}}:\ L_{\theta}\in S\}.\]

Note that each \(S_{n}\) is obtained from \(S\) by a rotation about the origin through an angle \(2\pi r_{n}\), where \(r_{n}\in[0,1)\) is a sequence \(r_{1},r_{2},\dots\) of rationals. Note that the \(S_{n}\)’s are disjoint and

\[B^{\prime}=\bigcup_{n}S_{n}.\]

Now we split \(B^{\prime}\) into two sets as follows:

\[C=\bigcup_{n}S_{2n}\quad\text{and}\quad D=\bigcup_{n}S_{2n+1}.\]

Observe that each \(S_{2n}\) can be rotated to \(S_{n}\), and the union of \(S_{n}\) give \(B^{\prime}\). Similarly, each \(S_{2n+1}\) can be rotated to \(S_{n}\) and the union of \(S_{n}\) gives \(B^{\prime}\) again. Thus \(B^{\prime}\) can be split into a countable set of disjoint pieces which can be rotated and translated to form disjoint sets whose union is two copies of \(B^{\prime}\).

Although the Banach-Tarski Decomposition clearly relies on the Axiom of Choice, in 1991, Pawlikowski [39] showed that the Banach-Tarski Theorem is implied by the famous the Hahn-Banach Theorem of functional analysis. Thus a weakened version of the Axiom of Choice will do. If you are asking the question what makes the Axiom of Choice true or false, notice that when we accept the Axiom of Choice, we declare our intention to treat certain mathematical objects as they exist regardless of if we can find examples of these objects. Of course, one can always refer to the inscription written on the tombstone of David Hilbert: “Wir Mussen Wissen, Wir Werden Wissen” (We Must Know, We Will Know).

## Chapter 3 Complex Analysis

Neither the true nor the false roots are always real; sometimes they are imaginary; that is, while we can always imagine as many roots for each equation as I have assigned, yet there is not always a definite quantity corresponding to each root we have imagined.

_Descartes_ 1637

### 3.1 The Complex Plane

#### Complex Numbers

Our idea of what a number is has evolved considerably throughout history. The ancients thought of numbers as extensions in space; we have looked at cardinalities, ratios, and invented negative numbers which are useful for describing debts, backward motion, etc. One more modern concept of what numbers are involves functions known as _polynomials_.

A _polynomial_ is a sum of multiples of powers of a variable \(x\), that is

\[p(x)=a_{0}+a_{1}x+\cdots+a_{n}x^{n}.\]

A number \(\alpha\) is a _root_ of a polynomial \(p(x)\) if \(p(\alpha)=0\), meaning if

\[a_{0}+a_{1}\alpha+\cdots+a_{n}\alpha^{n}=0.\]

Every real number \(\alpha\) is a root of a polynomial, namely \(p(x)=x-\alpha\). Thus, one idea for what it means to be a number is being “a root of a polynomial.” However, not every polynomial has a real root. For example, there is no real number \(x\) such that \(x^{2}+1=0\). It turns out that we can fix this by inventing a single new number \(i\) with the property that \(i^{2}=-1\).

A _complex number_ is a sum \(z=a+ib\) of a real number \(a\) and \(i\) times a real number \(b\); \(a\) is called the _real part_ of \(z\) denoted by \(\operatorname{Re}z\), while \(b\) is called the_imaginary part_ of \(z\) and denoted by \(\operatorname{Im}z\). The term “imaginary” is unfortunate since it misleadingly suggests that these numbers are somehow fake or unreal.

Where real numbers can be understood as lengths of segments, complex numbers can be identified with points of a plane (Figure 3.1):

* We can identify a complex number either by using Cartesian coordinates (Figure 3.2) \(z=a+ib\) or by using _polar coordinates_ \[z=re^{i\theta}=r\cos\theta+ir\sin\theta\] We write \(e^{i\theta}\) as shorthand for \(\cos\theta+i\sin\theta\) and \(z=re^{i\theta}\) is called the polar form of a complex number. Here \(r=|z|=\sqrt{x^{2}+y^{2}}\) and for \(z=0\), we can choose \(\theta\) arbitrarily; for \(z\neq 0\), the angle \(\theta\) is not unique. Because of the periodic nature of \(\sin\theta\) and \(\cos\theta\), \(\theta\) is only determined up to an integer multiple of \(2\pi\). We call any value of \(\theta\) with \(z=re^{i\theta}\)_an argument_ of \(z\). In Section 3.4 we will examine the series expansion of the exponential function \(e^{z}\) that properly justifies the notation \(e^{i\theta}=\cos\theta+i\sin\theta\). Note that if we extend the series definition of the exponential function to complex numbers, we have for any real \(\theta\), \[e^{i\theta} =1+i\theta+\frac{1}{2!}(i\theta)^{2}+\frac{1}{3!}(i\theta)^{3}+ \frac{1}{4!}(i\theta)^{4}+\frac{1}{5!}(i\theta)^{5}+\cdots\] \[=\left(1-\frac{1}{2!}\theta^{2}+\frac{1}{4!}\theta^{4}-\cdots \right)+i\left(\theta-\frac{1}{3!}\theta^{3}+\frac{1}{5!}\theta^{5}-\cdots\right)\] \[=\cos\theta+i\sin\theta.\]
* The following are some special complex numbers (Figure 3.3). For \(\theta\in\mathbb{R}\), \[e^{i0}=1\quad\text{and}\quad e^{i\theta}=1\Longleftrightarrow\theta=2k\pi,\ \ (k \in\mathbb{Z}),\]

Figure 3.1: Cartesian representation

Figure 3.2: Polar representation

\[e^{i\pi}=-1\quad\text{and}\quad e^{i\theta}=-1\Longleftrightarrow\theta=(2k+1)\pi, \ (k\in\mathbb{Z}),\] \[e^{i\pi/2}=i\quad\text{and}\quad e^{i\theta}=i\Longleftrightarrow \theta=\frac{1}{2}(2k+1)\pi,\ (k\in\mathbb{Z}).\]