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\[f(n)=0.a_{n1}a_{n2}a_{n3}a_{n4}\dots\]
If we write down the \(1-1\) correspondence between \(\mathbb{N}\) and \((0,1)\) for \(n=1,2,3,4,\dots\), we get the following array:
\[\begin{array}{rcl}1\leftrightarrow f(1)&=&0.\boldsymbol{a_{11}}a_{12}a_{13} a_{14}\dots\\ 2\leftrightarrow f(2)&=&0.a_{21}\boldsymbol{a_{22}}a_{23}a_{24}\dots\\ 3\leftrightarrow f(3)&=&0.a_{31}a_{32}\boldsymbol{a_{33}}a_{34}\dots\\ 4\leftrightarrow f(4)&=&0.a_{41}\;a_{42}\;a_{43}\;\boldsymbol{a_{44}}\;\dots\\ \vdots\;\;\vdots&&\vdots\;\;\vdots\;\;\vdots\end{array}\]
We are assuming here that every real number you can think of in \((0,1)\) appears somewhere on the above list. Now we are going to define a real number \(x\in(0,1)\) which is not in the above list. Let \(x=0.b_{1}b_{2}b_{3}\dots\). To choose the digit \(b_{1}\), look at the digit \(a_{11}\) in the upper left-hand corner of the above array and choose \(b_{1}\neq a_{11}\). For example, if \(a_{11}=3\), then choose \(b_{1}=2\), and if \(a_{11}=2\), then choose \(b_{1}=3\). Notice that the real number \(x=0.b_{1}b_{2}b_{3}\dots\) cannot be equal to \(f(1)\). Similarly choose \(b_{2}\neq a_{22}\), and thus \(x\neq f(2)\). Continuing this process we see that \(x\neq f(n)\) for any \(n\in\mathbb{N}\). In other words the decimal \(x=0.b_{1}b_{2}b_{3}\dots\) is a real number, and it does not end in all nines but cannot be in the above list since it differs from each number we list in at least one digit. So we reach a contradiction. Consequently, the real numbers between \(0\) and \(1\) are not countable.
**Remark 4**.: Since \((0,1)\) is uncountable if and only if \(\mathbb{R}\) is uncountable (why? see Exercise 11), the above theorem also shows \(\mathbb{R}\) is uncountable. Cantor initially published his discovery that \(\mathbb{R}\) is uncountable and later offered this same fact with an amazing simple proof technique called Cantor’s diagonalization method as discussed in the proof of Theorem 2 above. Note that \(|\mathbb{N}|=\aleph_{0}\) and
\[c=|(0,1)|=|\mathbb{R}|,\]
from the inequality \(\aleph_{0} ### Equivalence Relations Sometimes in mathematics, we find that we need a customized mathematical structure with certain desired properties. In other cases, we find that the things we are studying have too much information, with distracting details making it difficult to see what is important and what is irrelevant. In these and other situations, structures known as _relations on a set_ can make the difference. Let \(X\) be a set. As we have seen, a function \(f:X\to X\) can be understood as a subset of the Cartesian product \(X^{2}\) with the property that every element of \(X\) appears exactly once in the first component, e.g., the function \(f:\{1,2,3\}\to\{1,2,3\}\) defined by \(f(1)=3\), \(f(2)=1\) and \(f(3)=3\) is really the subset \[\{(1,3),(2,1),(3,3)\}\subset\{1,2,3\}\times\{1,2,3\}.\] What about other subsets? More generally, any subset \(R\subset X^{2}\) of the Cartesian product of a set with itself defines a _relation_ on \(X\). We can think of a relation as a way of comparing two elements of \(X\); we will write \(xRy\) if the ordered pair \((x,y)\) belongs to \(R\). A relation \(\sim\) on \(X\) is an _equivalence relation_ if \(\sim\) satisfies the following three properties: * For all \(x\in X\), we have \(x\sim x\) (this is called the _reflexive_ property). If \(\sim\) is an equivalence relation on \(X\) and \(x\sim y\), we say that \(x\) is _equivalent_ to \(y\). **Example 8**.: The most familiar example of an equivalence relation is _equality_, generally denoted by “=.” We can easily verify the properties: * \(x=x\), so \(x\) is reflexive. * If \(x=y\) and \(y=z\), then \(x=z\), so \(=\) is transitive. As a subset of \(X\times X\), the relation \(=\) consists of ordered pairs where both elements of the pair are the same, i.e., \[\{(x,x)\mid x\in X\}.\] **Example 9**.: Let \(V\) be a real vector space, and say \(\vec{u}\sim\vec{v}\) if there is a nonzero scalar \(\alpha\in\mathbb{R}\) so that \(\vec{u}=\alpha\vec{v}\). Then \(\sim\) is an equivalence relation: * For any \(\vec{u}\in V\), \(\vec{u}=1\vec{u}\), so \(\vec{u}\sim\vec{u}\) and \(\sim\) is reflexive. **Example 10**.: Let \(X=\mathbb{Z}\times\mathbb{Z}\), the set of ordered pairs of integers, and say \((x,y)\sim(x^{\prime},y^{\prime})\) if there are nonzero integers \(m\) and \(n\) such that \((mx,my)=(nx^{\prime},ny^{\prime})\). Let us show that \(\sim\) is an equivalence relation: * \((1x,1y)=(1x,1y)\), so \(\sim\) is reflexive. **Example 11**.: Let \(X=\mathbb{Z}\) and fix a positive integer \(n\). Say \(x\sim y\) if \(x-y=nm\) for some integer \(m\). Then \(\sim\) is an equivalence relation: * \(x-x=0=0n\) so \(\sim\) is reflexive. This equivalence relation is known as _congruence modulo_\(n\), often denoted by \(x\equiv y\ (n)\). **Example 12**.: Let \(X=\mathbb{R}^{2}\), and define \(\vec{u}\sim\vec{v}\) if \(\vec{u}-\vec{v}\in\mathbb{Z}^{2}\), i.e., two vectors in the plane are equivalent if their components differ by integers. Then \(\sim\) is an equivalence relation; see Exercise 13 below. Given \(x\in X\), the _equivalence class_ of \(x\), denoted \([x]\), is the set of all elements of \(X\) which are equivalent to \(x\): \[[x]=\{y\in X\ |\ x\sim y\}.\] We can notice a few interesting things right away. First, equivalence classes are disjoint: Given \(x\neq y\in X\), we have either \([x]=[y]\) or \([x]\cap[y]=\emptyset\), since if \(z\in[x]\cap[y]\), then \(x\sim z\), \(y\sim z\), and transitivity says \(x\sim y\). In particular, an equivalence relation \(\sim\) partitions \(X\) into disjoint sets \([x]\). The set of these equivalence classes is called the _quotient space_ of \(X\)_modulo_\(\sim\), written \[X/_{\sim}=\{[x]\ |\ x\in X\}.\] For example, setting \(n=2\) in Example 11, we have an equivalence relation on \(\mathbb{Z}\); \(x\sim y\iff x-y=2m\) for some \(m\in\mathbb{Z}\). What are the equivalence classes? Well, for any number \(y\in\mathbb{Z}\), we can find its equivalence class \[[y]=\{x\in X\ |\ x\sim y\}.\] For instance, \[[0]=\{y\in\mathbb{Z}\ |\ y-0=2m\}=\{y\in\mathbb{Z}\ |\ y=2m\},\] so in fact the class of \(0\) is all even integers. Moreover \[[1]=\{y\in\mathbb{Z}\ |\ y-1=2m\}=\{y\in\mathbb{Z}\ |\ y=2m+1\},\] and the class of \(1\) is the set of all odd integers. Thus, the equivalence relation \(x\sim y\iff x-y=2m\) partitions the integers into the sets of even and odd integers. An equivalence relation on \(X\) determines a partition of \(X\) into disjoint subsets. It turns out that it works the other way too: If we start with a division of \(X\) into disjoint subsets \(X=X_{1}\cup\cdots\cup X_{n}\), we can define \(x\sim y\) if \(x\) and \(y\) are in the same \(X_{k}\). We claim that this defines an equivalence relation: * Clearly, an element \(x\) is in the same \(X_{k}\) as itself, so \(\sim\) is Reflexive. Hence, partitions and equivalence relations are really the same thing. Third, there is a map \(f:X\to X/_{\sim}\) called the _projection_ map defined by \[f(x)=[x],\] i.e., sending \(x\) to its equivalence class. Given any set \(X\) and equivalence relation \(\sim\), this projection map is surjective. This also works in the other direction: Given any surjective map \(f:X\to Y\), we can think of \(f\) as a projection map and identify \(Y\) with \(X/_{\sim}\), where \(x\sim x^{\prime}\) means \(f(x)=f(x^{\prime})\). Thus, equivalence relations, partitions, and surjective maps are all really the same thing. **Example 13**.: Many important objects in mathematics arise naturally as quotient sets. For instance, consider the equivalence relation in Example 12. What is the quotient set? Points in the quotient set are equivalence classes in \(X\), so to identify the quotient set we need to keep only one point from each equivalence class. Then every point in the plane is equivalent to a point in the unit square \([0,1]\times[0,1]\), so we only need the square. But points along the top edge are equivalent to points on the bottom edge, so we can picture gluing the top edge to the bottom edge to obtain a cylinder. The same holds for the sides, so gluing the sides together, we obtain a _torus_, the doughnut as in Figure 1.10. The quotient sets in Examples 10 and 11 are \(\mathbb{Q}\) and \(\mathbb{Z}_{n}\), where \(\mathbb{Z}_{n}\) denotes the integers modulo \(n\). We will give the definition of \(\mathbb{Z}_{n}\) in Section 1.2 when we explain modular arithmetic. The real numbers, as we will see, can be understood as equivalence classes of certain infinite sequences of rational numbers known as _Cauchy_ sequences where two sequences are equivalent if the difference between them approaches zero as \(n\) gets large. Even the natural numbers can be understood as equivalence classes of finite sets where two sets are equivalent if there is a bijection between them. ### Axiom of Choice The _axiom of choice_ is one of the several rules that one can use for building sets out of other sets. Roughly speaking, the axiom of choice says that we are allowed to make an arbitrary number of unspecified choices to form a set. It explicitly or implicitly appears in many proofs in mathematics. For example, let us examine the well-known proof that the countable union of countable sets is Figure 1.10: Torus as a quotient set
* For all \(x,y\in X\), \(x\sim y\Rightarrow y\sim x\) (this is called the _symmetric_ property).
* For all \(x,y,z\in X\), \(x\sim y\) and \(y\sim z\) imply \(x\sim z\) (this is called the _transitive_ property).
* If \(x=y\), then \(y=x\), so = is symmetric.
* For any \(\vec{u},\vec{v}\in V\), if \(\vec{u}\sim\vec{v}\), then \(\vec{u}=\alpha\vec{v}\) for some \(\alpha\neq 0\), so \(\vec{v}=\alpha^{-1}\vec{u}\) and \(\vec{v}\sim\vec{u}\). Thus, \(\sim\) is symmetric.
* For any \(\vec{u},\vec{v},\vec{w}\in V\), if \(\vec{u}\sim\vec{v}\) and \(\vec{v}\sim\vec{w}\), then we have \(\vec{u}=\alpha\vec{v}\) and \(\vec{v}=\beta\vec{w}\) for \(\alpha,\beta\neq 0\), so \(\vec{u}=\alpha\beta\vec{w}\) and \(\vec{u}\sim\vec{w}\).
* If \((x,y)\sim(x^{\prime},y^{\prime})\), then we have \((mx,my)=(nx^{\prime},ny^{\prime})\), so \((nx^{\prime},ny^{\prime})=(mx,my)\) and \((x^{\prime},y^{\prime})\sim(x,y)\); hence, \(\sim\) is symmetric.
* If \((x,y)\sim(x^{\prime},y^{\prime})\) and \((x^{\prime},y^{\prime})\sim(x^{\prime\prime},y^{\prime\prime})\), then there are integers \(n,m\) and \(j,k\) such that \[(mx,my)=(nx^{\prime},ny^{\prime})\quad\text{and}\quad(jx^{\prime},jy^{\prime}) =(kx^{\prime\prime},ky^{\prime\prime}).\] Then we have \[(mjx,mjy)=(njx^{\prime},njy^{\prime})=(nkx^{\prime\prime},nky^{\prime\prime}),\] and \(\sim\) is transitive.
* If \(x\sim y\), then \(x-y=nm\) for some \(m\); therefore \(y-x=-nm=n(-m)\), so \(y\sim x\) and \(\sim\) is symmetric.
* If \(x\sim y\) and \(y\sim z\), then \(x-y=nm\) and \(y-z=nk\); therefore \[x-z=x-y+y-z=nm+nk=n(m+k),\] and \(\sim\) is transitive.
* If \(x\) and \(y\) are in the same \(X_{k}\), then switching their order does not change that they are in the same \(X_{k}\), so \(x\sim y\) implies \(y\sim x\) and \(\sim\) is symmetric.
* If \(x\) and \(y\) are in the same \(X_{k}\) and \(y\) and \(z\) are in the same \(X_{j}\), then the fact that the \(X_{k}\)’s are disjoint means \(y\) is in only one subset, and \(x\) and \(z\) must both belong to the same subset. Thus, \(\sim\) is transitive.
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