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content/sets-functions-relations/sets/sets.tex
24 lines; SHA-256 eea34d38bb52811468a0025d348457a0f4d3f44aae4b7cabb28551d6328e2785. Open the preserved source file.
% Part: sets-functions-relations% Chapter: sets\documentclass[../../../include/open-logic-chapter]{subfiles}\begin{document}\olchapter{sfr}{set}{Sets}\olimport{basics}\olimport{subsets}\olimport{important-sets}\olimport{unions-and-intersections}\olimport{pairs-and-products}\olimport{russells-paradox}\OLEndChapterHook\end{document}
content/sets-functions-relations/sets/basics.tex
103 lines; SHA-256 8bc9151af0985e6e20c374aa38cddd1add7a8dffbabccc94b89f324f62c40f8c. Open the preserved source file.
% Part: sets-functions-relations% Chapter: sets% Section: basics\documentclass[../../../include/open-logic-section]{subfiles}\begin{document}\olfileid{sfr}{set}{bas}\olsection{Extensionality}A \emph{set} is a collection of objects, considered {as} a singleobject. The objects making up the set are called \emph{elements} or\emph{members} of the set. If $x$ is !!a{element} of a set~$A$, wewrite $x \in A$; if not, we write $x \notin A$. The set which has no!!{element}s is called the \emph{empty} set anddenoted~``$\emptyset$''.\begin{explain}It does not matter how we \emph{specify} the set, or how we\emph{order} its !!{element}s, or indeed how \emph{many times} wecount its !!{element}s. All that matters are what its !!{element}sare. We codify this in the following principle.\end{explain}\begin{defn}[Extensionality]If $A$ and $B$ are sets, then $A = B$ iffevery !!{element} of~$A$ is also !!a{element} of~$B$, and viceversa.\end{defn}Extensionality licenses some notation. In general, when we have someobjects $a_{1}$, \dots, $a_{n}$, then $\{a_{1}, \dots, a_{n}\}$ is\emph{the} set whose !!{element}s are $a_1, \ldots, a_n$. We emphasisethe word ``\emph{the}'', since extensionality tells us that there canbe only \emph{one} such set. Indeed, extensionality also licenses thefollowing:\[\{a, a, b\} = \{a, b\} = \{b,a\}.\]This delivers on the point that, when we consider sets, we don't careabout the order of their !!{element}s, or how many times they arespecified.\begin{tagblock}{novice}\begin{ex}Whenever you have a bunch of objects, you can collect them together ina set. The set of Richard's siblings, for instance, is a set thatcontains one person, and we could write it as $S=\{\textrm{Ruth}\}$.The set of positive integers less than $4$ is $\{1, 2, 3\}$, but itcan also be written as $\{3, 2, 1\}$ or even as $\{1, 2, 1, 2, 3\}$.These are all the same set, by extensionality. For every !!{element}of $\{1, 2, 3\}$ is also !!a{element} of $\{3, 2, 1\}$ (and of $\{1,2, 1, 2, 3\}$), and vice versa.\end{ex}\end{tagblock}Frequently we'll specify a set by some property that its !!{element}sshare. We'll use the following shorthand notation for that:$\Setabs{x}{\phi(x)}$, where the $\phi(x)$ stands for the propertythat~$x$ has to have in order to be counted among the !!{element}s ofthe set.\begin{tagblock}{novice}\begin{ex}In our example, we could have specified $S$ also as\[S = \Setabs{x}{x \text{ is a sibling of Richard}}.\]\end{ex}\end{tagblock}\begin{tagblock}{math}\begin{ex}A number is called \emph{perfect} iff it is equal to the sum of itsproper divisors (i.e., numbers that evenly divide it but aren'tidentical to the number). For instance, $6$ is perfect because itsproper divisors are $1$, $2$, and~$3$, and $6 = 1 + 2 + 3$. In fact,$6$ is the only positive integer less than $10$ that is perfect. So,using extensionality, we can say:\[\{6\} = \Setabs{x}{x\text{ is perfect and }0 \leq x \leq 10}\]We read the notation on the right as ``the set of $x$'s such that $x$is perfect and $0 \leq x \leq 10$''. The identity here confirms that,when we consider sets, we don't care about how they are specified.And, more generally, extensionality guarantees that there is alwaysonly one set of $x$'s such that $\phi(x)$.So, extensionality justifies calling$\Setabs{x}{\phi(x)}$ \emph{the} set of $x$'s such that~$\phi(x)$.\end{ex}\end{tagblock}Extensionality gives us a way for showing that sets are identical: toshow that $A = B$, show that whenever $x \in A$ then also $x \in B$,and whenever $y \in B$ then also $y \in A$.\begin{prob}Prove that there is at most one empty set, i.e., show that if $A$ and $B$are sets without !!{element}s, then $A = B$.\end{prob}\end{document}
content/sets-functions-relations/sets/subsets.tex
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% Part: sets-functions-relations% Chapter: sets% Section: subsets\documentclass[../../../include/open-logic-section]{subfiles}\begin{document}\olfileid{sfr}{set}{sub}\olsection{Subsets and Power Sets}\begin{explain}We will often want to compare sets. And one obvious kind of comparisonone might make is as follows: \emph{everything in one set is in theother too}. This situation is sufficiently important for us tointroduce some new notation.\end{explain}\begin{defn}[Subset]If every !!{element} of a set $A$ is also !!a{element} of~$B$, then wesay that $A$ is a \emph{subset} of~$B$, and write $A \subseteq B$. If$A$ is not a subset of~$B$ we write $A \not\subseteq B$.If $A \subseteq B$ but $A \neq B$, we write $A \subsetneq B$ and saythat $A$ is a \emph{proper subset} of $B$.\end{defn}\begin{ex}Every set is a subset of itself, and $\emptyset$ is a subset of everyset. The set of natural even numbers is a subset of the set of naturalnumbers. Also, $\{ a, b \} \subseteq \{ a, b, c \}$. But $\{ a, b, e\}$ is not a subset of $\{ a, b, c \}$.\end{ex}\begin{ex}The number $2$ is an !!{element} of the set of integers, whereas theset of even numbers is a subset of the set of integers. However, a setmay happen to \emph{both} be !!a{element} and a subset of some otherset, e.g., $\{0\} \in \{0, \{0\}\}$ and also $\{0\} \subseteq \{0,\{0\}\}$.\end{ex}Extensionality gives a criterion of identity for sets: $A = B$ iffevery !!{element} of~$A$ is also !!a{element} of~$B$ and vice versa.The definition of ``subset'' defines $A \subseteq B$ precisely as thefirst half of this criterion: every !!{element} of~$A$ is also!!a{element} of~$B$. Of course the definition also applies if weswitch $A$ and $B$: that is, $B \subseteq A$ iff every !!{element}of~$B$ is also !!a{element} of~$A$. And that, in turn, is exactly the``vice versa'' part of extensionality. In other words, extensionalityentails that sets are equal iff they are subsets of one another.\begin{prop}$A = B$ iff both $A \subseteq B$ and $B \subseteq A$.\end{prop}Now is also a good opportunity to introduce some further bits ofhelpful notation. In defining when $A$ is a subset of~$B$ we said that``every !!{element} of~$A$ is \dots,'' and filled the ``$\dots$'' with``!!a{element} of $B$''. But this is such a common \emph{shape} ofexpression that it will be helpful to introduce some formal notationfor it.\begin{defn}\ollabel{forallxina}$(\forall x \in A)\phi$ abbreviates $\forall x(x \in A \lif\phi)$. Similarly, $(\exists x \in A)\phi$ abbreviates $\exists x(x\in A \land \phi)$.\end{defn}Using this notation, we can say that $A \subseteq B$ iff $(\forallx \in A)x \in B$.Now we move on to considering a certain kind of set: the set of allsubsets of a given set.\begin{defn}[Power Set]The set consisting of all subsets of a set~$A$ is called the\emph{power set of}~$A$, written $\Pow{A}$.\[\Pow{A} = \Setabs{B}{B \subseteq A}\]\end{defn}\begin{ex}What are all the possible subsets of $\{ a, b, c \}$? They are:$\emptyset$, $\{a \}$, $\{b\}$, $\{c\}$, $\{a, b\}$, $\{a, c\}$, $\{b,c\}$, $\{a, b, c\}$. The set of all these subsets is$\Pow{\{a,b,c\}}$:\[\Pow{\{ a, b, c \}} = \{\emptyset, \{a \}, \{b\}, \{c\}, \{a, b\},\{b, c\}, \{a, c\}, \{a, b, c\}\}\]\end{ex}\begin{prob}List all subsets of $\{a, b, c, d\}$.\end{prob}\begin{prob}Show that if $A$ has $n$ !!{element}s, then $\Pow{A}$ has $2^n$!!{element}s.\end{prob}\end{document}
content/sets-functions-relations/sets/important-sets.tex
69 lines; SHA-256 b1b998cad3fa5aef48670f755245f86efe00b239d332a50e764147602405fb32. Open the preserved source file.
% Part: sets-functions-relations% Chapter: sets% Section: important-sets\documentclass[../../../include/open-logic-section]{subfiles}\begin{document}\olfileid{sfr}{set}{imp}\olsection{Some Important Sets}\begin{ex}We will mostly be dealing with sets whose !!{element}s aremathematical objects. Four such sets are important enough to havespecific names:\begin{multline*}\Nat = \{0, 1, 2, 3, \ldots\} \\\shoveright{\text{the set of natural numbers}}\\\shoveleft{\Int = \{\ldots, -2, -1, 0, 1, 2, \ldots\}} \\\shoveright{\text{the set of integers}}\\\shoveleft{\Rat = \Setabs{\nicefrac{m}{n}}{m, n \in \Int\text{ and }n \neq 0}}\\\shoveright{\text{the set of rationals}}\\\shoveleft{\Real = (-\infty, \infty)}\\\text{the set of real numbers (the continuum)}\end{multline*}These are all \emph{infinite} sets, that is, they each haveinfinitely many !!{element}s.As we move through these sets, we are adding \emph{more} numbers toour stock. Indeed, it should be clear that $\Nat \subseteq \Int\subseteq \Rat \subseteq \Real$: after all, every natural number is aninteger; every integer is a rational; and every rational is a real.Equally, it should be clear that $\Nat \subsetneq \Int \subsetneq\Rat$, since $-1$ is an integer but not a natural number, and$\nicefrac{1}{2}$ is rational but not integer. It is less obviousthat $\Rat \subsetneq \Real$, i.e., that there are some real numberswhich are not rational\oliflabeldef{sfr:arith:real:realline}{, but we'llreturn to this in \olref[arith][real]{realline}}{}.We'll sometimes also use the set of positive integers $\PosInt = \{1,2, 3, \dots\}$ and the set containing just the first two naturalnumbers $\Bin = \{0, 1\}$.\end{ex}\begin{tagblock}{compsci}\begin{ex}[Strings]Another interesting example is the set $A^{*}$ of \emph{finitestrings} over an alphabet $A$: any finite sequence of elements of~$A$is a string over $A$. We include the \emph{empty string $\Lambda$}among the strings over~$A$, for every alphabet~$A$. For instance,\begin{multline*}\Bin^*=\{\Lambda,0,1,00,01,10,11,\\000,001,010,011,100,101,110,111,0000,\ldots\}.\end{multline*}If $x=x_{1}\ldots x_{n}\in A^{*}$is a string consisting of $n$``letters'' from $A$, then we say \emph{length} of the string is~$n$and write $\len{x}=n$.\end{ex}\end{tagblock}\begin{ex}[Infinite sequences]For any set $A$ we may also consider the set~$A^\omega$ of infinitesequences of !!{element}s of~$A$. An infinite sequence$a_1a_2a_3a_4\dots$ consists of a one-way infinite list of objects,each one of which is !!a{element} of~$A$.\end{ex}\end{document}
content/sets-functions-relations/sets/unions-and-intersections.tex
174 lines; SHA-256 2ad0eec70308ceeff2a4158b2dee60e59a9b1cf9268fc95c4bb1da05f65ad60d. Open the preserved source file.
% Part: sets-functions-relations% Chapter: sets% Section: unions-and-intersections\documentclass[../../../include/open-logic-section]{subfiles}\begin{document}\olfileid{sfr}{set}{uni}\olsection{Unions and Intersections}\begin{explain}In \olref[sfr][set][bas]{sec}, we introduced definitions of sets byabstraction, i.e., definitions of the form $\Setabs{x}{\phi(x)}$.Here, we invoke some property~$\phi$, and this property can mentionsets we've already defined. So for instance, if $A$ and~$B$ are sets,the set $\Setabs{x}{x \in A \lor x \in B}$ consists of all thoseobjects which are !!{element}s of either $A$ or~$B$, i.e., it's theset that combines the !!{element}s of $A$ and~$B$. We can visualizethis as in \olref{fig:union}, where the highlighted area indicates the!!{element}s of the two sets $A$ and~$B$ together.\begin{figure}\olasset{assets/diagrams/union.tikz}\caption{The union $A \cup B$ of two sets is set of !!{element}s of$A$ together with those of~$B$.}\ollabel{fig:union}\end{figure}This operation on sets---combining them---is very useful and common,and so we give it a formal name and a symbol.\end{explain}\begin{defn}[Union]The \emph{union} of two sets $A$ and $B$, written $A \cup B$, is theset of all things which are !!{element}s of $A$, $B$, or both.\[A \cup B = \Setabs{x}{x \in A \lor x \in B}\]\end{defn}\begin{ex}Since the multiplicity of !!{element}s doesn't matter, the union of twosets which have !!a{element} in common contains that !!{element} only once,e.g., $\{ a, b, c\} \cup \{ a, 0, 1\} = \{a, b, c, 0, 1\}$.The union of a set and one of its subsets is just the bigger set: $\{a,b, c \} \cup \{a \} = \{a, b, c\}$.The union of a set with the empty set is identical to the set: $\{a,b, c \} \cup \emptyset = \{a, b, c \}$.\end{ex}\begin{prob}Prove that if $A \subseteq B$, then $A \cup B = B$.\end{prob}\begin{explain}We can also consider a ``dual'' operation to union. This is theoperation that forms the set of all !!{element}s that are !!{element}sof~$A$ and are also !!{element}s of~$B$. This operation is called\emph{intersection}, and can be depicted as in \olref{fig:intersection}.\begin{figure}\olasset{assets/diagrams/intersection.tikz}\caption{The intersection $A \cap B$ of two sets is the set of!!{element}s they have in common.}\ollabel{fig:intersection}\end{figure}\end{explain}\begin{defn}[Intersection]The \emph{intersection} of two sets $A$ and $B$, written $A \cap B$, isthe set of all things which are !!{element}s of both $A$ and~$B$.\[A \cap B = \Setabs{x}{x \in A \land x \in B}\]Two sets are called \emph{disjoint} if their intersection isempty. This means they have no !!{element}s in common.\end{defn}\begin{ex}If two sets have no !!{element}s in common, their intersection is empty:$\{ a, b, c\} \cap \{ 0, 1\} = \emptyset$.If two sets do have !!{element}s in common, their intersection is the set ofall those: $\{a, b, c \} \cap \{a, b, d \} = \{a, b\}$.The intersection of a set with one of its subsets is just the smallerset: $\{a, b, c\} \cap \{a, b\} = \{a, b\}$.The intersection of any set with the empty set is empty: $\{a, b, c \}\cap \emptyset = \emptyset$.\end{ex}\begin{prob}Prove rigorously that if $A \subseteq B$, then $A \cap B = A$.\end{prob}\begin{explain}We can also form the union or intersection of more than twosets. An elegant way of dealing with this in general is thefollowing: suppose you collect all the sets you want to form the union(or intersection) of into a single set. Then we can define the unionof all our original sets as the set of all objects which belong to atleast one !!{element} of the set, and the intersection as the set ofall objects which belong to every !!{element} of the set.\end{explain}\begin{defn}If $A$ is a set of sets, then $\bigcup A$ is the set of !!{element}s of!!{element}s of~$A$:\begin{align*}\bigcup A & = \Setabs{x}{x \text{ belongs to !!a{element} of } A},\text{ i.e.,}\\& = \Setabs{x}{\text{there is a } B \in A\text{ so that } x \in B}\end{align*}\end{defn}\begin{defn}If $A$ is a set of sets, then $\bigcap A$ is the set of objects whichall elements of~$A$ have in common:\begin{align*}\bigcap A & = \Setabs{x}{x \text{ belongs to every !!{element} of } A},\text{ i.e.,}\\& = \Setabs{x}{\text{for all } B \in A, x \in B}\end{align*}\end{defn}\begin{ex}Suppose $A = \{ \{ a, b \}, \{ a, d, e \}, \{ a, d \} \}$.Then $\bigcup A = \{ a, b, d, e \}$ and $\bigcap A = \{ a \}$.\end{ex}\begin{prob}Show that if $A$ is a set and $A \in B$, then $A \subseteq \bigcup B$.\end{prob}We could also do the same for a sequence of sets $A_1$, $A_2$, \dots\begin{align*}\bigcup_i A_i & = \Setabs{x}{x \text{ belongs to one of the } A_i}\\\bigcap_i A_i & = \Setabs{x}{x \text{ belongs to every } A_i}.\end{align*}When we have an \emph{index} of sets, i.e., some set $I$ such that weare considering $A_i$ for each $i \in I$, we may also use theseabbreviations:\begin{align*}\bigcup_{i \in I} A_i & = \bigcup \Setabs{A_i }{i \in I}\\\bigcap_{i \in I} A_i & = \bigcap\Setabs{A_i}{i \in I}\end{align*}Finally, we may want to think about the set of all !!{element}s in~$A$which are not in~$B$. We can depict this as in \olref{difference}.\begin{figure}\olasset{assets/diagrams/difference.tikz}\caption{The difference $A \setminus B$ of two sets is the set ofthose !!{element}s of~$A$ which are not also !!{element}s of~$B$.}\ollabel{difference}\end{figure}\begin{defn}[Difference]The \emph{set difference}~$A \setminus B$ is the set of all !!{element}s of$A$ which are not also !!{element}s of~$B$, i.e.,\[A\setminus B = \Setabs{x}{x\in A \text{ and } x \notin B}.\]\end{defn}\begin{prob}Prove that if $A \subsetneq B$, then $B \setminus A \neq \emptyset$.\end{prob}\end{document}
content/sets-functions-relations/sets/pairs-and-products.tex
129 lines; SHA-256 3db7f0241d387b49488f70e78062c24192b813873618f761565d97bd7249431c. Open the preserved source file.
% Part: sets-functions-relations% Chapter: sets% Section: pairs-and-products\documentclass[../../../include/open-logic-section]{subfiles}\begin{document}\olfileid{sfr}{set}{pai}\olsection{Pairs, Tuples, Cartesian Products}\begin{explain}It follows from extensionality that sets have no order to theirelements. So if we want to represent order, we use \emph{orderedpairs} $\tuple{x, y}$. In an unordered pair $\{x, y\}$, the order doesnot matter: $\{x, y\} = \{y, x\}$. In an ordered pair, it does: if $x\neq y$, then $\tuple{x, y} \neq \tuple{y, x}$.How should we think about ordered pairs in set theory? Crucially, wewant to preserve the idea that ordered pairs are identical iff theyshare the same first element and share the same second element, i.e.:\[\tuple{a, b}= \tuple{c, d}\text{ iff both }a = c \text{ and }b=d.\]We can define ordered pairs in set theory using the Wiener--Kuratowskidefinition.\end{explain}\begin{defn}[Ordered pair]\ollabel{wienerkuratowski}$\tuple{a, b} = \{\{a\}, \{a, b\}\}$.\end{defn}\begin{prob}Using \olref[sfr][set][pai]{wienerkuratowski}, prove that $\tuple{a,b}= \tuple{c, d}$ iff both $a = c$ and $b=d$.\end{prob}\begin{explain}Having fixed a definition of an ordered pair, we can use it to definefurther sets. For example, sometimes we also want ordered sequences ofmore than two objects, e.g., \emph{triples} $\tuple{x, y, z}$,\emph{quadruples} $\tuple{x, y, z, u}$, and so on. We can think oftriples as special ordered pairs, where the first element is itself anordered pair: $\tuple{x, y, z}$ is $\tuple{\tuple{x, y},z}$. The sameis true for quadruples: $\tuple{x,y,z,u}$ is$\tuple{\tuple{\tuple{x,y},z},u}$, and so on. In general, we talk of\emph{ordered $n$-tuples} $\tuple{x_1, \dots, x_n}$.Certain sets of ordered pairs, or other ordered $n$-tuples, will be useful.\end{explain}\begin{defn}[Cartesian product]Given sets $A$ and $B$, their \emph{Cartesian product} $A \times B$ isdefined by\[A \times B = \Setabs{\tuple{x, y}}{x \in A \text{ and } y \in B}.\]\end{defn}\begin{ex}If $A = \{0, 1\}$, and $B = \{1, a, b\}$, then their product is\[A \times B = \{ \tuple{0, 1}, \tuple{0, a}, \tuple{0, b},\tuple{1, 1}, \tuple{1, a}, \tuple{1, b} \}.\]\end{ex}\begin{ex}If $A$ is a set, the product of $A$ with itself, $A \times A$, is alsowritten~$A^2$. It is the set of \emph{all} pairs $\tuple{x, y}$ with$x, y \in A$. The set of all triples $\tuple{x, y, z}$ is $A^3$, andso on. We can give a recursive definition:\begin{align*}A^1 & = A\\A^{k+1} & = A^k \times A\end{align*}\end{ex}\begin{prob}List all !!{element}s of $\{1, 2, 3\}^3$.\end{prob}\begin{prop}\ollabel{cardnmprod}If $A$ has $n$ !!{element}s and $B$ has $m$ !!{element}s, then $A\times B$ has $n\cdot m$ elements.\end{prop}\begin{proof}For every !!{element}~$x$ in~$A$, there are $m$ !!{element}s of theform $\tuple{x, y} \in A \times B$. Let $B_x = \Setabs{\tuple{x, y}}{y\in B}$. Since whenever $x_1 \neq x_2$, $\tuple{x_1, y} \neq\tuple{x_2, y}$, $B_{x_1} \cap B_{x_2} = \emptyset$. But if $A = \{x_1,\dots, x_n\}$, then $A \times B = B_{x_1} \cup \dots \cup B_{x_n}$, and so has$n\cdot m$ !!{element}s.To visualize this, arrange the !!{element}s of~$A \times B$ in a grid:\[\begin{array}{rcccc}B_{x_1} = & \{\tuple{x_1, y_1} & \tuple{x_1, y_2} & \dots & \tuple{x_1, y_m}\}\\B_{x_2} = & \{\tuple{x_2, y_1} & \tuple{x_2, y_2} & \dots & \tuple{x_2, y_m}\}\\\vdots & & \vdots\\B_{x_n} = & \{\tuple{x_n, y_1} & \tuple{x_n, y_2} & \dots & \tuple{x_n, y_m}\}\end{array}\]Since the $x_i$ are all different, and the $y_j$ are all different, notwo of the pairs in this grid are the same, and there are $n\cdot m$of them.\end{proof}\begin{prob}Show, by induction on~$k$, that for all $k \ge 1$, if $A$ has $n$!!{element}s, then $A^k$ has $n^k$ !!{element}s.\end{prob}\begin{ex}If $A$ is a set, a \emph{word} over~$A$ is any sequence of!!{element}s of~$A$. A sequence can be thought of as an $n$-tuple of!!{element}s of~$A$. For instance, if $A = \{a, b, c\}$, then thesequence ``$bac$'' can be thought of as the triple~$\tuple{b, a, c}$.Words, i.e., sequences of symbols, are of crucial importance incomputer science. By convention, we count !!{element}s of~$A$ assequences of length~$1$, and $\emptyset$ as the sequence of length~$0$.The set of \emph{all} words over~$A$ then is\[A^* = \{\emptyset\} \cup A \cup A^2 \cup A^3 \cup \dots\]\end{ex}\end{document}
content/sets-functions-relations/sets/russells-paradox.tex
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% Part: sets-functions-relations% Chapter: sets% Section: russells-paradox\documentclass[../../../include/open-logic-section]{subfiles}\begin{document}\olfileid{sfr}{set}{rus}\olsection{Russell's Paradox}Extensionality licenses the notation $\Setabs{x}{\phi(x)}$, for\emph{the} set of $x$'s such that~$\phi(x)$. However, all thatextensionality \emph{really} licenses is the following thought.\emph{If} there is a set whose members are all and only the $\phi$'s,\emph{then} there is only one such set. Otherwise put: having fixedsome~$\phi$, the set $\Setabs{x}{\phi(x)}$ is unique, \emph{if itexists}.But this conditional is important!{} Crucially, not every propertylends itself to \emph{comprehension}. That is, some properties do\emph{not} define sets. If they all did, then we would run intooutright contradictions. The most famous example of this is Russell'sParadox.Sets may be !!{element}s of other sets---for instance, the power setof a set~$A$ is made up of sets. And so it makes sense to ask orinvestigate whether a set is !!a{element} of another set. Can a set bea member of itself? Nothing about the idea of a set seems to rulethis out. For instance, if \emph{all} sets form a collection ofobjects, one might think that they can be collected into a singleset---the set of all sets. And it, being a set, would be !!a{element}of the set of all sets.Russell's Paradox arises when we consider the property of not havingitself as !!a{element}, of being \emph{non-self-membered}. What if wesuppose that there is a set of all sets that do not have themselves as!!a{element}? Does\[R = \Setabs{x}{x \notin x}\]exist? It turns out that we can prove that it does not.\begin{thm}[Russell's Paradox]\ollabel{thm:russells-paradox}There is no set $R = \Setabs{x}{x \notin x}$.\end{thm}\begin{proof}If $R = \Setabs{x}{x \notin x}$ exists, then$R \in R$ iff $R \notin R$, which is a contradiction.\end{proof}\begin{tagblock}{novice}\begin{explain}Let's run through this proof more slowly. If $R$ exists, it makes sense to ask whether $R \inR$ or not. Suppose that indeed $R \in R$. Now, $R$~was defined as the set of allsets that are not !!{element}s of themselves. So, if $R \in R$,then $R$ does not itself have $R$'s defining property. But only setsthat have this property are in~$R$, hence, $R$ cannot be !!a{element}of~$R$, i.e., $R \notin R$. But $R$ can't both be and not be!!a{element} of~$R$, so we have a contradiction.Since the assumption that $R \in R$ leads to a contradiction, we have$R \notin R$. But this also leads to a contradiction!{} For if $R\notin R$, then $R$ itself does have $R$'s defining property, and so $R$ would be!!a{element} of $R$ just like all the other non-self-membered sets.And again, it can't both not be and be !!a{element} of~$R$.\end{explain}\end{tagblock}\begin{digress}How do we set up a set theory which avoids falling intoRussell's Paradox, i.e., which avoids making the \emph{inconsistent}claim that $R = \Setabs{x}{x \notin x}$ exists? Well, we would need tolay down axioms which give us very precise conditions for stating whensets exist (and when they don't).The set theory sketched in this chapter doesn't do this. It's\emph{genuinely na\"ive}. It tells you only that sets obeyextensionality and that, if you have some sets, you can form theirunion, intersection, etc. It is possible to develop set theory morerigorously than this. \oliflabeldef{cumul:::part}{That rigour will bereserved for Part \olref[cumul][][]{part}. For now, we will proceedna\"ively, and carefully try to sidestep contradictions.}{}\end{digress}\end{document}