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Sets and membership

Set theory is all about sets. For now, we can say that sets are some specific collections of things. These things can be of any sort, including sets themselves. Going back to file systems, in some important aspects, sets are like folders, which may contain other folders, which in turn may contain other folders, and so on. In fact, sets are denoted just as we denoted folders in our language f\mathcal{L}_f; using curly braces. Thus, the following is a set: {Peter,Leo,Taylor,Curtis}\{\text{Peter}, \text{Leo}, \text{Taylor}, \text{Curtis}\}

Sets have members. The members of a set are those things that are in the set. So the members of the set {Peter,Leo,Taylor,Curtis}\{\text{Peter}, \text{Leo}, \text{Taylor}, \text{Curtis}\} are Peter, Leo, Taylor, and Curtis. Set membership is denoted by \in. So for example, Peter{Peter,Leo,Taylor,Curtis}\text{Peter} \in \{\text{Peter}, \text{Leo}, \text{Taylor}, \text{Curtis}\} We can denote sets by using capital letters. The first choice is usually SS (you can guess why…). This makes it easier to talk about them. So for example: S={Peter,Leo,Taylor,Curtis}PeterS\begin{gathered} S=\{\text{Peter}, \text{Leo}, \text{Taylor}, \text{Curtis}\}\\ \text{Peter} \in S \end{gathered} Similarly, we can denote something not being in the set by using \notin. So for example: HenryS\text{Henry} \notin S So far, this is rather simple. Now let’s look at some complications.

Sets are extensional entities. This is a fancy word to say that all a set depends on is what its members are. So in a set, there is no ordering between its members, and it doesn’t matter whether the set is represented as having some members twice (or however many times). Each member is only counted once, in whatever order. So for example: {Peter,Leo,Taylor,Curtis}={Peter,Peter,Leo,Taylor,Curtis}{Peter,Leo,Taylor,Curtis}={Curtis,Leo,Peter,Taylor}\begin{aligned} \{\text{Peter}, \text{Leo}, \text{Taylor}, \text{Curtis}\}&=\{\text{Peter}, \text{Peter}, \text{Leo}, \text{Taylor}, \text{Curtis}\}\\ \{\text{Peter}, \text{Leo}, \text{Taylor}, \text{Curtis}\}&= \{\text{Curtis}, \text{Leo}, \text{Peter}, \text{Taylor}\} \end{aligned}

Definition 3.1 (Set extensionality). Two sets SS, QQ, are identical if they have exactly the same members. That is, for all xx, xSx \in S if, and only if, xQx \in Q. If SS and QQ are identical sets, we write S=QS=Q.

Exercise 3.1. Determine whether the following sentences are true or false. In each case, explain your reasoning.

  1. a{b,g,h,a}a \in \{b, g, h, a\};

  2. if {b,g,h,a}=S\{b, g, h, a\}=S, then bSb \in S;

  3. the set {b,b,b}\{b, b, b\} has exactly one member;

  4. if Q={3,5,7,7}Q=\{3, 5, 7, 7\}, then the sum of its members is 2222;

  5. if S={a,h,q,r}S=\{a, h, q, r\}, then PeterS\text{Peter} \notin S.

There is one set that is unlike others, denoted \emptyset. This is the empty set. The empty set is so-called because it is, you guessed it, empty. In other words, it has no members. So for any candidate member xx, xx \notin \emptyset.

Now sets can be members of other sets, and it is very important to be clear whether something is in a set, or it is in another set that is part of another set, and so on. Note: since \emptyset is a set, it can also be a member of other sets. Let’s look at an example: S={a,b,c,{a,d},d,}Q={{a,b,c},{d},d,}\begin{aligned} S&=\{a, b, c, \{a, d\}, d, \emptyset\}\\ Q&=\{\{a, b, c\}, \{d\}, d, \emptyset\} \end{aligned}

Now SS and QQ are not the same set. They do share some elements, but they differ in others. Since they do not have the same elements, they are not the same set. In particular, S,Q\emptyset \in S, Q, and dS,Qd \in S, Q. But note that a,b,cQa, b, c \notin Q. What is in QQ is the set {a, b, c}. Conversely, a,b,cSa, b, c \in S, but not {a,b,c}\{a, b, c\}. Similarly, {d}Q\{d\} \in Q but {d}S\{d\} \notin S, though again, dS,Qd \in S, Q.

Exercise 3.2. Determine whether the following expressions are true or false. In each case, explain your reasoning.

  1. S={}S=\{\emptyset\} has no members;

  2. \emptyset has exactly one member, \emptyset;

  3. S={a,b,{b},{{a,b}},b,{}}S=\{a, b, \{b\}, \{\{a, b\}\},b, \{\emptyset\}\} has exactly 55 members;

  4. the sets \emptyset, {}\{\emptyset\}, and {,}\{\emptyset, \emptyset\} are pairwise distinct (that is, no two of them are the same set);

  5. if S={a,{a},{a,{a}}}S=\{a, \{a\}, \{a, \{a\}\}\} and Q={a,{a,{a}}}Q=\{a, \{a, \{a\}\}\}, then every member of QQ is a member of SS;

  6. if S={a,{a},{a,{a}}}S=\{a, \{a\}, \{a, \{a\}\}\} and Q={a,{a,{a}}}Q=\{a, \{a, \{a\}\}\}, then every member of SS is a member of QQ.

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Learning Logic Backwards Copyright © by Peter Susanszky is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License, except where otherwise noted.