Listing the members versus stating the property they share

0:00Every set answers one question: is this object in, or not? But there are two completely different ways to write a set down, and they answer that question by different means. The first hands you a list. You look the object up. If it is on the list it is in, and if it is not, it is not. The second hands you a condition. You do not look anything up; you test the object against the condition and see whether it passes.

0:31Those two feel interchangeable on small examples. They are not. One of them stops working the moment the list gets long, and the other one never stops working at all. Start with the list. Write the members inside curly braces, separated by commas. The even positive integers below seven: two, four, six. That is the whole set, in braces. This is called roster form, and the braces are doing real work. They say: what is inside is the set, and the set is nothing else.

1:10Here is a bigger one. The natural numbers that divide forty-two. Forty-two is two times three times seven. So build the divisors from those primes: one, two, three, six, seven, fourteen, twenty-one, forty-two. Eight members. And notice what you had to do to write that roster. You had to know all of them. A roster is a promise that the list is complete. Leave one out and you have not written a shorter description; you have written a different set.

1:45Now two things about rosters that look like rules to memorise and are really just consequences. The first is order. Take those eight divisors and scramble them. Seven, twenty-one, one, forty-two, three, fourteen, two, six. That is a different sequence. It is the same set. And you can see why from the membership question alone. The question was: is this object in? Nothing in that question mentions position. Nothing asks where in the list an object sits.

2:21So there is nothing for the order to change. Both lists give the same answer to every object you hand them. The second is repetition, and it follows from exactly the same place. Take the letters of the word SCHOOL. As written, that is six letters: S, C, H, O, O, L. As a set, it is five members: S, C, H, O and L. The O is written twice, and being written twice does not put it in twice.

2:53Because again: is O in? Yes. That is the entire answer. There is no count attached to it. So writing a member a second time adds nothing, and the set does not grow. Try it on BETTER. Six letters written, four members: B, E, T and R. And on LOYAL. Five written, four members: L, O, Y and A. Every one of those shrinks, and always by exactly the number of extra writings.

3:25Now the hard case. What do you do when the list does not stop? The odd natural numbers. One, three, five, and then three dots. Those dots are not laziness. They are a promise, and it is a strong one. The promise is: you can work out the next member yourself. And you can. One, three, five. The gap is two every time, so the next one is seven, and after that nine.

3:52That is what makes the dots honest. There is a rule sitting under the first few members that regenerates them and then keeps going. Same with the even numbers: two, four, six, dots. Gap of two, next one is eight. Same with two, four, eight, sixteen. That one is not a gap, it is a doubling, but it is still a rule you can carry on. So the test for the dots is simple. Can a reader name the next member?

4:24Which means there are sets the dots cannot handle at all, and it is worth seeing why. Try to write the real numbers with dots. Start anywhere. Say zero. What comes next? Name the very next real number after zero. You might try one. But a half is between them, so a half comes first. Try a half. A quarter is between zero and a half. Try a quarter. An eighth is between them. And this never ends.

4:54Whatever you name, there is always another one between it and zero. So there is no next real number. Not a hard one to find - there is not one. And the dots promise a next member. A promise you cannot keep is not a description. So the real numbers get no roster. Not a long one, not a clever one. None. Which is a problem, because they are obviously a perfectly good set. Every object either is a real number or is not.

5:27The list has run out before the mathematics has. So we need the other way of writing a set, and here it is. Instead of listing the members, state the property that picks them out. Let V be the set of vowels. In this form you write: V equals, open brace, x, colon, x is a vowel, close brace. Read that out loud and it is a sentence. V is the set of all x, such that x is a vowel.

5:58This is set-builder form, and it does not need the list at all. Hand it any letter and it decides. Is a a vowel? Yes, so a is in. Is b? No, so b is out. It never asks how many members there are. It never needs them written down anywhere. And that is exactly the thing the roster could not do. Let us slow down on the notation, because every piece of it has a job.

6:27The braces mean the same as before: this is a set. The letter, x, is a stand-in. It is not a particular number; it is whatever object you are currently asking about. The colon is read as such that. Some people write a vertical bar instead. It means the same thing. And everything after the colon is the condition the object has to satisfy. So the whole expression reads: the set of all x, such that x has this property.

7:00One warning. The letter is arbitrary. The set of all x such that x is a vowel, and the set of all y such that y is a vowel, are the same set. The variable is a placeholder, not a member. Changing its name changes nothing. Now the two translations, and they are the actual work of this topic. Property to list first. Take the set of x such that x squared plus x minus two equals zero.

7:33You do not guess the members. You solve. The left side factorises as x minus one, times x plus two. So the roster is one and minus two. And you check by substituting. One plus one minus two is zero. Four minus two minus two is zero. Both hold. Here is a second one. The positive integers whose square stays under forty. Again, derive. One, four, nine, sixteen, twenty-five, thirty-six - all under forty.

8:08Six squared is thirty-six, which is under. Seven squared is forty-nine, which is over. So the set stops at six. The roster is one to six. Notice the shape of both. The condition was given; the list was produced. The list is the output. The other direction is harder, and it is harder for a reason worth naming. Take the set one, four, nine, sixteen, twenty-five, dots. You can see it. They are the squares. So one description is: the set of x such that x is the square of a natural number.

8:48Here is a second description that is just as correct. The set of x such that x equals n squared, where n is a natural number. Two different sentences. Same set. And that is not a contradiction; it is the normal situation. What is fixed is the set. The sentence naming it is not fixed, and there are usually many. But going this way you are not reading something off. You are proposing a test.

9:20You are claiming your condition lets in every member and nothing else. That claim can be wrong. Here is a case where it goes wrong quietly, so watch the size of the thing. A half, two thirds, three quarters, four fifths, five sixths, six sevenths. Look at the pattern. Every numerator is exactly one below its denominator. So a description suggests itself: the set of n over n plus one, where n is a natural number.

9:53And every one of the six fits it. So that is right, surely? No. Because that condition also lets in seven eighths, and eight ninths, and it never stops. The description is not wrong about the six. It is wrong about everything else. The fix is one clause. n over n plus one, where n runs from one to six. That bound is not tidying-up. It is part of the set. Delete it and six members become endlessly many.

10:26Which gives us the check you should be running every single time, and it has two halves. First half: does every member of the set satisfy the condition? Second half: does anything outside the set satisfy the condition? You need both. The first half alone is the trap, because it is so easy to pass. Here is a set: one, two, three, six, nine, eighteen. The divisors of eighteen. Now propose a description: a positive integer up to eighteen.

10:59Run the first half. Is every member a positive integer up to eighteen? Yes, all six of them. It passes. And it is completely wrong. Because the second half catches it. Four is a positive integer up to eighteen, and four is not in the set. So is five, so is seven, so is eight, and eight more besides. Twelve extra objects the description lets in. A description that admits an extra element does not describe your set. It describes a different one.

11:35Run both halves. Every time. Let us put the machinery to work on a few that catch people out. The two-digit numbers whose digits add to eight. The temptation is to start at eight itself. But eight is not two-digit, so it is out. Work the tens digit instead, one through eight: seventeen, twenty-six, thirty-five, forty-four, fifty-three, sixty-two, seventy-one, eighty. Eight members. And notice eighty-nine is not one of them - eight and nine add to seventeen.

12:13Next. The prime numbers that divide sixty. Sixty is two times two times three times five. So the primes dividing it are two, three and five. Three of them. Not twelve. Sixty has twelve divisors altogether, but the condition said prime. Next. The letters of TRIGONOMETRY. Twelve letters written, nine distinct: T, R, I, G, O, N, M, E, Y. And MATHEMATICS. Eleven written, eight distinct: M, A, T, H, E, I, C, S.

12:50One more, because it is a nice reminder that members need not be numbers. The months that do not have thirty-one days: February, April, June, September, November. Five of them, out of twelve. So which form should you use? The honest answer is that they are not equal partners. A roster only exists when you can either finish the list or show a pattern that finishes it for you. The set-builder form has no such condition. It works for the six divisors of eighteen and it works for the real numbers, unchanged.

13:30It also works for sets nobody has ever listed and nobody ever will. So the property is the primary thing, and the roster is a report of it - convenient when it is available, and simply unavailable otherwise. Which is why the translation exercises are never busywork. Writing a list as a property is proposing a test and claiming it admits exactly those members and no others. That is a real claim, and you now have the two-way check that decides it.

14:06Here is one to try. The set of x such that x is a letter in the word PRINCIPAL. Write the roster. Watch what happens to the two letters that appear twice, and tell me how many members you get.