1 1
1 2
1 3
1 4
1 5
1 6
1
7
2 1
2 2
2 3
2 4
2 5
2 6
2
7
3 1
3 2
3 3
3 4
3 5
3 6
3
7
4 1
4 2
4 3
4 4
4 5
4 6
4
7
::: ::: ::: ::: ::: ::: :::
Imagining the positive rational numbers arranged as above, we can show that the natural numbers can be paired with them. That is, we will define a one-to-one functionf takingNontoQC. As we define the function, you should keep looking back at the array to see the pattern that we are using.
Definef .1/ D 11 andf .2/ D 12. (We can’t continue byf .3/ D 13,f .4/ D 14, . . . , for thenf would only map onto those rational numbers with numerator 1.) Definef .3/ D 21 andf .4/ D 31. We can’t just keep going down in our array; we must include the numbers above as well. We need not include 22 however, since
2
2 D 11, which is already paired with 1. Thus, we let f .5/ D 13, f .6/ D 14, f .7/ D 23,f .8/ D 32,f .9/ D 41, andf .10/ D 51. We need not consider 42, since
4
2 D 21, and we need not consider 33 D 11 or 24 D 12. Thus, f .11/is defined to be 15 andf .12/ D 16. It is apparent that a pairing of the natural numbers and the positive rational numbers is indicated by continuing to label rational numbers with natural numbers in this manner, “zigzagging,” you might say, through the above
array. Therefore,jQCj D jNj. ut
10.2 Countable Sets and Uncountable Sets
You may be wondering whether or not every infinite set can be paired with the set of natural numbers. If the elements of a set can be paired with the natural numbers, then the elements can be listed in a sequence. For example, if we let s1 be the element of the set corresponding to the natural number 1,s2be the element of the set corresponding to the natural number 2,s3 to 3, and so on, then the set could be displayed:
fs1; s2; s3; : : :g
Pairing the elements of a set with the set of natural numbers is, in a sense, “counting the elements of the set.”
90 10 Sizes of Infinite Sets
Definition 10.2.1. A set is countable (sometimes called denumerable, or enumerable) if it is either finite or has the same cardinality as the set of natural numbers. A set is said to beuncountableif it is not countable.
One example of an uncountable set is the following.
Theorem 10.2.2. The set of all real numbers between 0 and 1 is uncountable.
Proof. We must show that there is no way of pairing the set of natural numbers with the set of real numbers between 0 and 1. LetSdenote the set of real numbers between 0 and 1:S D fxW0x1g. We will show that every pairing of natural numbers with elements ofS fails to include some members ofS. In other words, we will show that there does not exist any function that mapsNontoS.
Note that the elements ofS can be written as infinite decimals; that is, in the form :c1c2c3: : :, where each ci is a digit between 0 and 9. Some numbers have two different such representations. For example,:9999 : : :is the same number as 1:0000 : : :, and :19999 : : :is the same number as :20000 : : :. For the rest of this proof, let us agree that we choose the representation involving an infinite string of 9’s rather than the representation involving an infinite string of 0’s for all numbers that have two different representations.
Suppose, then, thatf is any function takingNtoS. To show thatf cannot be onto, we imagine writing out all the values off in a list, as follows:
f .1/D:a11a12a13a14a15: : : f .2/D:a21a22a23a24a25: : : f .3/D:a31a32a33a34a35: : : f .4/D:a41a42a43a44a45: : : f .5/D:a51a52a53a54a55: : :
:::
We now construct a number inSthat is not in the range of the functionf. We do that by showing how to choose digitsbj so that the numberxD:b1b2b3b4: : :is not in the range off. Begin by choosingb1 D3ifa11 6D3andb1 D4ifa11 D3. No matter what digits we choose for thebj forj 2, the numberxwill be different fromf .1/since its first digit is different from the first digit off .1/. Then choose b2 D3ifa22 6D3andb2 D4ifa22 D3. This insures thatx6Df .2/. We continue in this manner, choosingbj D 3 if ajj 6D 3 andbj D 4 ifajj D 3, for every natural numberj. The numberxthat is so constructed differs fromf .j /in itsjth digit. Therefore,f .j / 6D xfor allj, soxis not in the range off. Thus, we have proven that there is no function (one-to-one or otherwise) takingNontoS, so we conclude thatShas cardinality different from that ofN. ut
10.2 Countable Sets and Uncountable Sets 91
Of course, any given functionf in the above proof could be modified so as to produce a function whose range does include the specific number x that we constructed in the course of the proof. For example, given anyf, define the function gWN!Sby definingg.1/Dxandg.n/Df .n1/, forn2. The range ofg includesxand also includes the range off. However,gdoes not mapNontoS, for the above proof could be used to produce a differentxthat is not in the range ofg.
Definition 10.2.3. Foraandbreal numbers withab, theclosed interval froma tobis the set of all real numbers betweenaandb, includingaandb. It is denoted Œa; b. That is,Œa; bD fxWaxbg.
The theorem we have just proven asserts that the closed unit interval,Œ0; 1, is uncountable. How does the cardinality of other closed intervals compare to that of Œ0; 1?
Theorem 10.2.4. Ifaandbare real numbers anda < b, thenŒa; bandŒ0; 1have the same cardinality.
Proof. The theorem will be established if we can construct a functionf WŒ0; 1! Œa; bthat is one-to-one and onto. That is easy to do. Simply definef byf .x/ D aC.ba/x. Thenf .0/D aandf .1/D b. Moreover, the functionf increases fromatobasx increases from 0 to 1. Ify 2 Œa; b, letx D ybaa. Thenx 2Œ0; 1 andf .x/ D y. This shows thatf is onto. To show thatf is one-to-one, assume thataC.ba/x1DaC.ba/x2. Subtractingafrom both sides of this equation and then dividing both sides bybayieldsx1 D x2. This shows thatf is one- to-one. Thus,f is a pairing of the elements ofŒ0; 1with the elements ofŒa; b, so
ˇˇŒ0; 1ˇˇDˇˇŒa; bˇˇ. ut
There are other intervals that frequently arise in mathematics.
Definition 10.2.5. Ifaandbare real numbers anda < b, then theopen interval betweenaandb, denoted.a; b/, is defined by
.a; b/D fxWa < x < bg Thehalf-open intervalsare defined by
.a; bD fxWa < xbg and Œa; b/D fxWax < bg
How does the size of a half-open interval compare to the size of the correspond- ing closed interval?
Theorem 10.2.6. The intervalsŒ0; 1and.0; 1have the same cardinality.
Proof. We want to construct a one-to-one functionf takingŒ0; 1onto.0; 1. We will definef .x/ D x for mostx inŒ0; 1, but we need to make a place for 0 to go to in the half-open interval. For each natural numbern, the rational number 1n is in both intervals. Definef on those numbers byf 1
n
D nC11 , forn 2 N. In particular,f .1/D 12. Note that the number 1, which is in.0; 1, is not in the range
92 10 Sizes of Infinite Sets
off as defined so far. We definef .0/to be 1. We definef on the rest ofŒ0; 1by f .x/D x. That is,f .x/ D x for thosex other than 0 that are not of the form 1n withna natural number. It is straightforward to check that we have constructed a
one-to-one function mappingŒ0; 1onto.0; 1. ut
Suppose thatjSj D jTjandjTj D jUj; mustjSj D jUj? If this was not the case, we would be using the “equals” sign in a very peculiar way.
Theorem 10.2.7. IfjSj D jTjandjTj D jUj, thenjSj D jUj.
Proof. By hypothesis, there exist one-to-one functionsf andgmappingSontoT andT ontoU, respectively. That is,f WS !T andg WT !U. Leth Dgıf be the composition ofg andf. In other words,his the function defined onSby h.s/Dg.f .s//. We must show thathis a one-to-one function takingSontoU. Let ube any element ofU. Sincegis onto, there exists atinT such thatg.t /Du. Since f is onto, there is ansinSsuch thatf .s/Dt. Thenh.s/Dg.f .s//Dg.t /Du.
Thus,his onto.
To see thath is one-to-one, suppose that h.s1/ D h.s2/; we must show that s1 D s2. Nowg.f .s1//D g.f .s2//, sof .s1/D f .s2/sincegis one-to-one. But f is also one-to-one, and sos1Ds2. We have shown thathis one-to-one and onto,
from which it follows thatjSj D jUj. ut
Theorem 10.2.8. Ifa,b,c, andd are real numbers witha < b andc < d, then the half-open intervals.a; band.c; d have the same cardinality.
Proof. The functionf defined byf .x/ D aC.ba/x is a one-to-one function mapping.0; 1 onto.a; b, as can be seen by a proof almost exactly the same as that in Theorem10.2.4. Hence,ˇˇ.0; 1ˇˇDˇˇ.a; bˇˇ. Similarly the functiongdefined byg.x/ D cC.d c/x is a one-to-one function mapping.0; 1 onto.c; d , so ˇˇ.0; 1ˇˇDˇˇ.c; d ˇˇ. It follows from Theorem10.2.7thatˇˇ.a; bˇˇDˇˇ.c; d ˇˇ. ut Are there more positive real numbers than there are numbers in Œ0; 1? The, perhaps surprising, answer is “no.”
Theorem 10.2.9. The cardinality of the set of nonnegative real numbers is the same as the cardinality of the unit intervalŒ0; 1.
Proof. We begin by showing that the setSD fx Wx 1ghas the same cardinality as.0; 1. Note that the functionf defined byf .x/D x1 mapsS into.0; 1. For if x 1, then 1x 1. Also,f mapsSonto.0; 1. For ify 2 .0; 1, then y1 1and f
1 y
Dy. To see thatf is one-to-one, suppose thatf .x1/Df .x2/. Then x1
1 D
1
x2, sox1Dx2. Hence,f is one-to-one and onto, and it follows thatjSj Dˇˇ.0; 1ˇˇ. Now letT D fx W x 0g. Define the functiong byg.x/ D x1. Thengis obviously a one-to-one function mappingSontoT. HencejTj D jSj. Therefore, by Theorem10.2.7,jTj D ˇˇ.0; 1ˇˇ. But, by Theorem10.2.6,ˇˇŒ0; 1ˇˇ D ˇˇ.0; 1ˇˇ. It
follows thatjTj DˇˇŒ0; 1ˇˇ. ut
Must the union of two countable sets be countable? A much stronger result is true.