Intervals and the size of R

In document Basic Analysis: Introduction to Real Analysis (Page 38-41)

Real Numbers

1.4 Intervals and the size of R

Note: 0.5–1 lecture (proof of uncountability ofRcan be optional)

You surely saw the notation for intervals before, but let us give a formal definition here. For a,b∈Rsuch thata<bwe define

[a,b]:={x∈R:a≤x≤b},

(a,b):={x∈R:a<x<b},

(a,b]:={xR:a<x≤b},

[a,b):={xR:a≤x<b}.

The interval[a,b]is called aclosed intervaland(a,b)is called anopen interval. The intervals of

the form(a,b]and[a,b)are calledhalf-open intervals.

The above intervals were allbounded intervals, since bothaandbwere real numbers. We define unbounded intervals,

[a,∞):={xR:a≤x},

(a,∞):={xR:a<x},

(−∞,b]:={x∈R:x≤b},

(−∞,b):={x∈R:x<b}.

For completeness, we define (−∞,∞):=R. The intervals[a,∞),(−∞,b], and Rare sometimes

calledunbounded closed intervals, and (a,∞),(∞,b), andRare sometimes calledunbounded

open intervals.

In short, an interval is a set with at least two points that contains all points between any two points. The proof of the following proposition is left as an exercise.

Proposition 1.4.1. Let IRbe a set. Then I is an interval if and only if I contains at least 2 points, and whenever a<b<c and a,cI, then bI.

We have already seen that any open interval(a,b)(wherea<bof course) must be nonempty.

For example, it contains the number a+b

2 . An unexpected fact is that from a set-theoretic perspective, all intervals have the same “size,” that is, they all have the same cardinality. For example the map

f(x):=2xtakes the interval[0,1]bijectively to the interval[0,2].

Maybe more interestingly, the function f(x):=tan(x)is a bijective map from(π/2,π/2) to R. Hence the bounded interval (−π/2,π/2) has the same cardinality asR. It is not completely

straightforward to construct a bijective map from[0,1]to(0,1), but it is possible.

And do not worry, there does exist a way to measure the “size” of subsets of real numbers that “sees” the difference between[0,1]and[0,2]. However, its proper definition requires much more

machinery than we have right now.

Let us say more about the cardinality of intervals and hence about the cardinality of R. We

have seen that there exist irrational numbers, that is R\Qis nonempty. The question is: How ∗Sometimes single point sets and the empty set are also called intervals, but in this book, intervals have at least 2 points.

many irrational numbers are there? It turns out there are a lot more irrational numbers than rational numbers. We have seen thatQis countable, and we will show thatRis uncountable. In fact, the

cardinality ofRis the same as the cardinality ofP(N), although we will not prove this claim here.

Theorem 1.4.2(Cantor). Ris uncountable.

We give a modified version of Cantor’s original proof from 1874 as this proof requires the least setup. Normally this proof is stated as a contradiction proof, but a proof by contrapositive is easier to understand.

Proof. LetXRbe a countably infinite subset such that for any two real numbersa<b, there is

anx∈X such thata<x<b. WereRcountable, then we could takeX =R. If we show thatX is

necessarily a proper subset, thenX cannot equalR, andRmust be uncountable.

As X is countably infinite, there is a bijection from Nto X. Consequently, we write X as a

sequence of real numbersx1,x2,x3, . . ., such that each number inX is given byxjfor some j∈N.

Let us inductively construct two sequences of real numbersa1,a2,a3, . . .andb1,b2,b3, . . .. Let a1:=x1andb1:=x1+1. Note thata1<b1andx1∈/(a1,b1). Fork>1, supposeak−1andbk−1 have been defined. Let us also suppose(ak−1,bk−1)does not contain anyxjfor any j=1, . . . ,k−1.

(i) Defineak:=xj, where jis the smallest j∈Nsuch thatxj∈(ak−1,bk−1). Such anxjexists by our assumption onX.

(ii) Next, definebk:=xj where jis the smallest j∈Nsuch thatxj∈(ak,bk−1).

Notice thatak<bk andak−1<ak<bk<bk−1. Also notice that(ak,bk)does not containxk and hence does not contain anyxj for j=1, . . . ,k.

Claim: aj<bkfor all j and k inN.Let us first assume j<k. Thenaj<aj+1<···<ak−1< ak<bk. Similarly for j>k. The claim follows.

LetA={aj: j∈N}andB={bj: j∈N}. By and the claim above we have

supAinfB.

Definey:=supA. The numberycannot be a member ofA. Ify=ajfor some j, theny<aj+1, which is impossible. Similarly, ycannot be a member ofB. Therefore,aj <yfor all j∈Nand

y<bjfor all j∈N. In other wordsy∈(aj,bj)for all j∈N.

Finally, we must show thaty/X. If we do so, then we will have constructed a real number not

inX showing thatX must have been a proper subset. Take anyxk∈X. By the above construction xk∈/(ak,bk), soxk6=yasy∈(ak,bk).

Therefore, the sequencex1,x2, . . .cannot contain all elements ofRand thusRis uncountable.

1.4.1

Exercises

Exercise1.4.1: For a<b, construct an explicit bijection from(a,b]to(0,1].

Exercise1.4.2: Suppose f:[0,1](0,1)is a bijection. Using f , construct a bijection from[1,1]toR. Exercise1.4.3: Prove . That is, suppose IRis a subset with at least 2 elements such that

if a<b<c and a,c∈I, then b∈I. Prove that I is one of the nine types of intervals explicitly given in this section. Furthermore, prove that the intervals given in this section all satisfy this property.

Exercise1.4.4(Hard): Construct an explicit bijection from(0,1]to(0,1). Hint: One approach is as follows:

First map(1/2,1]to(0,1/2], then map(1/4,1/2]to(1/2,3/4], etc. Write down the map explicitly, that is, write

down an algorithm that tells you exactly what number goes where. Then prove that the map is a bijection.

Exercise1.4.5(Hard): Construct an explicit bijection from[0,1]to(0,1).

Exercise1.4.6:

a) Show that every closed interval[a,b]is the intersection of countably many open intervals.

b) Show that every open interval(a,b)is a countable union of closed intervals.

c) Show that an intersection of a possibly infinite family of bounded closed intervals, T

λ∈I

[aλ,bλ], is either

empty, a single point, or a bounded closed interval.

Exercise1.4.7: Suppose S is a set of disjoint open intervals inR. That is, if(a,b)∈S and(c,d)∈S, then

either(a,b) = (c,d)or(a,b)(c,d) =/0. Prove S is a countable set.

Exercise1.4.8: Prove that the cardinality of[0,1]is the same as the cardinality of(0,1)by showing that

|[0,1]| ≤ |(0,1)|and|(0,1)| ≤ |[0,1]|. See . This proof requires the Cantor–Bernstein–

Schröder theorem we stated without proof. Note that this proof does not give you an explicit bijection.

Exercise1.4.9(Challenging): A number x isalgebraicif x is a root of a polynomial with integer coefficients,

in other words, anxn+an−1xn−1+···+a1x+a0=0where all an∈Z.

a) Show that there are only countably many algebraic numbers.

b) Show that there exist non-algebraic numbers (follow in the footsteps of Cantor, use uncountability ofR).

Hint: Feel free to use the fact that a polynomial of degree n has at most n real roots.

Exercise1.4.10(Challenging): Let F be the set of all functions f:R→R. Prove|R|<|F|using Cantor’s

.

Interestingly, ifCis the set of continuous functions then

1.5

Decimal representation of the reals

In document Basic Analysis: Introduction to Real Analysis (Page 38-41)