You have seen the notation for intervals before, but let us give a formal definition here. For
a;b 2 R such that a < b we define
[a;b] := fx 2 R j a x bg;
(a;b) := fx 2 R j a < x < bg;
(a;b] := fx 2 R j a < x bg;
[a;b) := fx 2 R j a x < bg:
The interval [a;b] is called a closed interval and (a;b) is called an open interval. The intervals of
the form (a;b] and [a;b) are called half-open intervals.
The above intervals were all bounded intervals, since both a and b were real numbers. We define
unbounded intervals,
[a;¥) := fx 2 R j a xg;
(a;¥) := fx 2 R j a < xg;
(¥;b] := fx 2 R j x bg;
(¥;b) := fx 2 R j x < bg:
For completeness we define (¥;¥) := R.
We have already seen that any open interval (a;b) (where a < b of 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) := 2x takes the interval [0;1] bijectively to the interval [0;2].
Or, maybe more interestingly, the function f (x) := tan(x) is a bijective map from (p;p)
to R, hence the bounded interval (p;p) has the same cardinality as R. It is not completely
straightforward to construct a bijective map from [0;1] to say (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 RnQ is nonempty. The question is, how
many irrational numbers are there. It turns out there are a lot more irrational numbers than rational
numbers. We have seen that Q is countable, and we will show in a little bit that R is uncountable.
In fact, the cardinality of R is the same as the cardinality ofP(N), although we will not prove this
claim.
Theorem 1.4.1 (Cantor). R is uncountable.
36 CHAPTER 1. REAL NUMBERS
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 always
easier to understand.
Proof. Let X R be a countable subset such that for any two numbers a < b, there is an x 2 X such
that a < x < b. If R were countable, then we could take X = R. If we can show that X must be a
proper subset, then X cannot equal to R and R must be uncountable.
As X is countable, there is a bijection from N to X. Consequently, we can write X as a sequence
of real numbers x1;x2;x3; : : :, such that each number in X is given by some x j for some j 2 N.
Let us construct two other sequences of real numbers a1;a2;a3; : : : and b1;b2;b3; : : :. Let a1 := 0
and b1 := 1. Next, for each k > 1:
(i) Define ak := x j, where j is the smallest j 2 N such that x j 2 (ak1;bk1). As an open interval
is nonempty, we know that such an x j always exists by our assumption on X.
(ii) Next, define bk := x j where j is the smallest j 2 N such that x j 2 (ak;bk1).
Claim: aj < bk for all j and k in N. This is because aj < aj+1 for all j and bk > bk+1 for all k.
If there did exist a j and a k such that aj bk, then there is an n such that an bn (why?), which is
not possible by definition.
Let A = faj j j 2 Ng and B = fbj j j 2 Ng. We have seen before that
sup A inf B:
Define y = sup A. The number y cannot be a member of A. If y = aj for some j, then y < aj+1,
which is impossible. Similarly y cannot be a member of B.
If y =2 X, then we are done; we have shown that X is a proper subset of R. If y 2 X, then there
exists some k such that y = xk. Notice however that y 2 (am;bm) and y 2 (am;bm1) for all m 2 N.
We claim that this means that y would be picked for am or bm in one of the steps, which would be a
contradiction. To see the claim note that the smallest j such that x j is in (ak1;bk1) or (ak;bk1)
always becomes larger in every step. Hence eventually we will reach a point where x j = y. In this
case we would make either ak = y or bk = y, which is a contradiction.
Therefore, the sequence x1;x2; : : : cannot contain all elements of R and thus R is uncountable.
a;b 2 R such that a < b we define
[a;b] := fx 2 R j a x bg;
(a;b) := fx 2 R j a < x < bg;
(a;b] := fx 2 R j a < x bg;
[a;b) := fx 2 R j a x < bg:
The interval [a;b] is called a closed interval and (a;b) is called an open interval. The intervals of
the form (a;b] and [a;b) are called half-open intervals.
The above intervals were all bounded intervals, since both a and b were real numbers. We define
unbounded intervals,
[a;¥) := fx 2 R j a xg;
(a;¥) := fx 2 R j a < xg;
(¥;b] := fx 2 R j x bg;
(¥;b) := fx 2 R j x < bg:
For completeness we define (¥;¥) := R.
We have already seen that any open interval (a;b) (where a < b of 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) := 2x takes the interval [0;1] bijectively to the interval [0;2].
Or, maybe more interestingly, the function f (x) := tan(x) is a bijective map from (p;p)
to R, hence the bounded interval (p;p) has the same cardinality as R. It is not completely
straightforward to construct a bijective map from [0;1] to say (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 RnQ is nonempty. The question is, how
many irrational numbers are there. It turns out there are a lot more irrational numbers than rational
numbers. We have seen that Q is countable, and we will show in a little bit that R is uncountable.
In fact, the cardinality of R is the same as the cardinality ofP(N), although we will not prove this
claim.
Theorem 1.4.1 (Cantor). R is uncountable.
36 CHAPTER 1. REAL NUMBERS
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 always
easier to understand.
Proof. Let X R be a countable subset such that for any two numbers a < b, there is an x 2 X such
that a < x < b. If R were countable, then we could take X = R. If we can show that X must be a
proper subset, then X cannot equal to R and R must be uncountable.
As X is countable, there is a bijection from N to X. Consequently, we can write X as a sequence
of real numbers x1;x2;x3; : : :, such that each number in X is given by some x j for some j 2 N.
Let us construct two other sequences of real numbers a1;a2;a3; : : : and b1;b2;b3; : : :. Let a1 := 0
and b1 := 1. Next, for each k > 1:
(i) Define ak := x j, where j is the smallest j 2 N such that x j 2 (ak1;bk1). As an open interval
is nonempty, we know that such an x j always exists by our assumption on X.
(ii) Next, define bk := x j where j is the smallest j 2 N such that x j 2 (ak;bk1).
Claim: aj < bk for all j and k in N. This is because aj < aj+1 for all j and bk > bk+1 for all k.
If there did exist a j and a k such that aj bk, then there is an n such that an bn (why?), which is
not possible by definition.
Let A = faj j j 2 Ng and B = fbj j j 2 Ng. We have seen before that
sup A inf B:
Define y = sup A. The number y cannot be a member of A. If y = aj for some j, then y < aj+1,
which is impossible. Similarly y cannot be a member of B.
If y =2 X, then we are done; we have shown that X is a proper subset of R. If y 2 X, then there
exists some k such that y = xk. Notice however that y 2 (am;bm) and y 2 (am;bm1) for all m 2 N.
We claim that this means that y would be picked for am or bm in one of the steps, which would be a
contradiction. To see the claim note that the smallest j such that x j is in (ak1;bk1) or (ak;bk1)
always becomes larger in every step. Hence eventually we will reach a point where x j = y. In this
case we would make either ak = y or bk = y, which is a contradiction.
Therefore, the sequence x1;x2; : : : cannot contain all elements of R and thus R is uncountable.
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