CSIR UGC NET Real Analysis-Basics

  View Differently-Enjoy Pure Maths

M. Velrajan

This illustrates how viewing pure mathematics differently and asking related questions lead to more interesting results.

We start with

Theorem (Cantor) Cardinality of the set of real numbers ℝ, Card(ℝ) is not the same as Card(ℕ), in fact Card(ℕ) < Card(ℝ) and ℝ is uncountable.

Proof  Since the map x → x from ℕ to ℝ is 1-1, card(ℕ) ≤ card(ℝ).

We prove Card (ℝ) is not the same as Card (ℕ) by contrapositive way, i.e. prove by contradiction, by a variation of Cantor’s diagonalization argument.

Suppose there is a one-one correspondence between ℕ and ℝ. Then the real numbers can be arranged in a sequence as x1 , x2, . . . , xn, . . .  .

Consider all the real numbers  x1 , x2, . . . , xn, . . .  in decimal form.

x1 :  x10 . x11 x12 . . . x1n . . . . .

x2 :  x20 . x21 x22 . . . x2n . . . . .

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xn :  xn0 . xn1 xn2 . . . xnn. . . . .

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Let a0 be any integer, for each n ≥ 1, let an ∈ {0, 1, 2, . . . , 9} and an ≠ xnn, the nth decimal place of xn. Then a = a0.  a1 a2 . . . an . . . is a real number. 

Since for each n ≥ 1, the nth decimal place of xn and a are not equal,  a ≠ xn, for all n ≥ 1.

Hence x1, x2, . . . , xn, . . . is not the complete list of real numbers and hence there is no one-one correspondence between ℕ and ℝ.

Therefore, Card (ℝ) is not the same as Card (ℕ).

Hence Card (ℕ) < Card (ℝ) and ℝ is uncountable.

Note : If we understand the proof of the above theorem, we can easily observe that Cantor's diagonalization argument part is the crux of the above proof and that also the integral part of  the decimal representations of real numbers plays no role. Hence, if the integral parts are omitted, we can consider them  as sequences of digits {0, 1, 2, . . . , 9}, hence we get that the set of all sequences of digits {0, 1, 2, . . . , 9} is uncountable. 

x1 :  x11 x12 . . . x1n . . . . .

x2 :  x21 x22 . . . x2n . . . . .

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xn :  xn1 xn2 . . . xnn. . . . .

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For each n ≥ 1, let an ∈ {0, 1, 2, . . . , 9} and an ≠ xnn, the nth term of xn. Then the sequence

a = ( a1 , a2, . . . , an, . . .) ≠ xn, for all n ≥ 1. 

Also we note that the digits {0, 1, 2, . . . , 9} can be replaced with any finite set S. Hence, in general, the set of all sequences of elements of any finite set S is uncountable and in particular the set of all sequences of digits {0, 1} is uncountable. 

But we know that a sequence on a set S is a function from ℕ into S.

Hence the set of all functions from ℕ into any finite set S is uncountable.

Also, if An, n = 1, 2, 3, . . . are finite sets then A =  A1 × A2 × . . . × An × . . .  

is uncountable.(Since any element of  A1 × A2 × . . . × An × . . .  can be considered as sequence

( a1 , a2, . . . , an, . . .), where an ∈ An.)

2. Since the set of all functions from the set of natural numbers ℕ to {0, 1} is uncountable, the set of all functions from any countably infinite set S to {0, 1} is uncountable.

Think in another way:

The sequence                                         0,  1,  1,  0,  1,  1,  1,   0,  0,  1,  0,  0,  1, …………..

can be considered as the

subset of {x1, x2, … , xn, …} :            {       x2,  x3,      x5, x6,  x7,            x10,          x13, … … } and  

subsequence of (xn)               :                    x2,  x3,      x5, x6,  x7,            x10,          x13, … …

3. Let A be a subset of a countably infinite set S = {x1, x2, . . . , xn, . . .}. Then A is determined by whether each element of S is in A or not in A. Hence A can be considered as a function from ℕ into {0,1}, where A(n) = 1, if xn ∈ A, A(n) = 0, if xn ∉ A.

i.e. A can be considered as a sequence with terms 0 or 1. And there is a bijection between the set of all subsets of S  and the set of all sequences with terms 0 or 1. 

Hence the number of subsets of the countably infinite set S is the same  as the number of sequences with terms 0 or 1.

Therefore, the number of subsets of the countably infinite set S is uncountable.

Aliter :

Let S = {s1 , s2 , . . . , sk , . . . } be a countably infinite set.

Suppose the power set, the set of all subsets of S, 𝓟(S) is countable and

𝓟(S) = { A1 , A2 , . . . , Ak , . . . }.

Let A be the subset of S consisting of all sk which are not in Ak.

i.e. For each k, sk ∈ A  if and only if sk ∉ Ak. Then A ∉ 𝓟(S) but A ≠ Ak , k = 1, 2, 3, . . . , which is a contradiction. Hence 𝓟(S) is uncountable.

In general,

Theorem (Cantor's Theorem) If A is any set, then Card(A) < Card(𝓟(A)). 

Proof. Since the map x → {x} from A in to 𝓟(A) is 1-1, Card(A) ≤ Card(𝓟(A)). Now we show that there is no bijection from A onto 𝓟(A).

Suppose g : A → 𝓟(A) is a bijection. Let S = {a ∈ A : a ∉ g(a)}. Then S ∈ 𝓟(A), hence S = g(x), for some x ∈ A, because g is onto. There are two possibilities: x ∈ S and x ∉ S. 

1. If x ∈ S, then x ∉ g(x) = S, i.e. x ∉ S, a contradiction.
2. If x ∉ S, then x ∈ g(x) = S, i.e. x ∈ S, a contradiction.

Therefore, there is no bijection g : A → 𝓟(A). Hence Card(A) < Card(𝓟(A)).

Cantor's theorem implies that there are infinitely many infinite cardinal numbers, and that there is no largest cardinal number. 

Card(A) < Card(𝓟(A)) < Card(𝓟(𝓟(A))) < Card(𝓟(𝓟(𝓟(A)))) < Card(𝓟(𝓟(𝓟(𝓟(A))))) < . .  . .  

It also has the following interesting consequence :

There is no such thing as the "set of all sets''.

For,

Suppose A is the set of all sets. Since every element of 𝓟(A) is a set, we have 

𝓟(A) ⊆ A. Hence Card(𝓟(A)) ≤ Card(A) ≤ Card(𝓟(A)). 

By the Schröder–Bernstein Theorem, 

Card(𝓟(A)) = Card(A), but this contradicts Cantor's Theorem.

3.1. The number of subsets of any infinite set S is uncountable.

3.2. The number of subsets of any finite set S is finite and hence countable.

The number of subsets of the empty set is one. So, by 3.1.there is no set which has countably infinite subsets.

But,.

4. The set of all sequences with terms 0 or 1 with only finite number of 1s is countable.

For, The number of sequences with no 1 and only one 1 are clearly countable.

For any k, in a sequence with exactly k 1s by replacing any one of the terms with 0 by 1 we get a sequence with exactly k+1 1s. Since the replacement of 0 by 1 can be done in a countably infinite number of terms, from a sequence with exactly k 1s we can get only a countably infinite number of sequences with exactly k+1 1s.

And any sequence with exactly k+1 1s can be obtained by this way from a sequence with exactly k 1s.

Hence if the set of all sequences with exactly k 1s is countable then the set of all sequences with exactly k+1 1s is also countable, since a countable union of countable sets is countable.

Therefore, by induction for all n = 1, 2, …., the set of all sequences with exactly n 1s is countably infinite.

Hence the set of all sequences with terms 0 or 1 with only finite number of 1s is countably infinite(including the sequence with no 1s).

4.1. And hence the set of all finite subsets of a countably infinite set is countably infinite.

4.2. The set of all infinite subsets of a countably infinite set is uncountable(since the set of all subsets of a countably infinite set is uncountable).

4.3. The set of all sequences with terms 0 or 1 with an infinite number of 1s is uncountable.

5. The sequence 1, 2, 1, 2, 1, 2, …….  has the following infinitely many subsequences :  

1, 2, 2, 2, 2, . . . ;           2, 1, 2, 2, 2, . . . ;                2, 2, 2, 1, 2, . . . ;  

2, 2, 2, 2, 1, 2, 2, 2,. . . ; and so on.     And all these subsequences are convergent. 

But 1, 2, 1, 2, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, ……. has only finitely many distinct subsequences, since after a finite stage all the terms are 2. 

1, 2, 1, 2, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, ……. ;   2, 1, 2, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, ……. ;

1, 2, 2, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, ……. ;       1, 2, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, …….  ;

2, 2, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, ……. ;           2, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, ……. ;

1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, ……. ;           2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, ……. .

In general, if all the terms of a sequence are a constant,  after a finite number of terms then it has only a finite number of distinct subsequences and in all other cases the sequence has uncountable distinct subsequences.  

For, A subsequence of a sequence can be considered as a sequence with terms 0 or 1 with an infinite number of 1s and vice-versa and this correspondence is one - one. Hence the set of all subsequences of a sequence which is not a constant after a finite number of terms is uncountable.

(if (ank) is a subsequence of a sequence (an) then (ank) can be identified with the sequence with terms 0 and 1 with 1 at n1, n2, . . . , nk, . . . . th terms and 0 in all other terms, and vice-versa)    

6. In this way by observing, viewing differently and asking related questions we can get more interesting results. This is a way of enjoying the beauty of pure mathematics.

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