The Uncountability of R

There are many ways to prove that RR, the set of real numbers, is uncountable, i.e., it has no bijection with a subset (may be proper or improper) of NN, the set of naturals. Here I will provide a simple proof. The proof is basically done by assuming [0,1] to be countable (its elements can be listed) and then devising a method to exclude all the elements of [0,1] one by one and show that still at least one real remains which is not in the list

First we shall state the result that a superset of an uncountable set is uncountable. So, if we can prove a subset of RR to be uncountable, then it follows that RR is uncountable. We choose the closed interval [0,1][0,1]. We start by assuming that this interval is countable, or in other words, it has a bijection with a subset of NN. Further see that [0,1][0,1] being an infinite set (this is quite trivial, for, if xx belongs to [0,1][0,1], then xn[0,1]nN\frac { x }{ n } \in [0,1]\forall n\in N and thus there are infinite elements in [0,1][0,1] ), if such a bijection exists then such a bijection has to be with NN itself and not any proper subset of NN.

Thus NN can be used as an index set of [0,1][0,1], or in other words the elements of [0,1][0,1] can be indexed as x1,x2,x3,...{ x }_{ 1 },{ x }_{ 2 },{ x }_{ 3 },... Thus we can write [0,1]={xnnN}[0,1]=\left\{ { { x }_{ n } }|{ n\in N } \right\} Now we shall construct a sequence of nested closed intervals, such that the ii-th interval excludes all the above elements till the ii-th element. How? Simply using the bisection method:

i) Bisect [0,1][0,1] into the two closed intervals [0,12][0,\frac { 1 }{ 2 } ] and [12,1][\frac { 1 }{ 2 } ,1].

ii) If x1{ x }_{ 1 } belongs to [12,1][\frac { 1 }{ 2 } ,1], choose I1=[0,12]{ I }_{ 1 }=[0,\frac { 1 }{ 2 } ].

iii) If x1{ x }_{ 1 } belongs to [0,12][0,\frac { 1 }{ 2 } ], then choose I1=[12,1]{ I }_{ 1 }=[\frac { 1 }{ 2 } ,1]

iv) If x1=12{ x }_{ 1 }=\frac { 1 }{ 2 } then bisect [0,12][0,\frac { 1 }{ 2 } ] into similar closed intervals and choose I1{ I }_{ 1 } to be the left interval.

Continue this process ad infinitum. (It may happen that at the ii-th step, xi{ x }_{ i } does not belong to any of the two closed sub-intervals of the bisected interval. Then we can choose any one of the two sub-intervals as Ii{ I }_{ i }, though, for making a definiteness in our choice I prefer the left one always.)

By this process we get a sequence of closed nested intervals which have a non-empty intersection by the Nested Interval Property of RR. See that since each interval belongs to [0,1][0,1], so, their non-empty intersection belongs to [0,1][0,1]. Also see that if li{ l }_{ i } is the length of the interval Ii{ I }_{ i }, then 0<li12iiN0<{ l }_{ i }\le \frac { 1 }{ { 2 }^{ i } } \forall i\in NThus by sandwich property of sequences,lim(li)=0\lim _{ }{ \left( { l }_{ i } \right) } =0. Thus i=1Ii\bigcap _{ i=1 }^{ \infty }{ { I }_{ i } } is a singleton set containing one point only, say xx. Now we shall argue that xxiiNx\neq { x }_{ i }\forall i\in NFor this let us assume that x=xix={ x }_{ i } for some iNi\in N. Then xii=1Ii{ x }_{ i }\in \bigcap _{ i=1 }^{ \infty }{ { I }_{ i } } which is a contradiction since the element xi{ x }_{ i } was eliminated at the ii-th step of our bisection.

Hence we see that x\exists x such that x[0,1]x\in [0,1] but x{xnnN}x\notin \left\{ { { x }_{ n } }|{ n\in N } \right\}. Thus [0,1]{xnnN}[0,1]\neq \left\{ { { x }_{ n } }|{ n\in N } \right\} which is a contradiction. Thus our assumption is wrong and RR is uncountable. [Q.E.D.][Q.E.D.]

Note by Kuldeep Guha Mazumder
5 years, 6 months ago

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In the second paragraph you write " if such a bijection exists then such a bijection has to be with N itself and not any proper subset of N ." I did not understand this part. Why can't the bijection be assumed to be with a proper subset of N.

Max Jerry - 3 years, 1 month ago

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