# The Uncountability of R

There are many ways to prove that $R$, the set of real numbers, is uncountable, i.e., it has no bijection with a subset (may be proper or improper) of $N$, 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 $R$ to be uncountable, then it follows that $R$ is uncountable. We choose the closed interval $[0,1]$. We start by assuming that this interval is countable, or in other words, it has a bijection with a subset of $N$. Further see that $[0,1]$ being an infinite set (this is quite trivial, for, if $x$ belongs to $[0,1]$, then $\frac { x }{ n } \in [0,1]\forall n\in N$ and thus there are infinite elements in $[0,1]$ ), if such a bijection exists then such a bijection has to be with $N$ itself and not any proper subset of $N$.

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

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

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

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

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

Continue this process ad infinitum. (It may happen that at the $i$-th step, ${ x }_{ i }$ 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 ${ I }_{ i }$, 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 $R$. See that since each interval belongs to $[0,1]$, so, their non-empty intersection belongs to $[0,1]$. Also see that if ${ l }_{ i }$ is the length of the interval ${ I }_{ i }$, then $0<{ l }_{ i }\le \frac { 1 }{ { 2 }^{ i } } \forall i\in N$Thus by sandwich property of sequences,$\lim _{ }{ \left( { l }_{ i } \right) } =0$. Thus $\bigcap _{ i=1 }^{ \infty }{ { I }_{ i } }$ is a singleton set containing one point only, say $x$. Now we shall argue that $x\neq { x }_{ i }\forall i\in N$For this let us assume that $x={ x }_{ i }$ for some $i\in N$. Then ${ x }_{ i }\in \bigcap _{ i=1 }^{ \infty }{ { I }_{ i } }$which is a contradiction since the element ${ x }_{ i }$ was eliminated at the $i$-th step of our bisection.

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

- 3 years, 1 month ago