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In the post Not enough names for numbers, we prove that there are more real numbers between 0 and 1 than natural numbers, technically we say that the set of real numbers is uncountable. We now apply the same argument to prove that there are uncountably many real numbers made up entirely by 1 and 2 in the unit interval.

Consider a list of real numbers made up entirely by 1 and 2 in the unit interval:

0.α

_{11}α

_{12}α

_{13}α

_{14}α

_{15}…

0.α

_{21}α

_{22}α

_{23}α

_{24}α

_{25}…

0.α

_{31}α

_{32}α

_{33}α

_{34}α

_{35}…

0.α

_{41}α

_{42}α

_{43}α

_{44}α

_{45}…

……………

construct a real number B=0.β

_{1}β

_{2}β

_{3}β

_{4}β

_{5}… such that β

_{j}=1 if α

_{jj}=2; 2 if α

_{jj}=1.

Again this number B is not in the list. Hence there is never a complete list of such real numbers.

Lets consider a somewhat different problem. The set of natural numbers is {1,2,3,…}. Each subset

*S*of natural numbers is corresponding to a unique real number α

_{S}=0.α

_{1}α

_{2}α

_{3}α

_{4}α

_{5}… in a way that α

_{j}=1 if j∈

*S*; 2 if j∉

*S*, for example

α

_{{1,2}}=0.1122…

α

_{{1,3,4}}=0.121122…

α

_{{2,4,6,8,…}}=0.2121212…

Suppose there is a given list of subsets of natural numbers, then we convert it to a list of numbers in the unit interval made up entirely by 1 and 2, then we have a B which is not in the list, and finally we convert B to a subset S

_{B}={j β

_{j}=1} which is of course not in the given list of subsets of natural numbers.

Can we construct S

_{B}directly from the given list? Yes, we can!

S

_{B}={j : j∉ the j-th subset in the list}

In the first part of the previous section, we have a correspondence that each subset of natural numbers

*S*maps to a real number α

_{S}in (0,1). Different subsets map to different real numbers in (0,1), and hence there is at least as many real numbers in (0,1) as the subsets of natural numbers.

Lets consider another correspondence. For each real number in (0,1), expressed in binary number system, for example 5/8=0.101000…. If α=0.α

_{1}α

_{2}α

_{3}α

_{4}α

_{5}…, then let S

_{α}={j : α

_{j}=1}, for example S

_{0.101000…}={1,3}. Different real numbers in (0,1) maps to different subsets of natural numbers, and hence there is at least as many subsets of natural numbers as real numbers in (0,1).

Combining, we see that there are as many subsets of natural numbers as real numbers.