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Is the following proof for the uncountability of $\Bbb{R}$ sufficient? We first assume that the interval $(0,1)$ is countable. So we can define a bijection $f:\Bbb{N}\rightarrow(0,1)$

$$x_{1}=x_{11}x_{12}x_{13}\\x_{2}=x_{21}x_{22}x_{23}\\x_3=x_{31}x_{32}x_{33}\\.\\.\\.$$

Where $x_{ij}$ is the digit in the $jth$ decimal place of the $ith$ number in the list. Now construct some number $y$ whose $jth$ decimal place $y_{j}=x_{ii}+1$ when $x_{ii}\neq9$ and $0$ otherwise. But $y$, while clearly in $(0,1)$, is not in the list, for it differs from $x_{1}$ in the first decimal, $x_2$ in the second, and so on. So $f:\Bbb{N}\rightarrow(0,1)$ is not surjective, and so not a bijection. $(0,1)$ is therefore not countable, and so neither is $\Bbb{R}$.

Ali Lodhi
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    As Reinhard Meier points out, this proof is flawed. But you can easily fix it by using a different mapping. For example, $y_i=x_{ii+5}$ if $x_{ii}\le 4$, and $y_i=x_{ii}-5$ if $x_{ii}\ge 5$. Then $y_i$ and $x_{ii}$ can't represent the same number. – TonyK Mar 28 '19 at 10:14

2 Answers2

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The problem with this proof is that it assumes there is a one-to-one mapping between the real numbers in $(0,1)$ and the numbers that can be represented as $0.x_1x_2x_3\ldots$ But this is wrong. Let's say $x_1 = \frac{1}{2},$ but you choose the unusual representation $0.49999\ldots$ for it. Now it happens by accident that $x_{22}=x_{33}=x_{44}=\ldots=9$ (you have not excluded this case from consideration, and you can obviously find a sequence of different real numbers $x_2,x_3,\ldots$ with this property.) Then your $y$ will be $0.50000$ which represents exactly the same real number as $x_1$, and you haven't proven anything.

Sure, all those drawbacks originating from the usage of the decimal representation of real numbers can be addressed somehow. But there are much simpler proofs which do not have to deal with the ambiguity of decimal representation.

If you want to stay with the "diagonal argument", you have to ensure at least that the $y$ cannot be a different decimal representation of a real number that is already in the list.

Reinhard Meier
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  • +1 for this. But the problem is easy to fix $-$ see my comment to the OP. – TonyK Mar 28 '19 at 10:16
  • I proceeded by proof by contradiction, and so that is why I assumed that there is a one-to-one mapping between the real numbers in (0,1) and the numbers that can be represented as 0.x1x2x3.... – Ali Lodhi Mar 28 '19 at 15:51
  • @AliLodhi But your proof is not guaranteed to produce a contradiction. You have to construct the $y$ differently, as TonyK pointed out. Alternatively, make up a proof that does not depend on the representation of the real numbers, but on their construction (Dedekind cuts, Cauchy sequences, nested intervals, etc.) This would be a better solution. As soon as you use the decimal representation, you are already far away from the definition/construction of the real numbers. It is not a trivial fact that we have such thing as a representation of the real numbers as decimal numbers. – Reinhard Meier Mar 28 '19 at 16:52
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You can simplify the diagonal argument considerably by considering the binary representation of real numbers. Then you simply go along the diagonal flipping 0s to 1s and 1s to 0s.

The binary version goes like this:

Consider all binary sequences. Each of these sequences maps to a number in $[0,1]$ via:

$$\{a_n\} \mapsto \sum_{j=1}^{\infty} \frac{a_j}{2^j}.$$

Note that because of the geometric series result, this sum is in $[0,1]$, and hence we have mapped each of these sequences to a number in $[0,1]$. Now suppose the set of these sequences were countable. Then we could simply order the sequences. But, if we can do this, then I can create a new sequence where the $j$th entry is the negation of the $j$th sequence's $j$th entry. By construction, this sequence is different from all the others. This contradicts the assumption that this set of sequences is countable. Hence, $[0,1]$ must be uncountable, and therefore $\mathbb{R}$ must be uncountable.

MatthewPeter
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  • This proof has the same flaw as the OP's proof, and it is even more difficult to work around it, because you do not have enough one-digit numbers to use TonyK's approach. – Reinhard Meier Mar 28 '19 at 10:35
  • Oh geez. Thanks for pointing that out. You are correct, but if you consider the subset of these sequences such that the limit of the sequence does not exist then the argument works, although this tidbit does introduce some undesirable complexity. – MatthewPeter Mar 28 '19 at 10:42