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If we try to map the rational numbers that terminate $\left[0,1\right)$ with the irrational numbers $\left[0,1\right)$, we could do it like this.

$$ 0.1 \to 0.\mathbf1271828182845 \dots $$ $$ 0.14 \to 0.1\mathbf41421356237 \dots $$ $$ 0.16 \to 0.16\mathbf1803398874 \dots $$ $$ 0.13 \to 0.131\mathbf4159265359 \dots $$ $$ 0.101 \to 0.1010\mathbf0100010 \dots $$ $$ 0.2 \to 0.20200\mathbf20002000 \dots $$ The idea is to match each irrational with the shortest prefix from each irrational number that has not been used yet. Since the numbers are irrational we can always find some rational number to pair with each irrational.

The problem is that we can't list all the irrational numbers, because we can diagonalize a new number. In this case we can take the diagonal $0.141402 \dots$ and change every digit to something else $0.251513 \dots$then we will know that this number can't be on the list.

But, we can still use our idea to find a rational number to match with the new diagonal number, in this case the rational number $0.25$, and since the new irrational number will always be different from all the other numbers, at some finite point, we can always find a rational number to pair with it.

My question is "How can we prove that the set of irrational numbers is uncountable, without invoking Cantor's diagonal argument?"

Ivan Hieno
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    The problem here is that the diagonalisation argument doesn't work as smoothly as it does for the real numbers. You have to ensure that the new number you produce is irrational, which means you have to take special care exactly how it is supposed to be different from each number on the list. – Arthur Mar 14 '17 at 10:40
  • We can use the Baire Category Theorem insetad, as shown in the duplicate. – Dietrich Burde Mar 14 '17 at 10:43
  • Do we know that $\mathbb R$ is uncountable? If we do, then result is immediate. – AlvinL Mar 14 '17 at 10:43
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    @AlvinLepik Do we know that $\mathbb{R}$ is uncountable without invoking Cantor's diagonal argument, because that's the question. – Dietrich Burde Mar 14 '17 at 10:45
  • @DietrichBurde If we have nested intervals $\ldots\supset [a_j,b_j]\supset [a_{j+1},b_{j+1}]\supset\ldots $, with $a_j<b_j$ for each $j$, it's standard analysis result that their countable intersection is non-empty [ i.e the metric space $\mathbb{R}$ is complete]. One concludes from this fact, that any open interval in $\mathbb{R}$ is uncountable, therefore $\mathbb{R}$ is uncountable. – AlvinL Mar 14 '17 at 10:47
  • @ Arthur, if the number you produce is not irrational, then you do not need to map anything to that number. You need only map some unique rational to each irrational, so no special care is needed. – Ivan Hieno Mar 14 '17 at 12:23

2 Answers2

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The usual proof goes:

  • The rationals are countable
  • The reals are uncountable
  • Therefore, the irrationals are uncountable

If this proof form is not already known, one mechanic for deducing it is to assume the irrationals are countable. Then we could count the reals by:

  • Let $f_n$ be an enumeration of the rationals
  • Let $g_n$ be an enumeration of the irrationals
  • Then $h_n$ is an enumeration of the reals, where

$$ h_n = \begin{cases} f_m & n = 2m \\ g_m & n = 2m + 1 \end{cases} $$

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Cantor's argument works by contradiction, because proving something to non-exist is difficult.

It works by showing that whatever enumeration you can think of, there is an element which will not be enumerated. And Cantor gives an explicit process to build that missing element.

I guess that it is uneasy to work in other way than by contradiction and by exhibiting an element which differs from all the enumerated ones. So a variant of the diagonal argument seems hard to avoid.

Of course, another proof technique is to establish a bijection between the set at hand and another already known to be uncountable.