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Cantor's diagonal argument shows that there is no bijection between the real segment [0; 1] and natural numbers. Does that mean that we can't enumerate this set ?

If we take an application f(n) from N to R that, for any number, reverse it and place it after the comma, like it:

f(0) = 0.0
f(1) = 0.1
...
f(1234) = 0.4321
...

We clearly see that the application enumerate any given finite real in [0; 1]. What's wrong ?

rafoo
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  • Yes; it is uncountably infinite – Mauro ALLEGRANZA Jan 27 '21 at 15:49
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    Your bijection is between every natural and every real number with a finite decimal expansion. However, most reals have infinite decimal expansion. For example, $\frac13$. – Rushabh Mehta Jan 27 '21 at 15:49
  • What maps to, say, $1/\sqrt{2}$? – Amit Rajaraman Jan 27 '21 at 15:49
  • This is not a bijection $\mathbb{N} \rightarrow [0,1]$, since for exmaple nothing maps to $1/\pi$. It is not even a bijection between $\mathbb{N}$ and $\mathbb{Q} \cap [0,1]$ (which is indeed countable) since many rational numbers have an infinite decimal expantion as @DonThousand points out. – Enforce Jan 27 '21 at 15:51
  • "We clearly see that there is a bijection between N and any given finite real" What is a "finite real"? Can you define what is meant by a finite real? This may help you to answer the question. – Adam Rubinson Jan 27 '21 at 15:52
  • @AdamRubinson In some schools, finite reals are used to refer to reals with finite decimal expansion. But generally, it's only a useful notion if you are considering multiple bases, since any rational base will have all finite reals be rational. – Rushabh Mehta Jan 27 '21 at 15:55
  • "In some schools, finite reals are used to refer to reals with finite decimal expansion." I suspected that, but I wanted OP to answer and have an "ahah!" moment: by realising then that no integer would map to $1/3$. "Bases" has nothing to do with this. – Adam Rubinson Jan 27 '21 at 15:58
  • So the problem is about infinity. Can't we say it will be reached "at infinity" ? – rafoo Jan 27 '21 at 16:01
  • @rafoo No, since the reals are literally "at" infinity, while the naturals "approach" infinity (I mean both terms very very loosely). – Rushabh Mehta Jan 27 '21 at 16:06

2 Answers2

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None of your inputs map to $\frac13 = 0.\overset{.}{3}.$

Adam Rubinson
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Well it makes sense that you can attempt to "construct" such a bijection since those are all rational numbers (ones of the form: $x_Nx_{N-1}...x_0.y_1y_2...y_M$) and the field of rational numbers $\mathbb{Q}$ is countable, so $\mathbb{Q}\cap I$ is countable for any interval $I\subset \mathbb{R}$. You can give an ordering to any interval by just taking the well ordering theorem (by assuming the axiom of choice), but that ordering will not be indexed by the natural numbers (since that would contradict the fact that intervals are uncountable, unless the interval is empty or a singleton).

Here is an intuition for what is going on: any subset of an interval I indexed by the natural numbers, is a sequence in I, this sequence is visually a (one-dimensional) "lattice" of points in I, and the actual interval I itself is too "large" to be "filled in" by a sequence.

More clearly, $[0,1]$ has the same cardinality as $J:= (-\pi/2, \pi/2)$ (take a dilation bijection from J to the unit interval $(0,1)$ and then another bijection to $[0,1]$), but $tan(x)\mid _{(-\pi/2, \pi/2)}$ is a bijection from J to the real line (perhaps it's more intuitive to see that $\mathbb{R}$ does not have an enumeration by $\mathbb{N}$ than the closed unit interval).

Stephen Landsittel
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