5

After watching a proof of the set of computer programs being countable, I thought of the following argument: Consider the sets of real numbers (of the form $0.xxx...$) with $0$, $1$, $2$, $3$, $...$ significant decimal places. Each of these sets is finite as it has $0$, $10^{1}$, $10^{2}$, $10^{3}$, $...$ elements respectively.

Now, there are countably many of these sets, as I can list them using the number of significant decimals as index.

Finally, I have seen a proof (using a zig-zag argument) that the countable union of at most countable sets is countable. But, we just showed (seemingly) that the reals can be written as such an union, so they must be countable!? What may have I done wrong?

Andrés E. Caicedo
  • 81,687
guillefix
  • 510
  • 2
  • 15
  • To put it more glibly: "Why can't I describe the real numbers as a countable union of countable sets?" – Semiclassical Aug 04 '14 at 19:19
  • 2
    @Semiclassical It is not the same. Without the axiom of choice, it is consistent that the reals are a countable union of countable sets (though the set is still uncountable, of course). The argument in the question is still wrong in that case. – Andrés E. Caicedo Aug 04 '14 at 19:20
  • Hmm. Would it have been valid had I said "...of the countable sets of finite decimals?" (Lesson probably being, physicists shouldn't weigh in on set theory) @AndresCaicedo – Semiclassical Aug 04 '14 at 19:21
  • 2
    @Semiclassical One could simply say: "Why can't I describe the reals is a way that excludes $\pi,17/9,\sqrt2,$ and just about anything interesting from existing?" – Andrés E. Caicedo Aug 04 '14 at 19:25
  • 1
    You never get $1/9$ in your set. $1/9$ is a limit point of your enumerated values, but it isn't one of the enumerated values. – Thomas Andrews Aug 04 '14 at 19:26
  • Your definition of "reals" excludes numbers with $\infty$ significant digits, not to mention reals < 0 or > 1. – Tim S. Aug 04 '14 at 19:54

2 Answers2

22

Because the real numbers having finite decimal expansions are all rational, and your enumeration doesn't capture any irrational numbers (and misses out some rationals as well).

Mark Bennet
  • 101,769
  • Hm, I think I see. So the problem is that any particular of the subsets in my list have finitely many decimal places? So even though the natural numbers themselves has the same size than the number of decimal places in an infinite decimal (countable), I never get to a natural number that big? It's tough to get the head around that :P – guillefix Aug 04 '14 at 19:32
  • @guillefix No, all the elements of your list have only finitely many non-zero decimal places. All of the numbers in your list can be written as $\frac{M}{10^n}$ where $M$ is an integer and $n$ is a positive integer. Your list simply does not contain any other numbers. It has nothing to do with subsets. – Thomas Andrews Aug 04 '14 at 19:39
  • So just to confirm, every natural number is finite? – guillefix Aug 04 '14 at 19:42
  • 4
    @guillefix Can you think of an infinite natural number? – Mark Bennet Aug 04 '14 at 19:44
  • No, I can not :P – guillefix Aug 04 '14 at 19:45
  • Induction wouldn't be our intuition if we allowed infinite natural numbers. @guillefix – Thomas Andrews Aug 04 '14 at 20:31
10

Your proof only tells you that the collection of numbers with finitely long decimal expansions is countable, which it is. But what about real numbers whose decimal expansions don't terminate (where, for example, would $\pi - 3$ come in your list of decimals)?

Peter Smith
  • 56,527