Given the irrationals are an uncountable infinity, (on, say, (0,1),) the intervals between the irrationals which partition the interval (0,1) must also be an uncountable infinity. However, on each interval between any two irrationals, we know that there lies at least a countable infinity of rationals.

Since there are an uncountable number of intervals, each with at least a countable infinity of rationals on them, how can there only be only a countable infinity of rationals?


marked as duplicate by Asaf Karagila cardinals Jan 18 '18 at 4:50

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    $\begingroup$ Why can't one rational be used for uncountably many intervals? $\endgroup$ – Asaf Karagila Jan 18 '18 at 4:49
  • $\begingroup$ Those intervals aren't all disjoint. $\endgroup$ – Lord Shark the Unknown Jan 18 '18 at 4:53
  • $\begingroup$ Now when you require the intervals are disjoint, there are only countably many of them. $\endgroup$ – Asaf Karagila Jan 18 '18 at 4:59
  • $\begingroup$ (Also, there is no partition of an open interval to more than one open interval. But let's ignore that and assume you mean half open intervals...) $\endgroup$ – Asaf Karagila Jan 18 '18 at 5:01
  • $\begingroup$ Thanks, Lord Shark. The proof that the interval (0,1) can only be countably partitioned by disjoint intervals relies on the countability of the rationals. And that's well established. Hmm. So it is impossible to partition the interval into an uncountable number of disjoint intervals on the irrationals. And I would think that similarly true of any (bounded?) space. But. Does this mean we have to invoke the rationals to establish distance? And Thanks, Asaf. And I was a little casual with the word 'partition,' since I was concerned with the interval between the reals. $\endgroup$ – Charles St Pierre Jan 18 '18 at 6:33