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The problem is to prove that any set of disjoint non-overlapping intervals that contain more than a single point ("non-singular intervals) is countable. The text seems to want me to offer the following "proof:"

$1$. Start with the first interval and associate it with the number one.

$2.$ For each interval, assign its successor to $n+1$

$3.$ This is an injection from the set of intervals to the set of natural numbers, so the set is countable.

The problem I have with this "proof" is that is that for any interval in the set, there is no guarantee that it has a unique predecessor. Consider the union of two sets of intervals

First Set :

$(1/(n+1),1/(n+2))$ for $n$ all natural numbers $n$

Second set :

An unspecified set of intervals to the right.

The first interval in the second set has have no unique predecessor since there is no maximum value for $n$

I thought about trying to show that if an infinite set of intervals contained in $(0,1)$ were uncountable then it would have to include at least one singular interval.

My difficulty with thinking about this approach is that I don't have any examples of uncountable sets other than the reals, the rationals, etc. I would like to construct an uncountable set of intervals as an example.

I will be asking on meta how to embed markdown and LaTex in my next post.

Anna Naden
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    The set of intervals of the form $(a,a+1)$ is in obvious bijection with $\Bbb R$, so how should it be countable? – Hagen von Eitzen Jul 30 '20 at 05:46
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    Have you missed a condition that the intervals must be disjoint? – badjohn Jul 30 '20 at 05:54
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    Yes they have to be disjoint. – Anna Naden Jul 30 '20 at 07:46
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    Each interval must contain a rational number. – badjohn Jul 30 '20 at 07:53
  • Allright, each interval contains a least rational number, so there is a bijection from the set of intervals to a subset (not necessarily proper) of the rational numbers. This implies that the set is countable, is that right? – Anna Naden Jul 30 '20 at 09:16
  • @AnnaNaden Yes, this is right. Apart from the part "not necessarily proper". Since every interval contains infinite many rational numbers and we assign every interval to one such rational number, the bijection goes necessarily to a proper subset of $\mathbb Q$. This does not matter for the proof, I just wanted to point that out. – Peter Aug 04 '20 at 13:33

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