# uncountable nested intervals

When we use the axiom of completeness to prove the nested intervals theorem we use a countable quantity of nested intervals,but i don't know why this restriction, where is the error taking a uncountable quantity of nested intervals and prove that the intersection is not empty?

• What exactly is this "nested intervals theorem" that you're referring to? Do you mean that the countable intersection of nested closed intervals is non-empty? – Omnomnomnom Jul 14 '14 at 16:22
• yes, this theorem here. personal.bgsu.edu/~carother/cantor/Nested.html – Victor Rafael Jul 14 '14 at 16:23

The problem is that an uncountable set can no longer be described in terms of sequences. If we were to replicate this proof in the case of an uncountable collection of intervals, we would need the notion of a net.

In particular, let's say we defined $I_x = [-x,x]$ for $x \in (0,1)$. Then we could indeed conclude that $$\bigcap_{x \in (0,1)} I_x = \{0\}$$ but we could no longer take the set $\{a_x:x \in (0,1)\} = \{-x: x \in (0,1)\}$ to be indexed as a sequence of real numbers.

• What would our inability to index the set as a sequence of real numbers contradict? Would it invalidate the nested interval property? If so, what step in the proof would it invalidate? – dts Jul 31 at 1:26
• @dts There are examples (involving linear orders over different underlying sets, not the reals) when one could find strictly decreasing transfinite sequences of nested closed intervals, and where an analog of the nested closed interval theorem would hold. I posted an answer that includes some such suggestions. – Mirko Aug 28 at 15:17

There is no uncountable well-ordered increasing (or decreasing) sequence of reals, since you can chose a rational between any two consecutive points. The endpoints of a uncountable nested sequence must therefore after some countable stage be constant.

• I just posted an answer, to complement yours, where one could come up with some other complete linear orders (different underlying sets) where one could find strictly decreasing transfinite sequences of closed intervals (that never stabilize) and an analog of the closed intervals theorem would hold. – Mirko Aug 28 at 15:08

There is no error, why should there be? We just need not bother with uncountable collections in that context. (Countable ones are easier to work with.) Besides, as Rene Schipperus points out in his answer, every uncountable nested family must stabilize (i.e. the intervals become constant).

Just to complement that answer (and the answer by Omnomnomnom, which are both correct in the context of the reals), you could come up with a different linear order (and the underlying set is different, not the reals), where there will be uncountable nested family of intervals, and one could state and prove a suitable analog of the nested interval theorem.

One such order would be $$\omega_1+1=\{\alpha:\alpha\le\omega_1\}$$ where $$\omega_1$$ is the first uncountable ordinal. Then the family $$\{[\alpha,\omega_1]:\alpha<\omega_1\}$$ is a nested decreasing (of length $$\omega_1$$) family of closed intervals, and its intersection is singleton $$\{\omega_1\}$$. This order is complete, though not connected (or, more precisely, the topology generated by this linear order is not connected). One could make it connected, by inserting a copy of the interval of reals $$(0,1)$$ between $$\alpha$$ and $$\alpha+1$$ for each $$\alpha<\omega_1$$ (obtaining the one point compactification of the so-called long ray).

As a different example one could take all transfinite sequences of $$0$$'s and $$1$$'s (or of arbitrary integers), of length $$\omega_1$$, ordered lexicographically (dictionary order), take completion if necessary (and add two endpoints if desired, when working with the integers, to get a compact space, but I think this order is already complete, at least when working with $$0$$'s and $$1$$'s), and then again there will be a completeness theorem and there will also be uncountable strictly decreasing transfinite sequences of nested closed intervals (that do never stabilize). The intersection of such a transfinite sequence will be non-empty (by compactness of the largest element of such a sequence).