My question concerns the interpretation of the nested interval property in Understanding Analysis by Abbott (2015).

I would appreciate some input from members of the community here as to whether my understanding of it is appropriate. That is because it forms part of a proof that I am struggling to understand, and which I will ask in a separate question.


The statement of the nested interval property I am using from Abbott is:

Theorem 1.4.1. (Nested Interval Property). For each $n \in \mathbb{N}$, assume we are given a closed interval $I_n = [a_n, b_n] = \{x \in \mathbb{R} : a_n \leq x \leq b_n \}$. Assume also that each $I_n$ contains $I_{n+1}$. Then the resulting nested sequence of closed intervals

$$I_1 \supseteq I_2 \supseteq I_3 \supseteq I_4 \supseteq \cdots$$ has a nonempty intersection; that is, $\bigcap^{\infty}_{\space n=1} I_n \neq \emptyset$.

The proof uses the axiom of completeness to show that there exists an $x \in \bigcap^{\infty}_{n=1} I_n$, where $x \in \mathbb{R}$.


Does this theorem admit the interpretation that any real number $x \in \mathbb{R}$, as an entity in a continuum, can be characterised by the property that it belongs to the intersection of a nested infinite sequence of closed non-empty bounded intervals of real numbers?

I am not experiencing difficulty with understanding the proof or the theorem, in the sense that it is a statement about the intersection of a nested infinite sequence of closed non-empty bounded intervals of real numbers being non-empty.

And to be clear, I am not asking whether the interpretation that I have outlined is useful for characterising the reals $\mathbb{R}$ in any way, as I have heard that modern definitions do this using Dedekind cuts, rather, whether it is a faulty intepretation. And if the question seems trivial, it might be helpful to bear in mind that I have a pre-analysis (i.e. high school/16th century possibly earlier) understanding of real numbers, where real numbers are these point-like entities whose existence I take for granted.

  • 2
    $\begingroup$ To make "is characterized" work in your question, you want to additionally include in "the property" that the lengths of the intervals approach $0$ as $n \rightarrow \infty.$ Also, to help clarify my previous comment, maybe observe that $[0,\frac{1}{2}]$ "belongs to" the intersection. (There is a slight phrasing issue involved also, since the intersection of these intervals is a set of real numbers, not a real number, so what you really want to be asking is whether the intersection is a set consisting of a single real number.) $\endgroup$ Jun 9, 2021 at 17:53
  • 1
    $\begingroup$ I think it is a fine question in that it points to many issues that are easily confused. Sometimes books try to side-step certain technicalities, such as a set consisting of one real number vs. a real number, in the informal discussion stuff to avoid drowning the reader in all sorts of minutiae (at least as far as the "big picture" is concerned). I also think your question could easily motivate someone (who has time, which I don't right now) to discuss this topic (and thus add to the usefulness of this site), and so in that sense I also think your question is fine, so I've upvoted it. $\endgroup$ Jun 9, 2021 at 18:02
  • 1
    $\begingroup$ I enjoyed reading your question. I found it interesting, useful and clear. To declare something "Trivial" is very dependent on the person making the judgement. +1 $\endgroup$
    – 311411
    Jun 9, 2021 at 18:12
  • 1
    $\begingroup$ Two quick googled results of possible interest: Foundations of Mathematical Analysis by Fabio Bagagiolo (see page 61 = .pdf page 20) and An Analysis of the First Proofs of the Heine-Borel Theorem (from MAA web pages). $\endgroup$ Jun 9, 2021 at 18:15
  • 1
    $\begingroup$ You have to be very cautious with sentences like where real numbers are these point-like entities whose existence I take for granted.! With a naive view, the real line can only contain rational numbers. In which case the answer to your query maybe negative. It is only by constructing properly the reals that you can have a definitive answer to your query. $\endgroup$ Jun 9, 2021 at 18:32

1 Answer 1


The notion of bounded intervals $I_n\,=\,\{x: a_n \leq x \leq b_n, -\infty <a_n \leq b_n < \infty \}$, each one a subset of the last, and such that $d_n\,=\,b_n-a_n\,\to\,0,$ can be the basis of a characterisation of real numbers. Here I follow the development in Knopp's book Theory and Application of Infinite Series.

Define a nest of intervals to be a sequence of $I_n=[a_n,b_n]$ with rational endpoints, always $I_{n+1}\subset I_n,$ and $d_n \to 0$ as $n \to \infty.$

Then the following can be shown.

  1. There is at most one rational $r$ belonging to every $I_n$ of a given nest, and if there is one, we can say $r$ is defined by the nest, and we may view the nest $(I_n)$ as "another symbol for the number $r$."

  2. There exist nests with no rational $r$ as in the previous item.

  3. Since a nest with no $r$ is in a sense unoccupied, we can try to give some meaning to the idea that this given nest is a number.

As in the case of Dedekind cuts, there is much to check to make certain that this outline yields a complete ordered field, containing and extending the rational field. This verification is shown in some detail within the first chapter of Knopp's book.

Let's again go to your query. You specifically asked about intervals with real end-points. By completeness, a nest of real-endpoint intervals (again with $d_n \to 0$ as $n \to \infty$) strikes a unique real number. The characterization you proposed is a bit trivial, because any interval is already a subset of the real numbers, and we have already taken the real numbers for granted. But it is true, if we add the shrinking-to-nothing mentioned by Renfro in the comments. If $\xi$ is real, there exists such a nest. If a nest is given, it contains a real.

Your theorem 1.4.1 is not the place to seek a characterization of an arbitrary real. But this theorem shares the idea of nested intervals with a valid construction of the real number system.

  • $\begingroup$ +1. Thank you again for taking the time to assist. I hope you will understand that it would be an act of bad faith on my part to accept this without taking the time to sit down and thoroughly parse the answer you have written, as well as @DaveL.Renfro's comments and resources, to ascertain where my confusion lies. In particular, he has made me aware of a number of issues concerning 1) the semantics of characterisation 2) the distinction between a set containing a real number and a real number 3) the length of the intervals approaching 0 as $n \rightarrow 0$, which you've also raised. $\endgroup$
    – microhaus
    Jun 9, 2021 at 19:22
  • 1
    $\begingroup$ Of course, no problem, it is my pleasure to think about numbers. It is common on Math Stack for people to accept too hastily. You may get other responses over the next few days. $\endgroup$
    – 311411
    Jun 9, 2021 at 19:27
  • 1
    $\begingroup$ By the way, to me a "characterisation" is (usually? always?) an if-and-only-if theorem. So a characterisation (in the positive integers) of the set $\{6,12,18,\dots\}$ is $\{n:n\, \text{is divisible by 2 and 3}\}$. (en.wikipedia.org/wiki/Logical_equivalence#In_mathematics) $\endgroup$
    – 311411
    Jun 9, 2021 at 19:41
  • $\begingroup$ The way you spelled out and formalised some of David L. Renfro's comments was clear and extremely instructive for me. I'm not sure I would have been able to appreciate his comment about requiring the additional property that $d_n \rightarrow 0$ without it being spelt out. $\endgroup$
    – microhaus
    Jun 17, 2021 at 16:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.