This is a theorem from Abott's Real Analysis textbook.

Theorem 1.4.1 (Nested Interval Property)

For each $n\in N,$ assume we are given a closed interval $I_n = [a_n,b_n] = \{x\in R : a_n\leq x\leq b_n \}$. Assume also that each $I_n$ contains $I_n{_+}{_1}$. Then the resulting sequence of closed intervals

$I_1\subseteq I_2\subseteq I_3\subseteq I_4\subseteq \cdots$

has a nonempty intersection; that is, $\bigcap\limits_{n=1}^{\infty} I_{n} \neq \emptyset$


Wikipedia says that an given an interval $I_n$ and $I_n{_+}{_1}$, $I_n{_+}{_1}$ is a subset of $I_n$ which makes total sense. But in the theorem it reads that the "resulting sequence of closed intervals" and below it says $I_1$ is a subset of $I_2$, so on and so forth, which is not true. Am I misreading this?

  • 7
    $\begingroup$ If you have quoted the theorem exactly, you are right and Abbot's book has a typo in that all the containment symbols should be reversed to agree with what is said verbally (and with other statements I've seen of nested interval theorem). $\endgroup$
    – coffeemath
    May 18, 2021 at 23:32
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    $\begingroup$ @coffeemath The OP's quote was actually not exact: see screenshot below. $\endgroup$ Feb 12 at 14:29

1 Answer 1


The book says $$I_1 ⊇ I_2 ⊇ I_3 ⊇ I_4 ⊇···$$ (you have it flipped)

enter image description here

  • 3
    $\begingroup$ Although this remark is very useful, you'd better write it as a comment, not as an answer. $\endgroup$
    – Dominique
    Feb 12 at 12:49
  • $\begingroup$ This does not provide an answer to the question. Once you have sufficient reputation you will be able to comment on any post; instead, provide answers that don't require clarification from the asker. - From Review $\endgroup$
    – Dominique
    Feb 12 at 12:50
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    $\begingroup$ @Dominique After having voted like you, I now think this answers exactly the OP's question, so it was better to edit it as an answer than as a comment. I shall improve it with a screenshot of the book. $\endgroup$ Feb 12 at 14:25

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