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Abbot's Understanding Analysis Problem 1.4.8 (d)

If the following statement is possible, give an example; if it is not, provide a compelling argument as to why it is not possible.

A sequence of closed bounded (not necessarily nested) intervals $I_1, I_2, I_3,...$ with the property that

(1) $$\cap_{n=1}^{N} I_n \neq \emptyset, \quad \forall N \in \mathbb{N},$$

and

(2)$$\cup_{n=1}^{\infty} I_n = \emptyset.$$

It seems to me that the statement is not possible. I considered the sequence of nested intervals $(0,1/n]$ for all $n \in \mathbb{N}.$ I know that this does not satisfy the given conditions (it is not closed), but I went through it simply as an exercise. This sequence of open intervals would satisfy both conditions. Since the sequence of intervals must all include zero (as each interval must be closed), it would not otherwise satisfy the properties. Does the falsity of the statement have anything to do with the properties of open and closed intervals?

I tried considering sequences of intervals that were not nested, yet was unable to produce such a sequence that would satisfy both properties.

If the statement is not true, would you give an complete explanation as to why it is impossible? It would seem to me (if it is not true) that it has something to do with the fact that the intervals are closed.

Moreover, it seems to me that a sequence that fulfills both properties cannot be nested, as for all nested, closed sequences of intervals, $$\cup_{n=1}^{\infty} I_n \neq \emptyset.$$

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    $\begingroup$ Sure you do not mean $\cap_n^\infty I_n$? $\endgroup$
    – Rgkpdx
    Jun 10 '18 at 21:37
  • $\begingroup$ Yes! I just doubled check to make sure! $\endgroup$ Jun 10 '18 at 21:57
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    $\begingroup$ You should check that your (1) and (2) can never hold together. $\endgroup$
    – Rgkpdx
    Jun 10 '18 at 22:00
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    $\begingroup$ $\cup_{n=1}^{\infty} I_n = \emptyset$ would imply $\forall n, I_n=\emptyset$. $\endgroup$ Jun 10 '18 at 22:06
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Here is a proof that such a sequence is impossible, without assuming knowledge of open covers and compact sets.

Let $(I_n)_{n=1}^\infty$ be a sequence of closed, bounded intervals (not necessarily nested) such that $\bigcap_{n=1}^NI_n\not=\emptyset$ for all $N\in\mathbb N$. We show that $\bigcap_{n=1}^\infty I_n\not=\emptyset$.

Proof: Define the following sequence:

$$\left(\bigcap_{n=1}^NI_n\right)_{N=1}^\infty=\left(I_1,I_1\cap I_2,I_1\cap I_2\cap I_3,...\right)\text.$$

Note that these intervals are nested, since $A\cap B\subset A$ for any sets $A$ and $B$. By the Nested Interval Property, we have the following:

$$\bigcap_{n=1}^\infty I_n=\bigcap_{N=1}^\infty\left(\bigcap_{n=1}^NI_n\right)\not=\emptyset\text,$$

where the first equality is true because

$$I_1\cap I_2\cap I_3\cap...=I_1\cap(I_1\cap I_2)\cap(I_1\cap I_2\cap I_3)\cap...\text.$$

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  • $\begingroup$ The subset of a closed, bounded interval can be open, so I would also confirm or mention that $\cap^N_{n=1}I_n$ is closed (and bounded) for all $N\in \mathbb{N}$. This way we know all the conditions for NIP are met. $\endgroup$
    – hiroshin
    Apr 12 '20 at 3:10
  • $\begingroup$ The logic is sound but appears to be so random. How did you come up with this solution? I can never come up with this in a million years for it is so hard to connect the dots with just reading the first four sections of Steven Aboit's Understanding Analysis. $\endgroup$
    – Andes Lam
    Jan 12 at 7:39
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    $\begingroup$ I came up with this solution a year after taking analysis, when I was given the same problem again in a different course. It's funny how math works like that! $\endgroup$
    – Kyle
    Jan 13 at 13:37
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It is not possible. Assume $\bigcap_{n=1}^N I_n \ne \emptyset, \forall N \in\mathbb{N}$ but $\bigcap_{n=1}^\infty I_n = \emptyset$.

Notice that $\bigcap_{n=1}^\infty I_n \subseteq I_1$ so we have

$$I_1 = I_1 \setminus \emptyset = I_1 \setminus \left(\bigcap_{n=1}^\infty I_n\right) = \bigcup_{n=1}^\infty (I_1 \setminus I_n)$$

The sets $I_1 \setminus I_n$ are open in $I_1$ so $(I_1 \setminus I_n)_{n=1}^\infty$ is an open cover of the compact set $I_1$.

Therefore, there exists $N \in \mathbb{N}$ such that $I_1 = \bigcup_{n=1}^N(I_1 \setminus I_n)$. Again taking the complement in $I_1$ gives $\bigcap_{n=1}^N I_n = \emptyset$ which is a contradiction with the assumption.


Note that it is possible if $I_n$ are not assumed to be closed:

$$I_n = \left\langle0, \frac1n\right]$$

or bounded: $$I_n = \left[n, +\infty\right\rangle$$

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Hint (assuming the exercise is motivated by a version of Cantor's intersection Theorem): If $(2) \cap_{n=1}^\infty I_n=\emptyset$, then consider $A_N:=\cap_{n=1}^NI_n, $ for $ N\in \mathbb N$, which are non empty nested, closed and bounded sets, and $\cap_{N=1}^M A_N=\cap_{n=1}^MI_n$ for all $M$.

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