Dtyjlui

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, $$cup_{n=1}^{infty} I_n neq emptyset.$$

It is not possible. Assume $bigcap_{n=1}^N I_n ne emptyset, forall N inmathbb{N}$ but $bigcap_{n=1}^infty I_n = emptyset$.

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

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

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:

$$I_n = leftlangle0, frac1nright]$$

or bounded:
$$I_n = left[n, +inftyrightrangle$$

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 $ Nin 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 $M$.

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 $bigcap_{n=1}^NI_nnot=emptyset$ for all $Ninmathbb N$. We show that $bigcap_{n=1}^infty I_nnot=emptyset$.

Proof: Define the following sequence:

$$left(bigcap_{n=1}^NI_nright)_{N=1}^infty=left(I_1,I_1cap I_2,I_1cap I_2cap I_3,...right)text.$$

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

$$bigcap_{n=1}^infty I_n=bigcap_{N=1}^inftyleft(bigcap_{n=1}^NI_nright)not=emptysettext,$$

where the first equality is true because

$$I_1cap I_2cap I_3cap...=I_1cap(I_1cap I_2)cap(I_1cap I_2cap I_3)cap...text.$$