Vtyjkyuk

For all $n in mathbb N$, let $D_n = [a_n, b_n]$ be a closed interval in $ mathbb R$ with $b_n - a_n > 0$. Suppose that
$$D_1 supset D_2 supset ....$$
Moreover, suppose that $lim_nrightarrowinfty(b_n - a_n) = 0$. Prove that there exists precisely one $p in mathbb R$ such that $p in bigcap_n=1^infty D_n$. (Then I will continue to check if this is also valid everywhere in $mathbb Q$.

Intuitively, it's very simple to draw a picture and see what is going on, and I have concluded that I have to work with the supremum of $(a_n)_ninmathbb N$ but I have no idea how to get started.

This is general preparation for a test, so hints would be preferred, thanks a lot.

(EDIT) My progress:

Define $p = sup$$(a_n)_n in mathbb N$.

It goes without saying that $pin D_n = [a_n, b_n]$ for all $ninmathbb N$ given by the properties of a supremum and how $b_n - a_n > 0$. Moreover by hypothesis; $e>0$, $ |b_n-p| le |b_n-a_n| le e$ (since $p ge a_n$) hence $b_n$ converges to p. I guess I need to show that the infimum of $a_n = p$ or something and thus by properties of infimum and supremum, there lies nothing inbetween.

Left to prove: p is the only thing in $bigcap_n=1^infty D_n$

Hint

• Prove that the sequence $(a_n)$ is Cauchy.


• Prove that its limit $p$ belongs to all $D_n$.


• Prove that there cannot exists two distinct points belonging to
  $displaystyle cap_n in mathbb N D_n$ by considering the
  distance between those potentially two distinct points.

Hint for uniqueness:

The (possibly infinite) intersection of closed intervals is a closed interval. Consider the width of the interval $cap D_n$

Just a thought - here is how I would try to attack the problem:

Let $varepsilon>0$. Since $undersetnrightarrowinftylim(b_n-a_n)=0,$ there exist $N_k in mathbbN$ such that for all $n geq N_k$, $|b_n-a_n|< frac12^k+1$. Then we see that by your definition of $mathcalD_n$,
$$bigcaplimits_n geq N_kmathcalD_nsubseteq bigcaplimits_n geq N_kbigg[a_n,a_n+frac12^k+1bigg).$$
What happens if we union over $N_k in mathbbN$, and then intersect over all $k geq 1$?

If you use similar inequalities and the convergence of $a_n-b_n$ you can show that $a_n$ Cauchy, and by completeness of $mathbbR$ conclude that it converges to some $p in mathbbR$. How does this impact the set-theoretic containments you obtained?

Somebody cited compactness, I'll briefly state here: in a compact space, any collection of closed sets such that every finite subcollection has nonempty intersection (your nested sequence of closed intervals works), the intersection of the entire collection is not empty.
This basically comes from the very definition of compactness, suitably manipulated. Of course, closed intervals in $BbbR $ are compact.

From the linearity of the limit:
$$lim_nto infty b_n = lim_nto infty a_n $$
Call this limit $p$. We show that if $xneq p$, there is a closed interval that does not contain it; first of all, note that for every two distinct points in $BbbR $ you can find two disjoint open intervals that contain them. Let $U$ and $V$ be such intervals, for $p$ and $x$ respectively.

Since $a_n to p$, there is $N$ such that $a_i in U$ for every $i geq N$. Similarly, since $b_n to p$, there is $N'$ such that $b_i in U$ for every $i geq N'$. Let $barN = max (N, N')$.

Then the closed interval $[a_barN , b_barN] = D_barN$ is entirely contained in $U$; thus, it is disjoint from $V$. Since there is a closed interval that does not contain it, $x$ does not belong to the intersection of all $D_n$.

Here is a late answer from a Rudin fan. As you correctly suspected, you should want to consider
$$ x = sup_n geq 1 a_n, $$
which exists finitely since $a_n leq b_1 < infty$ for all $n geq 1$. Then since we know that
$$ a_n leq a_n+m leq b_n+m leq b_m $$
for all $n,m geq 1$, we can conclude that
$$a_m leq sup a_n = x leq b_m$$
for all $m geq 1$. Hence $x in [a_m, b_m]$ for all $m geq 1$, and
$$ x in bigcap_m geq 1 [a_m, b_m] = X. $$

Now assume that $X$ contains another point $y not= x$. Then $y$ is at a positive distance from $x$. Put $r = |x-y| > 0.$ Then since $b_n-a_n to 0$, as $n to infty,$ we can choose $Ngeq 1$ so that
$$ b_n - a_n < r, $$
for all $n geq N$, and then the distance between any two points of $[a_n, b_n]$ is strictly less than $r$ for all $n geq N$. Hence $y$ cannot belong to $X$, and $X = x.$

Thanks for all the responses, here is what I came up with in the end:

Choose $x = sup$$(a_n)$ and $y = inf$$(b_n)$.

It's easy to check $xin D_n = [a_n, b_n]$ $forall n in mathbb N$ (by definition of $x$ and hypothesis $b_n - a_n > 0$). Similarly with $y$.

Following from the properties of supremum of $x$ (or properties of infimum $y$, you choose), we briskly conclude that $xle y$. Hence $[x,y]subseteq D_n$ for all $ninmathbb N$ which means $[x,y]subseteq bigcap_n=1^inftyD_n$. Now suppose $z in bigcap_n=1^inftyD_n$ then $a_n le x le z le y le b_n$ for all $n in mathbb N$, hence $bigcap_n=1^inftyD_n subseteq [x,y]$.

Hence $$bigcap_n=1^inftyD_n = [x,y]$$

Suppose with a view to contradiction that $y ne x$, then set $epsilon = y-x > 0$. But... by hypothesis; $|b_n-a_n| lt epsilon$.

And so it must be the case that, $x=y$.