# Prove that there exists precisely one $p \in \mathbb R$ such that $p \in \bigcap_{n=1}^\infty D_n$

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 $\lim_{n\rightarrow\infty}(b_n - a_n) = 0$. 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 {$(a_n)_{n\in\mathbb N}$} 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 $p\in D_n = [a_n, b_n]$ for all $n\in\mathbb N$ 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 $\bigcap_{n=1}^\infty D_n$

• Sorry! Supremum of $a_n$ I mean! – Florian Suess Sep 8 '18 at 6:53
• Have you already heard of compacteness ? – Suzet Sep 8 '18 at 6:56
• Not by that name at the very least. Looking at this math.stackexchange.com/questions/607889/…, this seems to be not familiar to me at all. – Florian Suess Sep 8 '18 at 7:00
• This should work indeed. Note that $\sup (a_n)$ also is the limit of the sequence $(a_n)$ (why does it converge ?). This may help proving that it lies in every $D_n$. Then, showing that the intersection can not contain more than one element should be easier. – Suzet Sep 8 '18 at 7:18
• The notation $(a_n)_{n\in \mathbb{N}}$ is usually defined, in ZFC set theory, as a function from the natural numbers into the reals. As such, you can write it $a : \mathbb{N} \rightarrow \mathbb {R}$ and then the notation $\{{(a_n)}_{n \in \mathbb{N}}\}$ just means the image / range of $a$ which is definitely a valid set, since a function is simply a collection of tuples so that the range is projection onto the second component. – Mark Sep 8 '18 at 7:35

## 6 Answers

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.
• To justify the existence of $\lim (a_n)$, wouldn't it be easier to look at monotony ? – Suzet Sep 8 '18 at 7:19
• @Suzet That is an alternate option. I wouldn’t say easier as the fact that an increasing sequence converges in $\mathbb R$ uses completeness of $\mathbb R$. – user532133 Sep 8 '18 at 7:23
• @j314159 Nice point, thank you ! – Suzet Sep 8 '18 at 7:23
• @Mark For sure! The result doesn’t necessarily hold in a non complete space. – user532133 Sep 8 '18 at 7:28
• Oh yes and this also clarifies OP's question about whether this is true in $\mathbb {Q}$ – Mark Sep 8 '18 at 7:41

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 $\underset{n\rightarrow\infty}{\lim}(b_{n}-a_{n})=0,$ there exist $N_{k} \in \mathbb{N}$ such that for all $n \geq N_{k}$, $|b_{n}-a_{n}|< \frac{1}{2^{k+1}}$. Then we see that by your definition of $\mathcal{D}_{n}$, $$\bigcap\limits_{n \geq N_{k}}\mathcal{D}_{n}\subseteq \bigcap\limits_{n \geq N_{k}}\bigg[a_{n},a_{n}+\frac{1}{2^{k+1}}\bigg).$$ What happens if we union over $N_{k} \in \mathbb{N}$, 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 $\mathbb{R}$ conclude that it converges to some $p \in \mathbb{R}$. How does this impact the set-theoretic containments you obtained?

• Thanks for the help! I'm a bit slow so I was wondering, how does the $|b_{n}-a_{n}|< \frac{1}{2^{k+1}}$ inequality hold? I understand $|b_{n}-a_{n}|<\epsilon$ – Florian Suess Sep 8 '18 at 7:06
• Since the $\varepsilon>0$ in the definition of convergence is arbitrary, for each $k \in \mathbb{N}$ I can pick $\varepsilon_{k}=\frac{1}{2^{k+1}}$. Then to each $\varepsilon_{k}$ corresponds an $N_{k} \in \mathbb{N}$ such that for all $n \geq N_{k}$, $|a_{n}-b_{n}|<\varepsilon_{k}=\frac{1}{2^{k+1}}$. – Dan Sep 8 '18 at 7:09

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 $\Bbb{R}$ are compact.

From the linearity of the limit: $$\lim_{n\to \infty} b_n = \lim_{n\to \infty} a_n$$ Call this limit $p$. We show that if $x\neq p$, there is a closed interval that does not contain it; first of all, note that for every two distinct points in $\Bbb{R}$ 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 $\bar{N} = \max (N, N')$.

Then the closed interval $[a_\bar{N} , b_\bar{N}] = D_\bar{N}$ 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 $N\geq 1$ 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 $x\in D_n = [a_n, b_n]$ $\forall n \in \mathbb N$ (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 $x\le y$. Hence $[x,y]\subseteq D_n$ for all $n\in\mathbb N$ which means $[x,y]\subseteq \bigcap_{n=1}^{\infty}D_n$. Now suppose $z \in \bigcap_{n=1}^{\infty}D_n$ then $a_n \le x \le z \le y \le b_n$ for all $n \in \mathbb N$, hence $\bigcap_{n=1}^{\infty}D_n \subseteq [x,y]$.

Hence $$\bigcap_{n=1}^{\infty}D_n = [x,y]$$

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, $x=y$.

• You proved that $[x,y]\subseteq \cap_{n \in \mathbb N} D_n$. You need to explain why the reverse inclusion holds. – mathcounterexamples.net Sep 8 '18 at 9:03
• Yeah, I've actually been working on it. I think a massive auxiliary is required, showing convergence of $a_n$ and $b_n$ and stuff and then some ugly counter example and stuff. It's in the works. – Florian Suess Sep 8 '18 at 9:07
• Only simple arguments are required! If $z \in \cap_{n \in \mathbb N} D_n$ then $a_n \le z \le b_n$ for all $n$. What can you conclude from there? – mathcounterexamples.net Sep 8 '18 at 9:13
• Oh true! $a_n \le x \le z \le y \le b_n$ by definition of x and y! Hence $z \in [x,y]$ and so $[x,y] = \cap_{n \in \mathbb N} D_n$ – Florian Suess Sep 8 '18 at 9:16
• Thanks so much! – Florian Suess Sep 8 '18 at 9:18