Constructing a nest of intervals with rational end points given $x\in\mathbf{R}$ 
Let $(I_n)_{n\in\mathbf{N}}$ be a family of nonempty closed intervals
in $\mathbf{R}$. Suppose the following hold:

*

*$I_{n+1}\subset I_n$, for all $n\in\mathbf{N}$;

*For each $\varepsilon>0$, there is some $n\in\mathbf{N}$ such that $\sup(I_n)-\inf(I_n)<\varepsilon$.

Then $(I_n)_{n\in\mathbf{N}}$ is called a nest of intervals.

Suppose we are given $x\in\mathbf{R}$. We want to construct a nest of intervals $(I_n)_{n\in\mathbf{N}}$ with rational endpoints such that $\{x\}=\bigcap_{n\in\mathbf{N}}I_n$.
I am not allowed to use limits, accumulation points, etc; only the basic properties of real numbers, supremums, etc.
For $x=0$, it is easy to find such family of intervals. Let $0<x$. I am not sure how to think about this. Can someone please give me a hint to get me started?
 A: If your real $r$ is actually rational the task is rather trivial.
If $r\in\mathbb{R}\setminus\mathbb{Q}$ is defined by some Dedekind cut $A\mid B$ pick sequences $s_1<s_2<s_3<\cdots$ in $A$ and $t_1>t_2>t_3>...$ in $B$ so that $t_n-s_n>0$ becomes arbitrarily small as $n\to\infty$.
Then $I_n=(s_n,t_n)$ does it.
A: Okay basic properties of the reals.  What are they?
One of them is that between any two unequal reals, $x,y$ so that $x < y$ there is a rational $r$ so that $x < r < y$.  (Proof at at end.)
So if $x$ is not rational (let's assume it isn't) then let $l_k = x - \frac 1k < x$ and $u_k = x+\frac 1k$.  And we for every $k$ then $l_k < l_{k+1} < x < u_{k+1} < u_k$.
For each $k$ let $r_k$ be a rational number between $l_k$ and $l_{k+1}$ and let $s_k$ be a rational number between $u_{k+1}$ and $u_k$.
Let $I_k = [r_k, s_k]$ and we are done.
......
Alternatively, just let $r_0, s_0$ be any rational so that $r_0 < x < s_0$ and find a sequence of  rational $r_n, s_n$ so that $r_k < r_{k+1} < x < s_{k+1} < s_k$ and let  $I_k = [r_k, s_k]$.
...
This result was meant to be obvious and almost trivial.  The purpose of the exercise was to get you used to the complicated but precise language that is needed to justify abstractly and formally the basic and obvious notions of the nature of real numbers.
One thing I didn't realize when I took it about Real analysis was that you will not in the first several months learn anything you didn't already know, but you will develop a formal and justifiable basis to describe real nubers and concepts to apply to abstract and other systems that do not superficially resemble real numbers at all.
In this case you are only being asked to to take a series of smaller and smaller intervals with rational numbers as their endpoints. You should have been able to do that in the thirde grad.  $[0,1],[\frac 1{10}, \frac 9{10}], [\frac 1{100}, \frac {99}{100}]....$etc.   and you can make these intervals around any point you want.
That should be obvious.
What is new and non-trivial is to justify the reasons you can do this and how to formally describe them.
===== Addenda:  Proof that between any two real numbers there is a rational number; and a proof that the rationals are not bounded below or above====
Proof that between $x,y; x< y$ there is a rational number $r$ so that $x < r < y$.
Let $M=\{q\in \mathbb Q| q \le x\}$ this is set that is bounded above by $x$ and presumably not empty.  (because the rationals are not bounded; proof to follow). So $\sup M$ exist.
Likewise if $N = \{q\in \mathbb Q| q\ge y\}$ this set is bounded below by $y$ and not empty.  So $\inf N$ exists.
Let $d = y-x > 0$ and we know that $\inf M -d$ is not an upper bound of $M$ and so there is a rational $r_1$ so that $\sup M - d < r_1 \le \sup M$.
And likewise there is an $r_2$ so that $\inf N \le r_2 < \inf N + d$.
Now consider $q= \frac {r_1 + r_2}2$.  Simple algebra shows that $\sup M < q < \inf N$ ($r_1> \sup M-d$ and $r_2 > \inf N$ so $r_1 + r_2 > \sup M +(\inf N -d)$ but $\sup M \le x < x+d = y \le \inf N $) so $\inf N-d \ge \sup M$ so $r_1 + r_2 > 2\sup M$ and $q > \sup M$.  Similarly $q < \inf M$.)
So $q \not \in M$ and $q\not \in N$ so $q > x$ and $q < y$.
.......
Now a proof that the rationals aren't bound above:
If the were there'd be a $\sup \mathbb Q$ and therefor there would be a rational $q$ so that $\sup \mathbb Q - 1 < q \le \sup \mathbb Q$.  But that would mean $\sup \mathbb Q < q + 1$.  But $q + 1$ is rational and that contradicts $\sup \mathbb Q$ being bounded above.
To prove the aren't bounded below is done with the same method.
