Show that $\exists x \in I$ such that: $\forall n \in \mathbb{N}$, the set $\left\{s \in S \colon |s-x|<\frac{1}{n}\right\}$ is infinite I'm trying to solve this homework:
Let $I = [a,b]$, real closed bounded interval where $a<b$. Let $S \subset I$ s.t $S$ is infinite. Show that $\exists x \in I$ such that: $\forall n \in \mathbb{N}$, the set $\left\{s \in S \colon |s-x|<\frac{1}{n}\right\}$ is infinite.
I'm planning to answer by using definition of density. Since $I$ is dense in $\mathbb{R}$, one can affirm that $\forall X \subset I$, $I \cap X \neq \emptyset$. But how can I show that exists a specific x that solves the question? And must I demonstrate $I$ is dense, isn't it dense by construction?
Edit: I cannot use definitions from sequences, series nor topology. I can use until theorems and definitions from Field, Nested Interval Theorem, Interval, Supremum/Infimum, Archimedean property...
 A: As $S$ is infinite in the compact segment $I=[a,b]$, $S$ has at least a limit point $x \in I$. $x$ answers the question.
Other proof using Nested Interval Theorem:

*

*As $S$ is infinite, $[a, \frac{a+b}{2}] \cap S$ or $[\frac{a+b}{2}, b] \cap S$ is infinite. Select for $I_1$ the interval for which the intersection is infinite.

*Proceed in a similar way to define from $I_n$ an interval $I_{n+1}$which length is the half of the one of $I_n$ and such that $I_{n+1} \cap S$ is infinite.

*Select $x \in \cap_{n \in \mathbb N} I_n$ which exists according to the Nested Interval Theorem.

Some additional remarks:

*

*$I$ is not dense in $\mathbb R$. For example $2\sup(\vert a \vert , \vert b \vert)$ is not a limit point of $I$.

*$S$ may not be dense in $I$. For example if $I=[-1,1]$ and $S = \{1/n \mid n \in \mathbb N\}$.

A: We shall use Bolzano-Weierstrass compactness.
Let $X$ be a metric space.Then $X$ is compact iff every infinite subset of $X$ has a limit point.
$I=[a,b]$ is a compact interval and $S\subset I$ is an infinite subset.So,$S$ has a limit point in $I$ .Let $x\in I$ be the point.
Then $x$ is a limit point of $S$ and hence for each $n\in \mathbb N$,we have $B(x,\frac{1}{n})\cap S$ infinite i.e. $\{x\in S: |x-s|<\frac{1}{n}\}$ is infinite for each $n\in \mathbb N$.
