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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...

paulmuaddib
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2 Answers2

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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 $S = \{1/n \mid n \in \mathbb N\}$.
mathcounterexamples.net
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  • Is there a way to answer without using definitions from Topology? Maybe by using Nested Interval Theorem on $S$ and showing that $S$ has at least an element? – paulmuaddib Sep 08 '20 at 12:39
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    See updated answer using Nested Interval theorem. – mathcounterexamples.net Sep 08 '20 at 12:53
  • Hey, thank you! And would you be so kind to give me some guideline to how to prove that this $x \in \cap_{n \in \mathbb N} I_n$ is the $x$ with the property that $\forall n \in \mathbb{N}$ $\left{s \in S \colon |s-x|<\frac{1}{n}\right}$ is infinite? Without using limit point definition. Kept trying solve this since yesterday but I'm not sure. – paulmuaddib Sep 09 '20 at 20:26
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    Take an interval $I_k$ that contains $x$ and that has a length less than $1/n$. Remember that $I_k$ contains an infinite number of points in $S$. – mathcounterexamples.net Sep 10 '20 at 06:23
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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 $n\in \mathbb N$.

Kishalay Sarkar
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