Existence of a set with precisely the limit points $\{\frac{1}{n}:n\in\mathbb{N}\}$

Question

The book I'm reading Real Analysis by Jay Cummings contains the following question: Can a set with precisely the limit points $\left\{1,\frac{1}{2},\frac{1}{3},\cdots\right\}$ exist?

My immediate answer was/is no, one can easily prove that any such set also contains a limit point for 0. But that made me think, what if the set did contain 0? That is, what is the answer to the question:

Can a set with precisely the limit points $S=\{0,1,\frac{1}{2},\frac{1}{3},\cdots\}$ exist?

I think I have a contruction, $$P=\bigcup^\infty_{n=1}\bigcup^\infty_{j=1}\left\{\frac{1}{n}+\frac{1}{10^j}\right\}$$

It can be easily proven that $P$ contains all the limit points in $S$, but how does one go about proving that $P$ contains only the limit points in $S$?

Answer 1

As Jakobian indicated, my initial answer was wrong because I was providing a satisfying sequence, rather than a satisfying set.

Answer corrected.

For each $~n \in \Bbb{Z^+},~$
let $~S_n~$ denote the set $\displaystyle \left\{ ~\frac{1}{1} + \frac{\sqrt{2}}{n}, ~\frac{1}{2} + \frac{\sqrt{2}}{n}, ~\frac{1}{3} + \frac{\sqrt{2}}{n}, ~\cdots, ~\frac{1}{n} + \frac{\sqrt{2}}{n} ~\right\}.$

Let $~T~$ denote the set $\displaystyle \bigcup_{n\in\Bbb{Z^+}} S_n.$

Let $~A~$ denote the set $\displaystyle \left\{ ~0, ~\frac{1}{1}, ~\frac{1}{2}, ~\frac{1}{3}, ~\cdots ~\right\}.$

I need to show that:

• For $~i,j \in \Bbb{Z^+} ~: ~i \neq j,~$ you have that $~S_i, ~S_j~$ are disjoint.

• Every element of $~A~$ is a limit point of $~T.$

• There are no limit points of $~T~$ that are not elements of $~A.~$

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Given $~i,j \in \Bbb{Z^+} ~: ~i \neq j,~$
let $~a \in S_i, ~b \in S_j.~$
Suppose that $~a = b.~$

Then, there exists $~k_a \in \{1,2,3,\cdots,i\}~$ and $~k_b \in \{1,2,3,\cdots,j\},~$ such that

$\displaystyle \frac{1}{k_a} + \frac{\sqrt{2}}{i} = \frac{1}{k_b} + \frac{\sqrt{2}}{j} \implies $ $\displaystyle \frac{1}{k_a} - \frac{1}{k_b} = \sqrt{2} \times \left[ ~\frac{1}{j} - \frac{1}{i} ~\right] \implies $

$\displaystyle \sqrt{2} = \left[ ~\frac{1}{k_a} - \frac{1}{k_b} ~\right] \times \left\{ ~\left[ ~\frac{1}{j} - \frac{1}{i} ~\right]^{-1} ~\right\}.$

With $~i,j~$ assumed distinct, this yields a contradiction, because $~\sqrt{2}~$ is not rational, and the RHS above is rational.

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$~0~$ is a limit point of the following subset of $~T~$:

$$\left\{ ~\frac{1}{1} + \frac{\sqrt{2}}{1}, ~\frac{1}{2} + \frac{\sqrt{2}}{2}, ~\frac{1}{3} + \frac{\sqrt{2}}{3}, ~\frac{1}{4} + \frac{\sqrt{2}}{4}, ~\cdots ~\right\}.$$

For any $~n \in \Bbb{Z^+},~~\dfrac{1}{n}~$ is a limit point of the following subset of $~T~$ :

$$\left\{ ~\frac{1}{n} + \frac{\sqrt{2}}{n}, ~\frac{1}{n} + \frac{\sqrt{2}}{n+1}, ~\frac{1}{n} + \frac{\sqrt{2}}{n+2}, ~\frac{1}{n} + \frac{\sqrt{2}}{n+3}, ~\cdots ~\right\}.$$

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Suppose $~r \not\in A.~$ I need to prove that $~r~$ can not be a limit point of $~T.$

I will do this by constructing a specific $~\epsilon,~$ and then proving that there can only be a finite number of elements $~t \in T,~$ such that $~|r - t| < \epsilon.~$ This will imply that $~r~$ can not be a limit point of $~T.~$

If $~r < 0,~$ then set $~\epsilon = \dfrac{1}{2}\,|r|.~$
Since every element of $~T~$ is positive, this implies that $~\forall ~t \in T, ~|r - t| > \epsilon.~$

Therefore, if $~r < 0,~$ then $~r~$ can not be a limit point of $~T.~$ Therefore, without loss of generality, $~r > 0.~$

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If $~r > 1,~$ set $~M = 0.~$
Otherwise, set $~M~$ as the largest positive integer such that $~r < \dfrac{1}{M}.~$

This implies that regardless of whether $~r > 1,~$ you will have that $~\dfrac{1}{M+1} < r.~$

If $~M = 0,~$ then set $~\displaystyle \epsilon = \left[ ~\frac{~r - \frac{1}{M+1}}{2} ~\right].$

Otherwise, set $~\displaystyle \epsilon = \min\left[ ~\frac{~r - \frac{1}{M+1}}{2}, ~\frac{1}{M} - r ~\right].$

When $~M > 0,~$ and $~t \in T~$ has form $~\dfrac{1}{d} + \dfrac{\sqrt{2}}{e} ~: ~d \leq M,~$
then you will have that
$~t > \dfrac{1}{M} \implies |r - t| = t - r > \dfrac{1}{M} - r \geq \epsilon.~$

Consequently, when $~M > 0,~$ it is impossible for there to be an infinite number of elements $~t \in T~$ of form $~\dfrac{1}{d} + \dfrac{\sqrt{2}}{e} ~: ~d \leq M,~$
such that $~|r - t| < \epsilon.~$

Therefore, it only remains to show that it is impossible for there to be an infinite number of elements $~t \in T~$ of form $~\dfrac{1}{d} + \dfrac{\sqrt{2}}{e} ~: ~d \geq M+1,~$
such that $~|r - t| < \epsilon.~$

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From the specification for $~\epsilon,~$ you have that
$\dfrac{1}{M+1} \leq r - 2\epsilon.$

Choose $~E \in \Bbb{Z^+}~$ such that $~\dfrac{\sqrt{2}}{E} < \epsilon.$

Then, for every element $~t \in T,~$ of form
$~\dfrac{1}{d} + \dfrac{\sqrt{2}}{e} ~: ~d \geq M+1, ~e \geq d, ~e \geq E,~$

you will have that $~t \leq \dfrac{1}{M+1} + \dfrac{\sqrt{2}}{E} < (r - 2\epsilon) + \epsilon.~$

So, this will imply that $~t < r - \epsilon \implies |r - t| > \epsilon.~$

Therefore, the only elements $~t \in T~$ of form $~\dfrac{1}{d} + \dfrac{\sqrt{2}}{e}~$
where $~d \geq M+1,~$ and $~|r - t| < \epsilon,~$ will have to be in the following subset:

$$\bigcup_{n\in\Bbb{Z^+}, ~M+1 \leq n < E} S_n. \tag1 $$

However, the subset of $~T~$ referred to in (1) above only has a finite number of elements. Thus, it is impossible for there to be an infinite number of elements $~t~$ of form $~\dfrac{1}{d} + \dfrac{\sqrt{2}}{e},~$ such that $~d \geq M+1~$ and $~|r - t| < \epsilon.~$

Thus, when $~r > 0,~$ and $~r \not\in A,~$ it is also impossible for $~r~$ to be a limit point of $~T.$

Answer 2

For a decreasing sequence $a_n$ convergent to $0$ let $$S_n=\{a_n+2^{-k}(a_{n-1}-a_n)\,:\, k\in \mathbb{N}\}$$ The number $a_n$ is the only accumulation point of $S_n.$ Therefore the set of accumulation points of $S=\bigcup_{n=1}^\infty S_n$ coincides with $\{0\}\cup \{a_n\}_{n=1}^\infty.$

Indeed, assume $x_m\in S$ and $x_m\to x.$ If infinitely many terms of $x_m$ belong to $S_n$ for some $n,$ then $x=a_n.$ Otherwise $x_m$ is not bounded below by a positive constant, hence $x=0.$