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Can someone give me an example or a hint to come up with a countable compact set in the real line with infinitely many accumulation points? Thank you in advance!

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  • $\begingroup$ @carmichael561: I think that you’re right. Careless reading on my part. $\endgroup$ Sep 2, 2016 at 23:41
  • $\begingroup$ Every point in the set is an accumulation point. So, you just need a set that is countably infinite and compact. $\endgroup$
    – Doug M
    Sep 2, 2016 at 23:46
  • $\begingroup$ @DougM I think here "accumulation point" is the same as "limit point" (see here), while what you have in mind is an "adherent point" (it always bugs me too). $\endgroup$
    – user228113
    Sep 3, 2016 at 11:05

4 Answers 4

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What about if we define $H = \{ \frac{1}{n} : n\in \mathbb{N}\} \cup \{0\}$, a sort of standard countable compact set with 0 at its sole limit point, then define your countable compact set to be:

$$S = \{ x + y \mid x, y \in H\}.$$

To unpack the thought behind this definition:

  1. This set $S$ is countable.
  2. This set $S$ is bounded, because its smallest member is 0 and its largest member is 1+1 = 2.
  3. Each $x \in H$ is a limit point of $S$. If we fix $x$ and vary $y\in H$ in the definition of $S$, we can see that $x$ is a limit point of $S$.
  4. This set $S$ is closed (and therefore compact, because it is a bounded subset of the real line.) ($S$ is closed because it is the sum of two compact subsets of $\mathbb{R}$ and is therefore closed in $\mathbb{R}$.)
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    $\begingroup$ $S$ is closed since $S= H+H$ and $H$ is compact. See this post for a proof that the Minkowski sum of compact sets is closed. $\endgroup$ Sep 3, 2016 at 0:00
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HINT: Start with $\{0\}\cup\left\{\frac1n:n\in\Bbb Z^+\right\}$, and figure out how to add for each $n\in\Bbb Z^+$ a sequence between $\frac1{n+1}$ and $\frac1n$ converging downwards to $\frac1{n+1}$.

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First consider the set:

$$A=\left\{ \frac{1}{n}:n\in\mathbb{Z}^+\right\}\cup\{0\}$$

For each $n>1$, take a sequence $\{\varepsilon_k^n\}_{k=1}^\infty$ such that $\frac{1}{n}<\varepsilon^n_k<\frac{1}{n-1}$, and $\varepsilon_k^1>1$ such that

$$\lim_{k\to\infty}\varepsilon_k^n=\frac{1}{n}$$

Put $$K=A\cup \{\varepsilon_k^n:n,k\in\mathbb{Z}^+\}$$

Notice that this $K$ works.

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Let $a_0=0$ and let $(a_n)_{n\in N}$ be a strictly increasing sequence of members of $(0,1)$ with $\lim_{n\to \infty}a_n=1.$

For each $n\in N$ (i.e. $n>0$) let $(b_{n,j})_{j\in N}$ be a strictly increasing sequence of members of $[a_{n-1},a_n)$ with $b_{n,1}=a_{n-1}$ and $\lim_{j\to \infty}b_{n,j}=a_n.$

Let $S=\{b_{n,j}:n,j\in N\}\cup \{1\}.$ Then S is closed and bounded, hence compact, and the set of accumulation points of $S$ is $\{a_n:n\in N\}\cup \{1\}.$

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