10
$\begingroup$

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!

$\endgroup$

closed as off-topic by T. Bongers, Claude Leibovici, user91500, Mike Haskel, Watson Sep 3 '16 at 8:57

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – T. Bongers, Claude Leibovici, user91500, Mike Haskel, Watson
If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ @carmichael561: I think that you’re right. Careless reading on my part. $\endgroup$ – Brian M. Scott Sep 2 '16 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 '16 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 '16 at 11:05
16
$\begingroup$

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 $\mathbb{R}$.)
$\endgroup$
  • 3
    $\begingroup$ $S$ 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. $\endgroup$ – Dominique R.F. Sep 3 '16 at 0:00
7
$\begingroup$

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 $\frac1{n+1}$.

$\endgroup$
5
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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 $\{a_n:n\in N\}\cup \{1\}.$

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.