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I came up with the following example of a set that I believe is compact and whose limit points form a countable set.

Let $E_k=\cup_{n=2}^\infty \{\frac{1}{n}+k\}\cup \{k\}. $ Note that for each $k$, $E_k$ is a compact set and the set of limit points of $E_k$ is precisely $\{k\}$.

Then $E=\cup_{k=0} ^\infty E_k$ is a compact set (I want to say this is definitely true) whose limit points are the set of nonnegative integers, which is a countable set.

Does this example work?? Thank you!!

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    $\begingroup$ $E$ is not compact: it’s unbounded. $\endgroup$ – Brian M. Scott Oct 3 '16 at 23:14
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Your set $E$ is not compact, because it’s unbounded. However, a modification of it will work. For $k\ge 2$ let

$$E_k=\left\{\frac1k\right\}\cup\left\{\frac1k+\frac1m:m>k(k-1)\right\}\;,$$

and let $E=\{0\}\cup\bigcup_{k\ge 2}E_k$; I’ll leave it to you to show that this works.

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  • $\begingroup$ Ahh, yes, my set is unbounded. Thank you. Quick (and probably silly) question. WHy do you require that $m>k(k-1)$? $\endgroup$ – Michael Oct 3 '16 at 23:54
  • $\begingroup$ @Michael: You’re welcome. That restriction isn’t actually necessary, but it makes it easier to see why the example works: if you work it out, you’ll see that it’s exactly what’s needed to ensure that $\frac1k+\frac1m<\frac1{k-1}$. This means that the sets $E_k$ really do line up end to end instead of overlapping in any way. $\endgroup$ – Brian M. Scott Oct 3 '16 at 23:56
  • $\begingroup$ Is the intention here to get a compact set whose set of limit points is countably infinite? (The question just says "countable".) $\endgroup$ – Mitchell Spector Oct 4 '16 at 0:34
  • $\begingroup$ @Mitchell: I assume so, in view of the OP’s original attempt and the widespread (and unfortunate) use of countable to mean countably infinite. $\endgroup$ – Brian M. Scott Oct 4 '16 at 0:36

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