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By definition,

Perfect set $E_1$ is closed set without isolated points.

Compact set $E_2$ is bounded and closed set in Euclidean space; $\mathbb{R}^n$.

Is the following equation true? $$E_2 \subseteq E_1$$

In my understanding, examples of perfect set; $$ \mathbb{R}^n, [10, \infty), [-10, 20] \cup [30, 40], \mbox{Cantor set}, [(0,0) (20, \infty)], \cdots $$

and examples of compact set; $$ [10, 20], [-7, -2]\cup[0, 5], [(10, 20), (15, 23)], \cdots $$

In a nutshell, my question is whether every compact set is perfect.

In addition, are there any problems in my examples?

Asaf Karagila
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Danny_Kim
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1 Answers1

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If $p\in\Bbb R^n$, the set $\{p\}$ is compact but not perfect, since $p$ is an isolated point. The same is true of any finite set. There are also infinite compact sets that are not perfect; a simple example in $\Bbb R$ is $\{0\}\cup\left\{\frac1n:n\in\Bbb Z^+\right\}$, in which $0$ is the only non-isolated point.

Brian M. Scott
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  • Ah, isolated points could be compact!!!! Except isolated point cases, every compact set is perfect? – Danny_Kim Mar 12 '16 at 06:03
  • In my opinion, $\left{\cdots, -\frac{1}{2}, -1, 0, 1, \frac{1}{2}, \cdots\right}$ is not closed and not bounded. So this set is not compact by the *Heine-Borel theorem*, isn't it? – Danny_Kim Mar 12 '16 at 06:09
  • @Danny: It’s more complicated than that. Take a look at my answer to this question, The diagram shows a compact set in $\Bbb R^2$. If you remove the isolated points, you have another compact set in $\Bbb R^2$, but it still has isolated points. Do this two more times, and you end up with a one-point space. What is true is that every compact set either is countable or has a perfect subset. – Brian M. Scott Mar 12 '16 at 06:11
  • @Danny: Assuming that you mean the set $$\left{-1,-\frac12,-\frac13,\ldots,0,\ldots,\frac13,\frac12,1\right};,$$ you’re set is bounded — it’s a subset of $[-1,1]$ — and closed — its only limit point is $0$, which is in the set — so it’s compact. – Brian M. Scott Mar 12 '16 at 06:13
  • Ah, thank you!!! I got my fault, now I totally agree with that the set is bounded. And now, I am pondering about the set is closed or not. I agree with the union of isolated points are closed, but I want to check near zero point. – Danny_Kim Mar 12 '16 at 06:25
  • @Danny_Kim The point is that the only "weird" point - precisely, the only point which is a limit of a sequence of points from the set - is zero, and since zero is in the set, that means the set is closed. – Noah Schweber Mar 12 '16 at 07:00
  • The set {1/n} has 1 limit point. That limit point is 0. The set {1/n} U {0} has the same limit point. 0 is in {1/}U {0} so it is closed. BTW don't say "in my opinion" in math. Either you are right or you are wrong. Your opinion has nothing to do with it. – fleablood Mar 12 '16 at 07:18