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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?

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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.

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  • $\begingroup$ Ah, isolated points could be compact!!!! Except isolated point cases, every compact set is perfect? $\endgroup$
    – Danny_Kim
    Mar 12 '16 at 6:03
  • $\begingroup$ 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? $\endgroup$
    – Danny_Kim
    Mar 12 '16 at 6:09
  • $\begingroup$ @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. $\endgroup$ Mar 12 '16 at 6:11
  • $\begingroup$ @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. $\endgroup$ Mar 12 '16 at 6:13
  • $\begingroup$ 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. $\endgroup$
    – Danny_Kim
    Mar 12 '16 at 6:25

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