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From 'baby' Rudin.

I've seen that a set is closed iff it contains all of its limit points. In Rudin, $(d)$ says if every limit point of E is a point of E, then $E$ is closed. He also says $(h)$: $E$ is perfect if $E$ is closed and if every point of $E$ is a limit point of $E$.

But Closed $\implies$ contains all of its limit points. So, is every closed set a perfect set?

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    $\begingroup$ No. Closed = all limit points are in E. Perfect = all points in E are limit points. Those are completely different statements. $\endgroup$
    – fleablood
    Mar 4, 2016 at 4:09
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    $\begingroup$ Consider the set {2}. It has no limit points. All limit points are in {2} so it is closed. 2 is no a limit point so it is not perfect. Any closed set with so much as a single point that isn't a limit point, won't be perfect. A less trivial example is {a + 1/n| a an integer and n a positive integer}. It's limit points are all the integers. All integers are in the set (a+1/1) so it is closed. But if n>1, a + 1/n is not a limit point. So the set isn't perfect. $\endgroup$
    – fleablood
    Mar 4, 2016 at 4:26
  • $\begingroup$ Re-read (b); What is a limit point? $\endgroup$ Mar 4, 2016 at 4:33
  • $\begingroup$ Even though this is an old post - I've been extensively working out of this book and work into Rudin's Green book ($Real~And~Complex~Analysis$, W. Rudin, 3rd Ed. -- I've heard this book is coined "Father" Rudin similar to, and maybe not as popular, "Baby" Rudin). The way I've come to distinguish the two definitions is, in an arbitrarily fixed metric space $(X,d_{X})$, we let $E\subseteq X$. Define $E':=\big\{\ell\in X:\ell\text{ is a limit point of }E\big\}$. If $E'\subseteq E$, then $E$ is closed in $X$; conversely, if $E$ is closed and $E\subseteq E'$, then $E$ is perfect in $X$. $\endgroup$
    – Procore
    Jun 3, 2017 at 23:53

4 Answers 4

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No finite set is perfect but every finite set is closed; a finite set has no limit points and thus all of its limit points (all zero of them) belong to it, so it's closed.

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    $\begingroup$ Nitpick: the empty set is perfect . $\endgroup$
    – chi
    Mar 4, 2016 at 9:57
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    $\begingroup$ Nitpick pick; it is common to require perfect sets to be nonempty, for similar reasons to requiring prime numbers to not be one. For example, spaces are scattered if they contain no (nonempty!) perfect sets. But this is definitional and not important. $\endgroup$ Mar 4, 2016 at 11:26
  • $\begingroup$ @chi is entirely correct here, using the definitions quoted by OP. And since the question is entirely about these definitions, chi is being pleasantly modest to call it a nitpick. If you did require perfect sets to be nonempty then the question would have an immediate and unhelpful answer: the empty set is closed by everyone's definition and not perfect by this definition. $\endgroup$ Mar 4, 2016 at 12:24
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Closed means all limit points are in $E$. But that doesn't mean all points in $E$ are limit points. Any closed set with a point that is not a limit point will not be perfect.

The easiest counter example is a set with a single point. That set is closed but its one point isn't a limit point.

Less trivial and less contrived is $D = \{a + 1/n| a \in \mathbb Z,n \in \mathbb N\} $. Every integer is a limit point. No other point is a limit point. All integers are in $D$ (because $a + 1/1$ is an integer) so $D$ is closed. But for all $n > 1$ then $a + 1/n $ is in D but is not a limit point. So $D$ is not perfect.

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  • $\begingroup$ Many closed sets are perfect. [a,b] is perfect. For point not to be a limit point it must be an isolated point. $\endgroup$
    – fleablood
    Mar 4, 2016 at 4:59
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Not quite. $[0, 1]\cup \{2\}$ is a closed set but not perfect.

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No. Consider the set $[0,1] \cup \{2\} \subseteq \mathbb{R}$ with the usual topology. By the definition of limit point, if $x_0$ is a limit point of the set $S \subseteq \mathbb{R}$, $$ \forall r > 0, \exists x \in \mathbb{R}\mid x \neq x_0, |x-x_0| < r $$ Clearly the point $2$ does not satisfy this condition(taking $r = 1/2$), and thus is not a limit point.

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