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Given a set $A \subseteq \mathbf{R}$, let $L$ be the set of all limit points of $A$. I recently worked through the proof that $L$ is closed. However, I have a few queries thinking deeper about this result.

Since $L$ is closed, this implies that if $x$ is a limit point of $L$ then $x \in L$, which further implies that $x$ must be a limit point of $A$. Therefore, if $x$ is a limit point of $L$ then it must also be a limit point of $A$. But consider the set $$A = \{0\} \cup \{\frac{1}{n}: n \in \mathbf{N}\}$$

Clearly, $L = \{0\}$ but does not have any limit point itself, i.e., the set of all limit points of $L$ is $\emptyset$, but clearly $\emptyset$ is not a limit point of $A$, so isn't this a counter example to the above bolded statement?

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    $\begingroup$ Of course $\varnothing$ isn't a limit point of $A$. That isn't what the bolded statement says at all. $\endgroup$ – Kenny Lau Jun 4 '17 at 5:21
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    $\begingroup$ The bolded statement claims that every point in $\varnothing$ is a limit point of $A$, which is vacuously true. $\endgroup$ – Kenny Lau Jun 4 '17 at 5:21
  • $\begingroup$ $\emptyset$ is both closed and open in any topology, as is the whole space. $\endgroup$ – Araske Jun 4 '17 at 6:53
  • $\begingroup$ $\emptyset$ isn't even a point, but every element in it (although there is none) is a point. $\endgroup$ – md2perpe Jun 4 '17 at 6:54
  • $\begingroup$ @siTTmo when you say "the set of all limit points of $L$ is $\emptyset$", it means that $L$ has no limit points. Therefore, the sentence "every limit point of $L$ is also a limit point of $A$" (equivalent to the sentence "if $x$ is a limit point of $L$ then $x$ is a limit point of $A$) is vacuously true. Hence, the bolded statement is true, just easily true in this particular case as there are no points to check. $\endgroup$ – Darío G Jun 4 '17 at 6:58
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Consider. If $A = [0,1]$ then $L = [0,1] = A$ and set of limit points of $L = [0,1] = A$. But $[0,1]$ is not a limit point of $A$! Contradiction? No. A set is not a member of itself (by ZFC that can never happen) so the set of limit points is not a limit point.

Clearly, L={0} but does not have any limit point itself, i.e., the set of all limit points of L is ∅, but clearly ∅ is not a limit point of A

Which doesn't matter because $\emptyset \not \in \emptyset = \{$limit points of $L\}$.

But since there aren't any $x \in \{$limit points of $L\} = \emptyset$ that aren't limit points (because there aren't any $x \in \emptyset$ period) the all $x \in \emptyset$ (all ZERO of them) are limit points. They are also all french pigs farting green sausages. (Because there are any $x \in \emptyset$ that arent french pigs farting green sausages.)

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From the comments above.


When you say that "$\emptyset$ is not a limit point of $A$", there seems to be some confusion: sets can anyway never be limit points, only points (in the metric space) can be limit points.

What you instead need to verify is that every element of $\emptyset$ is a limit point of $A$, and this is vacuously true.

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