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Since the intersection of any set with $\emptyset$ is $\emptyset$, it does not seem like $\phi$ has any limit point. Is my reasoning correct?

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    $\begingroup$ Yes, it is correct. No matter what topology you put, $\emptyset$ has no limit points. $\endgroup$ – Jonas Gomes Nov 1 '14 at 15:10
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    $\begingroup$ For $\emptyset$, use \emptyset; for $\varnothing$ use \varnothing. $\endgroup$ – MJD Nov 1 '14 at 15:14
  • $\begingroup$ the empty set has a unique topology, and of course it has no limit points since $\emptyset$ has no points at all. $\endgroup$ – matiasdata Nov 1 '14 at 15:17
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Empty set has no limit points since no set can have non-empty intersection with $\varnothing$.

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Suppose $\emptyset$ has at least one limit point, say $x$. Then there exist a sequence $(x_n)$ in $\emptyset$ which converge to $x.$ That is each $x_n\in\emptyset.$ This gives if empty set has a limit point it is not empty which contradicts to its definition. Hence??

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    $\begingroup$ That's not the definition of limit point. A limit point of a set $S$ is a point $x$ such that $U\cap (S\setminus \{x\}) \neq \varnothing$ for every neighbourhood $U$ of $x$. It does not follow that there is a sequence in $S$ converging to $x$ - and sometimes, such a sequence does not exist. If we were dealing with metric spaces, such a sequence would necessarily exist. $\endgroup$ – Daniel Fischer Nov 1 '14 at 15:48
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    $\begingroup$ After a long time now I can understand your comment :) $\endgroup$ – Bumblebee Dec 30 '16 at 20:05
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Yes you would appear to be correct. And since it contains all its limit points (vacuously), it's closed. Additionally, this appears to be true in any topology.

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