# In an indiscrete topology, does the empty set have any limit points?

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

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

## 3 Answers

Empty set has no limit points since no set can have non-empty intersection with $$\varnothing$$.

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??

• 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. – Daniel Fischer Nov 1 '14 at 15:48
• After a long time now I can understand your comment :) – Bumblebee Dec 30 '16 at 20:05

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.