- We have the following characterisation of limit points:
`x`is a limit point of`S`if and only if it is in the closure of`S`\\ {`x`}.*Proof*: We assume the fact that a point is in the closure of a set if and only if every neighbourhood of the point meets the set. Now,*x*is a limit point of*S*, iff every neighbourhood of*x*contains a point of*S*other than*x*, iff every neighbourhood of*x*contains a point of*S*\\ {*x*}, iff*x*is in the closure of*S*\\ {*x*}.

- If we use L(
*S*) to denote the set of limit points of*S*, then we have the following characterisation of the closure of*S*: The closure of*S*is equal to the union of*S*and L(*S*).*Proof*: ("Left subset") Suppose*x*is in the closure of*S*. If*x*is in*S*, we are done. If*x*is not in*S*, then every neighbourhood of*x*contains a point of*S*, and this point cannot be*x*. In other words,*x*is a limit point of*S*and*x*is in L(*S*). ("Right subset") If*x*is in*S*, then every neighbourhood of*x*clearly meets*S*, so*x*is in the closure of*S*. If*x*is in L(*S*), then every neighbourhood of*x*contains a point of*S*(other than*x*), so*x*is again in the closure of*S*. This completes the proof.

- A corollary of this result gives us a characterisation of closed sets: A set
*S*is closed if and only if it contains all of its limit points.*Proof*:*S*is closed iff*S*is equal to its closure iff*S*=*S*∪ L(*S*) iff L(*S*) is contained in*S*.*Another proof*: Let*S*be a closed set and*x*a limit point of*S*. Then*x*must be in*S*, for otherwise the complement of*S*would be an open neighborhood of*x*that does not intersect*S*. Conversely, assume*S*contains all its limit points. We shall show that the complement of*S*is an open set. Let*x*be a point in the complement of*S*. By assumption,*x*is not a limit point, and hence there exists an open neighborhood*U*of*x*that does not intersect*S*, and so*U*lies entirely in the complement of*S*. Hence the complement of*S*is open.

- No isolated point is a limit point of any set.
*Proof'\': If*x*is an isolated point, then {*x*} is a neighbourhood of*x*that contains no points other than*x''.

- A space
*X*is discrete if and only if no subset of*X*has a limit point.*Proof*: If*X*is discrete, then every point is isolated and cannot be a limit point of any set. Conversely, if*X*is not discrete, then there is a singleton {*x*} that is not open. Hence, every open neighbourhood of {*x*} contains a point*y*≠*x*, and so*x*is a limit point of*X*.

- If a space
*X*has the trivial topology and*S*is a subset of*X*with more than one element, then all elements of*X*are limit points of*S*. If*S*is a singleton, then every point of*X*\\*S*is still a limit point of*S*.*Proof*: As long as*S*\\ {*x*} is nonempty, its closure will be*X*. It's only empty when*S*is empty or*x*is the unique element of*S*.