Limit Point - Some Facts

Some Facts

• 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 use 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, if and only if every neighbourhood of x contains a point of S other than x, if and only if every neighbourhood of x contains a point of S \ {x}, if and only if 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 if and only if S is equal to its closure if and only if S = S ∪ L(S) if and only if L(S) is contained in S.
  • Another proof: Let S be a closed set and x a limit point of S. If x is not in S, then we can find an open set around x contained entirely in the complement of S. But then this set contains no point in S, so x is not a limit point, which contradicts our original assumption. 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. Since this argument holds for arbitrary x in the complement of S, the complement of S can be expressed as a union of open neighborhoods of the points 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.
• By definition, every limit point is an adherent point.