Limit point

Informally speaking, a limit point (or cluster point) of a set S in a topological space X is a point x in X that can be "approximated" by points of S other than x as well as one pleases. This concept profitably generalizes the notion of a limit and is the underpinning of concepts such as closed set and topological closure. Indeed, a set is closed if and only if it contains all of its limit points, and the topological closure operation can be thought of as an operation that enriches a set by adding its limit points.

Table of contents

1 Definition 2 Relationship to accumulation point 3 Some facts

Definition

Let S be a subset of a topological space X. We say that a point x in X is a limit point of S if every open set containing x also contains a point of S other than x itself. This is equivalent to requiring that every neighbourhood of x contains a point of S other than x itself. (It is often convenient to use the "open neighbourhood" form of the definition to show that a point is a limit point and to use the "general neighbourhood" form of the definition to derive facts from a known limit point.)

Relationship to accumulation point

Some authors require that every open neighbourhood of x contains infinitely many points of S. Clearly this condition is stronger than our definition, but the following result shows that it is not equivalent.

• Let X be a topological space. Then X is a T₁ space (i.e. singletons are closed in X) if and only if for every point x in X and every subset S of X, x is a limit point of S if and only if every open neighbourhood of x contains infinitely many points of S.
  • Proof: Suppose singletons are closed in X. Let S be a subset of X and x a limit point of S. Suppose there is an open neighbourhood U of x that contains only finitely many points of S. Then U \\ (S \\ {x}) is an open neighbourhood of x that does not contain any points of S other than x. (Here is where we use the fact that singletons are closed.) This contradicts the fact that x is a limit point of S. Thus, every open neighbourhood of x contains infinitely many points of S. Conversely, suppose there is a point x in X such that the singleton {x} is not closed. Then there is a point y ≠ x in the closure of {x}. We claim that any open neighbourhood U of y contains x. For suppose not; then the complement of U in X would be a closed set containing x, and the closure of {x} would be contained in the complement of U. Since y is in the closure of {x}, this would force y not to be in U, contradicting the fact that U is a neighbourhood of y. We have shown that y is a limit point of S = {x}. But it is clear that X is a neighbourhood of y that does not contain infinitely many points of S. This completes the proof.

Let S be a subset of a topological space X. We say that a point x in X is an accumulation point of S if every open set containing x contains infinitely many points of S. This is equivalent to requiring that every neighbourhood of x contains infinitely many points of S. (Again, it is often convenient to use the "open neighbourhood" form of the definition to show that a point is an accumulation point and to use the "general neighbourhood" form of the definition to derive facts from a known accumulation point.)

Many authors, especially those who only work with T₁ spaces, use the term "accumulation point" to mean what we have defined as a "limit point". Of course, when one is only working in T₁ spaces, there is no confusion and either of the terms may be freely used. However, there are many articles and examples in Wikipedia that involve non-T₁ spaces. For this reason, Wikipedia makes the above distinction between limit points and accumulation points, with the understanding that whenever an article discusses only T₁ spaces, either term may be freely used.

The simplest example of a non-T₁ space is Sierpinski space X = {x,y} with topology {{}, {x}, X}. Here, y is a limit point, but not an accumulation point, of {x}.

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 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 has more than one point, then its topology is trivial if and only if every point is a limit point of every nonempty subset.
  • Proof: In the trivial topology on X, the only neighbourhood of a point x in X is X, which necessarily intersects any other nonempty subset of X. On the other hand, with a non-trivial topology, X has a nonempty closed proper subset Y. Then the limit points of Y are elements of Y and so not every point of X is a limit point of Y.