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# Limit point

(Redirected from Accumulation point)

In mathematics, 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.

## 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.)

## 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 yx, 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.

03-10-2013 05:06:04
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