Limit Point

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 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.

Anagrams: Limit Point

Scrabble® Enable2K-Verified Anagrams
	Words within the letters "i-i-i-l-m-n-o-p-t-t"
	-3 letters: pinitol.
	-4 letters: pontil.
	-5 letters: imino, limit, lipin, pilot, pinot, pinto, piton, point.
	Words containing the letters "i-i-i-l-m-n-o-p-t-t"
	+4 letters: multiplication.
	+5 letters: contemptibility, incompatibility, multiplications.
Source: compiled by the editor from various references; see credits. SCRABBLE® is a registered trademark. All intellectual property rights in and to the game are owned in the U.S.A and Canada by Hasbro Inc., and throughout the rest of the world by J.W. Spear & Sons Limited of Maidenhead, Berkshire, England, a subsidiary of Mattel Inc. Mattel and Spear are not affiliated with Hasbro.