Open main menu
Home
Random
Recent changes
Special pages
Community portal
Preferences
About Wikipedia
Disclaimers
Incubator escapee wiki
Search
User menu
Talk
Dark mode
Contributions
Create account
Log in
Editing
Accumulation point
(section)
Warning:
You are not logged in. Your IP address will be publicly visible if you make any edits. If you
log in
or
create an account
, your edits will be attributed to your username, along with other benefits.
Anti-spam check. Do
not
fill this in!
==Properties== Every [[Limit of a sequence#Topological spaces|limit]] of a non-constant sequence is an accumulation point of the sequence. And by definition, every limit point is an [[adherent point]]. The closure <math>\operatorname{cl}(S)</math> of a set <math>S</math> is a [[disjoint union]] of its limit points <math>L(S)</math> and isolated points <math>I(S)</math>; that is, <math display="block">\operatorname{cl} (S) = L(S) \cup I(S)\quad\text{and}\quad L(S) \cap I(S) = \emptyset.</math> A point <math>x \in X</math> is a limit point of <math>S \subseteq X</math> if and only if it is in the [[Closure (topology)|closure]] of <math>S \setminus \{ x \}.</math> {{math proof | proof = We use the fact that a point is in the closure of a set if and only if every neighborhood of the point meets the set. Now, <math>x</math> is a limit point of <math>S,</math> if and only if every neighborhood of <math>x</math> contains a point of <math>S</math> other than <math>x,</math> if and only if every neighborhood of <math>x</math> contains a point of <math>S \setminus \{x\},</math> if and only if <math>x</math> is in the closure of <math>S \setminus \{x\}.</math> }} If we use <math>L(S)</math> to denote the set of limit points of <math>S,</math> then we have the following characterization of the closure of <math>S</math>: The closure of <math>S</math> is equal to the union of <math>S</math> and <math>L(S).</math> This fact is sometimes taken as the {{em|definition}} of [[Closure (topology)|closure]]. {{math proof | proof = ("Left subset") Suppose <math>x</math> is in the closure of <math>S.</math> If <math>x</math> is in <math>S,</math> we are done. If <math>x</math> is not in <math>S,</math> then every neighbourhood of <math>x</math> contains a point of <math>S,</math> and this point cannot be <math>x.</math> In other words, <math>x</math> is a limit point of <math>S</math> and <math>x</math> is in <math>L(S).</math> ("Right subset") If <math>x</math> is in <math>S,</math> then every neighbourhood of <math>x</math> clearly meets <math>S,</math> so <math>x</math> is in the closure of <math>S.</math> If <math>x</math> is in <math>L(S),</math> then every neighbourhood of <math>x</math> contains a point of <math>S</math> (other than <math>x</math>), so <math>x</math> is again in the closure of <math>S.</math> This completes the proof. }} A corollary of this result gives us a characterisation of closed sets: A set <math>S</math> is closed if and only if it contains all of its limit points. {{math proof | proof = ''Proof'' 1: <math>S</math> is closed if and only if <math>S</math> is equal to its closure if and only if <math>S=S\cup L(S)</math> if and only if <math>L(S)</math> is contained in <math>S.</math> ''Proof'' 2: Let <math>S</math> be a closed set and <math>x</math> a limit point of <math>S.</math> If <math>x</math> is not in <math>S,</math> then the complement to <math>S</math> comprises an open neighbourhood of <math>x.</math> Since <math>x</math> is a limit point of <math>S,</math> any open neighbourhood of <math>x</math> should have a non-trivial intersection with <math>S.</math> However, a set can not have a non-trivial intersection with its complement. Conversely, assume <math>S</math> contains all its limit points. We shall show that the complement of <math>S</math> is an open set. Let <math>x</math> be a point in the complement of <math>S.</math> By assumption, <math>x</math> is not a limit point, and hence there exists an open neighbourhood <math>U</math> of <math>x</math> that does not intersect <math>S,</math> and so <math>U</math> lies entirely in the complement of <math>S.</math> Since this argument holds for arbitrary <math>x</math> in the complement of <math>S,</math> the complement of <math>S</math> can be expressed as a union of open neighbourhoods of the points in the complement of <math>S.</math> Hence the complement of <math>S</math> is open. }} No [[isolated point]] is a limit point of any set. {{math proof | proof = If <math>x</math> is an isolated point, then <math>\{x\}</math> is a neighbourhood of <math>x</math> that contains no points other than <math>x.</math> }} A space <math>X</math> is [[discrete space|discrete]] if and only if no subset of <math>X</math> has a limit point. {{math proof | proof = If <math>X</math> is discrete, then every point is isolated and cannot be a limit point of any set. Conversely, if <math>X</math> is not discrete, then there is a singleton <math>\{ x \}</math> that is not open. Hence, every open neighbourhood of <math>\{ x \}</math> contains a point <math>y \neq x,</math> and so <math>x</math> is a limit point of <math>X.</math> }} If a space <math>X</math> has the [[trivial topology]] and <math>S</math> is a subset of <math>X</math> with more than one element, then all elements of <math>X</math> are limit points of <math>S.</math> If <math>S</math> is a singleton, then every point of <math>X \setminus S</math> is a limit point of <math>S.</math> {{math proof | proof = As long as <math>S \setminus \{ x \}</math> is nonempty, its closure will be <math>X.</math> It is only empty when <math>S</math> is empty or <math>x</math> is the unique element of <math>S.</math> }}
Edit summary
(Briefly describe your changes)
By publishing changes, you agree to the
Terms of Use
, and you irrevocably agree to release your contribution under the
CC BY-SA 4.0 License
and the
GFDL
. You agree that a hyperlink or URL is sufficient attribution under the Creative Commons license.
Cancel
Editing help
(opens in new window)