Talk:Closure (topology)

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Dense set

I phrased the bit about density as a fact relating it to closure (expressed with "iff") rather than as a definition in terms of closure (expressed with "if"), since there are alternative definitions of dense sets. (And someday I'll probably edit the article on dense sets to mention them.) -- Toby Bartels 2002/05/08

Closed set

A separate article is titled closed set. Should these two articles be merged? -- Michael Hardy, 2003 Aug 12

I don't think so; conceptually they are quite different, since one is a property of sets, while the other is an operation on them. To be sure, they are closely related and should be interlinked; but the concepts of open set, dense set, and the like are also closely related, yet separate. We can afford to have an article on each of these, eventually expanding them all like Open set is now. -- Toby Bartels 05:11, 25 Sep 2003 (UTC)

Alternative characterization

I have found this characterization of a closure useful. For any S in topological space X

1. Cl(S) is in topological space X

2. Cl(S) is closed

3. For any closed A in topological space X, A ⊇ Cl(S) iff A ⊇ S

The Interior can be defined similarly.

This seems to me a much less clumsy way of doing things than using the standard definition. Many results follow immediately by substituting various terms for A. Since CL(S) ⊇ CL(S) always holds, we also have CL(S) ⊇ S, CL(S) being closed by assumption. For closed S, we similarly have S ⊇ CL(S), hence S = CL(S). Jamie Oglethorpe 8 July 2004

Yes, this seems to be another way of saying that Cl(S) is the smallest closed set containing S.

Adherent point

Is an adherent point of a topological space the same thing as a closure point? Surely a neighbourhood gives you an open set, and vice versa? Xanthoxyl 14:11, 12 March 2007 (UTC)[reply]

Yes. Paul August ☎ 23:24, 9 July 2008 (UTC)[reply]

Point of Closure

Hi,

Regarding the present statement :

"For S a subset of a Euclidean space, x is a point of closure of S if every open ball centered at x contains a point of S (this point may be x itself)."

Would it be OK to clarify this in the following (or similar) way (OR would this be making the paragraph too long?):

"For S a subset of a Euclidean space, x is a point of closure of S if every open ball centered at x contains a point of S (this point may be x itself). Whether x is or is not a member of S depends on the space in which the closure is taken."

The motivation for this was the statement lower down that :

"The closure of a set also depends upon in which space we are taking the closure. For example, if X is the set of rational numbers, with the usual subspace topology induced by the Euclidean space R, and if S = {q in Q : q2 > 2}, then S is closed in Q, and the closure of S in Q is S; however, the closure of S in the Euclidean space R is the set of all real numbers greater than or equal to \sqrt2."

Please let me know your thoughts,

Thanks

Arjun r acharya (talk) 11:05, 22 January 2009 (UTC)[reply]