Talk:Clopen set

message: This is trivial: since {x} is connected (in any Hausdorff space), then any clopen set is the union of all its elements: A = \bigcup_{x \in A} { \{ x \} } \,.
Should the text be Any clopen set is a union of (possibly infinitely many) connected clopen components?
I don't think this is true: \mathbb{Q}\, a likely counterexample. Albmont 15:56, 8 December 2006 (UTC)
thread_title:  Any clopen set is a union of (possibly infinitely many) connected components 

message: This is not a commen term.  I think that it is one the terms rlm introduced.  —Preceding unsigned comment added by 130.164.67.108 (talk) 09:49, 22 April 2008 (UTC)
thread_title:  Do we really need this section? 

message: This isn't about clopen groups, but surely the subset of rational numbers whose square is more than 2 is not a closed subset. (1+1)^1/2 = 1 + 1/2 - 1/8 + 1/16 - 5/128 + 7/256 -... more rational terms which get smaller (cos of binomial expansion) so by saying
x0= 1 + 1/2
x1= 1 + 1/2 - 1/8 + 1/16
x2= 1 + 1/2 - 1/8 + 1/16 - 5/128 + 7/256
and so on, we have that all xn are in the subset, but the square root of 2 isn't.  —Preceding unsigned comment added by Timcrinion (talk • contribs) 16:54, 1 October 2008 (UTC)
replies: text: As noted in the article, it's not closed as a subset of R, but it is as a subset of Q. The complement of the set (sqrt(2), infinity) is (-infinity, sqrt(2)], but because sqrt(2) isn't rational, this is the same interval as (-infinity, sqrt(2)). Dcoetzee 18:56, 1 October 2008 (UTC)

text: There's no need for a proof. It's a consequence of the definitions. — Fly by Night (talk) 02:36, 3 March 2011 (UTC)
replies: text: Short proofs are still proofs. I guess I was actually asking for some intuition. That is is open makes sense. Pick a rational number, you can find a number larger or smaller than it that is still in this subset. As for it being closed, I think the intuition is this: there is no intuition when the space is not the reals, or some subset of the reals. 128.8.211.24 (talk) 19:42, 3 March 2011 (UTC)
replies: text: A subset  is open if there exists an open set  such that . That's the definition of an open set in Q. (It has the induced topology from R.) Let  be given by  Clearly S is open and so  is, by definition, also open. All we need to do is to show that A is also closed. Well, a set is closed if, and only if, it contains all of its limit points. For  to be a limit point of a set it must not belong to the set, but every open neighbourhood of the point must contain at least one point of the set. Clearly √2 is a limit point of , but it isn't a limit point of A because  So what about the other  with  Well, none of these are limit points of A either. To see this, for each rational  let  By definition  are open in Q, and also  That means that A contains all of its limit points and so A must also be closed. This implies that A is a clopen set. — Fly by Night (talk) 18:02, 14 March 2011 (UTC)
replies: text: There is a very definitional proof, and I think you avoided that with your proof that the set is open. Why not put this on the page itself? I think it would help.  Is it possible for a non-null proper subset of the reals to be clopen? Either way, this would be worth mentioning. 151.200.117.207 (talk) 18:18, 21 March 2011 (UTC)
replies: text: Feel free to add or copy-and-paste whatever you like to the article. As for the question: Is it possible for a non-null proper subset of the reals to be clopen?, that's a very interesting question. Such properties depend not only on the underlying space, but on the topology you choose. To answer your question: yes there are non-empty proper subsets of the reals that are clopen. The discrete topology, on any set X is given by declaring all subsets of X to be open. Since the complement of a set is again a set, this means that all subsets of X are also closed. Thus; every subset of any set X is clopen with respect to the discrete topology. The nest question to ask, to which I don't yet know the answer, is: What are some examples of the coarsest topology on R for which there exist a non-empty, clopen, proper subset. — Fly by Night (talk) 20:01, 21 March 2011 (UTC)
replies: text: I was thinking of limiting it to topologies where the union of all the proper subsets is the reals (or, i.e. a portion of the real line). But, yeah, your "next question" is another way of thinking about this. 151.200.190.212 (talk) 21:29, 23 March 2011 (UTC)
replies: text: I started a thread on the maths reference desk and the question was answered. For a fixed real number x, we define a topology on R where the open sets are R itself, the empty set ∅, the singleton set { x } and the complement R \ { x }. This also fits in with your requirement that the union of all the proper subsets is the reals. The proper subsets are { x } and R \ { x } while the union is . The same hold if we replace { x } by any subset, say U, of R. The discreet topology comes from setting . — Fly by Night (talk) 18:46, 27 March 2011 (UTC)
thread_title:  Square root of 2 example 

message: Let me just say that Clopen is my least favorite terminology in all of mathematics. I really can't stand it. If I ever become a mathematician, I will make it my priority to get this changed. (Along with doing important mathematical things.) 74.192.194.29 (talk) 01:29, 1 October 2009 (UTC)
thread_title:  Etymology 

message: Is my spell check making mistakes of it clopen spelled wrong? It gives alternatives cl-open, cl open, clop-en, and clop en. 66.183.58.62 (talk) 05:13, 27 November 2010 (UTC)
replies: text: It's making mistakes. Algebraist 22:30, 27 November 2010 (UTC)
replies: text: Spell check doesn't understand a lot of math terms.--75.80.43.80 (talk) 13:52, 8 April 2011 (UTC)
replies: text: Spell checks dont understand biology terms either 190.60.93.218 (talk) 17:09, 9 September 2011 (UTC)
thread_title:  Spell check