Talk:Clopen set

untitled

This doesn't make much sense. Perhaps an example of a clopen set might make the idea comprehensible ? Theresa knott 06:46 25 May 2003 (UTC)

If you look at the articles on Open sets and Closed sets, you will find examples of Clopen sets. Perhaps some of them could be added to this article but if so it would only be repeating what exists elsewhere. -- Derek Ross 07:07 25 May 2003 (UTC)

Any clopen set is a union of (possibly infinitely many) connected components

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)

Standard terminology ?

I haven't heard of the term clopen set before; that's why I put the reference tag in the article. Can anyone state a reference ? Thanks. MP ^{(talk•contribs)} 20:21, 14 January 2008 (UTC)

Here are just a few of the textbooks mentioning this term:

http://www.amazon.com/s/ref=nb_ss_/002-0295115-4340819?url=search-alias%3Daps&field-keywords=clopen

I don't know where the term originated, but it pops up all over the place. Dcoetzee 23:17, 14 January 2008 (UTC)

I have the books cited for the "clopen" naming convention, and I believe these books do not mention the term "clopen". In fact [1] is quite confusing, while [2] does not seem to give a special treatment to "clopen" sets. Furthermore, the link points to MP3 music. I would love to use such term, however, it does not seem standard. --VitorAMJorge (talk) 05:52, 28 October 2015 (UTC)

Do we really need this section?

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)

Square root of 2 example

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)

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)

"one can show quite easily that A is a clopen subset of Q." There isn't a single proof, despite clopen sets being mentioned in open sets, closed sets and, obviously here. Could someone do this "quite eas[y]" exercise? 64.202.138.67 (talk) 18:15, 2 March 2011 (UTC)

There's no need for a proof. It's a consequence of the definitions. — Fly by Night (talk) 02:36, 3 March 2011 (UTC)

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)

A subset A ⊆ Q is open if there exists an open set S ⊆ R such that Q ∩ S = A. That's the definition of an open set in Q. (It has the induced topology from R.) Let S ⊆ R be given by { x ∈ R : x > √2 }. Clearly S is open and so Q ∩ S = { q ∈ Q : q > √2 } = A 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 q ∈ Q 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 S ⊆ R, but it isn't a limit point of A because √2 ∉ Q. So what about the other q ∈ Q with q < √2? Well, none of these are limit points of A either. To see this, for each rational q < √2 let S_q = { x ∈ R : x < ½(√2 − q) }. By definition Q ∩ S_q are open in Q, and also A ∩ S_q = ∅. 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)

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)

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)

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)

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 R \ { x } ∪ { x } = R. The same hold if we replace { x } by any subset, say U, of R. The discreet topology comes from setting U = R. — Fly by Night (talk) 18:46, 27 March 2011 (UTC)

Etymology

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)

Spell check

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)

It's making mistakes. Algebraist 22:30, 27 November 2010 (UTC)

Spell check doesn't understand a lot of math terms.--75.80.43.80 (talk) 13:52, 8 April 2011 (UTC)

Spell checks dont understand biology terms either 190.60.93.218 (talk) 17:09, 9 September 2011 (UTC)

Graph

Graph

Can somebody explain in which sense the shown graph defines a topological space? What are the points? What are the open sets? --Jobu0101 (talk) 14:19, 28 November 2014 (UTC)

I have no clue what the graph is trying to say, so I'm removing it. I think it is the result of some beginner confusion or possibly OR ("original research") 67.198.37.16 (talk) 20:48, 17 November 2023 (UTC)

Well, OK, If i have to guess, the green dots are the points. Points are closed. I'm guessing that any subgraph of the graph is meant to be a closed set as well. I'm guessing that the complement of a subgraph $A$ is what you get if you erase the edges connecting that subgraph to the outside world. The complement is thus again a subgraph, and it is disjoint (disconnected) from starting subgraph $A$. Since complements are open (by definition) but are still subgraphs, they are also closed. Thus they are clopen, and thus everything is clopen in this example. This is the only plausible explanation I can think of for what this graph was trying to say. I don't know if there is a name for this kind of topology ... is it a subgraph topology or a graph topology? Clearly not the last thing.

Here's another way of describing this. A topology is just .. a collection of sets, with some rules about intersection and union. If a set consists of points, then, for any two points in the same set, draw an edge between them. Else not. The result is a graph. Does this graph accurately represent the topology $T$? Well, lets see. If $U\in T$, $V\in T$ are two open sets, then one must have that $U\cup V\in T$ and $U\cap V\in T$ which would seem to correctly capture the topology of $T$. Well, except not: look at finite topological space or even Sierpiński space -- we cannot take Sierpiński space and represent it with this graph topology construction.

Anyway, I've written a bunch of "original research" in these last two paragraphs. I don't know of anything in Wikipedia that would explain this accurately. Nor have I seen this as an example in any of the books on general topology I've read (and I've read 3 or 4 or 5). If you are reading this, and can clarify the situation here, that would be great. 67.198.37.16 (talk) 21:16, 17 November 2023 (UTC)

You have been mislead by the caption. There would not have been any confusion if the first sentence of the caption would have been changed from "A graph with several clopen sets" to "A figure with several clopen sets (for the topology induced by the topology of the plane)".

So, I suggest to restore the image with a new caption.

By the way, graph connectivity is defined from a path-connectedness that involves implicitly this topology. It would probably be useful to add to this article a paragraph or a section on connected components of graphs. D.Lazard (talk) 09:51, 18 November 2023 (UTC)