Talk:Cover (topology)

Latest comment: 8 years ago by Baking Soda in topic vanchor

Definition edit

Is the given definition adequate to force the correct answer to the Nightmare Freshman's questions: "Can't I cover the real number line with itself?" and "Since every set is a subset of itself, isn't every cover a subcover?"Halfb1t (talk) 17:05, 30 April 2012 (UTC)Reply

Jargon edit

I know that such a term is deeply embedded in mathematical jargon, however I do believe that something in layman's terms should be added for these types of terms, which need to be understood by any who wish to try to follow the subsequent articles that such terms relate to, such as Lebesgue covering dimension or fractal dimensions. Regardless of one's familiarity with mathematical terms and algebraic expressions, some concrete definition should be able to be applied to give the reader some basic information to build on, in order to understand the whole picture. Livingston 14:51, 1 December 2009 (UTC)Reply

Link to Heine-Borel edit

What are thoughts on providing a link to "Heine–Borel theorem" in the "See Also" section. Is this OK or is it too specialised a topic? Thanks. —Preceding unsigned comment added by Arjun r acharya (talkcontribs) 11:57, 8 August 2009 (UTC)Reply

The Heine–Borel theorem provides criteria for compactness, so a link would be appropriate in the compact space article rather than here. Now, in fact, the Heine–Borel theorem already is extensively discussed in said article. — Emil J. 10:20, 10 August 2009 (UTC)Reply


finite edit

The article says about locally finite, then what is globally finite?

does the definition of cover requires disjoint between sets?

Concerning the compact set, it says, "finite subcover", what does a finite subcover mean? since there is finite subcover, so there must be infinite subcover, am I right? This is not explained in the article. Jackzhp (talk) 18:03, 11 July 2010 (UTC)Reply

  • There is no such thing as "globally finite", locally finite is a local version of the property of being finite.
  • No, sets in a cover do not have to be disjoint (otherwise it would be mentioned in the definition).
  • "Finite subcover" is not a term on its own, it just means a subcover which is finite. A subcover which is not finite is, obviously, an infinite subcover. As for more explanation, the "Compactness" section is just an overview list, you are supposed to click the links if you want to know more details.—Emil J. 12:11, 12 July 2010 (UTC)Reply
Thank for the information! Still when you say a cover is finite. What does it mean? I did click on the finite set, but it is not easy to get the point. A set with finite number of elements? A few years ago, when I was studying this kind of shit. I thought I get it. But with time, I don't remember and I don't understand, and I don't want to look for my old notes. I think that we need a simple example to elaborate this, or it is good to put a link linking to a well explained article available online. It is not that complicated, but just not well presented, so people like me have difficulty to understand it. Jackzhp (talk) 21:02, 16 August 2010 (UTC)Reply
Yes, a set is finite if it has finitely many elements. In other words, a finite cover of X is a set of the form C = {U1,U2,...,Un} for some natural number n, such that  . (It is allowed to have n = 0 here, in which case the notation should be understood in such a way that C and X are both empty.)—Emil J. 10:51, 17 August 2010 (UTC)Reply
Thank you for the response. Let's work out an example, and put it on the article page.
Let's use X=[0,1] which is compact,   is an open cover of X, and   is a subcover of C, since it has 2 elements, so it is finite.
Another example, let X=(0,1), it is not compact, how to construct an open cover which does not have finite subcover? Jackzhp (talk) 22:21, 17 August 2010 (UTC)Reply

vanchor edit

@Paul August:, open cover vanchored for Open covering. Just updated Open cover. Feel free to update if you mind direct link.. Baking Soda (talk) 16:54, 17 April 2016 (UTC)Reply