Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I know it's probably very basic. Please enlighten me with a simple example.

share|improve this question

closed as not a real question by Jasper Loy, tomasz, Ayman Hourieh, Henry T. Horton, Pete L. Clark Dec 22 '12 at 1:28

It's difficult to tell what is being asked here. This question is ambiguous, vague, incomplete, overly broad, or rhetorical and cannot be reasonably answered in its current form. For help clarifying this question so that it can be reopened, visit the help center.If this question can be reworded to fit the rules in the help center, please edit the question.

Both the space and the null set are closed. You will have to elaborate if you are asking something else. –  peoplepower Dec 22 '12 at 0:10
I don't understand the question. –  Qiaochu Yuan Dec 22 '12 at 0:16
I think the OP means "a" topology, as in topological space. As such, I don't think this is a bad question. –  Martin Argerami Dec 22 '12 at 0:53
@arkadiy: First of all, do you know the definition of ‘topology’? –  Haskell Curry Dec 22 '12 at 1:00
The question has been closed. It doesn't make sense as written: indeed there are closed sets in topology. Your question has already received multiple answers, all guessing at what you might mean. Please feel free to make edits which clarify your question: it could then be reopened. –  Pete L. Clark Dec 22 '12 at 1:30

4 Answers 4

up vote 14 down vote accepted

In topology, sets are not like doors: Doors are either open or closed, not both. In contrast, sets can be one of the following:

  • open and closed,
  • closed but not open,
  • open but not closed, or
  • neither open nor closed.

If you are worried about why topological spaces are often defined (in terms of open sets):

There are many other equivalent ways to define a topological space $X$: e.g., the concepts of neighborhood or of open sets can be reconstructed from the other starting points which satisfy the correct axioms. For example, using De Morgan's laws, the axioms defining open sets become axioms defining closed sets:

The empty set and topological space X are closed.
The intersection of any collection of closed sets is also closed.
The union of any pair of closed sets is also closed.
share|improve this answer
Are sets Windows? Or are they Linux? :-) –  Asaf Karagila Dec 22 '12 at 0:16
@Asaf Hahahaha...! Maybe Macs? –  amWhy Dec 22 '12 at 0:44
That's kind of Unix-y, borderline Linux. But I am willing to agree these are only cousin operating systems, and not really siblings. Maybe even second-cousin. –  Asaf Karagila Dec 22 '12 at 0:46
@AsafKaragila: I'd say they're neither, as Windows(es) are closed and Linux(es) are open. ;) –  tomasz Dec 22 '12 at 0:56
@tomasz: You went all the way to the source of the joke and made it better! :-) –  Asaf Karagila Dec 22 '12 at 0:58

The definition of a topology is simply the collection of open sets. However there is no axiom stating that an open set cannot be closed.

For example in the discrete topology every set is open, therefore every set is closed as well.

share|improve this answer

While I think you should clarify your question, here are two possible answers, depending on whether you referred to restating topology in terms of closed sets or having a set that's both closed and open.

share|improve this answer
I'm thinking about the following statement: "inverse images of open sets are open".Suppose $(X,\tau_X)$ and $Y,\tau_Y$ be a topological spaces. Then $f:X\to Y$ is continuous if $f^{-1}(U)\in\tau_X,\forall U\in \tau_Y$. –  arkadiy Dec 22 '12 at 0:23
So you're asking why aren't or why can't the notion of continuous functions be stated in terms of closed sets rather than open ones? Well, it could: the preimage of an open set is open if and only if the preimage of the corresponding closed set is closed, because for any (set-theoretical) function $f:A \rightarrow B$ with $U \subseteq B$ you get the identity $f^{-1}(U^c)=(f^{-1}(U))^c$. You should rewrite the thread's question properly. –  DoomMuffins Dec 22 '12 at 0:32

Let me gently go through the basics with you again. Let $ X $ be a set. A topology on $ X $ is a collection of subsets of $ X $ such that: (i) $ X $ and $ \varnothing $ belong to the collection, (ii) the union of sets in any subcollection belongs to the collection itself, and (iii) the intersection of any two sets in the collection belongs to the collection itself. These three conditions are the axioms of a topology.

A set $ X $ can have many topologies, because there can be many collections of subsets of $ X $ that satisfy the axioms of a topology. For a fixed topology $ \tau $, we call the sets belonging to $ \tau $, “$ \tau $-open sets”. The complement of $ \tau $-open sets are called “$ \tau $-closed sets”. If a topology $ \tau $ is understood, i.e., everyone knows what $ \tau $ is, then by convention, we drop any mention of $ \tau $ and simply say “open sets” or “closed sets”.

Clearly, for any topology, $ X $ and $ \varnothing $ qualify as closed sets, because they are the complement of $ \varnothing $ and $ X $ respectively, which are already open sets by definition.

share|improve this answer

Not the answer you're looking for? Browse other questions tagged or ask your own question.