# Definition of open set in topology

1. May 17, 2009

### kof9595995

A topological space is a set X together with T, a collection of subsets of X, satisfying the following axioms:
1.The empty set and X are in T.
2.The union of any collection of sets in T is also in T.
3.The intersection of any finite collection of sets in T is also in T.
The sets in T are the open sets
I can't see how it can become the open set definition in metric space.
Say,in 2-D Euclidean space,let X be a circle with all its interior points, then a trivial topology is T={{}, X}, it seems not to contradict to the axioms, but X is a closed set , can anybody help me clarify this? Thanks in advance.

2. May 17, 2009

### matt grime

There is nothing that prevents a set being both open and closed. In particular the empty set, and the whole space are always both open and closed.

Last edited: May 17, 2009
3. May 17, 2009

### kof9595995

So any closed curve with its interior points can defined to be open? Then what's the difference between open and closed set? I still can't understand .

4. May 17, 2009

### matt grime

A topology defines the open sets, and they obey the rules of being 'open'. A set is closed if and only if its complement is open.

There are many ways of doing this to create different topologies - you seem to want there to be only one topology on a set.

Consider R the set of real numbers. The following are all topologies:

T:= { {}, R } the trivial topology. There are only two open sets, and correspondingly two closed sets.

T:= P(R), the power set or the set of all subsets of R. This is called the discrete topology - all sets are open and all sets are closed.

T, the metric topology - open sets are arbitrary unions of open intervals. And the closed sets are what you think they are - arbitrary intersections of closed intervals.

Define S:={ x : f(x)=0 for some real polynomial f} these form the closed sets of a topology on R, called the Zariski topology. In this case S is equivalent to the set of all finite subsets of R, so T, the set of complements of things in S, is the set of all sets whose complements are finite. This is also called the cofinite topology.

Now, take some set X in R. It does not make sense to just ask 'is X open?' it only makes sense to ask 'is X open in a particular topology?'

5. May 17, 2009

### kof9595995

Well, I think I'm starting to get it, thanks a lot for the detailed answer, impressive

6. May 17, 2009

### matt grime

I misdefined the zariski topology.

I should have said: given a real poly f, define zeros(f) to be the set of zeros of f, then

{ zeros(f) : f a real poly }

form the closed sets for a topology on R.

7. May 17, 2009

### HallsofIvy

Staff Emeritus
I am a bit confused by this. You seem to be asking about metric spaces but then give, as an example, "let X be a circle with all its interior points, then a trivial topology is T={{}, X}" which is NOT a metric topology. In any case, it is always the case that the empty set and the entire set itself are both open and closed.

In fact, if you were to use the "discrete topology" in which all subsets of X are in the topology, then all subsets of X would be both open and closed.

8. May 19, 2009

### kof9595995

You exactly pointed out where I got lost:When I tried to understand open set in "common sense"(metric space), I should have used the metric topology to define the space.I didn't realize that I must associate a metric topology with a space before calling it a metric space.
I'm really a newb in topology and I'm still struggling with some very basic questions.

9. May 24, 2009

### jetplan

Thanks everyone for the nice explanation.
Anyone care to point out if the following is correct or not ?

1. An open set in a metric space is NEVER a closed set in ANY metric space.[/quote]
Given any set X, d(x,y)= 0 if x= y, 1 if $x\ne y$ is a metric. ALL subsets of X are both open and closed in that metric.

It's not clear what you mean by this. Do you mean two different topologies on the same space? For a finite set, at least, given any topology, you can define a new topology by taking all closed set in the first topology to be the open sets in the new topology.

Yes. See your question 1. In fact, in ANY topological space, both the empty set (maze's suggestion) and the entire set itself are both open and closed.

Thanks a lot

Last edited by a moderator: May 25, 2009
10. May 25, 2009

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