# Confused by open sets

1. Jul 6, 2009

### symbol0

A class T of subsets of X is called a topology on X if it satisfies these 2 conditions:
1. The union of every class of sets in T is a set in T.
2. the intersection of every finite class of sets in T is a set in T.
The sets in the class T are called the open sets of the topological space (X,T).

I'm a little confused by what open sets mean here. Do they mean that every point in the set is an interior point?
Let's say X is the metric space $$\mathbb{R}$$ with the regular metric, and I chose T to be all the closed sets in X. Then T satisfies the 2 conditions mentioned above. So T is a topology on X.
So the sets in T are closed, but I call them the open sets of the topological space (X,T)?

Can anybody help me to clarify this?

2. Jul 7, 2009

### Office_Shredder

Staff Emeritus
The class of closed sets on R does not satisfy the properties since for example the union of all sets of the form

[1/n,1]

which are all closed gives us the set (0,1] which is not closed (and hence not 'open' in your inverted topology)

Notice that if you modify property 1 to be true for only finite classes, and 2 to be true for arbitrary classes, then you have the 'closed' topology, i.e. all the closed sets of your topological space.

3. Jul 7, 2009

### Moo Of Doom

The closed sets in the metric space $$\mathbb{R}$$ actually don't satisfy criterion 1 above, since, for example, singletons are closed sets, but the union $$\bigcup_{x \in \left( 0, 1 \right)} \!\!\! \left\{ x \right\} = \left( 0, 1 \right)$$ is not a closed set.

But you're not too far off. There are plenty of other topologies you can give to $$\mathbb{R}$$, where the open sets don't look at all like you're used to. For example, the trivial topology, where only the empty set and all of $$\mathbb{R}$$ are open, or the discrete topology, where all subsets are open sets. A less obvious example is the set of subsets of $$\mathbb{R}$$ that are unions of half-open intervals [a, b). In this topology, [a, b) is both open and closed.

4. Jul 7, 2009

### HallsofIvy

Staff Emeritus
What do you mean by "interior point"? An interior point is defined using a metric and you don't necessarily have a metric. For example, If we define the topology for X to be only X itself and the empty set, the two conditions above are satisfied but we cannot define a metric on the set which will give only those two sets as "open" sets. What "open set" means here is exactly what it says: sets in the topology, T. Of course, you could then define "interior point" from that: a point, p, is an interior point of a set, A, if and only if there exist some set, O, in T such that p is in O and O is a subset of A or is A itself. In that case, yes, if A is itself in T, then every point in p is an "interior" point of A because you can take O to be A itself. The point is, however, that open sets are simply defined as "members of T".

By the way, many texts include the additional conditions "X itself is in T" and "the empty set is in T". Those can, in fact, be proven from (1) and (2) but the proofs require some rather abstract reasoning, looking at the union and intersection of the empty class.

As OfficeShredder and Moo of Doom said, the union of an infinite class of closed sets, as defined in a metric space, is not necessarily closed.

The collection of closes sets DOES satisfy, in a sense, the "reverse" of those:
The intersection of every class of closed sets is closed and the union of any finite class of closed sets is closed.

Last edited: Jul 7, 2009
5. Jul 7, 2009

### George Jones

Staff Emeritus
Open sets are defined to be those sets in a class that satisfies the axioms.
Well, once a topology, and thus open subsets, are defined for a set X, then a subset A is open if and only if A = Int(A), the interior of A, where Int(A) is the largest open set contained in A. An element x is in Int(A) if and only if there is a neighbourhood of x contained in A.

Even though this is correct, your intuition with respect to "interior" can fail you for some topologies.

6. Jul 7, 2009

### symbol0

So what I understand is that I was wrong with my particular example, but I was right with the general point I was trying to make, which is that what we call open sets in a topological space could be closed sets of a metric space. I understand that you don't need a metric to have a topological space, but if you start with a metric space, and then you pick your topology, you could have what I just mentioned.
Let me try another example:
If we start with the metric space X=$$\mathbb{R}$$ with the regular metric, and we pick the topology T= all the subsets of X, then each real number is an open set of the topological space (X,T), even when each real is a closed set of the metric space X.

Is that right?

7. Jul 7, 2009

### Office_Shredder

Staff Emeritus
That's correct

8. Jul 8, 2009

### HallsofIvy

Staff Emeritus
Okay, yes. A requirement for a "topology" is that the union of any collection of open sets (sets in the topology) is also open (is also in the topology). That is NOT in general true of closed sets but if you pick a topology in which it is true of closed sets, then, yes, you can swap the "open" and "closed" sets.

However, you say "If we start with the metric space X=$$\mathbb{R}$$ with the regular metric, and we pick the topology T= all the subsets of X". You should be aware that as soon as you "pick the topology T= all the subsets of X", you have thown away the "regular metric"- and there was really no need to mention it. If you take any set X and "pick the topology T= all the subsets of X", then every subset of x is both open and closed. That does happen to be a metric topology. It is called the "discrete topology" and is the metric topology you get if you define the metric d(x,y)= 0 is x=y, 1 otherwise.

9. Jul 10, 2009

### symbol0

Thanks HallsofIvy. Very clarifying reply. Especially the part about throwing away the metric.