Why is Empty set open?

1. Mar 2, 2009

soopo

1. The problem statement, all variables and given/known data
If $$\emptyset$$ has no elements, then $$\emptyset$$ is open.

3. The attempt at a solution
If $$\emptyset$$ is closed, then $$\emptyset$$ has at least an element.

This is a contradiction, so $$\emptyset$$ must be open.

I am not sure about the validity of my attempt.

2. Mar 2, 2009

CompuChip

It is also closed.
There is nothing to prove, really. It's open by definition:
Actually it also follows from 2 (take an empty union) or 3 (take two disjoint open sets, if possible).

Or, you can take the "analysis" definition: S is open if for all x in S there is a neighborhood of x contained in S, which is vacuously true for the empty set.

Note that open and closed are not mutually exclusive: a set can be open, closed, neither or both. Also note that being closed does not imply being non-empty.

3. Mar 2, 2009

soopo

Do you mean that an empty union is open?

4. Mar 2, 2009

lanedance

a union of 2 disjoint sets, an empty union, is the empty set... so both closed & open by the definition given - clopen

5. Mar 3, 2009

CompuChip

By empty union, I mean the union of no sets at all. But the argument relies on "arbitrarily many elements" being interpreted as including "no elements at all".

The intersection of two disjoint sets is empty. The union of 2 disjoint sets is ... the union of two disjoint sets

6. Mar 3, 2009

lanedance

woops yeah good catch - wandered off there, cheers

7. Mar 3, 2009

HallsofIvy

Staff Emeritus
WHY the empty set is open depends on what your definition of "open" is. As CompuChip said, the most general definition of a topological space defines a "topology" for a set as being a collection of subsets satisying certains conditions- that among those conditions is that the it include the empty set- and any set in that "topology" is open.

But you may be thinking in terms of a "metric space" where we are given a "metric function", d(x,y) and use that to define the "neighborhood of p of radius $\delta$, $N_\delta(p)= {q| d(p, q)< \delta}$ and define an "interior point", p, of set A to be a point in A such that for some $\delta$, $N_\delta(p)$ is a subset of A.

Even then there are two ways to define open set. Most common is "a set, A, is open if every member of A is an interior point of A" which can be expressed more formally as "if p is in A, the p is an interior point of A". If A is empty then the "hypothesis", "if p is in A" is false and so, logically, the statement is true: A is an open set.

Another way to define "open set" is to define p to be an "exterior point" of set A if it is an interior point of the complement of A and define p to be a "boundary point" of set A if and only if it is neither an interior point nor an exterior point of A. Now we can define a set A to be open if it contains NONE of its boundary points. (Here we could also define a set to be "closed" if it contains ALL of its boundary points. Remember how in Pre-Calculus, we say that intervals are "open" or "closed" depending upon whether they include their endpoints?

It is easy to see that every point in the space is an exterior point of the empty set so it has NO boundary points. That given, the statement "it contains all of its boundary points" is true and so the empty set is open. Because it has no boundary points it is also true that the empty set contains all (= none) of its boundary points and so the empty set is both closed and open.

8. Mar 3, 2009

soopo

I will try to summarize the different ways to define an open set
1. By interior point and the neighborhood of p of radius $\delta$ (Calculus): If p is in A, p is an interior point of A.
2. A set is open if every member of A is an interior point of A: if p is in A,
then p is an interior point of A. If A is empty and if p is not in A, then A
is an open set
.
3. A set A is open if it contains NONE of its boundary points.

9. Mar 9, 2009

Focus

If you make a statement like $$\forall x \in \emptyset$$ then it is true (trivially). So when you say $$\forall x \in \emptyset \quad x<x$$ is true, just as $$\forall x \in \emptyset \exists B_{\epsilon} \subset \emptyset \text{ s.t. }x \in B_\epsilon$$. The empty set possess more properties than most people realise ;)

10. Mar 10, 2009

HallsofIvy

Staff Emeritus
This is assuming a metric topology.

11. Mar 10, 2009

Focus

It would be a bit circular to try and justify why the empty set is open from topology :shy:

12. Mar 10, 2009

HallsofIvy

Staff Emeritus
No, it wouldn't: the empty set is open (in a general topological space) because the definition of a topology requires that it include the empty set. That was what CompuChip said. Nothing circular about that.