Is the Empty Set Open or Closed in Topology?

[soopo]

Homework Statement

If $$\emptyset$$ has no elements, then $$\emptyset$$ is open.

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.

 

[CompuChip]

It is also closed.
There is nothing to prove, really. It's open by definition:

Let X be any set and let T be a family of subsets of X. Then T is a topology on X if
1. Both the empty set and X are elements of T.
2. Any union of arbitrarily many elements of T is an element of T.
3. Any intersection of finitely many elements of T is an element of T.

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.

 

[soopo]

Actually it also follows from 2 (take an empty union) or 3 (take two disjoint open sets, if possible).

Do you mean that an empty union is open?

 

[lanedance]

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

 

[CompuChip]

Do you mean that an empty union is open?

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".

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

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

 

[lanedance]

woops yeah good catch - wandered off there, cheers

 

[HallsofIvy]

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.

 

[soopo]

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.

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.

 

[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 possesses more properties than most people realize ;)

 

[HallsofIvy]

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 possesses more properties than most people realize ;)

This is assuming a metric topology.

 

[Focus]

This is assuming a metric topology.

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

 

[HallsofIvy]

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.