Is the Empty Set Considered a Metric Space?

Click For Summary
The discussion centers on whether the empty set can be considered a metric space. Participants argue that while the empty set satisfies the metric properties vacuously, it is not formally a metric space without an accompanying metric function. The empty set does have a unique metric function, the empty function, which leads some to conclude that it can be classified as a metric space. However, others emphasize the importance of defining both the set and the metric, arguing that without this, the statement lacks formal validity. Ultimately, the debate highlights the nuances of mathematical definitions and the balance between formalism and practical understanding.
julypraise
Messages
104
Reaction score
0

Homework Statement


Is empty set a metric space?


Homework Equations


None.


The Attempt at a Solution


It seems so because all the metric properties are vacuously satisfied. Mabe the question
had better be put like this: Does mathematicians tend to think empty set as a metric space for some reason such as convenience?
 
Physics news on Phys.org
Simple answer- no- because a "set" alone is not a metric space. A metric space consists of a set and a "metric" function. What metric function are you using?
 
Second simple answer.
The empty set combined the empty metric function, which maps nothing since its domain is empty, is indeed a metric space.

For mathematicians this is not so much convenience, but simply a logical consequence of the definition.
 
I like Serena said:
The empty set combined with any metric function

Note that there is exactly one metric function on the empty set. This is simply the "empty function"

d:\emptyset\times \emptyset\rightarrow \mathbb{R}

Many people find the empty function a bit awkward, and they are right. To understand it, you must realize the very definition of a function.
A function f:A\rightarrow B is actually defined as a subset S of A\times B which satisfies some properties: the property is that for all x in A, there must exist exactly one y in B such that (x,y) is in S.

Now, our function d should then be a subset of (\emptyset\times \emptyset)\times \mathbb{R}, but this is just the empty set again! So our function d coincides with the empty set. And since the empty set has only one subset, there is only one such function d.

The metric axiom is trivially satisfied.
 
Last edited:
Actually, I'm going to have to contradict HallsofIvy.
The empty set is a metric space.

As wiki says: "Often, d is omitted and one just writes M (the set) for a metric space if it is clear from the context what metric is used."

In this case it is clear from the context which metric is used, so you can say that the empty set is a metric space.
 
I like Serena said:
Actually, I'm going to have to contradict HallsofIvy.
The empty set is a metric space.

As wiki says: "Often, is omitted and one just writes for a metric space if it is clear from the context what metric is used."

In this case it is clear from the context which metric is used, so you can say that the empty set is a metric space.

Wiki is wrong.
A metric space is formally defined as a pair (X,d). The empty set is not such a pair, so it is not a metric space in itself.
Halls was entirely correct in saying that a metric has to be defined first.

I agree that sometimes the metric is not mentioned, but that's informal notation. Pure formally, the empty set is not a metric space. I don't care what wiki says.
 
micromass said:
I agree that sometimes the metric is not mentioned, but that's informal notation. Pure formally, the empty set is not a metric space. I don't care what wiki says.

That's like saying that Z/nZ is not a group, but it is (isn't it?).
 
I like Serena said:
That's like saying that Z/nZ is not a group, but it is (isn't it?).

Pure formally, it is not a group.
 
Hmm, this is a bit of hairsplitting, but...

If we say: "the empty set is a metric space", this means unambiguously "the ordered pair (∅,d) where d is the uniquely defined metric function ∅x∅→R, is a metric space".

As far as I know, that means that the original statement is true.
Even though it is a bit informal, it is still unambiguously defined and as such true.

I thought that in math we often do not extensively list all required definitions and conditions as long as they are unambiguously clear from the context.
Otherwise that would make math a bit cumbersome.
 
Last edited:
  • #10
I like Serena said:
Hmm, this is a bit of hairsplitting, but...

If we say: "the empty set is a metric space", this means unambiguously "the ordered pair (∅,d) where d is the uniquely defined metric function ∅x∅→R", which is indeed a metric space.

As far as I know, that means that the original statement is true.
Even though it is a bit informal, it is still unambiguously defined and as such true.

I thought that in math we often do not extensively list all required definitions and conditions as long as they are unambiguously clear from the context.
Otherwise that would make math a bit cumbersome.

I agree that in practice we would say that "the empty set is a metric space" without trouble. And people say \mathbb{Z}/n\mathbb{Z} is a group without troubles and they're not wrong.

The difference is that this privilege is only there for experienced people. People who just see the subject should be aware that there's a difference between the set and the metric space \emptyset.

I'm not arguing that people be hyper-formal, and if somebody writes "\emptyset is a metric space" on an exam, I would not mark it down at all. But you should always be aware of how things are done formally. And pure formally \emptyset is not a metric space.
 
  • #11
I am a bit black and white on this.
In math a statement is either true or false.
As such an ill defined statement is false.

IMO the statement "The empty set is a metric space" is true, since it is well defined and true.
Another true statement is that "The empty set, without any metric attached, is not a metric space", which is of course what you mean.

I do agree that students should be made aware of the distinction, but I think it is wrong to say, even formally, that the statement "The empty set is a metric space" is false.

micromass said:
And people say Z/nZ is a group without troubles and they're not wrong.
It's not exactly without trouble. :wink:
I've asked students countless times what the operation is belonging to the group Z/nZ, addition or multiplication?
Until now, I've only had blank stares and had to explain that it is addition.
And then get a second blank stare when I asked which related group has multiplication as its operation.
 
Last edited:
  • #12
@micromass: Sorry if I'm a bit of a pain in the butt.
 
  • #13
If we're going to split hairs, a function requires a specification of a domain and a codomain, along with a relation, so the class of metrics on the empty set is so large it's not even a set, but I don't see what we've gained
 
  • #14
Yes, I kind of assumed that the metric (function) is an empty set. I agree with both the first two replies, including the third. Metric space indeed, strictly speaking, is an ordered pair, not just a set, thus (empty set, empty set) is indeed a metric space according to you guys' answers.
 

Similar threads

Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
9
Views
3K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 15 ·
Replies
15
Views
2K