Metrics on a 2-point set

[noob]

Question: Suppose X consists of only 2 points. Find all metrics on X.

My first thought was that there are none, because there must be 3 points in X to satisfy the triangle inequality using the definition of a metric:

A metric on a set X is a function (called the distance function or simply distance)

d : X × X → R
(where R is the set of real numbers). For all x, y, z in X, this function is required to satisfy the following conditions:

1. d(x, y) ≥ 0
2. d(x, y) = 0 if and only if x = y
3. d(x, y) = d(y, x)
4. d(x, z) ≤ d(x, y) + d(y, z).

But then I was reading some books on metrics and looking online, and it seems that the empty set is a metric space with its only metric being the empty function. Also, a singleton set has a metric, the zero function. So now I am confused.

I guess I don't understand how the empty set or a singleton set can have a metric (how do they satisfy the triangle inequality?). But if they can, then couldn't a set consisting of 2 points? How would I find the metrics on a 2-point set?

 

[pka]

Question: Suppose X consists of only 2 points. Find all metrics on X.
I guess I don't understand how the empty set or a singleton set can have a metric (how do they satisfy the triangle inequality?). But if they can, then couldn't a set consisting of 2 points? How would I find the metrics on a 2-point set?

Do you understand that the statement
"If 1=2 then the empty set contains 2 elements." is true?
Do you understand why that is a true statement?

The triangle inequality says:
"If $\displaystyle x, y, \& z$ are three points then $\displaystyle d(x,y)\le d(x,z)+d(z,y).$
That statement is true in any space with exactly two points.

 

[noob]

Do you understand that the statement
"If 1=2 then the empty set contains 2 elements." is true?
Do you understand why that is a true statement?

I'm sorry, I don't understand. How does "If 1=2" imply that "the empty set contains two elements"? You might need to break that down for me a little more (if you wouldn't mind).

 

[pka]

I'm sorry, I don't understand. How does "If 1=2" imply that "the empty set contains two elements"? You might need to break that down for me a little more (if you wouldn't mind).

It is no wonder that you had your doubts about this.
It is a fundamental property of logic that a false statement implies any statement.
Example: $\displaystyle \text{If }1=2\text{ then }\underline{~~~~~~~~~~}.$
You can fill in that blank with any statement whatsoever and you will made a true statement.
The false statement $\displaystyle 1=2$ implies any statement whatsoever makes a true statement.

In a twp-point the statement that $\displaystyle x,~y,~\&~z$ are three points is a false statement.

Therefore, $\displaystyle \text{If }x,~y,~\&~z\text{ are three points then }$
$\displaystyle d(x,y)\le d(x,z)+d(z,y)$ is a true statement.

 

[noob]

OH, right! Ok, thanks for explaining.

 

[noob]

Ok, so we have established how the empty set, a singleton set, and 2 point sets can be metric spaces.

So, given (X,d) is a metric space and supposing X consist of 2 points, how then would one determine the metrics on X? The empty set and a singleton set each have a specific metric on them (the empty function, and the zero function, respectively). Would the metrics on a two point set then be a constant function (as well as the metrics formed my multiplying that function by a constant k, k being a positive real number)?

Sorry if I'm being thick. This one problem has just been causing me quite a bit of confusion, and I'm trying to understand.

 

[pka]

Ok, so we have established how the empty set, a singleton set, and 2 point sets can be metric spaces.
So, given (X,d) is a metric space and supposing X consist of 2 points, how then would one determine the metrics on X? The empty set and a singleton set each have a specific metric on them (the empty function, and the zero function, respectively). Would the metrics on a two point set then be a constant function (as well as the metrics formed my multiplying that function by a constant k, k being a positive real number)?

First some comments. There are many respected topologist who do not accept is idea of an empty metric space. Let’s leave that alone. You seem to have some fundamental misunderstanding about how metric spaces are defined. It is common to say: A non-empty set with a metric defined on it is a metric space.
Now, the tricky part is defining a metric on a given set.
Consider the function $\displaystyle d(x,y) = \left\{ {\begin{array}{lr} {1,} & {x \ne y} \\ {0,} & {x = y} \\\end{array}} \right.$. That function defines a metric on any set. It is called the discrete metric. Thus any set can be made a metric space.

However, there are many other metrics that depend upon the nature of the underlying set. Now to confuse matters more two seeming different metrics are said to be equivalent if they generate the same topology on the set.

Here is the punch line. On a two element metric space there can only be four open sets. Therefore, there is really only one metric in the sense of equivalence.

 

[noob]

First some comments. There are many respected topologist who do not accept is idea of an empty metric space. Let’s leave that alone.

Oh! Noted. I had just read in one of the books on metric spaces that I was flipping through at the library that the empty set is a metric space, so I was just going with that.

You seem to have some fundamental misunderstanding about how metric spaces are defined. It is common to say: A non-empty set with a metric defined on it is a metric space.

I'm familiar with the definition of a metric space (apart from the "non-empty" specification). I think I just haven't been wording my thoughts as well as I'd have liked.

Now, the tricky part is defining a metric on a given set.
Consider the function $\displaystyle d(x,y) = \left\{ {\begin{array}{lr} {1,} & {x \ne y} \\ {0,} & {x = y} \\\end{array}} \right.$. That function defines a metric on any set. It is called the discrete metric. Thus any set can be made a metric space.

However, there are many other metrics that depend upon the nature of the underlying set. Now to confuse matters more two seeming different metrics are said to be equivalent if they generate the same topology on the set.

Here is the punch line. On a two element metric space there can only be four open sets. Therefore, there is really only one metric in the sense of equivalence.

Ok. We haven't really gone over the definition of a topology in the class I'm in, but I do have a limited understanding of topologies. Enough to understand what you're saying at least.

Thanks for talking me through this.

 

[Deveno]

it might be helpful just to consider all possible functions f:{a,b}x{a,b} → R+ U {0}. then we can see which of them might be metrics.

to define such a function, we need to consider 4 possible values:

f(a,a)
f(a,b)
f(b,a)
f(b,b).

the definition of a metric imposes on us the condition that f(a,a) = f(b,b) = 0, and also that f(a,b) = f(b,a). which means that if we are to have a metric on {a,b}, it will be entirely determined by f(a,b).

so let's define:

d(x,y) = r, if x ≠ y
d(x,x) = 0

is this a metric?

the only sticking point is the triangle inequality. since we have only 2 points: a and b, there's only the following cases to check:

x = y = z = a
x = y = z = b
x = y = a, z = b
x = y = b, z = a <--see below
x = z = a, y = b
x = z = b, y = a
x = a, y = z = b
x = b, y = z = a

these are all straight-forward to verify: i'll show case 4, just for grins:

d(x,z) = d(b,a) = d(a,b) = r ≤ 0 + r = d(b,b) + d(a,b) = d(b,b) + d(b,a) = d(x,y) + d(y,z)

there's nothing special about "r" in these proofs, any non-negative real number would do, so 1 is usually used.

although there are an infinite number of metrics on a two element set, they are all "similar" in that they produce the "same" metric space (up to isomorphism). so, in a sense, the discrete metric is the "only" metric on a two element set.

 

[noob]

That was a very thorough explanation, thank you. :smile: