# why is there no order in metric spaces?

Why is there no abstract notion of order in the metric spaces?

Surely, if there is a notion of distance there must be a notion of different values. If there is a notion of different values, why can't we specify a well defined order in the metric space to reflect the lowest to greatest values in the space?

• smallest to largest. – Qwertford Sep 10 '15 at 11:18
• How would you define the smallest element in a discrete metric space? – user29123 Sep 10 '15 at 11:23
• Which one is bigger? $\sin(x)$ or $\cos(x)$ ? What about $\sin(\frac{\pi}{2}-x)$ or $\cos(\frac{\pi}2-x)$? – N. S. Sep 10 '15 at 11:40
• Pretty much anything can be an element of a set, including functions. – Sasho Nikolov Sep 10 '15 at 12:15
• you should at least ask for an order that induces the same typology as the metric. otherwise you can order any set (but it is rather counter intuitive) – user251257 Sep 10 '15 at 12:30

## 7 Answers

The set of complex numbers $\Bbb{C}$ is clearly a metric space under the usual notion of modulus, i.e., $(\Bbb{C},d)$ is a metric space where $d(w,z)=|w-z|=\sqrt{(w-z)(w-z)^{\ast}}$. However, at the same time, $\Bbb{C}$ has no total ordering relation that is 'meaningful'. This is especially intuitive if we think of $\Bbb{C}$ as a plane analogous to $\Bbb{R}^{2}$. This is a concrete example of a metric space which has no (meaningful) ordering relation.

As of now, I have failed to prove to myself that there does not exist an order that can be constructed on $\Bbb{C}$ using only the metric. But my point is that, if such an order exists and is a total order, then it is 'meaningless'. Hence, considering orders constructed just from a metric might be an interesting problem in some specific instances, but may not be a great use of time in the general case, as some metric spaces are important for other, incompatible, reasons.

• Sorry, what do you mean Analagous to $\Bbb{R}^2$? – Qwertford Sep 10 '15 at 11:34
• $\mathbb{C}$ can be totally ordered, but there is no total order that is compatible with the ring structure of $\mathbb{C}$. – Augustin Sep 10 '15 at 11:39
• @Augustin: +1 Yes, I suppose I should have mentioned that. In the end, all metric spaces can be totally ordered. I guess my answer is currently unsatisfactory in that I have not demonstrated that a (unique) total ordering cannot be constructed from the metric space axioms alone. – Will R Sep 10 '15 at 11:44
• @MrMachine: By "analogous to $\Bbb{R}^{2}$ I mean we can imagine the complex numbers as points in a plane, and the plane is typically thought of as the Cartesian product of the real line with itself, i.e., $\Bbb{R}\times\Bbb{R}=\Bbb{R}^{2}$. – Will R Sep 10 '15 at 11:50
• "All metric spaces can be totally ordered." I think you need the axiom of choice. Without choice, does $\mathcal P(\Bbb R)$ (discrete metric) have a total ordering? – Akiva Weinberger Sep 10 '15 at 14:39

There need not be a meaningful such relation on a metric space. The basic reason you omit it is because it by omitting requirements cover up more cases. When talking about metric space we restrict the requirements to only mean that there exists a distance between the elements and nothing more (we do not require that the space has an origin, we do not require that the elements can be added or scaled and so on).

For example apart from the normal cases $\mathbb R$ and $\mathbb C$, the concept covers a lot of different cases. For example the trivial metric space with only one element, hamming metrics (the number of symbols that differ between two strings), euclid geometrics (that doesn't rely on the existence of an origin) and so on.

If you try to define the order naively by size you will run into the fact that a metric space doesn't have the notion of size either, it's defined in terms of distance between elements. And even if there were (and then it's not merely a metric space, but maybe a normed vector space) you would end up with something that does not fulfill the requirements of an order (you don't have the property that if $a\le b$ and $b\le a$ then $a=b$).

If you want an ordered metric space, then you should require that and not hope for that all abstract "structures" should conform to your requirements.

Math consists of a zoo of abstract "structures" from the most generic ones to more specialized - you have to choose which one to use in every occation (but if you use more generic one, your conclusions will become more generic).

• But im saying that a notion of distance exists if and only if a notion of size exists then from that we can make a well defined order. – Qwertford Sep 10 '15 at 11:20
• No, if you have the metric space $\{a, b\}$, where $d(a,b)=1$, which of $a$ and $b$ is then the largest? – skyking Sep 10 '15 at 11:25
• Didn't you use the values of a and b to get d(a,b)=1? – Qwertford Sep 10 '15 at 11:26
• @MrMachine: In this example the metric is defined to take the value $1$ for any two elements in the space. The 'values' $a$ and $b$ are not used to 'calculate' the distance in any way. – Will R Sep 10 '15 at 11:28
• @MrMachine Perhaps it is useful to know the definition of a metric. – Servaes Sep 10 '15 at 11:42

We can certainly derive a metric from certain notions of size (norms), but need not have any notion of size at all to have a metric.

Let $X$ be any set, and for $x,y\in X,$ define $$d(x,y)=\begin{cases}1 & x\ne y\\0 & x=0.\end{cases}$$ This is a metric on $X$ that tells us when two elements are different, but nothing else.

Order in itself has little to do with distance. I can order a set of words by alphabetical order. What matters is the order relationship: transitive, asymmetric, etc.

The reals both have an order and a metric. That makes them awesome. I could impose an order on the unit square by applying a space filling curve to the unit interval. It would be like unraveling a sweater. This order would have nothing to do with distance in $R2$, or anything else that makes two dimensions interesting. The sweater, once unraveled, is no longer a sweater.

Consider any possible ordering on the points on a circle. You will find that:

1. it is not invariant under isometries;

2. some of the sets $\{x: x<a\}$ are not open.

This tells you that the ordering cannot have so much relation with the metric structure...

I think this question suggests an answer addressing whether or not every metric topology is also induced by a linear order. The answer is no. I'll try to sketch a proof of this by means of an example.

Consider the topological space given by three copies of the closed unit interval where we identify the zeros. This is a metric space. Namely let $$I_1=[0,1]\times\{1\}$$ and $$I_i=(0,1]\times\{i\}$$ for $$i\in \{2,3\}$$ and the metric $$d$$ on $$X=\cup_i I_i$$ be given by $$d((x,i),(y,i))=|x-y|$$ and $$d((x,i),(y,j))=|x+y|$$ and $$d((0,1),(y,j))=y$$ for any $$x,y\in (0,1]$$ and any $$i\neq j$$. See Hedgehog space.

Suppose that the topology $$\tau$$ induced by $$d$$ is also generated by some linear order $$<$$ on $$X$$. We denote the intervals in $$<$$ with endpoints $$x$$ and $$y$$ by $$(x,y)_<$$. Consider the point $$x_0=(0,1)$$. Note that there must be two distinct $$i,j\in\{1,2,3\}$$ such that either $$(x_0,y)_{<} \cap I_i \neq \emptyset$$ and $$(x_0,y)_{<} \cap I_j \neq \emptyset$$ for any $$y>x_0$$ or $$(y,x_0)_{<} \cap I_i \neq \emptyset$$ and $$(y,x_0)_{<} \cap I_j \neq \emptyset$$ for any $$y. Without loss of generality we consider the first case. Let $$y_i>x_0$$ be a point in $$I_i$$ and then let $$x_0 be a point in $$I_j$$. The open intervals $$(-\infty, y_j)$$ and $$(y_j,+\infty)$$ partition $$\{x_0\}\cup I_i$$ into two open subspaces. However $$\{x_0\}\cup I_i$$ is homeomorphic to $$[0,1]$$ and hence connected. Contradiction.

It is worth noting that the converse is also not true. Namely, not all topologies induced by a linear order and metrizable. For example the space $$[0,\omega_1]$$, where $$\omega_1$$ denotes the first uncountable ordinal, with the topology induced by ordinal order, is compact but not separable, so it is not a metric space.

Let's consider the set of all points on the surface of the Earth and measure the distance between two points by the length of the shortest path over the surface. This is a metric space.

How are the points ordered? Which point is the greatest and which is the least?

• Ordering points by their latitude works just fine. The north pole is the greatest point and the south pole is least point. If you insist on an strict order, you can compare points by their latitude and if they're the same, use the longitude. – lhf Sep 10 '15 at 11:58
• So these points on the surface you express as vectors from the origin? – Qwertford Sep 10 '15 at 11:58
• @lhf You could also order the points so that Stockholm was the best and Dubuque Iowa was the worst, but that would equally have nothing to do with the metric space. Nobody is claiming that an arbitrary set cannot be ordered in an arbitrary way. – MJD Sep 10 '15 at 12:03
• @MrMachine You could think of them as vectors form the origin, but why go to so much effort? When someone says "How far is it from Paris to Mount Everest" do you reply "Are you thinking of Paris as a vector from the origin?" – MJD Sep 10 '15 at 12:04
• @MrMachine: more to the point, the orders given in comments here have pretty much nothing to do with the metric. As such it's not in any way useful to require that every metric space must come equipped with an order (all the usual metric space theorems are true for metric spaces that don't have one), and so we don't. If for some purpose you want to consider a metric space and an order then you just say, "a metric space together with an order" or words to that effect. – Steve Jessop Sep 10 '15 at 16:53