# Metrics and topologies

1. Oct 9, 2013

### friend

I could use some insight in the form of a summary.

Can the topology of a manifold determine the metric use on it? Or is the metric completely independent of topology and coordinates systems?

2. Oct 9, 2013

### WannabeNewton

The concept of a coordinate system has no meaning for arbitrary metric spaces; that is something we work with on $C^n$ manifolds. Every metric space is a topological space; the metric naturally determines a topology on the space by generating one through the base of open balls of the metric. On the other hand, not every topological space is a metric space; this is where metrization theorems come in.

3. Oct 9, 2013

### pasmith

It is also possible for different metrics to generate the same topology.

4. Oct 9, 2013

### friend

Thank you. So I'm not sure, which came first a topology or the metric. It would seem to me that you cannot have a metric until you first have a set of points to form a distance function onto. But perhaps that doesn't require the points to be a topology.

5. Oct 9, 2013

### pasmith

Metrics and topologies are both structures which can be put on arbitrary non-empty sets in order to define a notion of "closeness". Having done that one can then define notions of convergence of sequences and continuity of functions.

There is a sense in which topological spaces are more general than metric spaces, in as much as every metric space is a topological space but not every topological space can be a metric space.

6. Oct 9, 2013

### friend

Do the separation axioms on a topology imply a metric? If there is a continuous function from one point or set to another, doesn't that continuous function imply a measure between points, namely the continuous function has a different value for one point than for another, perhaps only for two point very close to each other?

And more generally, can you have a function on points without a metric between them. I mean, if there is a continuous function, then df/dx can be defined, and dx is a distance, right?

7. Oct 9, 2013

### economicsnerd

Urysohn's metrization theorem does exactly this. It says that if a topology on $X$ is Hausdorff (a separation condition), regular (a separation condition), and second-countable (a condition stating that the topology isn't too rich), then there exists a metric on on $X$ giving rise to said topology.

8. Oct 9, 2013

### friend

When you say, "giving rise to said topology", that seems confusing to me. Your first statements make it sound as though the topology already exists, with certain properties such as Hausdorff, etc. So would it be sufficient to say, ",then there exists a metric on X", without the rest of the sentence?

Last edited: Oct 9, 2013
9. Oct 10, 2013

### pwsnafu

The phrase "giving rise to a topology" means there exists a metric, such that the topology induced by said metric, is equivalent to the original topology you started with.

So to be precise: you have a set $T$ with topology $\mathcal{T}$. Now $\mathcal{T}$ is metrisable if and only if there exists metric $d$ (which induces a metric $\mathcal{D}$ via open balls) satisfying $\mathcal{T}=\mathcal{D}$.

10. Oct 10, 2013

### WannabeNewton

A function is a purely set theoretic concept so no there is no need for a notion of distance whatsoever. Also, you need the notion of differentiability to define derivatives, continuity isn't enough. For continuity you just need a topology but for differentiability you need a $C^n$ differentiable structure ($n \geq 1$). Finally, $dx$ has absolutely no relation to distance.

11. Oct 13, 2013

### friend

"a metric", you say. Is there a way to discern what kind of metric it is for a given topology, or what signature the metric might have? Or does this theorem only state that a metric can be constructed. Or does this mean that you can construct a metric of your chosing?

12. Nov 20, 2013

### gufiguer

For no more cofusing arise :)