# A Physical meaning of open set on manifold

1. Jan 7, 2017

### AlephClo

I understand the definition of continuity on a manifold based on open sets. I was questionning myself about what is the corresponding physical meaning of an open set of a manifold (M, Power-set-of-M, Atlas). Is it a simple (maybe simplest) assumption in order to define mathematically continuity?

Sorry I cannot help not questionning everything :-)

I am ''open'' to any reading references that can help.

Thank you, AlephClo

2. Jan 7, 2017

### rubi

The definition of a topological space using open sets is the mathematically most convenient, but not the most intuitive definition. There is an equivalent definition using neighborhoods, which leads to a very intuitive definition of continuity, which closely resembles the definition in $\mathbb R^n$. The definitions in terms of open sets can then be derived.

However, in order to define a differentiable manifold, you don't even need the concept of a topological space or continuity. There is an older (but equivalent) definition, which uses parametrizations instead of coordinate charts. The topology on the manifold (and hence the open sets) is then induced naturally. You can look it up for instance in do Carmo's book "Riemannian Geometry". It is equivalent to Whitney's modern definition, which first specifies a topological space and then equips it with an atlas. Again, the modern definition is mathematically much more convenient.

3. Jan 7, 2017

### AlephClo

i) The definition of continuity that is used is:
The map F: M into N is continuous if for all V that belongs to Powersets(N) the preimage,f(V) is an open in Powersets(M). M and N are sets on which the differentiable manifolds are built.
ii) The particular application is General Relativity, if this can help to nail the physical meaning.

Thank you both.

4. Jan 7, 2017

### rubi

Well, the purpose of my post was to make you aware of some more intuitive definitions, which are equivalent to the less intuitive definitions that you are questioning. In terms of neighborhoods, continuity of $F$ at a point $x\in M$ just means that for every neighborhood $V$ of $f(x)$, there exists a neighborhood $U$ of $x$ such that $F(U)\subseteq V$. A function is then said to be continuous, if it is continuous everywhere. The similarity to the $\epsilon$-$\delta$ definition of continuity in $\mathbb R^n$ should be apparent. If you define an open set to be a set, which is a neighborhood of all of its points, then the standard topology definition of continuity follows automatically. (By the way, the set of open sets is usually not the whole power set. Otherwise, the space would only admit a $0$-dimensional manifold structure.)

However, you don't need to worry about this, if you just adopt the manifold definition given in do Carmo's book. It doesn't require any knowledge about general topology at all.

In GR, we use manifolds to generalize the idea of Minkowski spacetime to spacetimes that look like Minkowski spacetime only locally. The notion of open sets in such spacetimes just arises as a mathematical consequence of the definition. It doesn't have any physical significance, but if you study objects that look like Minkowski spacetime locally, you cannot not have open sets.

5. Jan 7, 2017

### robphy

If you are interested in "physically motivated" topologies for spacetimes in general relativity,
you might be interested in https://www.google.com/search?q=Fullwood-McCarthy+topology

Some old posts:
https://www.physicsforums.com/threa...dying-general-relativity.144202/#post-1167336

https://en.wikipedia.org/wiki/Causality_conditions#Strongly_causal (note the comment for Strongly Causal spacetimes)

When I was in graduate school, this was a topic of interest for me. It's now on the backburner.

6. Jan 9, 2017

### AlephClo

Rubi,
I have read the topolgical definition of neighborhoods and its equivalence to open sets, and this clarifies my question about physical meaning open sets. I will further read on Do Carmo.

Robphy,
You opened a new area of interest that I will explore.

Merci to both of you. AlephClo