# Contitunity on a Manifold

1. Jun 21, 2006

### pmb_phy

I've been reading the book "Geometrical Methods for Mathematical Physics" by Schutz. I can't understand/visualize the definition of contituity given on this page 7. I.e. where it states in the 3rd paragraph
I don't understand/can't vizualize this definition and reconcile it with the normal definition of contituity given in simple calculus. Can someone help me along here??

Pete

2. Jun 22, 2006

### AKG

If you have f : Rn -> Rm say, then f is continuous at x if for all e > 0, there exists d > 0 such that every point in the neighbourhood of x of radius d maps inside the neighbourhood of f(x) of radius e. Choosing an e is like choosing a neighbourhood of f(x). Choosing a d is like choosing an open set containing x whose image under f is inside that e-neighbourhood. In fact, choosing a d is not just "like" choosing an open set, it is indeed choosing an open set. So the basic e-d definition sort of says:

"A map f : Rn -> Rm is 'continuous' at x if any round open set of Rm centered at f(x) contains the image of a round open set centered at x"

It turns out that this is equivalent to:

"A map f : Rn -> Rm is 'continuous' at x if any open set of containing f(x) contains the image of an open set containing x"

This is not hard to believe, since any open set containing f(x) [or x] contains a round open set centered at f(x) [or x, respectively], and more obviously, every round open set centered at f(x) [or x] contains (in fact is) an open set containing f(x) [or x].

And this second formulation of the normal calculus definition is very much like the manifold definition you're looking at, so it should now be easier to reconcile the two.

3. Jun 23, 2006

### pmb_phy

Thanks for responding AKG. I have no problem understanding the definition as you have stated it. Inb fact if my text stated it like that then the meaning of the definition be intuitively obvious to me. Had my text defined continuity similarly to yours then I never would have posted this question because it would be rather trivial to understand/visualize it.

However, the definition given by Schutz does not require what the definition you posted requires. I.e. the definition
requires only that for any open set of N containing f(x) to contain an image in M. It could be any image including those images which are disjoint in M. Hence my problem with the definition as stated in Schutz

Pete

4. Jun 23, 2006

### AKG

The only significant difference between:

"A map f : Rn -> Rm is 'continuous' at x if any open set of containing f(x) contains the image of an open set containing x"

and

"A map f: M -> N is 'continuous' at x in M if any open set of N containing f(x) contains the image of an open set of M"

is that the second definition doesn't require that the open set of M in question need not contain x. However, this is wrong. Let M = N = R with the usual topology, and let f(x) = 0 if x is negative, and f(x) = 1 otherwise. You know that f is discontinuous at 0, however every open set (in N) containing f(0) = 1 does indeed contain the image under f of an open set (in M). Your definition, as stated, is wrong. If instead it said:

"A map f: M -> N is 'continuous' at x in M if any open set of N containing f(x) contains the image of an open set of M containing x"

then it would be correct, and easily reconcilable with the definition:

"A map f : Rn -> Rm is 'continuous' at x if any open set of containing f(x) contains the image of an open set containing x"

which is easily seen to be equivalent with the e-d definition.

5. Jun 23, 2006

### pmb_phy

I agree. If so then this is the problem I had understanding the definition given by Schutz.

By the way, why do you use "Rn -> Rm" rather than the one I used "M -> N"? These mean different things. E.g. "Rn -> Rm" cannot describe a map from a torus (M) to a sphere (N).

Pete

6. Jun 23, 2006

### mathwonk

from a simple minded point of view, to be an open set around x means to be "near" x. so the defn says that the values of f will be near f(p), if their inputs are near p.

and every point on a manifold has a nbhd that looks like R^n so there is essentially no loss of generality in defining it just for R^n.

7. Jun 23, 2006

### mathwonk

i.e. a point on a torus or a sphere has a disk like nbhjd so assume firat that there is some disk like nbhd of p that maps entirely into a given disklike nbhd of f(p), and then just apply the defn from R^2 to the restriction of the map to those discs.

or, easier, all manifolds can be embdded in some R^n and then jnust use the isial defon in R^n.

i.e. abstract treatments of manifolds often obscure the fact that a manifold can always be regarded as a certain nice subset of R^n. In fact essentially any compact manifold looks like just a level set of a smooth function from one R^n to another.

Last edited: Jun 23, 2006
8. Jun 23, 2006

### mathwonk

by the way pete, if you are a mathematical physicist, some posters in the thread "who wants to be a mathematician" under academic guidance, have been asking whether they should major in math or physicts to become one. what do you advise?

9. Jun 23, 2006

### pmb_phy

10. Jun 23, 2006

### pmb_phy

I had two majors in college, physics and math. Most of what I do when I'm working in physics is only mathematical so in that sense I guess you could say that I'm a mathematical physicist.

I recommend to your friend that he double major in physics and math as I did. This way if he wants to be a mathematician he can utilize his physics when he's working on mathematical problems. E.g. its nice to have solid examples of the math one is working with, especially in GR.

Pete

11. Jun 23, 2006

### pmb_phy

On the contrary, look at the definition. It states quite clearly that its speaking of "x in M".
I posted no definition. What I'm saying is that all maps do not map from Rn -> Rm for any two m, n.

E.g. the surface of a sphere has n = 2. This set is often labeled S2 because if you label it Rn then there are points in Rn which are not in S2.

Pete

12. Jun 23, 2006

### AKG

You were trying to reconcile the given definition of continuity with the basic one from elementary calculus. The "Rn -> Rm" is the basic definition.
That's not contrary to what I wrote. Compare:

"A map f: M -> N is 'continuous' at x in M if any open set of N containing f(x) contains the image of an open set of M"

which is wrong, with:

"A map f: M -> N is 'continuous' at x in M if any open set of N containing f(x) contains the image of an open set of M containing x"

which is right.
Yes you did, it's in your first post.
When did I ever say anything to the contrary? Do you not remember what your own thread is about? You posted the following definition:

"A map f:M -> N is 'continuous' at x in M if any open set of N containing f(x) contains the image of an open set of M."

and asked how to reconcile this with the "normal definition of continuity given in simple calculus", which I would have to assume is the definition using Rn and Rm.

I did the following:

a) showed that the definition you gave was wrong with an example (the function that is 0 for negative reals and 1 otherwise)
b) suggested the proper definition
A map f:M -> N is 'continuous' at x in M if any open set of N containing f(x) contains the image of an open set of M containing x
c) gave the simple e-d definition of continuity from simple calculus
d) gave an equivalent definition of this "simple continuity" equivalent to the e-d definition, but in terms of open sets
e) suggested that this revised definition of "simple continuity" in terms of open sets is easily reconcilable with the proper definition of continuity for a map of manifolds

13. Jun 23, 2006

### shmoe

According to googles book search, the 1980 version of Schultz gives the definition AKG wants:

"A map f:M->N is continuous at x in M if any open set of N containing f(x) contains the image of an open set of M containing x."

That's the full sentence. Is that not what your book has? If you still have a problem with this version, your exercise is to prove it's equivalent to the old epsilon/delta one.

14. Jun 24, 2006

### pmb_phy

That is not a definition.

Pete

15. Jun 24, 2006

### AKG

Wow, I've had people in philosophical discussions debate what the proper definition of something is. I don't know if I've had anyone asking a math help question tell me that the definition I'm giving isn't even a definition.

16. Jun 24, 2006

### pmb_phy

My appolgies. For some reason I thought that you were speaking about something else.

Pete

Last edited: Jun 24, 2006
17. Jun 24, 2006

### mathwonk

yes pete I was talking local topology in ym first suggestion using disc nbhds, but the point is that continuity is essentially a local concept, so that is actually ok.

my remark in post 7 however globalized it completely by observing that all manifolds are realizable as just subsets of R^n, and then one can give the same definitrion as in R^n. I.e. if M and N are embedded in R^n, then f:M-->N is contin. at a point p of M if for every epsilon>0 there is a delta>0 such that whenever q is a point of M closer to p than delta, the f(q) is closer to f(p) than epsilon.

ona can also make my local version "global" in the sense of referring it to all of R^2, since an open disc is homeomorphic to all of R^2, by sending (r,theta) to (tan(rpi/2),theta).

Last edited: Jun 24, 2006
18. Jun 24, 2006

### pmb_phy

That isn't true. Consider the manifold M which is the surface of a sphere. This is sometimes labeled S^2. However S^2 does not have the global structure of R^2 and therefore is not identical to it. E.g. geodesics on S^2 are closed while geodesics in R^2 are not. I see no reason to assume that all manifolds are just subsets of R^n. What is your justification of this assertion? Thanks.

Pete

19. Jun 24, 2006

### Hurkyl

Staff Emeritus
But S^2 is realizable as a subset of R^3. There is a theorem... I think I have it right... that any n-dimensional manifold can be embedded in R^(2n).

(Incidentally, we can't talk about geodesics until we put a bunch of extra structure on our manifolds)

20. Jun 24, 2006

### pmb_phy

That requires an embedding of S^2 in S^3 which is not identical to a mapping of R^3 into R^n.

Pete