Contitunity on a Manifold

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
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.
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

Related Differential Geometry News on Phys.org

AKG

Homework Helper
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.

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
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.
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

AKG

Homework Helper
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.

pmb_phy

AKG said:
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.
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

mathwonk

Homework Helper
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.

mathwonk

Homework Helper
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:

mathwonk

Homework Helper
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?

pmb_phy

mathwonk said:
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. That refers only to neighborhoods and not to the entire set. I was speaking of sets in there entirety. E.g. A disk is a subset of R2 but is not identical to R2. It appears from what you posted that you're refering to local topology rather than global topology, correct?

Pete

pmb_phy

mathwonk said:
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?
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

pmb_phy

AKG said:
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.
On the contrary, look at the definition. It states quite clearly that its speaking of "x in M".
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.
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

AKG

Homework Helper
pmb_phy said:
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
You were trying to reconcile the given definition of continuity with the basic one from elementary calculus. The "Rn -> Rm" is the basic definition.
On the contrary, look at the definition. It states quite clearly that its speaking of "x in M".
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.
I posted no definition.
Yes you did, it's in your first post.
What I'm saying is that all maps do not map from Rn -> Rm for any two m, n.
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

shmoe

Homework Helper
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.

pmb_phy

AKG 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"
...
which is right.
That is not a definition.

Pete

AKG

Homework Helper
pmb_phy said:
That is not a definition.
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.

pmb_phy

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

Pete

Last edited:

mathwonk

Homework Helper
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:

pmb_phy

mathwonk said:
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.
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

Hurkyl

Staff Emeritus
Gold Member
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)

pmb_phy

Hurkyl said:
But S^2 is realizable as a subset of R^3.
That requires an embedding of S^2 in S^3 which is not identical to a mapping of R^3 into R^n.

Pete

Perturbation

pmb_phy said:
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
If you want to note that the two do not have the same global topology without having to invoke some additional structure required for geodesics, namely a metric, you can turn to the Borsuk-Ulam theorem in two dimensions, which states there's no homeomorphism between S^2 and R^2, so they have different global topologies. Locally though, which I think he may have been referring to, any manifold is homeomorphic to a subset of R^n by definition.

Last edited:

mathwonk

Homework Helper
this is a famous theorem called whitneys embedding theorem proved probably in the 1940's. let me see if i can make up a proof for the compact case off the top of my head. take any point p of M, say a 2 manifold. Then p has a disc like nbhd so there is a map of that disc nbhd taking the interior onto the complement of the north pole in some sophere in R^3. Then just extend continuously so that all the rest of M goes to that northpole.

now we have a map from M to R^3 which embeds the interior of a nbhd of p, but does not embed the rest of M. Now do this for an open nbhd of every point and pick a finite subcover of M by the interiors of these nbhds.

That emans we have chosen N points pi of M and a nbhd of each one, and and N embeddings, one of each of these nbhds into R^3. Moreover these N nbhds cover M.

Then take the product of these N functions, getting a map from M to R^(3N). I claim this embeds M into R^3N, not of cousre as an open subset, but as a closed subet, much as the sphere is naturally embedded as a closed subset of R^3.

I have to prove the map is injective.

Given any two points x,y at all of M, if they lie in one nbhd of one of the points pi, then the ith corrdinate map fi separates them, since the ith map fi is injective on that whole nbhd.

If x is in the interior of some nbhd, and y is not, then the ith map sends x to the complement of the northpole on the ith sphere, and sends y to the north pole, thus x,y, are still separated by the ith coordinate fucntion of the map f = f1xf2x...xfN.

WE cfan also arrange for our disc nbhd to map with injective deriavtive on the interior of each nbhd, hence actually getting an embedding onto a differentiable submanifold of R^N.

Notice in my definition of continuity for a subset of R^n I did nort require the subset to be open, since I sneakily said:”for every point q WHICH IS ALSO IN M, and which is within delta of p...”

after embedding a 2 dimensional M into some large dimensional R^N, one can then project down easily until it is actually differentiably embedded in R^5. Whitney showed with more difficulty one can even get it into R^4.

In fact since one knows what all the compact two manifodls are, one easil;y see that all orientable ones actually embed in R^3, since they are just toruses with more holes.

pmb_phy

This thread has gone off on a tangent. I didn't mean it to. I was merely curious as to why AKG wrote what he did. I didn't want to debate this point. In any case I now understand why AKG did that so that ends it for me. Thanks everyone.

Pete

mathwonk

Homework Helper
Well I respect that you are satisfied, but t does not seem at all tangential to me. it is directly related to answering the question of the connection between continuity on a manifold and continuity in R^n. namely they are exactly the same for embedded manifolds, and essentially the same locally for non embeddded ones. and all manifolds can be considered as embedded.

you can make your life much easier by knowing that (paracompact) manifolds are all embeddable. so the point is that you have nothing new to learn as far as continuity on manifold, but some books may still make it look strange by trying hard.

the point some posters above were missing is that the embedded manifold does not embed in an R^n of the same dimension, but one of higher dimension, i.e. twice the dimension suffices. As I said much earlier, the general picture of an n manifold is essentially as a zero set of a smooth map from R^2n to R^n.

Last edited:

pmb_phy

mathwonk - In the past I used to continue to participate in conversations such as this one which is an offshoot of the original topic which arose from asking AKG what he meant. Those past conversations in other boards/newsgroups almost always ended up in flame wars or heated arguements. It almost always contributed to large increases in stress levels and made my life more miserable than it already is. I therefore tend to bow out when I see myself participating in such a capacity and not for lack of input.

One last comment for in this thread. When a chart is described using the notation f: M -> N it means that every point in M is mapped to a point in N. When the notation f: Rn -> Rm is used then the idea behind the notation is destroyed since not all of Rn is mapped to N, only a subset of Rn. This is probably the strongest reason I have for disagreeing with the notation f: Rn -> Rm as a substitute notation for f: M -> N.

Pete

Last edited:

Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving