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So, I'm working on Topology again! Oh boy! I feel bad about vanishing from the last thread I posted. I made a ladyfriend and she ate all my time. Anyways, me and a friend are stuck on this problem:

Deﬁnition. Let M be an m-manifold.
i) A quasi-linear path in M is a path f : [a, b] → M for which there is an open set U containing f([a, b])
and a homeomorphism h from U to a convex open set in R
m
such that h◦ f is linear (that is, each coordinate
of h ◦ f is a linear function).
ii) A piecewise quasi-linear path in M is a path f : [a, b] → M for which there is a ﬁnite partition of [a, b]
into subintervals such that the restriction of f to each subinterval of the partition is quasi-linear

C) Prove: a piecewise quasi-linear path in an m-manifold with m > 1 cannot be onto. (Hint: ﬁrst show that
a ﬁnite intersection of dense open sets in any topological space is dense.

We can prove that it doesn't work if the path is quasi-linear, but induction seems to fail. Currently we are looking at the hint. Any help would be greatly appreciated.
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Marbas

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Marbas wrote:We can prove that it doesn't work if the path is quasi-linear, but induction seems to fail. Currently we are looking at the hint. Any help would be greatly appreciated.

How are you proving it doesn't work if the path is quasi-linear? Can you prove that if f is quasi-linear, then the complement of f([a,b]) is dense-open? The hint seems to be suggesting the following strategy:

1) Prove that if f is quasi-linear, then the complement of f([a,b]) is dense-open.
2) Prove that a finite intersection of dense-open sets is dense-open.
3) Conclude that if is piecewise quasi-linear, then the complement of f([a,b]) is dense-open, and hence nonempty.

Do you know how to do 1? Do you know how to do 2? Do you know how to do 3, given 1&2?
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How are you proving it doesn't work if the path is quasi-linear? Can you prove that if f is quasi-linear, then the complement of f([a,b]) is dense-open? The hint seems to be suggesting the following strategy:

It was mostly my roommate's idea, it was some goofy thing that showed that M must then be globally homeomorphic to R^M. Or some such. We actually threw it out for something much more like what you suggested. We've made quite a bit of progress.

What we currently have is that we assume that f is onto and quasi-linear. Then, there exists a finite number, n, of U_i open in M such that h_i(U_i) is a homeomorphism, and the union of all these U_i is M, since f is onto. And also that, for each of these U_i, there exists [a_i, a_i+1] such that f( [a_i, a_i+1]) :=S_i is contained in U_i, and h_i(S_i) is a linear function. Then, since h_i(S_i) is a closed set with empty interior contained inside of an open set, every point of h_i(S_i) is a boundary point(Hand-wavey!), and so we have that X-h_i(S_i) is open and is dense in X, where X is the union of all h_i(U_i)'s.

Now I'm just sort of staring at the intersection of all X-h_i(S_i). This should be dense open in X right? So if it is, then what?

I'm kind of stuck now.

Actually, do I really need to do this by contradiction?

Edit: Looking at Skeptic's suggestion, I've got 1 & 2, now I just need 3.

Edit 2: So, I think maybe I got it. If f_i is the restriction of f to [a_i, a_i+1], then f_i is quasi-linear. Thus there exists some U_i and h_i s.t U_i is open in M and h_i is a homeomorphism with h_i(U_i) :=V_i open and convex, and h_i(f_i) := S_i is linear. Then, V_i - S_i is dense and open in V_i.Then, letting Y be the union of all U_i's, h_i^-1(V_i-S_i) := W_i is dense and open in U_i, and thus is dense and open in Y. Then, Y- (Union of all W_i) is the intersection of all Y-W_i, and thus is a finite intersection of open dense sets. And thus, it is an open dense set in Y, and is thus non-empty.

Maaaybe? Hmm...
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Marbas

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Marbas wrote:Then, letting Y be the union of all U_i's, h_i^-1(V_i-S_i) := W_i is dense and open in U_i, and thus is dense and open in Y.

The bolded statement is false. Draw some pictures! This is a topology class, so if you're not drawing pictures, you're doing something wrong.

A nice way to draw pictures is to put yourself in the simplest possible situation. Imagine M=R2, and f is actually piecewise linear (so certainly piecewise quasi-linear) with two linear pieces, f1 and f2. Then U1 and U2 might be overlapping circles (neither of which contains the other) such that Ui contains the range of fi. (Since M=R2, Ui is convex, and f is piecewise linear for our picture, we can let hi be the identity restricted to Ui.) In this picture, W1 is dense in U1 (and open), and similarly W2 is dense in U2, but W1 will not be dense in Y, as the closure of U1 will be a single circle, while Y is two overlapping circles.

...thus is a finite intersection of open dense sets. And thus, it is an open dense set in Y, and is thus non-empty.

In order to conclude this, you still need to prove the hint, that a finite intersection of open dense sets is dense.
Last edited by skeptical scientist on Mon Nov 21, 2011 2:53 am UTC, edited 1 time in total.
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skeptical scientist wrote:
Marbas wrote:Then, letting Y be the union of all U_i's, h_i^-1(V_i-S_i) := W_i is dense and open in U_i, and thus is dense and open in Y.

The bolded statement is false. Draw some pictures! Imagine M=R2, and f is actually piecewise linear (so certainly piecewise quasi-linear) with two linear pieces, f1 and f2. Then U1 and U2 might be overlapping circles (neither of which contains the other!) such that Ui contains the range of fi. (This situation is nice to understand, as in this case we can actually take the homeomorphisms hi to be the identity restricted to Ui, which won't be the case in general.) In this picture, W1 is dense in U1 (and open), and similarly W2 is dense in U2, but W1 will not be dense in Y, as the closure of U1 will be a single circle, while Y is two overlapping circles.

Very much my bad. I think that statement was actually supposed to be W_i is dense and open in U_i. Thus Union of all U_j s.t j != i, union with W_i is dense and open in Y.

In order to conclude this, you still need to prove the hint, that a finite intersection of open dense sets is dense.

Yeah, I feel that third part is far harder than that. So I've mainly been trying to get that out of the way.
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Marbas

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Marbas wrote:Very much my bad. I think that statement was actually supposed to be W_i is dense and open in U_i. Thus Union of all U_j s.t j != i, union with W_i is dense and open in Y.

This is true, but you don't want this, because you want this intersection of the sets for each i to give you Y - range(f), which may not be the case for the sets you just defined. What you really want is Y - range(fi), because the intersection over all i of Y - range(fi) is Y - range(f) by de Morgan's laws.

Marbas wrote:
In order to conclude this, you still need to prove the hint, that a finite intersection of open dense sets is dense.

Yeah, I feel that third part is far harder than that. So I've mainly been trying to get that out of the way.

That's fine, as long as you don't forget that part. Anyways, I feel that for the steps 1-3 I suggested, step 3 should be easy. However, you seem to be trying to combine 1 and 3, jumping directly to proving things about piecewise quasi-linear functions (rather than simply quasi-linear functions). This is fine, if you can do it, but following the steps I suggested might be simpler.
I'm looking forward to the day when the SNES emulator on my computer works by emulating the elementary particles in an actual, physical box with Nintendo stamped on the side.

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skeptical scientist wrote:
Marbas wrote:Very much my bad. I think that statement was actually supposed to be W_i is dense and open in U_i. Thus Union of all U_j s.t j != i, union with W_i is dense and open in Y.

This is true, but you don't want this, because you want this intersection of the sets for each i to give you Y - range(f), which may not be the case for the sets you just defined. What you really want is Y - range(fi), because the intersection over all i of Y - range(fi) is Y - range(f) by de Morgan's laws.

Crudmuffins. So it is not true that h_i^-1(V_i-S_i) = h_i^-1(V_i) - h_i^-1(S_i) = U_i - f_i?

However, you seem to be trying to combine 1 and 3, jumping directly to proving things about piecewise quasi-linear functions (rather than simply quasi-linear functions). This is fine, if you can do it, but following the steps I suggested might be simpler.

Hmm. Maybe I should go back to the drawing board then. I mean, I feel like I've almost got it. But this road-block seems a bit troubling.

Edit: Can I show that the image of a piecewise quasi-linear function under all these "h_i's" is homeomorphic to the image of a quasi-linear one under some "h"?

Actually I'm not sure how much that would help.

Edit 2: Oops, S_i was supposed to be f([a_i,a_i+1])
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Marbas

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Marbas wrote:Crudmuffins. So it is not true that h_i^-1(V_i-S_i) = h_i^-1(V_i) - h_i^-1(S_i) = U_i - f_i?

No, that part is true (since hi-1 is one-to-one). But (U1 - f1) U U2 may not be the same as (U1 U U2) - f1. The former is what you get when you take W1 U U2, and the latter is Y - f1, which is what you want.

Technically all of these instances of fi should really be range(fi), since fi is a function, not a set. But it's pretty standard in topology to confuse a path with its range.

Edit: Can I show that the image of a piecewise quasi-linear function under all these "h_i's" is homeomorphic to the image of a quasi-linear one under some "h"?

No. A triangle is piecewise quasi-linear (in fact, piecewise linear), but no homeomorphism will ever turn it into a single line segment.
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No, that part is true (since hi-1 is one-to-one). But (U1 - f1) U U2 may not be the same as (U1 U U2) - f1. The former is what you get when you take W1 U U2, and the latter is Y - f1, which is what you want.

Oooh. That is a problem then. I can't seem to find a way to somehow join up all the homeomorphisms to get one one-to-one function either.
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Marbas

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Just read about half of this thread and without putting too much thought into it... have you tried showing something slightly stronger, such as given a quasilinear path, there is some neighborhood that the path never enters?
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z4lis

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z4lis wrote:Just read about half of this thread and without putting too much thought into it... have you tried showing something slightly stronger, such as given a quasilinear path, there is some neighborhood that the path never enters?

I have not, actually. Although I don't know how I'd start doing that.
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Marbas

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Marbas wrote:I have not, actually. Although I don't know how I'd start doing that.

It might not even work. It's just where I'd probably start heading if I was given the problem. Not put much thought into it.
What they (mathematicians) define as interesting depends on their particular field of study; mathematical anaylsts find pain and extreme confusion interesting, whereas geometers are interested in beauty.

z4lis

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z4lis wrote:
Marbas wrote:I have not, actually. Although I don't know how I'd start doing that.

It might not even work. It's just where I'd probably start heading if I was given the problem. Not put much thought into it.

Ahh.

I wonder what happens if I subtract S_i from every V_i. I think that should work.
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Marbas

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The more I've thought about it, the more I think the idea will go through, and inductively, to boot. Try showing that given a quasilinear path, there is some open set U of the manifold such that the quasilinear path never enters it by induction.
What they (mathematicians) define as interesting depends on their particular field of study; mathematical anaylsts find pain and extreme confusion interesting, whereas geometers are interested in beauty.

z4lis

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z4lis wrote:Just read about half of this thread and without putting too much thought into it... have you tried showing something slightly stronger, such as given a quasilinear path, there is some neighborhood that the path never enters?

The proof sketched by my steps 1-3 (which is the one suggested by the hint given in the problem) already shows that the complement of a quasilinear path is dense-open.
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Aaaaaaaaaaaaand done. I think. Thanks a lot guys, you have been very helpful.

The proof sketched by my steps 1-3 (which is the one suggested by the hint given in the problem) already shows that the complement of a quasilinear path is dense-open.

After working through the proof, this does in fact become quite apparent
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Marbas

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