reading Analysis on Manifolds by Munkres


by Matthollyw00d
Tags: analysis, manifolds, munkres, reading
Matthollyw00d
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#1
Sep8-09, 08:13 PM
P: 92
Show that U(x0, ε) is an open set.

I'm reading Analysis on Manifolds by Munkres. This question is in the review on Topology section. And I've just recently been introduced to basic-basic topology from Principles of Mathematical Analysis by Rudin.

I'm not really certain where to begin on this, nor can I find a sufficient definition of open set to somewhat utilize. Any help on where to begin would be appreciated. This doesn't seem like it would be difficult or long, just kind of in unfamiliar ground here.

Thanks in advanced.
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slider142
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#2
Sep8-09, 08:17 PM
P: 876
What is the topology of the set that U is a subset of? A subset is open if and only if it is an element of the topology; without a topology, or a basis for the topology, the quality of being open is undefined.
╔(σ_σ)╝
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#3
Sep8-09, 09:21 PM
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P: 851
I'm not really certain where to begin on this, nor can I find a sufficient definition of open set to somewhat utilize. Any help on where to begin would be appreciated.
I can not really help with the question since I haven't even taking a formal course in Analysis.
But a definition I got from Introduction to Real Analysis ( a book i "tried" to read over the summer ) says if the epsilon neighborhood of a point Xnot is contained in a set S then S is the neighborhood of Xnot and Xnot is an interior point in S. If every point is S is an interior point, then S is open.

Basically if you can find the epsilon neighborhood of any point in the Set S then S is open.

I don't know what else to say that would help,partly because I'm unsure of your notation.

Is U the union sign?

Anyway I assume that with the definition provided above you can show that your set is open. I guess a prove by contradiction would be appropriate, you can show that there is no point is your set that does not have an epsilon neighborhood.

Matthollyw00d
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#4
Sep8-09, 09:47 PM
P: 92

reading Analysis on Manifolds by Munkres


Quote Quote by slider142 View Post
What is the topology of the set that U is a subset of? A subset is open if and only if it is an element of the topology; without a topology, or a basis for the topology, the quality of being open is undefined.
I believe it to be general for Rn. But there is nothing else in the question, except before the problems it states: "Throughout, let X be a metric space with metric d."
Quote Quote by ╔(σ_σ)╝ View Post
I don't know what else to say that would help,partly because I'm unsure of your notation.

Is U the union sign?

Anyway I assume that with the definition provided above you can show that your set is open. I guess a prove by contradiction would be appropriate, you can show that there is no point is your set that does not have an epsilon neighborhood.
U is not the union sign, U(x0, ε) = {x | d(x, x0) < ε}
snipez90
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#5
Sep9-09, 03:23 AM
P: 1,106
OK, so it seems like you're trying to prove that the open ball about x_0 of radius epsilon is open, where the open ball is a subset of (X, d). There are probably a few ways to do this, but the way that I've learned recently is as follows:

-show that the open ball is a neighborhood of each of its points
-an subset of a metric space can be defined as open if it is a neighborhood of each of its points
slider142
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#6
Sep9-09, 05:52 AM
P: 876
Quote Quote by Matthollyw00d View Post
I believe it to be general for Rn. But there is nothing else in the question, except before the problems it states: "Throughout, let X be a metric space with metric d."

U is not the union sign, U(x0, ε) = {x | d(x, x0) < ε}
That means it is assuming the metric topology on X with metric d, which means the basis for the topology is the set of all balls, each of the form B(x[sub]0[/xub], r) = {x | d(x, x0) < r}. Your task is then to show that it is either a union of elements of this basis, or if you have covered homeomorphisms, that it is homeomorphic to such a union.
Tac-Tics
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#7
Sep9-09, 10:05 AM
P: 810
Quote Quote by slider142 View Post
That means it is assuming the metric topology on X with metric d, which means the basis for the topology is the set of all balls, each of the form B(x[sub]0[/xub], r) = {x | d(x, x0) < r}. Your task is then to show that it is either a union of elements of this basis, or if you have covered homeomorphisms, that it is homeomorphic to such a union.
This seems a bit tautological, though. In the metric topology, open balls are open.... but we're asked to prove that an open by is open? It's a strange problem, if in fact that is how it is stated.
Matthollyw00d
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#8
Sep9-09, 10:31 AM
P: 92
Quote Quote by Tac-Tics View Post
This seems a bit tautological, though. In the metric topology, open balls are open.... but we're asked to prove that an open by is open? It's a strange problem, if in fact that is how it is stated.
That is all that is written in book, verbatim.

I'll work through it a bit today in my free time and see what I come up with after reading the previous posts.
Petek
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#9
Sep9-09, 12:01 PM
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Let y be any point in [itex]U(x_0, \epsilon)[/itex]. You need to show that there exists a [itex]\delta[/itex] > 0 such that [itex]U(y, \delta) \subset U(x_0, \epsilon) [/itex]. (This is just restating what snipez90 already said.)

Hint: The triangle inequality might come in handy here.

HTH

Petek
aPhilosopher
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#10
Sep9-09, 04:39 PM
P: 244
Quote Quote by Tac-Tics View Post
This seems a bit tautological, though. In the metric topology, open balls are open.... but we're asked to prove that an open by is open? It's a strange problem, if in fact that is how it is stated.
Munkres doesn't spend enough time on the topology to define a basis so, in this context, it's not quite as tautological as it seems. It really only addresses manifolds as subsets of Rn and only covers "real" manifolds with a sketchy last chapter, "Life outside Rn". On the whole though, It's probably the best introduction to the ideas that I've seen. It's a great book.

Petek gave good advice.
Matthollyw00d
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#11
Sep9-09, 04:52 PM
P: 92
Yah, I think I have it after further thinking about it. Once I formalize it on paper, I'll post it here and see if I have it well understood.

And I do agree that it is a great book. I'm using it along side Calculus on Manifolds by Spivak for my Differential Geometry course.
Matthollyw00d
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#12
Sep13-09, 03:12 PM
P: 92
U(x0,ε)={x|d(x,x0)<ε}
Let y be in U(x0,ε) such that U(y,ε'(y)) is contained in U(x0,ε).
Hence d(y,x0)+ε'(y)≤ε. Let z be in U(y,ε'(y)), so d(z,y)<ε'(y).
Hence d(z,x0)≤d(z,y)+d(y,x0)<ε'(y)+ε-ε'(y)=ε
Thus d(z,x0)<ε for all z in U(y,ε'(y)) that is contained in U(x0,ε)
Thus U(x0,ε) is open.
EndofProof.

This is what I got for the proof, I don't see anything wrong with it, just tossing it out here to make sure everything is correct, since I'm still not completely confident when dealing with open sets. Just looking for confirmation or pointing out something that is incorrect, thanks.
Petek
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#13
Sep14-09, 03:15 PM
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P: 360
Quote Quote by Matthollyw00d View Post
U(x0,ε)={x|d(x,x0)<ε}
Let y be in U(x0,ε) such that U(y,ε'(y)) is contained in U(x0,ε).
You are assuming that such a U(y,ε'(y)) exists. Instead, you need to show that there exists an ε'(y) > 0 such that U(y,ε'(y)) is contained in U(x0,ε). That's Munkres' definition of a set being open.

Hint: let ε'(y) = ε - d(x0, y). Show that if z is contained in U(y,ε'(y)), then z is contained in U(x0,ε). Conclude that U(y,ε'(y)) is contained in U(x0,ε).

Petek

Hence d(y,x0)+ε'(y)≤ε. Let z be in U(y,ε'(y)), so d(z,y)<ε'(y).
Hence d(z,x0)≤d(z,y)+d(y,x0)<ε'(y)+ε-ε'(y)=ε
Thus d(z,x0)<ε for all z in U(y,ε'(y)) that is contained in U(x0,ε)
Thus U(x0,ε) is open.
EndofProof.

This is what I got for the proof, I don't see anything wrong with it, just tossing it out here to make sure everything is correct, since I'm still not completely confident when dealing with open sets. Just looking for confirmation or pointing out something that is incorrect, thanks.
Matthollyw00d
Matthollyw00d is offline
#14
Sep15-09, 05:14 PM
P: 92
Prove that U(x0,ε) is an open set.

Proof:

U(x0,ε)={x|d(x,x0)<ε}
Let y be in U(x0,ε) so there exists an ε'(y)>0 such that ε'(y) + d(y,x0) = ε.
Let z be in U(y,ε'(y)), so d(z,y)<ε'(y).
Hence d(z,x0)≤d(z,y)+d(y,x0)<ε'(y)+ε-ε'(y)=ε
Thus d(z,x0)<ε for all z in U(y,ε'(y)).
ThusU(y,ε'(y)) that is contained in U(x0,ε).
Thus U(x0,ε) is open.
QED.

Thanks for the help Petek. Is this better now or have a failed to sufficiently prove something?
Petek
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#15
Sep15-09, 08:06 PM
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P: 360
Looks good!

Petek


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