
#1
Sep809, 08:13 PM

P: 92

Show that U(x_{0}, 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 basicbasic 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. 



#2
Sep809, 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.




#3
Sep809, 09:21 PM

P: 851

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. 



#4
Sep809, 09:47 PM

P: 92

reading Analysis on Manifolds by Munkres 



#5
Sep909, 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 



#6
Sep909, 05:52 AM

P: 876





#7
Sep909, 10:05 AM

P: 810





#8
Sep909, 10:31 AM

P: 92

I'll work through it a bit today in my free time and see what I come up with after reading the previous posts. 



#9
Sep909, 12:01 PM

P: 360

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 



#10
Sep909, 04:39 PM

P: 244

Petek gave good advice. 



#11
Sep909, 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. 



#12
Sep1309, 03:12 PM

P: 92

U(x_{0},ε)={xd(x,x_{0})<ε}
Let y be in U(x_{0},ε) such that U(y,ε'(y)) is contained in U(x_{0},ε). Hence d(y,x_{0})+ε'(y)≤ε. Let z be in U(y,ε'(y)), so d(z,y)<ε'(y). Hence d(z,x_{0})≤d(z,y)+d(y,x_{0})<ε'(y)+εε'(y)=ε Thus d(z,x_{0})<ε for all z in U(y,ε'(y)) that is contained in U(x_{0},ε) Thus U(x_{0},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. 



#13
Sep1409, 03:15 PM

P: 360

Hint: let ε'(y) = ε  d(x_{0}, y). Show that if z is contained in U(y,ε'(y)), then z is contained in U(x_{0},ε). Conclude that U(y,ε'(y)) is contained in U(x_{0},ε). Petek 



#14
Sep1509, 05:14 PM

P: 92

Prove that U(x_{0},ε) is an open set.
Proof: U(x_{0},ε)={xd(x,x_{0})<ε} Let y be in U(x_{0},ε) so there exists an ε'(y)>0 such that ε'(y) + d(y,x_{0}) = ε. Let z be in U(y,ε'(y)), so d(z,y)<ε'(y). Hence d(z,x_{0})≤d(z,y)+d(y,x_{0})<ε'(y)+εε'(y)=ε Thus d(z,x_{0})<ε for all z in U(y,ε'(y)). ThusU(y,ε'(y)) that is contained in U(x_{0},ε). Thus U(x_{0},ε) is open. QED. Thanks for the help Petek. Is this better now or have a failed to sufficiently prove something? 


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