How to Prove Compactness of Infinite Union of Line Segments in R^2?

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In summary, this conversation discusses two different mathematical problems and asks for help in solving them. The first problem involves proving that a set in R^n is open using a given definition of an open set. The second problem involves proving that an infinite union of line segments in R^n is compact using the Bolzano-Weierstrass Theorem. The conversation includes attempts at solving the problems and asks for clarification and help in understanding the steps and theorems needed to solve them.
  • #1
kingwinner
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1) Prove rigorously that S={(x,y) | 1< x^2 + y^2 <4} in R^n is open using the following definition of an open set:
A set S C R^n is "open" if for all x E S, there exists some r>0 s.t. all y E R^n satisfying |y-x|<r also belongs to S.


[My attempt:
Let x E S, r1 = 2 - |x|, r2= |x| - 1, r = min {r1,r2}
Let y E R^n s.t. |y-x|<r
=>|y-x|<r1 and |y-x|<r2

|y|<=|y-x|+|x| (triangle inequality) [<= means less than or equal to]
<r1+|x|=2-|x|+|x|=2
So |y|<2

Now if I can prove that |y|>1, then I am done (y belongs to S). However, I tried many different ways, but still unable to prove that |y|>1, what should I do? ]


2) Let Li denote the line segment in R^2 from the origin (0,0) to the point (1/i, sqrt(1/i) ) on the curve f(x)=sqrt x. Prove that the infinite union S=U(i=1 to infinity) Li is compact using the Bolzano-Weierstrass Theorem.

Bolzano-Weierstrass Theorem: Let S C R^n. Then S is compact (bounded and closed) iff every sequence of points in S has a convergent subsequence whose limit lies in S.


Any help, explanation, or hints? I am feeling totally blank on this question...


My textbook basically has no examples, so I don't know how to use the theorems and I am really frustrated when doing the exercises...

I hope that someone would be kind enough to help me out! Thank you!
 
Last edited:
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  • #2
kingwinner said:
1) Prove rigorously that S={(x,y) | 1< x^2 + y^2 <4} in R^n is open using the following definition of an open set:
A set S C R^n is "open" if for all x E S, there exists some r>0 s.t. all y E R^n satisfying |y-x|<r also belongs to S.


[My attempt:
Let x E S, r1 = 2 - |x|, r2= |x| - 1, r = min {r1,r2}
Let y E R^n s.t. |y-x|<r
=>|y-x|<r1 and |y-x|<r2

|y|<=|y-x|+|x| (triangle inequality) [<= means less than or equal to]
<r1+|x|=2-|x|+|x|=2
So |y|<2

Now if I can prove that |y|>1, then I am done (y belongs to S). However, I tried many different ways, but still unable to prove that |y|>1, what should I do? ]
Use the triangle inequality but the other way around.
Since |y- x|= |x- y|< r2, |x|< |x-y|+ |y|, |x|< r2+ |y|, |x|< |x|-1+ |y|, 1< |y|.


2) Let Li denote the line segment in R^n from the origin (0,0) to the point (1/i, sqrt(1/i) ) on the curve f(x)=sqrt x. Prove that the infinite union S=U(i=1 to infinity) Li is compact using the Bolzano-Weierstrass Theorem.

Bolzano-Weierstrass Theorem: Let S C R^n. Then S is compact (bounded and closed) iff every sequence of points in S has a convergent subsequence whose limit lies in S.


Any help, explanation, or hints? I am feeling totally blank on this question...


My textbook basically has no examples, so I don't know how to use the theorems and I am really frustrated when doing the exercises...

I hope that someone would be kind enough to help me out! Thank you!
Well, only the obvious. Since you are told to use the Bolzano-Weierstrasse theorem, Let {xn} be a sequence of points in S. Prove that it has a convergent subsequence! You might want to break it into parts: If there happen to be an infinite number of points on a single one of the line segments Li, can you prove that has a convergent subsequence? (Is each Li closed and bounded?) If not then there must be an infinite subsequnce on {ai} with each ai on a different Li. Can you prove that that sequence converges to 0?
 
  • #3
2) I actually have a worked example to a similar question, and they did the same as you suggested, but I don't understand the difference between the following 2 cases and I am not sure why we need 2 cases...can you please explain more on this part?

Case 1: If there happen to be an infinite number of points on a single one of the line segments L

Case 2: If not then there must be an infinite subsequnce on {ai} with each ai on a different Li

I have a test tomorrow and I am still stuck on these kinds of questions...

Thanks a lot for your help!
 
Last edited:
  • #4
kingwinner said:
2) I actually have a worked example to a similar question, and they did the same as you suggested, but I don't understand the difference between the following 2 cases and I am not sure why we need 2 cases...can you please explain more on this part?

Case 1: If there happen to be an infinite number of points on a single one of the line segments L
Yes, that's true. And what is their limit?

Case 2: If not then there must be an infinite subsequnce on {ai} with each ai on a different Li
No, I didn't say that. There must be an infinite sequence {ai} on ONE Li. What is its limit?

I have a test tomorrow and I am still stuck on these kinds of questions...

Thanks a lot for your help!
 
  • #5
Sorry, for question #2, I am completely lost now...

Can anyone explain this question from scratch and provide the steps to solve this problem? What theorems do I need to use in these steps?

Thanks!
 
  • #6
I am sure someone here knows how to solve this problem. Please help me...
 

Related to How to Prove Compactness of Infinite Union of Line Segments in R^2?

1. What is topology?

Topology is a branch of mathematics that studies the properties of geometric figures that are unchanged by stretching, bending, or twisting. It focuses on the spatial relationships and structures of objects rather than their precise measurements.

2. What is compactness in R^n?

Compactness is a property of a topological space that describes how closely packed and finite its points are. In R^n, a set is compact if it is both closed and bounded, meaning it contains all of its limit points and is contained within a finite region of space.

3. What is the difference between compact and connected sets?

A compact set is one that is both closed and bounded, while a connected set is one that cannot be separated into two non-empty disjoint sets. This means that a set can be compact without being connected, and vice versa.

4. How is compactness related to continuity?

In topology, a function is considered continuous if the preimage of any open set is an open set. A function is said to be compact if its image is a compact set. Therefore, compactness is related to continuity because a continuous function maps compact sets to compact sets.

5. What is the Heine-Borel theorem?

The Heine-Borel theorem states that in R^n, a set is compact if and only if it is closed and bounded. This theorem is fundamental in understanding the concept of compactness in topology and is often used in proofs and applications.

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