Rudin Thm 2.41: E Closed & Bounded if Every Inf Subset Has a Lmt Pt

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I am currently reading Principles of Mathematical Analysis by Walter Rudin. I am a bit confused with theorem 2.41. He is trying to show at one point that if E is a set in ℝk and if every infinite subset of E has a limit point in E, then E is closed and bounded.

The proof starts by assuming that E is not bounded. He then says that if this is the case, then E contains points xn such that |xn| > n for each positive integer n. He then constructs a set S that contains all these points xn. Next he says "The set S ... is infinite and clearly has no limit point in ℝk..."

I don't see how it is obvious that there is no limit point in ℝk.
 
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Dschumanji said:
I don't see how it is obvious that there is no limit point in ℝk.

Suppose S has a limit point and consider the ball of radius 1 about this point. Does it contain infinitely many elements of S?
 
jgens said:
Suppose S has a limit point and consider the ball of radius 1 about this point. Does it contain infinitely many elements of S?
I see where you are heading. I came up with a proof using a similar idea:

Let p \in ℝk and suppose that it is a limit point of S.

There exists a positive integer n such that n ≤ |p| < n+1. Let C be the set of all y \in ℝk such that |y| < n+1. C is an open set and p \in C, so there exists a neighborhood N around p such that N \subset C.

Due to the construction of S, there can be at most n points of S in C and thus there can be at most n points in N. Since n is a finite number, p cannot be a limit point (theorem 2.20 of Rudin).

This shows there are no limit points in ℝk and thus no limit points in E. Therefore E must be bounded.

I believe that this proof is correct, but it hardly seems obvious.
 
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Dschumanji said:
I believe that this proof is correct, but it hardly seems obvious.

Your proof is fine. Try not to worry about people like Rudin claiming that things are obvious. The more you learn and the more you get a feel for a particular subject, the more obvious things become.
 
jgens said:
Your proof is fine. Try not to worry about people like Rudin claiming that things are obvious. The more you learn and the more you get a feel for a particular subject, the more obvious things become.
Thanks for the support!

It seems Rudin likes to use proof by intimidation more than anything. :-p
 
Dschumanji said:
It seems Rudin likes to use proof by intimidation more than anything.

This is actually something I like about Rudin. I always took it as a challenge to understand the material well enough to get why Rudin claimed something was obvious and this helped me learn the material properly. Just keep filling in all the details he leaves out and it will all seem obvious to you soon enough.
 
A sphere as topological manifold can be defined by gluing together the boundary of two disk. Basically one starts assigning each disk the subspace topology from ##\mathbb R^2## and then taking the quotient topology obtained by gluing their boundaries. Starting from the above definition of 2-sphere as topological manifold, shows that it is homeomorphic to the "embedded" sphere understood as subset of ##\mathbb R^3## in the subspace topology.
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