January 6th, 2017, 12:41 PM  #1 
Member Joined: Oct 2016 From: Arizona Posts: 60 Thanks: 15 Math Focus: Still deciding!  Quick question
In reference to 2.34 Theorem is $V_q$ supposed to be $V_p$?

January 6th, 2017, 01:05 PM  #2 
Senior Member Joined: Sep 2015 From: USA Posts: 1,763 Thanks: 905 
This proof is almost identical but a bit clearer. 
January 6th, 2017, 01:22 PM  #3  
Member Joined: Dec 2016 From: USA Posts: 46 Thanks: 11  Quote:
There is no such thing as $V_p$. The sets $V_q,W_q$ are defined for points $q \in K$. The proof in your book is valid. Last edited by quasi; January 6th, 2017 at 01:30 PM.  
January 6th, 2017, 01:26 PM  #4 
Member Joined: Oct 2016 From: Arizona Posts: 60 Thanks: 15 Math Focus: Still deciding! 
I've just never seen him refer to a neighborhood of $p$ using $V_q$ Thanks . 
January 6th, 2017, 01:28 PM  #5 
Member Joined: Oct 2016 From: Arizona Posts: 60 Thanks: 15 Math Focus: Still deciding! 
I realize that there is no such thing as $V_p$ but thanks for pointing that out anyways. I was not questioning the validity of the proof, just the notation. Last edited by ProofOfALifetime; January 6th, 2017 at 01:30 PM. 
January 6th, 2017, 01:37 PM  #6  
Member Joined: Dec 2016 From: USA Posts: 46 Thanks: 11  Quote:
But they're not just any old neigborhoods. The radii are defined so as to force them to be disjoint. For the proof in question, the point $p$ is fixed; the points $q$ vary (over the points of $K$). The neighborhoods $V_q,W_q$ are defined based on the variable point $q \in K$. Last edited by quasi; January 6th, 2017 at 01:49 PM.  
January 6th, 2017, 01:54 PM  #7 
Member Joined: Oct 2016 From: Arizona Posts: 60 Thanks: 15 Math Focus: Still deciding! 
Okay I get it. Thanks quasi. I get the whole radii thing. I get the proof I guess the notation confused me a little at first.


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