
#1
Aug2011, 12:12 PM

P: 5

1. The problem statement, all variables and given/known data
I am not sure if I should post it under Calculus and Analysis but since it is for my Advanced Calculus class the I decided to do it anyways. If the metric space (S,d) is defined as S = R (set of real numbers) and d(x,y) = 0 if x=y and d(x,y) = 1 if x is not equal to y, find the limit points of A if A = Q (set of rational numbers). Based on my notes, the answer is Q but I don't seem to get it. Note: a point x element of S is a limit point of A if every open ball B(x,r) for r>0 (that is, an open ball or open interval that has center x and radius r) contains a point y element of A other than x. 2. Relevant equations 3. The attempt at a solution My answer is either {} null set or R. If I take a rational number x to be the center of an open interval, by property of real numbers, I can always find at least one rational number not equal to x that is within the open interval no matter how small or big the interval is. Therefore, all rational numbers are limit points and they are in the distance of 1 from each other provided that they are not equal. But following this logic also means that if the center is an irrational number say y, then I can always find a rational number within the open interval no matter how small or big the interval is. Therefore, R is the derived set or the set of limit points. On the other hand, I am also thinking that if the distance is 1 if the numbers are not equal and 0 if they are equal mean that all Q are isolated points and therefore the set of limit points is null set. 



#2
Aug2011, 12:51 PM

Mentor
P: 16,545

But here, the balls are a lot smaller. For example: B(x,1/2)={x}, so the singletons are open sets. What does that mean for the limit points? 



#3
Aug2011, 08:59 PM

P: 5

does that mean that the set of limit points of A is {} null set since the singletons do not have other elements other than x?



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