accumulation points

• Sep 23rd 2010, 09:10 PM
EmmWalfer
accumulation points
how can you prove this theorem from the book?

A finite set has no accumulation points.

I know that it's true, but fail to show it mathematically :(
• Sep 24th 2010, 06:13 AM
Ackbeet
Are you working in the real numbers? What ideas have you had so far?
• Sep 24th 2010, 11:03 AM
Danneedshelp
Quote:

Originally Posted by EmmWalfer
how can you prove this theorem from the book?

A finite set has no accumulation points.

I know that it's true, but fail to show it mathematically :(

Well, if you construct an epsilon neighborhood around a proposed limit point x in your finite set, then the intersection of your set with the epsilon neighborhood of x must contain points in your finite set other than x for x to be a limit point (accumulation point). If you can show this doesn't hold, then you are done.
• Sep 25th 2010, 05:26 AM
HallsofIvy
The definition of "accumulation point" is: p is an accumulation point of set A if and only if every neighborhood of p contains at least one point of A (other than p itself). Suppose A is finite. Then the set of all distances from p to points in A (other that p itself) is finite and so contains a smallest value. Take the radius of your neighborhood to be smaller than that value.

(I notice now, that is pretty much what Danneedshelp said!)
• Sep 25th 2010, 09:44 AM
Danneedshelp
Quote:

Originally Posted by HallsofIvy
The definition of "accumulation point" is: p is an accumulation point of set A if and only if every neighborhood of p contains at least one point of A (other than p itself). Suppose A is finite. Then the set of all distances from p to points in A (other that p itself) is finite and so contains a smallest value. Take the radius of your neighborhood to be smaller than that value.

(I notice now, that is pretty much what Danneedshelp said!)

I am not a 100% this correct, but I recall a lemma from my intro to real analysis book that stated something along to lines of: a point $x$ is a limit point (accumulation point) of a set $A$ iff for every $\epsilon>0$, the neighborhood $N_{\epsilon}(x)$ contains infinitly many points of the set $A$.

I think you would have to prove this to use it, but it answers your original question rather quickly.