1. ## 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

2. Are you working in the real numbers? What ideas have you had so far?

3. 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.

4. 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!)

5. 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 $\displaystyle x$ is a limit point (accumulation point) of a set $\displaystyle A$ iff for every $\displaystyle \epsilon>0$, the neighborhood $\displaystyle N_{\epsilon}(x)$ contains infinitly many points of the set $\displaystyle A$.

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