Statement: A non-empty set has no accumulation points iff all points are isolated.
So far I have said to Let $S= \{x_1,x_2,…,x_n\}$, such that $S$ has no accumulation points.
Then $\exists$ $\epsilon<|x-x_i|$ which by this I mean that since $S$ is finite, there will always be some $\epsilon$-neighborhood around an arbitrary $x\in S$ where $\epsilon <$ the smallest distance between $2$ points in $S$ which should show that all points are isolated.
If that portion is correct, how do I go about proving the other direction?
From my book, the definition of an isolated point just says "a point $x$ is an isolated point of $S$ if $x\in S$ and $x$ is not an accumulation point of $S$"
Is the answer then simply to say since all points are isolated, they are not accumulation points and therefore $S$ has no accumulation points?
Best Answer
Firstly in the statement given it is simply said "A non-empty set", therefore the set can be infinite or finite as well. So if you are indexing the elements of set up to $n$ elements it implies that set is finite. Then you may have to deal with infinite case separately.
Consider the set $S = \{\frac{1}{n}: n \in \mathbb{N}\}$. Each point in the set is an isolated point(meaning you can find a neighborhood around the point $x$ - isolated point, such that the neighborhood contains only $x$).
Reason: For every $x = 1/n \in S$, take $\epsilon < \frac{1}{n+1}$ then the open ball centered at $x$ with radius of $\epsilon$ - $B_{\epsilon} (x) \, \cap S = \{x\}$.
But $0$ is an accumulation point of the set $S$. To see this take any $\epsilon > 0, \exists \, n $ such that $1/n < \epsilon$ (By Archimedean property).
The above example illustrates that you can have a set $S$ where all of its points are isolated but has an accumulation point.