Determine all of the accumulation points of the following sets in $\mathbb{R}^1$ and decide whether the sets are open or closed or neither.
I have two problems with the following problems first of all consider the following set $S = \left\{\frac{1}{m}: m \in \mathbb{N}\right\}$ I have already proved $0$ is accumulation point.
My claim is that $0$ is the only accumulation point. I came with the following claim: suppose the only interval that could cause problems is $(0,1)$ (I have already proved that points outside of $(0,1)$ can't be accumulation points).
So, let $y \in (0,1)$ let $m$ be the smallest $m$ such that $y \geq \frac{1}{m}$. My claim is that $r := y – \frac{1}{m}$.
If we use such $r$ according to my intuition then $(B(y,r) – {y})\cap S = \emptyset$, however I can't prove it, so maybe it is wrong.
I am proving only using point-set topology.
So I have decided to edit the question to show the proof that I think is fully rigorous maybe it would benefit people in the future:
Proof:
Let m be the smallest integer such that $y > \frac{1}{m}$
Consider $r := min(y – \frac{1}{m},\frac{1}{m – 1} – y)$
One thing we must clarify here is that we can't have by definition the following that $\frac{1}{m – 1} < y$ according to how I defined my m to be.
Suppose $r = y – \frac{1}{m} \implies B(y,r) = (\frac{1}{m},2y – \frac{1}{m})$.
Now I will get a contradiction by supposing that $\frac{1}{q} \in (\frac{1}{m},2y – \frac{1}{m})$.
This means that $\frac{1}{q} > \frac{1}{m} \& \frac{1}{q} < 2y – \frac{1}{m – 1}$, however by hypothesis we have $min(y – \frac{1}{m},\frac{1}{m – 1} – y) = y – \frac{1}{m} \implies y – \frac{1}{m} < \frac{1}{m – 1} – y \implies 2y – \frac{1}{m} < \frac{1}{m – 1} \implies \frac{1}{q} < 2y – \frac{1}{m – 1} < \frac{1}{m – 1}$
So $q$ must be less than $\frac{1}{m}$ which is contradiction by construction that is from the fact that $m$ is the smallest integer such that $y > \frac{1}{m}$
Best Answer
One thing: when constructing a ball, make sure your radius is greater than 0. In your proof, you might want to say $y>\frac 1m$ instead of $y \geq \frac 1m$.
I don't think what you wrote is true, because although if $\frac 1 m$ is the largest element in $S$ that is less than $y$, then there are no elements of $S$ in the interval $(\frac 1 m,y)$, you cannot guarantee that to the right of $y$ within a distance of $y-\frac 1m$ there will be no elements of $S$.
For example, take $m=3$ and let $y = \frac{11}{24}$. Then the open ball $B(y, y-\frac 1m) = (\frac 13, \frac{7}{12})$, which contains $\frac 12$.
This happened because $y$ was closer to another element of $S$ than to $\frac 1m$, so to eliminate the mistake you can define $r$ to be the distance between $y$ and the element of $S$ that is closest to it.
Formally, let $r = $min$(\lvert y-\frac 1m\rvert,\lvert y-\frac{1}{m-1}\rvert)$, where $m>1$ because otherwise $y \notin (0,1)$.