Accumulation points of sets 
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 $y > \frac{1}{m}$
 A: Alternatively, you may use the fact: 

$a \in \mathbb R$ accumulation point of $S \iff \forall r>0:\; B(a,r) \cap S$ is infinite,

which clearly shows that $0$ is the only accumulation point of $S$



*

*If $x \in (0,1) \setminus S$, then you can always find a $n_x \in \mathbb N$, such that
$$\dfrac {1}{n_x+1} < x < \dfrac{1}{n_x}.$$


Consider any $r < \min\left\{ x-\dfrac{1}{n_x+1},\, \dfrac {1}{n_x}-x\right\}$. Thus, by definition we have:
$$\big(B(x,r)-\{x\}\big) \cap S = \emptyset.$$


*

*In case $x$ is of the form $x=\dfrac 1n,$ for some $n\in \mathbb N$, then you can choose $r <\dfrac 1 {n+1}$ and you will reach the same result.


So we proved that no point in $(0,1)$ can be an accumulation point of $S$.
A: 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 $y \notin (0,1)$. 
A: Observe that the sets $\left(\frac{1}{m+1}, \frac{1}{m}\right)$ are open and
$$\bigcup_{m=1}^\infty \left(\frac{1}{m+1}, \frac{1}{m}\right) = (0,1) \setminus S$$
Thus, for any point $y \in (0,1) \setminus S$, $y$ lies in one of the above open sets, say $\left(\frac{1}{m+1}, \frac{1}{m}\right)$.  By the definition of an open set, there is some $r$ so that $B(y,r) \subseteq \left(\frac{1}{m+1}, \frac{1}{m}\right)$, so such a $y$ cannot be an accumulation point.
To show that no point of $S$ is an accumulation point, apply the same reasoning with the sets $\left(\frac{1}{m+1}, \frac{1}{m-1}\right)$, each of which contains exactly one point of $S$.
A: The way to go is to guess the result and  prove $\{0\}\subset\text{Acc}(S),\:$ followed by $\:\text{Acc}(S)\subset\{0\}.$
We have $$S=\left\{\frac{1}{m}:m\in \mathbb N\right\}=\left\{1,\frac{1}{2},...,\frac{1}{m},\frac{1}{m+1},...\right\}\large\subset\normalsize\mathbb Q.$$
Assume that $0\in\text{Acc}(S)\iff\forall\delta_{>0}\:\exists\mathcal N_{\delta}(0)\large\cap \normalsize(S\setminus\{0\})\ne\emptyset.$
We can construct a sequence $\:\{x_m\}_{m\in\mathbb N}\:$ of terms of $\:\mathcal N_{\delta}(0)\ne\emptyset\:$ such that by the monotone subsequence theorem, $\:\exists\{x_{m_{k}}\}_{k\in\mathbb N}\:$ which is monotone and bounded on the $\:\delta$ - neighbourhood.
$$\implies\forall\epsilon_{>0}\:\exists M_{>0}:\forall m_k\ge M \:\:|x_{m_k}-\hat x|<\underbrace{\frac{1}{m_k}<\epsilon}_{\text{Archimedes}}$$
The key point is that our monotone subsequence $\:\{x_{m_{k}}\}_{k\in\mathbb N}\:$ yields a $\:k>0\:\:\text{s.t}\:\:\forall m\in\mathbb N,\:m_k\ge m$
By the squeeze theorem: $$-\epsilon<\frac{-1}{\:\:\:m}\le\frac{-1}{\:\:\:m_k}<-x_m\le -x_{m_k}\le \hat x\le x_{m_k}\le x_m<\frac{1}{m_k}\le\frac{1}{m}<\epsilon\iff \hat x=0.$$ 
By construction, $\:\{0\}\subset\text{Acc}(S).$
Now, let's assume the following: $\:\:p^*\in\text{Acc}(S)\large\cap\normalsize  S\ne\emptyset\:$ and in particular let $\:p^*\in S.$
Thus $$\text{let}\:\:D=\Large |\normalsize \frac{1}{m}-\frac{1}{m+1}\Large|\normalsize>\frac{1}{(m+1)^2}=\delta^*>0,\:\:\:\forall m\in\mathbb N \\ \implies\exists\delta^*<D:\:\forall \mathcal N_{\delta^*}(p^*)\large \cap \normalsize (S\setminus\{0\})=\emptyset.$$
Since $\:p^*\:$ is arbitrarily chosen in $\:S\:$ we have a contradiction, hence $\:\text{Acc}(S)\subset \{0\}.$
This proves that $\:\text{Acc}(S)=\{0\}$ 
