Prove that the set $S=\{\frac{1}{n} : n \in \mathbb{N}\}$ has exactly one limit point. Of course, it's obvious that $0$ is a limit point, because $(\frac{1}{n})_{n\in\mathbb{N}}$ is a sequence in $S \setminus \{0\}$ with  $\lim_{n\to\infty} \frac{1}{n}=0$. I think any other convergent sequence whose terms are elements of $S$ must either converge to $0$ or to a value that is a term of the sequence, but can't prove that formally.
 A: Hint: Around any non-zero value $L$, there exists a neighborhood $(L-\epsilon, L + \epsilon)$ small enough so that it contains at most one element of $S$.
A: This is one of those "follow your nose" proofs where you write one step down and see where you can go from there, and do it for each step.  Here is what I came up with:
We want to show $\{1/n \}_{n=1}^{\infty}$ has $0$ as its only limit point.  Assuming you can prove $0$ is a limit point, let's show that there are no other limit points.
Well, let $r$ be any nonzero real number. To be a limit point, we know for each $\epsilon > 0$, $(r - \epsilon, r + \epsilon)$ should contain an element of $\{1/n\}_{n=1}^{\infty}$ which is not equal to $r$.
Well, if $r = \frac{1}{n}$ for some $n$, we can choose our $\epsilon$ small enough so that $(\frac{1}{n} - \epsilon, \frac{1}{n} + \epsilon)$ doesn't contain any other elements in $\{ 1/n \}_{n=1}^{\infty}$.  For example, take $\epsilon = \frac{1}{n + 1000}$.
If $r \neq \frac{1}{n}$ for some $n$, then either $r > 1$, $r < 0$, or $r$ is in the set $(\frac{1}{2},1) \cup (\frac{1}{3},\frac{1}{2}) \cup (\frac{1}{4}, \frac{1}{3}) \cup (\frac{1}{5}, \frac{1}{4}) \cup \dots$.
If $r > 1$, choose $\epsilon$ small enough so that $(r - \epsilon, r + \epsilon)$ is contained in $(1,\infty)$ (which we can do since $(1,\infty)$ is open).  Then $(r - \epsilon, r + \epsilon)$ doesn't contain any elements of the form $\frac{1}{n}$.
If $r < 0$, choose $\epsilon$ small enough so that $(r - \epsilon, r + \epsilon)$ is contained in $(-\infty, 0)$ (which we can do since $(-\infty, 0)$ is open).  Then $(r - \epsilon, r + \epsilon)$ doesn't contain any elements of the form $\frac{1}{n}$.
If $r$ is in $(\frac{1}{n + 1}, \frac{1}{n})$ for some $n$, then since this is an open interval, we can find $\epsilon$ small enough so that $(r - \epsilon, r + \epsilon)$ is also contained in $(\frac{1}{n + 1}, \frac{1}{n})$, implying $(r - \epsilon, r + \epsilon)$ doesn't contain any elements of the form $\frac{1}{n}$.
So, in every case of $r$ being nonzero, we found that $r$ can't be a limit point of $\{ \frac{1}{n} \}_{n=1}^{\infty}$ since we are able to find a neighborhood around $r$ which doesn't contain elements of $\{\frac{1}{n} \}_{n=1}^{\infty}$.
A: One way to do it is to prove that if $x\neq 0$ then $x$ is not a limit point.
Try considering each of the following three cases:


*

*$x<0$

*$x>1$

*$0\leq1$


The last case is perhaps the trickiest.
We can divide it into two cases:


*

*$x$ is of the form $\frac{1}{n}$

*$x$ is between $\frac{1}{n-1}$ and $\frac{1}{n}$

