Problem proving "p", "k" are limit point. Good night, i have problem with this exercises:
$A=\left\{ 1-\frac{1}{n}:n=1,2,\ldots\right\} $
I make this:
$$ \lim_{n\rightarrow\infty} \left(1-\frac{1}{n}\right)=1$$

Prove:
Be $p=1$ and $r>0$, let '$s$ see "$p$" is a limit point.
If $p$ is a limit point then: $A\cap\left\{ \left(p-r,p+r\right)-\left\{ p \right\} \right\} \neq\emptyset$
Ovbiously $p$ is a limit point but i cannot prove that... 

$$B=\left\{ \left(1+\frac{1}{n}\right)^n :n=1,2,\ldots\right\}$$
I make this:
$$\lim_{n\rightarrow\infty}\left(1+\frac{1}{n}\right)^n=e\;$$

Prove: Be $k=e$ and $r>0$, let '$s$ see "$k$" be a limit point. If $k$ is a limit point then:$B\cap\{ (k-r,k+r)-\{ k\} \} \neq\emptyset$
But i cannot see $B\cap\{ (k-r,k+r)-\{ k\} \} \neq\emptyset$ Please, help.
 A: For $A$: 
Let $n$ be such that ${1\over n} < r$. Then 
$$ A\cap\left\{ \left(p-r,p+r\right)-\left\{ p \right\} \right\} =A\cap\left\{ \left(1-r,1+r\right)-\left\{ 1 \right\} \right\} $$
contains $1-{1\over n}$ and is therefore not empty. Since $r$ was arbitrary it follows that $p=1$ is a limit point of $A$. 
Try to do the same proof for $B$. 
A: You need more than the fact that $A\cap\left\{ \left(p-r,p+r\right)-\left\{ p \right\} \right\} \neq\emptyset$. This only says that the sequence is eventually within the punctured neighborhoods of $p$. You need to show that is eventually entirely within said neighborhoods. In symbols, you need to show:
$\exists N\in\mathbb{N}$ s.t. $\forall n>N$ you have $1-\frac{1}{n}\in\left\{ \left(p-r,p+r\right)-\left\{ p \right\} \right\}$. 
The reason you need that every $n>N$ is within the neighborhood is because of examples like $a_n=(-1^n)\left(1-\frac{1}{n}\right)$, which does not have a limit (as $n\rightarrow\infty$) because it has subsequences that converge to both $1$ and $-1$.
Other than that important detail, your approach is correct. Note that what you actually show in the course of the proof is the existence of a $N$ satisfying:
$\forall n>N$ that $1-\frac{1}{n}\in\left\{ \left(p-r,p+r\right)-\left\{ p \right\} \right\}$.
A: Well i work in this:
For first
Consider the open neighborhood $B(1,r)$. By the archimedean property, there exists some $r ∈ N$ so that  $r>\frac{1}{n}$. Then $B(1,\frac{1}{r})\subset B(1,r)$ and $A\cap\left\{ \left(p-r,p+r\right)-\left\{ p\right\} \right\} \neq\phi$ Then p it's limit point.
I'm working in the two.
