Accumulation points of a union of sequences Let $A$ be the subset of $\Bbb{R}^2$ defined by
$A = \{(1-\frac{1}{n}, (-1)^n(1-\frac{1}{n}) \vert \space n \in \Bbb{N}\} \cup \{(-1+\frac{1}{n}, (-1)^n(-1+\frac{1}{n}) \vert \space n \in \Bbb{N}\}$
Determine the accumulation points of $A$, $\text{cl}(A)$ and $\text{int}(A)$.
I'm having a difficult time with this because I'm not sure I understand what the accumulation points would look like for this. My understanding is that the accumulation point is basically the limit - someone please correct me if I'm misunderstanding.
So for the first point set in $A$, the limit is ($1$, and diverges), the second set would be ($-1$, and this diverges). So the Union seems to me like the $1$ and $-1$. Is this correct?
Help!
 A: The first thing to get straight is that $A$ is a subset of $\Bbb R^2$, the plane, so any accumulation points of $A$ will be points in the plane, not real numbers.
It might help to write out some elements of the set $A$ to get a better idea of what you’re dealing with. $A$ is the union of the sets
$$A_0=\left\{\langle 0,0\rangle,\left\langle\frac12,\frac12\right\rangle,\left\langle\frac23,-\frac23\right\rangle,\left\langle\frac34,\frac34\right\rangle,\left\langle\frac45,-\frac45\right\rangle,\ldots\right\}$$
and
$$A_1=\left\{\langle 0,0\rangle,\left\langle-\frac12,-\frac12\right\rangle,\left\langle-\frac23,\frac23\right\rangle,\left\langle-\frac34,-\frac34\right\rangle,\left\langle-\frac45,\frac45\right\rangle,\ldots\right\}\;.$$
It would probably be a good idea to sketch a graph of these points. It’s worth noticing that every point of $A$ lies on one or both of the lines $y=x$ and $y=-x$; this may help you with the graphing.
Apart from the origin, every point of $A_0$ lies in the right half-plane, and every point of $A_1$ lies in the left half-plane. (In fact, as you can easily verify algebraically, the points of $A_1$ can be obtained by multiplying the points of $A_0$ by $-1$: $A_1$ is the reflection of $A_0$ in the origin.) 
Look at the $x$-coordinates: for points in $A_0$ they ‘pile up’ towards $1$ and nowhere else, so you’d expect any limit points of $A_0$ to have $x$-coordinate $1$. Similarly, you’d expect any limit points of $A_1$ to have $x$-coordinate $-1$. What about the $y$-coordinates? They also ‘pile up’ only towards $1$ and $-1$. This gives you four good candidates for accumulation points of $A$: $\langle 1,1\rangle,\langle 1,-1\rangle,\langle -1,1\rangle$, and $\langle -1,-1\rangle$.
Take $\langle 1,1\rangle$, for instance. Can you find a sequence of points of $A$ converging to $\langle 1,1\rangle$? In fact you should be able to find a sequence of points of $A_0$ converging to it. That shows that every open nbhd of $\langle 1,1\rangle$ contains a point of $A$ different from $\langle 1,1\rangle$ (since $\langle 1,1\rangle\notin A$) and hence that $\langle 1,1\rangle$ is an accumulation point of $A$. You can apply the same reasoning to each of the other three points. Finally, if you’re supposed to prove your answer, you should show that every other point in $\Bbb R^2$ has an open nbhd that contains no point of $A$ (except itself, if it happens to be in $A$).
A: Let 
$\overline{A}$ is the closure of $A$
$A^o$ is the interior of $A$
$A'$ is the accumulation points of $A$.
We can get:
$A'=$ {$(1,1),(1,-1),(-1,1),(-1,-1)$} since for each point $a_i \in A'$ there is a series $a_{n, i} \in A$ approaching it. For example,
{$(1−\dfrac1n,(1−\dfrac1n)| n∈N$} $→(1,1), $ as $  n→∞$
{$(1−\dfrac1n,(−1)^{2n+1}(1−\dfrac1n)| n∈N$} $→(1,-1), $ as $  n→∞$
$\overline{A}=A'\cup A$
$A^o= \varnothing$ since $A$ is nowhere dense and does not have interior part. 
