Example of a compact set whose set of limit points is countably infinite. I need to find an example of a compact set whose set of limit points is countably infinite. 
 A: HINT: Start with the set $A=\{0\}\cup\left\{\frac1n:n\in\Bbb Z^+\right\}$; that’s a compact set with one limit point. Now add to $A$ a sequence converging to each of the points $\frac1n$. You can do this in $\Bbb R$, but it’s a little easier to describe and visualize if you first embed $A$ in $\Bbb R^2$ and then let the new sequence converge vertically to the points $\langle\frac1n,0\rangle$.
A: Hint: In $\mathbb{R}$ we want the set to be closed and bounded.  Make a set whose limit points are, say, $0$ and $1,1/2,1/3,1/4,1/5,\dots$. 
To make a set with the right limit points, you might start with making a set whose only limit points are $1$, $2$, $3$, and so on. (This is of course not compact.) Then produce the desired set by using the reciprocal, not forgetting the reciprocal of "$\infty$."
A: \begin{equation}
A = \{ (1/m, 1/n),(1/m,0),(0, 1/n) : m , n \in  \mathbb{N} \}
\end{equation}
The set of limit point of $ A $ is $ \{(1/m,0) : m  \in  \mathbb{N} \} \cup \{ (0, 1/n) : n \in    \mathbb{N} \}  $ is contable infinite and $ A  $ is compact because it is bounded and closed.
