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I need to find an example of a compact set whose set of limit points is countably infinite.

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up vote 6 down vote accepted

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 $\langle\frac1n,0\rangle$.

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Excellent, thanks a lot. – Jorge Dec 7 '12 at 19:51
@Jorge: You’re welcome. – Brian M. Scott Dec 7 '12 at 19:55

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 "$\infty$."

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Excellent, thank you. – Jorge Dec 7 '12 at 19:50

\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.

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