Show that any uncountable subset of the reals has uncountably many limit points.

Let $S\subseteq \mathbb R$ be uncountable and let $L$ be the set of all the limit points of $S$.

Assume on the contrary that $L$ is at most countable.

Now. For all $x\in S\setminus L$, there exist $a_x,b_x\in \mathbb Q$ be such that the open interval $U_x=(a_x,b_x)$ contains exactly one point of $S$.

Note that for all $x,y\in S\setminus L$, we have $x\neq y\iff U_x\neq U_y$. Write $\mathcal C=\{U_x:x\in S\setminus L\}$.

Note that $\mathcal C$ is in bijection with $S\setminus L$.

But $\mathcal C$ can be viewed as a subset of $\mathbb Q\times \mathbb Q$ and hence $\mathcal C$ is countable.

This forces $S$ to be countable which contradicts the hypothesis.

Hence we achieve the required contradiction and the proof is complete.

  • 2
    $\begingroup$ Possible duplicate with here and here $\endgroup$ – Golbez Aug 25 '14 at 4:33

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