Originally Posted by

**Drexel28** I was a little lazy. I think if I rephrase my argument more rigorously it will make more sense.

Assume that $\displaystyle \xi\in S'$, then $\displaystyle \forall\varepsilon>0$ there exists some $\displaystyle x\in S,x\ne\xi$ such that $\displaystyle x\in N_{\varepsilon}(\xi)$. Now if there exists some $\displaystyle \varepsilon'>0$ such that $\displaystyle N_{\varepsilon'}(\xi)\subseteq S$ then $\displaystyle \xi\in S^{\circ}$. If not, then $\displaystyle \forall\varepsilon>0$ it is true that $\displaystyle N_{\varepsilon}(\xi)\nsubseteq S$ which means there exists some point $\displaystyle y\in S^{c}$ such that $\displaystyle y\in N_{\varepsilon}(\xi)$. Consequently, every neighborhood of $\displaystyle \xi$ contains a point of $\displaystyle S^c$, but since $\displaystyle \xi\in S'$ it is true that every neighborhood also contains a point of $\displaystyle S$. This tells us that $\displaystyle \xi\in\partial S$.

$\displaystyle S^{\circ}$-interior

$\displaystyle \partial S$-boundary

$\displaystyle S'$-limit points

$\displaystyle S^{c}$-complement