Notes - Metric Spaces MT23, Limit points

Any open ball about $a$ contains a point of $S$ other than $a$ itself.

We instead show that $L(S)^C$ is open.

Suppose $a \in L(S)^C$. Then there is a ball $B(a, \varepsilon)$ whose intersection with $S$ is either empty or $\{a\}$.

We claim $B(a, \varepsilon/2) \subseteq L(S)^C$. Let $b \in B(a, \varepsilon / 2)$. If $b = a$, then clearly $b \in L(S)^C$. Otherwise if $b \ne a$, then there is some open ball about $b$ which is contained in $B(a, \varepsilon)$, but does not contain $a$ (and hence is in $L(S)^C$).

The ball $B(b, \delta)$ where $\delta = \min(\varepsilon / 2, d(a, b))$ has this property. The ball meets $S$ in the empty set, so $b \in L(S)^C$.

$S \cup L(S) \subseteq \overline S$: We have $S \subseteq \overline S$, so we need only to show that $L(S) \subseteq \overline S$. Suppose $a \in \overline S^C$. Since $\overline S^c$ is open, there is some ball $B(a, \varepsilon)$ which lies in $\overline S^C$, and hence also in $S^C$, and therefore $a$ cannot be a limit point of $S$.

$\overline S \subseteq S \cup L(S)$: If $a \in \overline S$, there is a sequence of elements $(x _ n)^\infty _ {n = 1}$ in $S$ such that $x _ n \to a$. If $x _ n = a$ for some $n$ we are done, since this implies that $a \in S$. Otherwise, suppose $x _ n \ne a$ for all $n$. Let $\varepsilon > 0$. Then for all the $x _ n$ where $n$ is sufficiently large, there are elements of $B(a, \varepsilon) \setminus \{a\}$ that also lie in $S$. It follows that $a$ is a limit point of $S$, and so we are done.

Todo.

Todo.