# Notes - Metric Spaces MT23, Limit points

> Source: https://ollybritton.com/notes/uni/part-a/mt23/metric-spaces/notes/limit-points/ · Updated: 2023-07-30 · Tags: uni, notes

- [Course - Metric Spaces MT23](https://ollybritton.com/notes/uni/part-a/mt23/metric-spaces/)

### Flashcards
What does it mean for $a \in X$ to be a limit point of $S \subseteq X$?::

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

What notation is used for the set of limit points of $S \subseteq X$?::

$$
L(S)
$$

What does it mean for $a \in S$ to be an isolated point?::

$$
\exists \varepsilon >0 \text{ s.t. } B(a, \varepsilon) \cap S = \\{a\\}
$$

Suppose:

- $X$ is a metric space
- $S \subseteq X$
- $L(S)$ is the set of limit points of $S$

Quickly show that $L(S)$ is closed.::

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 $b \in L(S)^C$.

Suppose:

- $X$ is a metric space
- $S \subseteq X$
- $L(S)$ is the set of limit points of $S$
- $\overline S$ is the closure

Quickly prove that
$$
\overline S = S \cup L(S)
$$
::

$S \cup L(S) \subseteq \overline S$: 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.

### Proofs
Prove that if $S \subseteq X$, then $\overline S = S \cup L(S)$.::Todo.

Prove that if $S \subseteq X$, then $S$ is closed if and only if it contains its limit points.::Todo.

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