Math4201 Topology I (Lecture 8)

Let $X=\mathbb{R}$ with standard topology.

Let $A=(0,1)$, then set of limit points of $A$ is $[0,1]$.

Let $A=\left\{\frac{1}{n}\right\}_{n\in \mathbb{N}}$, then set of limit points of $A$ is $\{0\}$.

Let $A=\{0\}\cup (1,2)$, then set of limit points of $A$ is $[1,2]$

Let $A=\mathbb{Z}$, then set of limit points of $A$ is $\emptyset$.

First we want to prove the theorem implies the proposition,

$\Rightarrow$

Let $A$ be a close set in $X$, then $\overline{A}=A$ because $A$ in the intersection of all closed subsets $Z\subseteq A$ in $X$ that contains $A$.

So $Z=A$ is such a closed subset of $X$ that contains $A$.

By the theorem, $\overline{A}=A\cup A'$. Combining this with the fact that $A$ is closed, we have $A=A\cup A'$.

So $A'\subseteq A$.

$\Leftarrow$

Suppose $A\subseteq X$ is a set that includes all its limit points, then $A'\subseteq A$.Then $A'\cup A=A$.

By the theorem, $\overline{A}=A\cup A'=A$.

Since $\overline{A}$ is the smallest closed subset of $X$ that contains $A$, we have $A$ is closed.

$\Leftarrow$

We proceed by contradiction.

Suppose $A\notin \overline{A}$, then $x\notin \overline{A}$.

Then $\overline{A}=\bigcap_{A\subseteq Z, Z\text{ is closed}} Z$

So, there is $A\subseteq Z\subset X$ and $Z$ is closed.

So this implies that $x\in X-Z\coloneq U$ and $U$ is open since it a complement of a closed set $Z$.

Since $A\subseteq Z$, we have $A\cap U= \emptyset$. (disjoint)

So $U$ and $A$ are disjoint. So $U$ is an open neighborhood of $x$ that is disjoint from $A$.

This contradicts the assumption that $x\in \overline{A}$.

$\Rightarrow$

Let $x\in \overline{A}$, and we want to show that any neighborhood of $U$ of $x$ non-trivial intersects $A$. ($A\cap U\neq \emptyset$)

By contradiction, suppose that there is an open neighborhood of $x$ that is disjoint from $A$. Then $Z\coloneq X-U$ is closed and $A\subseteq Z$ because $U\cap A= \emptyset$.

Also $x\notin Z$.

By the definition of closure, $\overline{A}=\subset Z$.

Since $x\notin Z$, we have $x\notin \overline{A}$.

This contradicts the assumption that $x\in \overline{A}$.

First we show $A\subseteq \overline{A}$.

If $x\in A'$, then any open neighborhood $U$ of $x$ has a non-trivial intersection with $A$ by the lemma.

So $x\in \overline{A}$.

We already know $A\subseteq \overline{A}$.

Therese two inductions implies $A\cup A'\subseteq \overline{A}$.

Next we show that $\overline{A}\subseteq A\cup A'$.

If $x\in \overline{A}$, then by the lemma, any open neighborhood $U$ of $x$ has a non-trivial intersection with $A$.

If $x\in A$, then $x\in A\cup A'$.

If $x\notin A$, then the intersection of any open neighborhood $U$ of $x$ with $A$ does not contain $x$.

This implies that this intersection has to include a point $y$ that is not $x$.

Since this holds for any open neighborhood $U$ of $x$, we have $x\in A'$. ($x$ is a limit point of $A$)

So $x\in A'$.

Therese two inductions implies $\overline{A}\subseteq A\cup A'$.