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content/Math4201/Math4201_L8.md

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+# Math4201 Lecture 8
+
+Recall from real analysis, a set is closed if and only if it has limit points.
+
+## New materials
+
+### Limit points
+
+Let $(X,\mathcal{T})$ be a topological space. $A$ is a subset of $X$, then we say $x\in X$ is a limit point of $A$ if any open set $U\subset X$ containing $x$ has another point $y\in A-\{x\}$.
+
+<details>
+<summary>Example of limit points</summary>
+
+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 $\emptyset$.
+
+</details>
+
+#### Proposition of limit points and closed sets
+
+A set is close if and only if it has limit points.
+
+Theorem: For any subset $A$ of a topological space $X$, the closure of $A$ is $\overline{A}=A\cup A'$.
+
+<details>
+<summary>Proof</summary>
+
+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.
+
+</details>
+
+#### Definition of neighborhood
+
+Let $(X,\mathcal{T})$ be a topological space. A neighborhood of a point $x\in X$ is an open set $U\in \mathcal{T}$ such that $x\in U$.
+
+#### Lemma of intersection of neighborhoods for closure of a set
+
+$x\in \overline{A}$ if and only if any neighborhood of $U$ of $x$ non-trivial intersects $A$. ($A\cap U\neq \emptyset$)
+
+<details>
+<summary>Proof of Lemma</summary>
+
+$\Leftarrow$
+
+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 $x\in \overline{A}$.
+
+</details>
+
+#### Proof of theorem
+
+For any subset $A$ of a topological space $X$, the closure of $A$ is $\overline{A}=A\cup A'$.
+
+<details>
+<summary>Proof</summary>
+
+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 $\overline{A}\subseteq A\cup A'$.
+
+</details>
+
+> [!TIP]
+>
+> Now the three definition of closure are equivalent.
+>
+> 1. The smallest closed subset of $X$ that contains $A$.
+> 2. $A\cup A'$.

content/Math4201/_meta.js

@@ -8,5 +8,7 @@ export default {
     Math4201_L3: "Topology I (Lecture 3)",
     Math4201_L4: "Topology I (Lecture 4)",
     Math4201_L5: "Topology I (Lecture 5) Bonus",
-    Math4201_L6: "Topology I (Lecture 6)"
+    Math4201_L6: "Topology I (Lecture 6)",
+    Math4201_L7: "Topology I (Lecture 7)",
+    Math4201_L8: "Topology I (Lecture 8)",
 }