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+# Math4201 Topology I (Lecture 29)
+
+## Compact and connected spaces
+
+### Compact spaces
+
+#### Theorem of uncountable compact Hausdorff spaces without isolated points
+
+Any non-empty compact Hausdorff space $X$ without any isolated points is uncountable.
+
+<details>
+<summary>Proof</summary>
+
+By contradiction, let $X=\{x_n\}_{n\in\mathbb{N}}$ be a countable set.
+
+We construct inductively a sequence of open non-empty subspaces $\{V_i\}$ of $X$ such that
+
+$$
+\overline{V_1}\supseteq \overline{V_2}\supseteq \overline{V_3}\supseteq \dots
+$$
+
+where $x_i$ not in $\overline{V_i}$.
+
+This could imply that for any $j\leq i$, $x_j$ is not in $\overline{V_j}$ but $\overline{V_j}\supseteq \overline{V_i}$. This contradicts the fact that $x_j$ not in $\overline{V_i}$, that is $x_1, ..., x_i$ are not in $\overline{V_i}$.
+
+This is a contradiction because $\{\overline{V_i}\}$ satisfies the finite intersection property.
+
+$$
+\bigcap_{i=1}^\infty \overline{V_i} \neq \emptyset
+$$
+
+But in this case, $\bigcap_{i=1}^\infty \overline{V_i} = \emptyset$ because $x_i$ not in $\overline{V_i}$.
+
+To construct such $\{V_i\}$, we can start with $V_1,\dots, V_{k-1}$ are constructed, then there is a point $y_k$ in $V_{k-1}$ which isn't same as $x_k$.
+
+So $x_k$ not in $\overline{V_k}$ because $U_k$ is an open neighborhood of $x_k$ that don't intersect with $V_k$.
+
+Since $X$ is Hausdorff, there exists an open neighborhood $U_k$ of $x_k$ and $U_y$ of $y_k$ such that $U_k\cap U_y=\emptyset$.
+
+Let $x_k\in U_k$ and $y_k\in W_k\subseteq X$ that is open, and $U_k\cap W_k=\emptyset$.
+
+Let $V_k=W_k\cap V_{k-1}$. Then this is open and is contained in $V_{k-1}$ since $\overline{V_k}\subseteq \overline{V_{k-1}}$.
+
+Case 1: $x_k\notin V_{k-1}$. There is such $y_k$ because $V_{k-1}$ is not empty.
+
+Case 2: $x_k\in V_{k-1}$. Since $x_k$ is not isolated points, any open neighborhood of $x_k$ including $V_{k-1}$ contains another point.
+
+Therefore, $X$ is uncountable.
+
+</details>
+
+#### Definition of limit point compact
+
+A space $X$ is limit point compact if any infinite subset of $X$ has a limit point.
+
+#### Definition of sequentially compact
+
+A space $X$ is sequentially compact if any sequence has a convergent subsequence. i.e. If $\{x_n\}_{n\in\mathbb{N}}$ is a sequence in $X$, then there are $n_1<n_2<\dots<n_k<\dots$ such that $\{y_i=x_{n_i}\}_{i\in\mathbb{N}}$ is convergent.
+
+#### Theorem of limit point compact spaces
+
+If $(X,d)$ is a metric space, then the following are equivalent:
+
+1. $X$ is compact.
+2. $X$ is limit point compact.
+3. $X$ is sequentially compact.
+
+<details>
+<summary>Example of limit point compact spaces but not compact</summary>
+
+Let $X'=\{a,b\}$ with trivial topology, and $X=\mathbb{N}\times X'$ with the product topology where we use the discrete topology on $\mathbb{N}$.
+
+$X$ isn't compact because $\{\{i\}\times X':i\in\mathbb{N}\}$ is an open cover of $X$ that doesn't have a finite subcover.
+
+because these open sets are disjoint, $X$ is limit point compact.
+
+Let $A\subseteq X$ be an infinite subset of $X$. In particular, it contains a point of the form $(i,a)$ or $(i,b)$ for $i\in\mathbb{N}$. Let $(i,a)\in A$. Then $(i,b)$ is a limit point of $A$, since any open neighborhood ($\{i\}\times X'$) of $(i,b)$ contains a point of the form $(i,a)$ or $(i,b)$.
+
+$X$ is not sequentially compact because the sequence $\{(n,a)\}_{n\in\mathbb{N}}$ has no convergent subsequence.
+
+</details>
+
+<details>
+<summary>Proof</summary>
+
+First, we show that 1. implies 2.
+
+Let $X$ be compact and $A\subseteq X$ be an infinite subset of $X$ that doesn't have any limit points.
+
+Then $X-A$ is open because any $x\in X-A$ isn't in the closure of $A$ otherwise it would be a limit point for $A$, and hence $x$ has an open neighborhood contained in the complement of $A$.
+
+Next, let $x\in A$. Since $x$ isn't a limit point of $A$, there is an open neighborhood $U_x$ of $x$ in $X$ that $U_x\cap A=\{x\}$. Now consider the open covering of $X$ given as
+
+$$
+\{X-A\}\cup \{U_x:x\in A\}
+$$
+
+This is an open cover because either $x\in X-A$ or $x\in A$ and in the latter case, $x\in U_x$ since $X$ is compact, this should have a finite subcover. Any such subcover should contain $U_x$ for any $x$ because $U_x$ is the element in the subcover for $x$.
+
+This implies that our finite cover contains infinite open sets, which is a contradiction.
+
+---
+
+Continue with the proof that 2. implies 3. next time.
+</details>
+
+#### Corollary of compact spaces
+
+1. If $X$ is a compact topological space, then it is limit point compact.
+2. If $X$ is a sequentially compact topological space, then it is limit point compact.
+
+<details>
+<summary>Proof</summary>
+
+Proof of 1. follows from the theorem of limit point compact spaces.
+
+
+</details>
diff --git a/content/Math4201/_meta.js b/content/Math4201/_meta.js
index 47a0510..308e7d9 100644
--- a/content/Math4201/_meta.js
+++ b/content/Math4201/_meta.js
@@ -32,4 +32,5 @@ export default {
     Math4201_L26: "Topology I (Lecture 26)",
     Math4201_L27: "Topology I (Lecture 27)",
     Math4201_L28: "Topology I (Lecture 28)",
+    Math4201_L29: "Topology I (Lecture 29)",
 }