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+# Math4201 Lecture 16 (Topology I)
+
+## Continuous maps
+
+The following maps are continuous:
+
+$$
+F_+:\mathbb{R}\times \mathbb{R}\to \mathbb{R}, (x,y)\to x+y
+$$
+
+$$
+F_-:\mathbb{R}\times \mathbb{R}\to \mathbb{R}, (x,y)\to x-y
+$$
+
+$$
+F_\times:\mathbb{R}\times \mathbb{R}\to \mathbb{R}, (x,y)\to x\times y
+$$
+
+$$
+F_\div:\mathbb{R}\times (\mathbb{R}\setminus \{0\})\to \mathbb{R}, (x,y)\to \frac{x}{y}
+$$
+
+### Composition of continuous functions is continuous
+
+Let $f,g:X\to \mathbb{R}$ be continuous functions. $X$ is topological space.
+
+Then the following functions are continuous:
+
+$$
+H:X\to \mathbb{R}\times \mathbb{R}, x\to (f(x),g(x))
+$$
+
+Since the composition of continuous functions is continuous, we have
+
+$$
+F_+\circ H:X\to \mathbb{R}, x\to f(x)+g(x)
+$$
+
+$$
+F_-\circ H:X\to \mathbb{R}, x\to f(x)-g(x)
+$$
+
+$$
+F_\times\circ H:X\to \mathbb{R}, x\to f(x)\times g(x)
+$$
+
+are all continuous.
+
+More over, if $g(x)\neq 0$ for all $x\in X$, then
+
+$$
+F_\div\circ H:X\to \mathbb{R}, x\to \frac{f(x)}{g(x)}
+$$
+
+is continuous following the similar argument.
+
+### Defining metric for functions
+
+#### Definition of bounded metric space
+
+A metric space $(Y,d)$ is **bounded** if there is $M\in\mathbb{R}^{\geq 0}$ such that
+
+$$
+\forall y,y'\in Y, d(y,y')<M
+$$
+
+<details>
+<summary>Example of bounded metric space</summary>
+
+If $(Y,d)$ is a bounded metric space, let $M$ be a positive constant, then $\overline{d}=\min\{M,d\}$ is a bounded metric space.
+
+In fact, the metric topology by $d$ and $\overline{d}$ are the same. (proved in homeworks)
+
+</details>
+
+Let $X$ be a topological space. and $(Y,d)$ be a **bounded** metric space.
+
+$$
+\operatorname{Map}(X,Y)\coloneq \{f:X\to Y|f \text{ is a map}\}
+$$
+
+Define $\rho:\operatorname{Map}(X,Y)\times \operatorname{Map}(X,Y)\to \mathbb{R}$ by
+
+$$
+\rho(f,g)=\sup_{x\in X} d(f(x),g(x))
+$$
+
+#### Lemma space of map with metric defined is a metric space
+
+$(\operatorname{Map}(X,Y),\rho)$ is a metric space.
+
+<details>
+<summary>Proof</summary>
+
+Proof is similar to showing that the square metric is a metric on $\mathbb{R}^n$.
+
+$\rho(f,g)=0\implies \sup_{x\in X}(d(f(x),g(x)))=0$
+
+Since $d(f(x),g(x))\geq 0$, this implies that $d(f(x),g(x))=0$ for all $x\in X$.
+
+The triangle inequality of being metric for $\rho$ follows from the similar properties for $d$.
+
+</details>
+
+#### Lemma continuous maps form a closed subset of the space of maps
+
+Let $(\operatorname{Map}(X,Y),\rho)$ be a metric space defined before.
+
+and 
+
+$$
+Z=\{f:X\to Y|f \text{ is a continuous map}\}
+$$
+
+Then $Z$ is a closed subset of $(\operatorname{Map}(X,Y),\rho)$.
+
+<details>
+<summary>Proof</summary>
+
+We need to show that $\overline{Z}=Z$.
+
+Since $\operatorname{Map}(X,Y)$ is a metric space, this is equivalent to showing that: $f_n:X\to Y\in Z$ continuous,
+
+Which is to prove the uniform convergence,
+
+$$
+f_n \to f \in \operatorname{Map}(X,Y)
+$$
+
+Then we want to show that $f$ is continuous.
+
+Let $B_r(y)$ be an arbitrary ball in $Y$, it suffices to show that $f^{-1}(B_r(y))$ is open in $X$.
+
+Take $N$ to be large enough such that for $n\geq N$, we have
+
+$$
+\rho(f_n(x), f(x)) < \frac{r}{3}
+$$
+
+In particular, this holds for $n=N$. So we have
+
+$$
+d(f_N(x), f(x)) < \frac{r}{3},\forall x\in X
+$$
+
+Take $x_0\in f^{-1}(B_r(y))$, we'd like to show that there is an open ball around $x_0$ in $f^{-1}(B_r(y))$.
+
+Since $f_N$ is continuous, $f^{-1}_N(B_{\frac{r}{3}}(y))$ is open in $X$.
+
+$d(f(x_0), f(x_0))<\frac{r}{3}$
+
+continue the proof in bonus video
+
+</details>
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diff --git a/content/Math4201/_meta.js b/content/Math4201/_meta.js
index 72ed2a9..bf60a58 100644
--- a/content/Math4201/_meta.js
+++ b/content/Math4201/_meta.js
@@ -18,4 +18,5 @@ export default {
     Math4201_L13: "Topology I (Lecture 13)",
     Math4201_L14: "Topology I (Lecture 14)",
     Math4201_L15: "Topology I (Lecture 15)",
+    Math4201_L16: "Topology I (Lecture 16)",
 }