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content/Math4201/Math4201_L15.md

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+# Math4201 Topology I (Lecture 15)
+
+## Continue on convergence of sequences
+
+### Closure and convergence
+
+#### Proposition in metric space, every convergent sequence converges to a point in the closure
+
+If $X$ is a metric space, $A\subset X$, and $x\in \overline{A}$, then there is a sequence $\{x_n\}_{n=1}^\infty\subseteq A$ such that $x_n\to x$.
+
+<details>
+<summary>Proof</summary>
+
+Let $U_n=B_{\frac{1}{n}}(x)$ be a sequence of open neighborhoods of $x$.
+
+Since $x\in \overline{A}$ and $U_n$ is an open neighborhood of $x$, then $U_n\cap A\neq \emptyset$. Take $x_n\in U_n\cap A$, we claim that $\{x_n\}_{n=1}^\infty\subseteq A$ that $x_n\to x$.
+
+Take an open neighborhood $U$ of $x$. Then by the definition of metric topology, there is $r>0$ such that $B_r(x)\subseteq U$.
+
+let $N$ be such that $\frac{1}{N}<r$. Then for an $n\geq N$, we have $\frac{1}{n}\leq \frac{1}{N}<r$. This implies that $B_{\frac{1}{n}}(x)\subseteq B_r(x)\subseteq U$.
+
+So $x_n\in B_{\frac{1}{n}}(x)\subseteq U$ for all $n\geq N$.
+
+</details>
+
+#### Corollary $\mathbb{R}^\omega$ with the box topology is not metrizable
+
+$\mathbb{R}^\omega$ is $\text{Map}(\mathbb{N},\mathbb{R})=\mathbb{R}\times \mathbb{R}\times \mathbb{R}\times \cdots$.
+
+> Note that $\mathbb{R}^\omega$ is Hausdorff. So not all Hausdorff spaces are metrizable.
+
+<details>
+<summary>Proof</summary>
+
+Otherwise, the last proposition holds for $\mathbb{R}^\omega$ with the box topology.
+
+Last time we showed that this is not the case.
+</details>
+
+#### Definition of first countability axiom
+
+A topological space $(X,\mathcal{T})$ satisfies the **first countability axiom** if for any point $x\in X$, there is a sequence of open neighborhoods of $x$, $\{V_n\}_{n=1}^\infty$ such that any open neighborhood $U$ of $x$ contains one of $V_n$.
+
+Note that if we take $U_n=V_1\cap V_2\cap \cdots \cap V_n$, then any open neighborhood $U$ that contains $V_N$, then it also contains $U_n$ for all $n\geq N$.
+
+> As the previous prof, for metric space, it is natural to try $V_n=B_{\frac{1}{n}}(x)$ for some $x\in X$.
+
+#### Proposition on first countability axiom
+
+Rewrite the [Proposition of metric space, every convergent sequence converges to a point in the closure](#proposition-in-metric-space-every-convergent-sequence-converges-to-a-point-in-the-closure)
+
+We can have the following:
+
+If $(X,\mathcal{T})$ satisfies the first countability axiom, then every convergent sequence converges to a point in the closure.
+
+We can easily prove this by takeing the sequence of open neighborhoods $\{V_n\}_{n=1}^\infty$ instead of $U_n=B_{\frac{1}{n}}(x)$.
+
+#### Proposition of continuous functions
+
+Let $f:X\to Y$ be a map between two topological spaces.
+
+1. If $f$ is continuous, then for any convergent sequence $\{x_n\}_{n=1}^\infty$ in $X$ converging to $x$, the sequence $\{f(x_n)\}_{n=1}^\infty$ converges to $f(x)$.
+2. If $X$ is equipped with the metric topology and for any convergent sequence $\{x_n\}_{n=1}^\infty\to x$ in $X$, the sequence $\{f(x_n)\}_{n=1}^\infty\to f(x)$ in $Y$, then $f$ is continuous.
+
+<details>
+<summary>Exercise</summary>
+
+Find an example of a function $f:X\to Y$ which is not continuous but for any convergent sequence in $X$, $\{x_n\}_{n=1}^\infty\to x$, the sequence $\{f(x_n)\}_{n=1}^\infty\to f(x)$.
+
+</details>
+
+<details>
+<summary>Solution</summary>
+
+Let $f:S^1\to [0,1)$ be defined by $f(x,y)=\sin^{-1}(\frac{y}{x})$.
+
+This is not continuous because $[0,1)$
+
+</details>
+
+<details>
+<summary>Proof</summary>
+
+Part 1:
+
+Let $f:X\to Y$ be a continuous map and
+
+$$
+\{x_n\}_{n=1}^\infty\subseteq X
+$$
+converges to $x$.
+
+Want to show that $\{f(x_n)\}_{n=1}^\infty$ converges to $f(x)$.
+
+i.e. for any open neighborhood $U$ of $f(x)$, we want to show $f(x_n)$ is eventually in $U$. Take $f^{-1}(U)$. This is an open neighborhood of $x$ since $x_n\to x$.
+
+There is $N$ such that $\forall n\geq N$, we have $x_n\in f^{-1}(U)$, $f(x_n)\in U$.
+
+This implies that $\{f(x_n)\}_{n=1}^\infty$ converges to $f(x)$.
+
+Part 2:
+
+Let $f:X\to Y$ be a map between two topological spaces with $X$ being metric such that for any convergent sequence in $X$, $\{x_n\}_{n=1}^\infty\to x$ in $X$, we have $f(x_n)\to f(x)$ in $Y$.
+
+Want to show that $f$ is continuous.
+
+Recall that it suffice to show that for any $A\subseteq X$, $f(\overline{A})\subseteq \overline{f(A)}$.
+
+Take $y\in f(\overline{A})$. Then $y=f(x)$ with $x\in \overline{A}$. By the previous proposition, there is a sequence $\{x_n\}_{n=1}^\infty\subseteq A$ (since $X$ is a metric space) such that $x_n\to x$.
+
+By our assumption,
+
+$$
+\{f(x_n)\}_{n=1}^\infty\to f(x) \tag{*}
+$$
+
+Note that $\{f(x_n)\}_{n=1}^\infty$ is a sequence in $f(A)$ and ($*$) implies that $y=f(x)\in f(A)$ (by the first part of the proposition). This gives us the claim.
+</details>
+
+> [!NOTE]
+>
+> The second part of the proposition is also true when $X$ is not a metric space but satisfies the first countability axiom.
+
+#### Equivalent formulation of continuity
+
+If $(X,d)$ and $(Y,d')$ are metric spaces and $f:X\to Y$ is a map, then $f$ is continuous if and only if for any $x_0\in X$ and any $\epsilon > 0$, there exists $\delta > 0$ such that if $\forall n\in X, d(x_n,x_0)<\delta$, then $d'(f(x_n),f(x_0))<\epsilon$.
+
+Proof similar to the $X=Y=\mathbb{R}$ case.

content/Math4201/_meta.js

@@ -17,4 +17,5 @@ export default {
     Math4201_L12: "Topology I (Lecture 12)",
     Math4201_L13: "Topology I (Lecture 13)",
     Math4201_L14: "Topology I (Lecture 14)",
+    Math4201_L15: "Topology I (Lecture 15)",
 }