133 lines
5.4 KiB
Markdown
133 lines
5.4 KiB
Markdown
# Math4201 Topology I (Lecture 15)
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## Continue on convergence of sequences
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### Closure and convergence
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#### Proposition in metric space, every convergent sequence converges to a point in the closure
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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 $x_n\to x$.
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<details>
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<summary>Proof</summary>
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Let $U_n=B_{\frac{1}{n}}(x)$ be a sequence of open neighborhoods of $x$.
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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 $x_n\to x$.
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Take an open neighborhood $U$ of $x$. Then by the definition of metric topology, there is $r>0$ such that $B_r(x)\subseteq U$.
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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 $B_{\frac{1}{n}}(x)\subseteq B_r(x)\subseteq U$.
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So $x_n\in B_{\frac{1}{n}}(x)\subseteq U$ for all $n\geq N$.
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</details>
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#### Corollary $\mathbb{R}^\omega$ with the box topology is not metrizable
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$\mathbb{R}^\omega$ is $\text{Map}(\mathbb{N},\mathbb{R})=\mathbb{R}\times \mathbb{R}\times \mathbb{R}\times \cdots$.
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> Note that $\mathbb{R}^\omega$ is Hausdorff. So not all Hausdorff spaces are metrizable.
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<details>
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<summary>Proof</summary>
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Otherwise, the last proposition holds for $\mathbb{R}^\omega$ with the box topology.
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Last time we showed that this is not the case.
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</details>
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#### Definition of first countability axiom
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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 $V_n$.
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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 $n\geq N$.
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> As the previous prof, for metric space, it is natural to try $V_n=B_{\frac{1}{n}}(x)$ for some $x\in X$.
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#### Proposition on first countability axiom
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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)
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We can have the following:
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If $(X,\mathcal{T})$ satisfies the first countability axiom, then every convergent sequence converges to a point in the closure.
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We can easily prove this by takeing the sequence of open neighborhoods $\{V_n\}_{n=1}^\infty$ instead of $U_n=B_{\frac{1}{n}}(x)$.
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#### Proposition of continuous functions
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Let $f:X\to Y$ be a map between two topological spaces.
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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 $f(x)$.
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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.
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<details>
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<summary>Exercise</summary>
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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 $\{f(x_n)\}_{n=1}^\infty\to f(x)$.
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</details>
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<details>
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<summary>Solution</summary>
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Consider $X=\mathbb{R}$ with complement finite topology and $Y=\mathbb{R}$ with the standard topology.
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Take identity function $f(x)=x$.
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This function is not continuous by trivially taking $(0,1)\subseteq \mathbb{R}$ and the complement of $(0,1)$ is not a finite set, so the function is not continuous.
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However, for every convergent sequence in $X$, $\{x_n\}_{n=1}^\infty\to x$, the sequence $\{f(x_n)\}_{n=1}^\infty\to f(x)$ trivially.
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</details>
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<details>
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<summary>Proof</summary>
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Part 1:
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Let $f:X\to Y$ be a continuous map and
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$$
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\{x_n\}_{n=1}^\infty\subseteq X
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$$
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converges to $x$.
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Want to show that $\{f(x_n)\}_{n=1}^\infty$ converges to $f(x)$.
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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 $x_n\to x$.
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There is $N$ such that $\forall n\geq N$, we have $x_n\in f^{-1}(U)$, $f(x_n)\in U$.
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This implies that $\{f(x_n)\}_{n=1}^\infty$ converges to $f(x)$.
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Part 2:
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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 $Y$.
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Want to show that $f$ is continuous.
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Recall that it suffice to show that for any $A\subseteq X$, $f(\overline{A})\subseteq \overline{f(A)}$.
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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 $x_n\to x$.
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By our assumption,
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$$
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\{f(x_n)\}_{n=1}^\infty\to f(x) \tag{*}
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$$
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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.
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</details>
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> [!NOTE]
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>
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> The second part of the proposition is also true when $X$ is not a metric space but satisfies the first countability axiom.
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#### Equivalent formulation of continuity
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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 $d'(f(x_n),f(x_0))<\epsilon$.
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Proof similar to the $X=Y=\mathbb{R}$ case.
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