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80 lines
2.6 KiB
Markdown
80 lines
2.6 KiB
Markdown
# Math4201 Topology I (Lecture 37)
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## Countable Axioms and Separation Axioms
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### Continue on Normal spaces
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#### Proposition of normal spaces
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A topological space $X$ is normal if and only if for all $A\subseteq U$ closed and $U$ is open in $X$, there exists $V$ open such that $A\subseteq V\subseteq \overline{V}\subseteq U$.
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#### Urysohn lemma
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Let $X$ be a normal space, $A,B$ be two closed and disjoint set in $X$, then there exists continuous function $f:X\to[0,1]$ such that $f(A)=\{0\}$ and $f(B)=\{1\}$.
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We say $f$ separates $A$ and $B$.
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> [!NOTE]
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>
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> We could replace $1,0$ to any $a<b$
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<details>
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<summary>Proof</summary>
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Step 1:
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Consider the rationals in $[0,1]$. Let $P=[0,1]\cap \mathbb{Q}$. This set is countable.
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First we want to prove that there exists a set of open sets $\{U_p\}_{p\in P}$ such that if $p<q$, then $\overline{U_p}\subseteq U_q$. ($U_p$ is open in $X$.)
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We prove the claim using countable induction.
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Define $U_1=X-B$, since $B$ is closed in $A,B$ are disjoint, $A\subseteq U_1$.
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Since $X$ is normal, then there exists $U_0$ such that $A\subseteq U_0\subseteq \overline{U_0}\subseteq X$.
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By induction step, for each $p\in P$, we have $U_0,U_1,\cdots,U_{p_n}$ such that if $p<q$, then $\overline{U_p}\subseteq U_q$.
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Choose $U_{p_{n+1}}$ as follows:
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$\exists p,q\in P_n\coloneqq\{1,0,p_3,\ldots,p_n\}$ and $p_{n+1}\in (p,q)$
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$\exists U_{p_{n+1}}$ such that $\overline{U_p}\subseteq U_{p_{n+1}}\subseteq \overline{U_{p_{n+1}}}\subseteq U_q$ by normality of $X$.
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This constructs the set satisfying the claim.
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Step 2:
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We can extend from $P$ to any rationals $\mathbb{Q}$.
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We set $\forall p<0$, $U_p=\emptyset$ and $\forall p>1$, $U_p=X$.
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Otherwise we use the $p$ in $P$.
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Step 3:
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$\forall x\in X$, set $\mathbb{Q}(x)=\{p\in \mathbb{Q}:x\in U_p\}\subseteq [0,\infty)$.
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This function has a lower bound and $f(x)=\inf\mathbb{Q}(x)$.
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Observe that $A\subseteq U_p,\forall p$ and $f(A)=\inf(0,\infty)=\{0\}$.
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$\forall b\in B$, $b\in U_p$ if and only if $p>1$, so $f(b)=\inf(1,\infty)=1$.
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Suppose $x\in \overline{U_r}$, then $x\in U_s,\forall s<r$, this implies that $f(x)\leq r$.
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Suppose $x\notin \overline{U_r}$, then $x\notin U_s,\forall s<r$, this implies that $f(x)\geq r$.
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If $x\in \overline{U_q}-U_p$, then $f(x)\in [p,q]$, $\exists p<q$.
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To show continuity of $f(x)$.
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Let $x_0\in X$ of $f(x_0)\in (c,d)$,
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our goal is to show that there exists open $U\subseteq X$ a neighborhood of $x_0\in U$ such that $f(U)\in (c,d)$.
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Choose $p,q\in \mathbb{Q}$ such that $c<p<f(x_0)<q<d$.
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By our construction, $x_0\in \overline{U_q}-U_p$, and $f(\overline{U_q}-U_p)\subseteq [p,q]\subset (c,d)$.
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</details> |