\import{topology/topological-space.tex} \import{topology/separation.tex} \import{topology/continuous.tex} \import{topology/basis.tex} \import{numbers.tex} \import{function.tex} \import{set.tex} \import{cardinal.tex} \import{relation.tex} \import{relation/uniqueness.tex} \import{set/cons.tex} \import{set/powerset.tex} \import{set/fixpoint.tex} \import{set/product.tex} \section{Urysohns Lemma} \begin{definition}\label{minimum} $\min{X} = \{x \in X \mid \forall y \in X. x \leq y \}$. \end{definition} \begin{definition}\label{maximum} $\max{X} = \{x \in X \mid \forall y \in X. x \geq y \}$. \end{definition} \begin{definition}\label{intervalclosed} $\intervalclosed{a}{b} = \{x \in \reals \mid a \leq x \leq b\}$. \end{definition} \begin{definition}\label{intervalopen} $\intervalopen{a}{b} = \{ x \in \reals \mid a < x < b\}$. \end{definition} \begin{definition}\label{one_to_n_set} $\seq{m}{n} = \{x \in \naturals \mid m \leq x \leq n\}$. \end{definition} \begin{definition}\label{sequence} $X$ is a sequence iff $X$ is a function and $\dom{X} \subseteq \naturals$. \end{definition} \begin{abbreviation}\label{urysohnspace} $X$ is a urysohn space iff $X$ is a topological space and for all $A,B \in \closeds{X}$ such that $A \inter B = \emptyset$ we have there exist $A',B' \in \opens[X]$ such that $A \subseteq A'$ and $B \subseteq B'$ and $A' \inter B' = \emptyset$. \end{abbreviation} \begin{abbreviation}\label{at} $\at{f}{n} = f(n)$. \end{abbreviation} \begin{definition}\label{chain_of_subsets} $X$ is a chain of subsets in $Y$ iff $X$ is a sequence and for all $n \in \dom{X}$ we have $\at{X}{n} \subseteq \carrier[Y]$ and for all $m \in \dom{X}$ such that $m > n$ we have $\at{X}{n} \subseteq \at{X}{m}$. \end{definition} \begin{definition}\label{urysohnchain}%<-- zulässig $X$ is a urysohnchain of $Y$ iff $X$ is a chain of subsets in $Y$ and for all $n,m \in \dom{X}$ such that $n < m$ we have $\closure{\at{X}{n}}{Y} \subseteq \interior{\at{X}{m}}{Y}$. \end{definition} \begin{definition}\label{finer} %<-- verfeinerung $X$ is finer then $Y$ in $U$ iff for all $n \in \dom{X}$ we have $\at{X}{n} \in \ran{Y}$ and for all $m \in \dom{X}$ such that $n < m$ we have there exist $k \in \dom{Y}$ such that $ \closure{\at{X}{n}}{U} \subseteq \interior{\at{Y}{k}}{U} \subseteq \closure{\at{Y}{k}}{U} \subseteq \interior{\at{X}{m}}{U}$. \end{definition} \begin{definition}\label{sequence_of_reals} $X$ is a sequence of reals iff $\ran{X} \subseteq \reals$. \end{definition} \begin{axiom}\label{abs_behavior1} If $x \geq \zero$ then $\abs{x} = x$. \end{axiom} \begin{axiom}\label{abs_behavior2} If $x < \zero$ then $\abs{x} = \neg{x}$. \end{axiom} \begin{definition}\label{realsminus} $\realsminus = \{r \in \reals \mid r < \zero\}$. \end{definition} \begin{definition}\label{realsplus} $\realsplus = \reals \setminus \realsminus$. \end{definition} \begin{definition}\label{epsilon_ball} $\epsBall{x}{\epsilon} = \intervalopen{x-\epsilon}{x+\epsilon}$. \end{definition} \begin{definition}\label{pointwise_convergence} $X$ converge to $x$ iff for all $\epsilon \in \realsplus$ there exist $N \in \dom{X}$ such that for all $n \in \dom{X}$ such that $n > N$ we have $\at{X}{n} \in \epsBall{x}{\epsilon}$. \end{definition} \begin{proposition}\label{iff_sequence} Suppose $X$ is a function. Suppose $\dom{X} \subseteq \naturals$. Then $X$ is a sequence. \end{proposition} \begin{theorem}\label{urysohnsetinbeetween} Let $X$ be a urysohn space. Suppose $A,B \in \closeds{X}$. Suppose $\closure{A}{X} \subseteq \interior{B}{X}$. Suppose $\carrier[X]$ is inhabited. There exist $U \subseteq \carrier[X]$ such that $U$ is closed in $X$ and $\closure{A}{X} \subseteq \interior{U}{X} \subseteq \closure{U}{X} \subseteq \interior{B}{X}$. \end{theorem} \begin{proof} Omitted. \end{proof} \begin{theorem}\label{urysohn} Let $X$ be a urysohn space. Suppose $A,B \in \closeds{X}$. Suppose $A \inter B$ is empty. Suppose $\carrier[X]$ is inhabited. There exist $f$ such that $f \in \funs{\carrier[X]}{\intervalclosed{\zero}{1}}$ and $f(A) = \zero$ and $f(B)= 1$ and $f$ is continuous. \end{theorem} \begin{proof} Let $X' = \carrier[X]$. Let $N = \{\zero, 1\}$. $1 = \suc{\zero}$. $1 \in \naturals$ and $\zero \in \naturals$. $N \subseteq \naturals$. Let $A' = (X' \setminus B)$. $B \subseteq X'$ by \cref{powerset_elim,closeds}. $A \subseteq X'$. Therefore $A \subseteq A'$. Define $U_0: N \to \{A, A'\}$ such that $U_0(n) =$ \begin{cases} &A &\text{if} n = \zero \\ &A' &\text{if} n = 1 \end{cases} $U_0$ is a function. $\dom{U_0} = N$. $\dom{U_0} \subseteq \naturals$ by \cref{ran_converse}. $U_0$ is a sequence. We show that $U_0$ is a chain of subsets in $X$. \begin{subproof} We have $\dom{U_0} \subseteq \naturals$. We have for all $n \in \dom{U_0}$ we have $\at{U_0}{n} \subseteq \carrier[X]$ by \cref{topological_prebasis_iff_covering_family,union_as_unions,union_absorb_subseteq_left,subset_transitive,setminus_subseteq}. We have $\dom{U_0} = \{\zero, 1\}$. It suffices to show that for all $n \in \dom{U_0}$ we have for all $m \in \dom{U_0}$ such that $m > n$ we have $\at{U_0}{n} \subseteq \at{U_0}{m}$. It suffices to show that $\at{U_0}{\zero} \subseteq \at{U_0}{1}$. Follows by \cref{one_in_reals,order_reals_lemma0,upair_elim,reals_one_bigger_zero,reals_order,reals_axiom_zero_in_reals,subseteq_refl,apply}. \end{subproof} $U_0$ is a urysohnchain of $X$. %We are done with the first step, now we want to prove that we have U a sequence of these chain with U_0 the first chain. \end{proof} \begin{theorem}\label{safe} Contradiction. \end{theorem} % %Ideen: %Eine folge ist ein Funktion mit domain \subseteq Natürlichenzahlen. als predicat % %zulässig und verfeinerung von ketten als predicat definieren. % %limits und punkt konvergenz als prädikat. % % %Vor dem Beweis vor dem eigentlichen Beweis. %die abgeleiteten Funktionen % %\derivedstiarcasefunction on A % %abbreviation: \at{f}{n} = f_{n} % % %TODO: %Reals ist ein topologischer Raum %