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-rw-r--r--library/topology/urysohn.tex44
1 files changed, 40 insertions, 4 deletions
diff --git a/library/topology/urysohn.tex b/library/topology/urysohn.tex
index 6662706..0c4052d 100644
--- a/library/topology/urysohn.tex
+++ b/library/topology/urysohn.tex
@@ -410,7 +410,7 @@ The first tept will be a formalisation of chain constructions.
Suppose $A,B \in \closeds{X}$.
Suppose $A \inter B$ is empty.
There exist $f$ such that $f \in \funs{\carrier[X]}{\intervalclosed{\zero}{1}}$
- and $f(A) = 1$ and $f(B)= 0$ and $f$ is continuous.
+ and $f(A) = 0$ and $f(B)= 1$ and $f$ is continuous.
\end{theorem}
\begin{proof}
@@ -442,8 +442,10 @@ The first tept will be a formalisation of chain constructions.
Omitted.
\end{subproof}
- For all $m \in \indexset[\zeta]$ we have $\pot(m) \neq \zero$.
-
+ We show that for all $m \in \indexset[\zeta]$ we have $\pot(m) \neq \zero$.
+ \begin{subproof}
+ Omitted.
+ \end{subproof}
%%------------- Maybe use Abstrect.hs line 368 "Local Function".
@@ -495,9 +497,43 @@ The first tept will be a formalisation of chain constructions.
\}$.
+ Let $\gamma(n) = \apply{\gamma}{n}$ for $n \in \naturals$.
+
+
+ We show that there exist $F$ such that
+ $F \in \funs{\carrier[X]}{\intervalclosed{\zero}{1}}$ and
+ for all $x \in \carrier[X]$ we have
+ there exist $k \in \intervalclosed{\zero}{1}$
+ such that $F(x) = k$ and
+ for all $\epsilon \in \reals$ such that $\epsilon > \zero$
+ we have there exist $N \in \naturals$ such that
+ for all $N' \in \naturals$ such that $N' > N$
+ we have $\abs{(k - \apply{\gamma(N')}{x})} \leq \epsilon$.
+ \begin{subproof}
+ Omitted.
+ \end{subproof}
+
-
+ We show that $F \in \funs{\carrier[X]}{\intervalclosed{\zero}{1}}$.
+ \begin{subproof}
+ Omitted.
+ \end{subproof}
+
+ We show that $F(A) = 0$.
+ \begin{subproof}
+ Omitted.
+ \end{subproof}
+
+ We show that $F(B) = 1$.
+ \begin{subproof}
+ Omitted.
+ \end{subproof}
+
+ We show that $F$ is continuous.
+ \begin{subproof}
+ Omitted.
+ \end{subproof}
%Suppose $\eta$ is a urysohnchain in $X$.
%Suppose $\carrier[\eta] =\{A, (X \setminus B)\}$