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authorSimon-Kor <52245124+Simon-Kor@users.noreply.github.com>2024-08-24 11:43:29 +0200
committerSimon-Kor <52245124+Simon-Kor@users.noreply.github.com>2024-08-24 11:43:29 +0200
commit29027c9d2cdbdfe59e48b5aa28eb2d32d1a4c1f7 (patch)
tree65a7689cc8ba001a7e914f8523fdc2c9e8c600c0 /library/topology/urysohn.tex
parentd5b31ee7dc5992687f214d77e795bab53d5fe65d (diff)
naproch sty extension
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diff --git a/library/topology/urysohn.tex b/library/topology/urysohn.tex
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--- a/library/topology/urysohn.tex
+++ b/library/topology/urysohn.tex
@@ -504,26 +504,31 @@ The first tept will be a formalisation of chain constructions.
-\begin{definition}\label{sequencetwo}
- $Z$ is a sequencetwo iff $Z = (N,f,B)$ and $N \subseteq \naturals$ and $f$ is a bijection from $N$ to $B$.
-\end{definition}
-
-\begin{proposition}\label{sequence_existence}
- Suppose $N \subseteq \naturals$.
- Suppose $M \subseteq \naturals$.
- Suppose $N = M$.
- Then there exist $Z,f$ such that $f$ is a bijection from $N$ to $M$ and $Z=(N,f,M)$ and $Z$ is a sequencetwo.
-\end{proposition}
-\begin{proof}
- Let $f(x) = x$ for $x \in N$.
- Let $Z=(N,f,M)$.
-\end{proof}
+%\begin{definition}\label{sequencetwo}
+% $Z$ is a sequencetwo iff $Z = (N,f,B)$ and $N \subseteq \naturals$ and $f$ is a bijection from $N$ to $B$.
+%\end{definition}
+%
+%\begin{proposition}\label{sequence_existence}
+% Suppose $N \subseteq \naturals$.
+% Suppose $M \subseteq \naturals$.
+% Suppose $N = M$.
+% Then there exist $Z,f$ such that $f$ is a bijection from $N$ to $M$ and $Z=(N,f,M)$ and $Z$ is a sequencetwo.
+%\end{proposition}
+%\begin{proof}
+% Let $f(x) = x$ for $x \in N$.
+% Let $Z=(N,f,M)$.
+%\end{proof}
%The proposition above and the definition prove false together with
% ordinal_subseteq_unions, omega_is_an_ordinal, powerset_intro, in_irrefl
+
+
+
+
+
\begin{theorem}\label{urysohn}
Let $X$ be a urysohn space.
Suppose $A,B \in \closeds{X}$.