proof update

spaces/S000150/properties/P000089.md

@@ -8,19 +8,9 @@ Observe that any increasing bijection
 $X\to [a,b]\cap\mathbb Q$, where $0\leq a<b\leq 1$,
 is a continuous self-map of $X$.
 
-
-Fix $\theta \in (0,1)\setminus \mathbb Q$ and two sequences of irrational numbers $(a_k)$ and $(b_k)$
-such that $1>a_k>a_{k+1}> \theta > b_{k+1}>b_k >0$
-and $a_k\to \theta$, $b_k\to \theta$.
-E.g. $\theta=1/\sqrt{2}$ and $a_k=\theta+\frac{1}{k+4}$, $b_k=\theta-\frac{1}{k+2}$.
-Pick also $\tilde a \in (a_1,a_0)\cap\mathbb Q$
-and $\tilde b\in (b_0,b_1)\cap\mathbb Q$.
-
-From now on, to simplify the notation,
-every interval is assumed to consist exclusively of rational numbers.
-
-We construct an increasing bijection $f:X\to [\tilde b,\tilde a]$
-by choosing arbitrary order isomorphism between the rational intervals:
-$[0,b_0)\to[\tilde b, b_1)$, $(a_0,1]\to(a_1,\tilde a]$,
-$(b_k,b_{k+1})\to(b_{k+1},b_{k+2})$, and $(a_{k+2},a_{k+1})\to(a_{k+1},a_k)$ for all $k\geq 0$.
-Function $f$ being union of those increasing bijections has no fixed point. 
+Consider the map $f(x):=\frac{x+1}{x+2}$.
+As a function on $\mathbb R$ it admits two irrational fixed points only.
+Restricted to $X$ it is an increasing bijection onto 
+$[\frac{1}{2},\frac{2}{3}]\cap \mathbb Q$
+(with inverse $f^{-1}(y)=\frac{1-2y}{y-1}$),
+therfore it is a continous self-map of $X$ with no fixed point.

spaces/S000151/properties/P000138.md

@@ -2,12 +2,21 @@
 space: S000151
 property: P000138
 value: false
+refs:
+ - doi:10.1007/978-93-80250-91-5_9
+   name: Rational Numbers (in "Notes on Infinite Permutation Groups")
 ---
 
 For every irrational $\theta\in[0,1]$ there exists
 a contiunuous self-map of $X$ such that its set
 of fixed points is exactly $[0,\theta)\cap\mathbb Q$.
 
-The construction of a map $(\theta,1]\cap X \to (\theta,1]\cap X$ with no fixed point is analogous as in the proof that
-{S150|P89}
-(use only the sequence $b_k$).
+To construct such a map for given
+$\theta\in[0,1]\setminus\mathbb Q$, consider the sequence
+$a_n:=\theta+\frac{1-\theta}{n+1}$
+and fix $b\in(a_2,a_1)\cap\mathbb Q$.
+Then pick increasing bijections between rational intervals (c.f. Cantor's isomorphism theorem, 9.3 in {{doi:10.1007/978-93-80250-91-5_9}})
+$(a_{n+1},a_n)\to(a_{n+2},a_{n+1})$ for $n\geq 1$ and $(a_1,1]\to(a_2,b]$.
+Together with the identity on $[0,\theta)\cap X$ we obtain an increasing bijection
+onto interval $[0,b]\cap\mathbb Q$.
+It is therefore a continuous self-map of $X$.