Update S132 (Duncan's space) (#1192)

spaces/S000132/README.md

@@ -5,27 +5,17 @@ counterexamples_id: 136
 refs:
   - doi: 10.1007/978-1-4612-6290-9
     name: Counterexamples in Topology
+  - zb: "0085.03002"
+    name: A topology for sequences of integers I (R. L. Duncan)
+  - zb: "0109.03302"
+    name: A topology for sequences of integers I (R. L. Duncan)
 ---
-Let $A$ be the set of all strictly increasing, infinite sequences of positive integers. For any $x\in A$, define $N(n,x)$ as the number of elements of $x$ less than $n$. Let $\delta (x) =\lim_{n\to\infty} \frac{N(n,x)}{n}$. Let $X$ be the subset of $A$ containing precisely the sequences, $x$, for which $\delta(x)$ exists. Now for any $x,y\in X$ define $k(x,y)$ to be the smallest index $i$ such that $x_i\neq y_i$. We define a metric $d(x,y)$ on $X$ as $$d(x,y)=\frac{1}{k(x,y)}+|\delta (x) - \delta (y)| \text{ with } d(x,x)=0.$$ Duncan's Space is the topology on $X$ induced by this metric.
+
+Let $A$ be the set of all strictly increasing, infinite sequences of positive integers. For any $x\in A$, define $N(n,x)$ as the number of elements of $x$ less than $n$. Let $\delta (x) =\lim_{n\to\infty} \frac{N(n,x)}{n}$. Let $X$ be the subset of $A$ containing precisely the sequences $x$ for which $\delta(x)$ exists. Now for any $x,y\in X$ define $k(x,y)$ to be the smallest index $i$ such that $x_i\neq y_i$. We define a metric $d(x,y)$ on $X$ as $$d(x,y)=\frac{1}{k(x,y)}+|\delta (x) - \delta (y)| \text{ with } d(x,x)=0.$$ Duncan's Space is the topology on $X$ induced by this metric.
 
 Defined as counterexample #136 ("Duncan's Space")
 in {{doi:10.1007/978-1-4612-6290-9}}.
 
-<!-- [[Proof of Topology]]
-To prove $d(x,y)$ is a metric on $X$, we must prove for every $x,y \in X$:
-$d(x,x)=0$,
-$d(x,y)>0$,
-$d(x,y)=d(y,x)$, and
-$d(x,y)+d(y,z)\geq d(x,z)$
-By definition, $d(x,x)=0$.
-Note that for all $x,y\in X$, $k(x,y)>0$, and clearly, $|\delta (x) - \delta (y)|>0$. It follows that $d(x,y)>0$.
-Note, $k(x,y)=k(y,x)$ and $|\delta (x) - \delta (y)| = |\delta (y)-\delta (x)|$. So, $$d(x,y)=\frac{1}{k(x,y)}+|\delta (x) - \delta (y)|=\frac{1}{k(y,x)}+|\delta (y)-\delta (x)|=d(y,x).$$
-For the final condition, we begin with the definition of our distance function. We must show $$\frac{1}{k(x,y)}+|\delta(x)-\delta(y)|+\frac{1}{k(y,z)}+|\delta(y)-\delta(z)|\geq \frac{1}{k(x,z)} +|\delta(x)-\delta(z)|.$$ It is sufficient to prove:
-$$|\delta(x)-\delta(y)|+|\delta(y)-\delta(z)|\geq | \delta(x)-\delta(z)|$$
-$$\frac{1}{k(x,y)}+\frac{1}{k(y,z)}\geq \frac{1}{k(x,z)}$$
-To prove the first, we utilize the triangle inequality $|a|+|b|\geq |a+b|$ with $a=\delta(x)-\delta(y)$ and $b=\delta(y)-\delta(z)$. So, $$|\delta(x)-\delta(y)|+|\delta(y)-\delta(z)|\geq |\delta(x)-\delta(y)+\delta(y)-\delta(z)| = |\delta(x)-\delta(z)|.$$
-Note that to prove the second, we will instead show that either $k(x,y) \text { or } k(y,z)$ is less than or equal to $k(x,z)$. It then will follow that the second holds. There are two cases:
-$$\text{ Case 1: } k(x,y)\leq k(x,z)$$
-This directly implies $$\frac{1}{k(x,y)}\geq\frac{1}{k(x,z)}.$$
-$$\text{Case 2:} k(x,y)>k(x,z)$$
-By definition, $x_n=y_n$ for $n<k(x,y)$, and $x_n=z_n$ for $n<k(x,z)$. Because $k(x,y)>k(x,z)$, it follows $x_n=y_n=z_n$ for $n<k(x,z)$. Furthermore, $x_{k(x,z)}\neq z_{k(x,z)}$, so $y_{k(x,z)}\neq z_{k(x,z)}$. Thus, $k(y,z)=k(x,z)$. -->
+Introduced and studied by R. L. Duncan in 
+{{zb:0085.03002}} (<https://www.jstor.org/stable/2309919>)
+and {{zb:0109.03302}} (<https://www.jstor.org/stable/2309171>).

spaces/S000132/properties/P000048.md

@@ -7,5 +7,4 @@ refs:
   name: Counterexamples in Topology
 ---
 
-Asserted in the General Reference Chart for space #136 in
-{{doi:10.1007/978-1-4612-6290-9_6}}.
+See item #6 for space #136 in {{doi:10.1007/978-1-4612-6290-9_6}}.

spaces/S000132/properties/P000056.md

@@ -0,0 +1,10 @@
+---
+space: S000132
+property: P000056
+value: true
+refs:
+- doi: 10.2307/2309171
+  name: A Topology for Sequences of Integers II (R. L. Duncan)
+---
+
+See Section 3 of {{doi:10.2307/2309171}}.

spaces/S000132/properties/P000065.md

@@ -0,0 +1,11 @@
+---
+space: S000132
+property: P000065
+value: true
+refs:
+- doi: 10.1007/978-1-4612-6290-9_6
+  name: Counterexamples in Topology
+---
+
+It is easily seen that for every real number $\alpha\in[0,1]$ there is an element $x\in X$ with asymptotic density $\delta(x)=\alpha$.  So $|X|\geq \mathfrak c$.
+And on the other hand, $|X|\leq\aleph_0^{\aleph_0}=2^{\aleph_0}=\mathfrak c$.

spaces/S000132/properties/P000066.md

@@ -0,0 +1,12 @@
+---
+space: S000132
+property: P000066
+value: false
+---
+
+Similar to the proof of {S28|P66}, it is easily seen that for each finite sequence $t$, $\left[ t \right] := \left\{ x \in X \mid x \text{ extends } t \right\}$ is open in $X$.
+
+Now define the open cover $\mathcal U_n = \left\{ \left[ t \right] \mid t \in \omega^n \right\}$.
+Given any finite subcollections $\mathcal F_n \subseteq \mathcal U_n$, we can choose $x_n$ such that $\left[ \left< x_1, \dots, x_n \right> \right] \notin \mathcal F_n$ and $x_n > 2 x_{n - 1}$.
+
+Then $x \in X$ since $\delta(x) = 0$ and $x \notin \bigcup_{n < \omega} \mathcal F_n$.

spaces/S000132/properties/P000184.md

@@ -0,0 +1,10 @@
+---
+space: S000132
+property: P000184
+value: true
+refs:
+- doi: 10.2307/2309171
+  name: A Topology for Sequences of Integers II (R. L. Duncan)
+---
+
+It is asserted in Section 4 of {{doi:10.2307/2309919}} that $S$ is homeomorphic to a certain subset of the plane, namely the graph of a function $\varphi(x)$.