Mysior plane

properties/P000022.md

@@ -15,3 +15,4 @@ Defined on page 20 of {{zb:0386.54001}}.
 
 - $X$ satisfies this property iff its Kolmogorov quotient $\mathrm{Kol}(X)$ does.
 - This property is preserved in any coarser topology.
+- This property is hereditary with respect to clopen sets.

spaces/S000215/README.md

@@ -0,0 +1,19 @@
+---
+uid: S000215
+name: Mysior plane
+refs:
+  - zb: "0469.54011"
+    name: A union of realcompact spaces (Mysior)
+  - zb: "1265.54111"
+    name: r-realcompact spaces (Bhattacharya & Dey)
+---
+
+Let $X=\mathbb R^2$ with each point $(x,y)$ with $y\ne 0$ isolated
+and each point $(x,0)$ having as local base the collection of open sets $U_n(x)$ ($n=1,2,\dots$),
+where $U_n(x)$ is the union of the three line segments
+* $\{(x, y): -1/n < y < 1/n\}$,
+* $\{(x+1+y, y): 0 < y < 1/n\}$,
+* $\{(x+\sqrt{2}+y, -y) : 0 < y < 1/n\}$.
+
+Introduced by Mysior in {{zb:0469.54011}}.
+Also described in Example 5 of {{zb:1265.54111}}.

spaces/S000215/properties/P000022.md

@@ -0,0 +1,7 @@
+---
+space: S000215
+property: P000022
+value: false
+---
+
+$U_1(x)$ is clopen and homeomorphic to {S133} and {S133|P22}.

spaces/S000215/properties/P000023.md

@@ -0,0 +1,8 @@
+---
+space: S000215
+property: P000023
+value: false
+---
+
+The closed set $\{0\}\times[0,1)\subseteq X$ is homeomorphic to {S133}
+and {S133|P23}.

spaces/S000215/properties/P000031.md

@@ -0,0 +1,7 @@
+---
+space: S000215
+property: P000031
+value: true
+---
+
+If $\mathcal{U}$ is an open cover of $X$, for each $x\in \mathbb{R}$ pick $n(x)$ such that $U_{n(x)}(x)\subseteq U$ for some $U\in\mathcal{U}$. Let $\mathcal{V} = \{U_{n(x)}(x) : x\in \mathbb{R}\}\cup \{\{y\} : y\in X\setminus \bigcup_{x\in \mathbb{R}} U_{n(x)}(x)\}$, then $\mathcal{V}$ is an open refinement of $\mathcal{U}$, and any point of $X$ is contained in at most two elements of $\mathcal{V}$.

spaces/S000215/properties/P000050.md

@@ -0,0 +1,7 @@
+---
+space: S000215
+property: P000050
+value: true
+---
+
+Each $U_n(x)$ is clopen, as is each singleton $\{(x,y)\}$ with $y\ne 0$.

spaces/S000215/properties/P000051.md

@@ -0,0 +1,8 @@
+---
+space: S000215
+property: P000051
+value: true
+---
+
+Let $Y\subseteq X$ be non-empty. If $Y$ contains a point $(x, y)$ with $y\neq 0$, then $(x, y)$ is isolated in $Y$.
+Otherwise, $Y\subseteq \mathbb{R}\times \{0\}$ and, since $U_n(x)\cap (\mathbb{R}\times \{0\}) = \{(x, 0)\}$, it follows that $Y$ is discrete.

spaces/S000215/properties/P000061.md

@@ -0,0 +1,10 @@
+---
+space: S000215
+property: P000061
+value: true
+---
+
+Note that if $V\subseteq X\setminus (\mathbb{R}\times \{0\})$ then $V = \bigcup_n V_n$ where $V_n = V\setminus(\mathbb{R}\times (-\frac{1}{n}, \frac{1}{n}))$ and each $V_n$ is clopen, so that $V$ is a cozero set as a countable union of cozero sets.
+
+If now $U\subseteq X$, let $V = X\setminus (U\cup (\mathbb{R}\times \{0\}))$.
+Then $V$ is a cozero set disjoint from $U$ and $U\cup V$ contains $X\setminus (\mathbb{R}\times \{0\})$ which is dense in $X$, so that $U\cup V$ is dense in $X$.

spaces/S000215/properties/P000062.md

@@ -0,0 +1,7 @@
+---
+space: S000215
+property: P000062
+value: false
+---
+
+The open cover $\mathcal{U} = \{\mathbb{R}\times (-1, 1)\}\cup \{\{x\} : x\in X\setminus (\mathbb{R}\times (-1, 1))\}$ is a partition of $X$, and if there is a subfamily $\mathcal{V}\subseteq \mathcal{U}$ such that $\bigcup \mathcal{V}$ is dense, then $\mathcal{V} = \mathcal{U}$. Since $\mathcal{U}$ is uncountable, $X$ is not {P62}.

spaces/S000215/properties/P000063.md

@@ -0,0 +1,13 @@
+---
+space: S000215
+property: P000063
+value: true
+---
+
+$X$ is {P50} and {P51}, hence {P6}
+[(Explore)](https://topology.pi-base.org/spaces?q=Zero+dimensional%2BScattered%2B%7E%24T_%7B3+%5Cfrac%7B1%7D%7B2%7D%7D%24).
+
+Let $\mathcal{U}_n = \{U_n(x) : x\in\mathbb{R}\} \cup \{\{y\} : y\in X\setminus \bigcup_{x\in \mathbb{R}} U_n(x)\}$. Suppose that $\mathcal{F}$ is a family of closed subsets of $X$ with finite intersection property, and such that for each $n$ there exists $F_n\in\mathcal{F}$ with $F_n\subseteq U$ for some $U\in\mathcal{U}_n$. If $U = \{y\}$, then $F_n = \{y\}$ and so $y\in \bigcap \mathcal{F}$. So we can assume that $F_n\subseteq U_n(x_n)$ where $x_n\in\mathbb{R}$. If $(x_n, 0)\notin F_n$, then $U_k(x_n)\cap F_n = \emptyset$ for some $k$, and so $F_n\subseteq X\setminus (\mathbb{R}\times (-\frac{1}{k}, \frac{1}{k}))$. And since $F_k\subseteq U_k(x_k)\subseteq \mathbb{R}\times (-\frac{1}{k}, \frac{1}{k})$, we must have $F_k\cap F_n = \emptyset$, which is a contradiction.
+So $(x_n,0)\in F_n$ for all $n$. Since $F_n\cap F_m\neq\emptyset$ it follows that $U_n(x_n)\cap U_m(x_m)\neq\emptyset$ and so $x_n = x_m$ or $1 < |x_n-x_m|\leq 1+\sqrt{2}$.
+But as the interval $[x_1-1-\sqrt{2}, x_1+1+\sqrt{2}]$ is bounded, the set $\{x_n : n\in\mathbb{N}\}$ must be finite, and so there is some $x\in\mathbb{R}$ such that $x_n = x$ for infinitely many $n$.
+If $F\in\mathcal{F}$, then $U_n(x)\cap F\supseteq F_n\cap F\neq\emptyset$ for infinitely many $n$, and so $U_n(x)\cap F\neq\emptyset$ for all $n$, which implies $(x, 0)\in F$ for all $F\in\mathcal{F}$ or in other words $(x, 0)\in\bigcap\mathcal{F}$.

spaces/S000215/properties/P000065.md

@@ -0,0 +1,7 @@
+---
+space: S000215
+property: P000065
+value: true
+---
+
+By definition.

spaces/S000215/properties/P000093.md

@@ -0,0 +1,7 @@
+---
+space: S000215
+property: P000093
+value: false
+---
+
+$U_n(x)$ is uncountable.

spaces/S000215/properties/P000105.md

@@ -0,0 +1,13 @@
+---
+space: S000215
+property: P000105
+value: false
+refs:
+  - mathse: 412625
+    name: Answer to "Every bounded non countable subset of $\mathbb{R}$ has a two-sided accumulation point."
+---
+
+Let $\mathcal{U} = \{U_1(x) : x\in\mathbb{R}\} \cup \{\{y\} : y\in X\setminus \bigcup_{x\in \mathbb{R}} U_1(x)\}$. If $X$ is {P105}, then by taking a locally countable open refinement of $\mathcal{U}$, for every $x\in \mathbb{R}$ there is $n(x)\in\mathbb{N}$ such that $\{U_{n(x)}(x): x\in \mathbb{R}\}$ is locally countable.
+Find $n$ such that $n = n(x)$ for uncountably many $x\in\mathbb{R}$, and let $C = \{x\in\mathbb{R} : n = n(x)\}$.
+Take a left-sided condensation point $x'$ of $C$; that is, for any $x < x'$ the set $(x, x')\cap C$ is uncountable (see {{mathse:412625}} for a proof that such a point exists).
+Then each basic neighborhood $U_m(x'+1)$ of $x'+1$ intersects uncountably many $U_n(y)$ with $y\in C$ and close enough to the left of $x$, which contradicts the local countability condition above.

spaces/S000215/properties/P000110.md

@@ -0,0 +1,10 @@
+---
+space: S000215
+property: P000110
+value: true
+---
+
+A development for $X$ is given by the open covers $\mathscr U_1,\mathscr U_2,\dots$ with
+
+$\quad\quad\mathscr U_n=\big\{U_n(x) : x\in\mathbb R\big\}
+\cup \big\{\{z\} : z\in X\setminus(\mathbb R\times\{0\}\big\}.$

spaces/S000215/properties/P000120.md

@@ -0,0 +1,7 @@
+---
+space: S000215
+property: P000120
+value: true
+---
+
+$U_1(x)$ is homeomorphic to {S133} and {S133|P133}.

spaces/S000215/properties/P000162.md

@@ -0,0 +1,10 @@
+---
+space: S000215
+property: P000162
+value: false
+refs:
+  - mathse: 4718866
+    name: Mysior plane is not realcompact
+---
+
+Proved in {{mathse:4718866}}.

spaces/S000215/properties/P000227.md

@@ -0,0 +1,7 @@
+---
+space: S000215
+property: P000227
+value: true
+---
+
+$\mathbb{R}\times \{0\}$ is a closed discrete subset of size $\mathfrak{c}$