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Completing S64 (pointed rational ext. of plane) (#1412)
Co-authored-by: yhx-12243 <yhx12243@gmail.com> Co-authored-by: Patrick Rabau <70125716+prabau@users.noreply.github.com>
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spaces/S000064/README.md

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---
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uid: S000064
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name: Pointed rational extension of the plane
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name: Rational extension of the plane
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counterexamples_id: 72
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refs:
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- doi: 10.1007/978-1-4612-6290-9
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- zb: "0386.54001"
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name: Counterexamples in Topology
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---
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If $(X,\tau)$ is the Euclidean plane, we define a new topology for $X$ by declaring open each point in the set $D = \{(x,y) | x \in Q, y \in Q\}$, and each set of the form $\{x\} \cup (D \cap U)$ where $x \in U \in \tau$.
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If $(X,\tau)$ is the {S176},
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we define a new topology for $X$ by declaring open each point in the set
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$D = \mathbb Q^2$, and each set of the form $\{x\} \cup (D \cap U)$
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where $x \in U \in \tau$.
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The topology is finer than that of {S176}.
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The construction combines the ideas from
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{S60} and {S62}.
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Defined as counterexample #72 ("Rational Extension in the Plane")
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in {{doi:10.1007/978-1-4612-6290-9}}.
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in {{zb:0386.54001}}.
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---
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space: S000064
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property: P000010
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value: false
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---
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Let $p\in X\setminus D$. Take $U$ be an open neighbourhood of $p$ such that $U\setminus D=\{p\}$ (any base for the topology has to contain such an open set).
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By definition, there exists a Euclidean-open $V$ such that $p\in V$ and $V\cap D\subset U$.
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Then $V\subset \mathrm{int}(\overline{V\cap D})\subset \mathrm{int}(\overline{U})$ and $V$ is not contained in $U$.
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This shows that $U$ is not a regular open set.

spaces/S000064/properties/P000011.md

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spaces/S000064/properties/P000018.md

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spaces/S000064/properties/P000022.md

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property: P000022
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value: false
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refs:
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- doi: 10.1007/978-1-4612-6290-9_6
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- zb: "0386.54001"
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name: Counterexamples in Topology
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---
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The function $f : X \to \mathbb{R}$ defined by $f(x,y) = x$ can be shown to be continuous, and is unbounded.
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Asserted in the General Reference Chart for space #72 in
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{{doi:10.1007/978-1-4612-6290-9_6}}.
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{{zb:0386.54001}}.

spaces/S000064/properties/P000026.md

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property: P000026
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value: true
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refs:
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- doi: 10.1007/978-1-4612-6290-9_6
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- zb: "0386.54001"
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name: Counterexamples in Topology
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---
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$D$ is countable and dense.
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See item #3 for space #72 in {{doi:10.1007/978-1-4612-6290-9_6}}.
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See item #3 for space #72 in {{zb:0386.54001}}.

spaces/S000064/properties/P000028.md

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spaces/S000064/properties/P000032.md

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property: P000032
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value: false
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refs:
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- doi: 10.1007/978-1-4612-6290-9_6
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- zb: "0386.54001"
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name: Counterexamples in Topology
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---
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Asserted in the General Reference Chart for space #72 in
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{{doi:10.1007/978-1-4612-6290-9_6}}.
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Consider an open cover consisting of $X\setminus (\mathbb Q\times\{\sqrt 2\})$ and $\{(q,\sqrt 2)\}\cup D$ for $q\in\mathbb Q$.
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Suppose there is a refinement $\mathscr V$ of the above cover. For each $q\in\mathbb Q$ there exists Euclidean-open $V_q$ such that $(q,\sqrt 2)\in V_q$ and
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$V_q\cap D$ is contained in an element of $\mathscr V$.
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Picking a convergent sequence of rationals $q_n \to q$ we
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observe that infinitely many $V_{q_n}$'s have to intersect any neighbourhood of $(q,\sqrt{2})$, therefore $\mathscr V$
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is not locally finite.

spaces/S000064/properties/P000033.md

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spaces/S000064/properties/P000048.md

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property: P000048
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value: false
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refs:
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- doi: 10.1007/978-1-4612-6290-9_6
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- zb: "0386.54001"
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name: Counterexamples in Topology
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---
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If $i$ is irrational, no two points $x_1 = (i, y_1), x_2 = (i, y_2)$ on the vertical line $x = i$ can be separated.
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See item #5 for space #72 in {{doi:10.1007/978-1-4612-6290-9_6}}.
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See item #5 for space #72 in {{zb:0386.54001}}.

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