Add some traits of S18

properties/P000162.md

@@ -6,13 +6,15 @@ refs:
     name: General Topology (Engelking, 1989)
   - doi: 10.1007/978-1-4615-7819-2
     name: Rings of Continuous Functions (Gillman and Jerison)
+  - wikipedia: Ultrafilter
+    name: Ultrafilter on Wikipedia
 ---
 
-A space that is homeomorphic to a closed subset of a (not necessarily finite) power of {S25}.
+A space $X$ that is homeomorphic to a closed subset of $\mathbb{R}^\kappa$ for some cardinal $\kappa$.
 
-Equivalently (see {{doi:10.1007/978-1-4615-7819-2}}), every real $z$-ultrafilter $\mathcal U$ on the space $X$ is fixed,
-i.e., we can find $x\in X$ such that $\mathcal U=\{Z\in Z(X):x\in Z\}$.
-Here, $Z(X)$ is the collection of all zero sets of X, a $z$-ultrafilter is an ultrafilter on the lattice $Z(X)$, and
-it is real when it is closed under countable intersections.
+Equivalently (see {{doi:10.1007/978-1-4615-7819-2}}), $X$ is {P6} and every real $z$-ultrafilter $\mathcal U$ on the space $X$ is fixed,
+that is, $\bigcap\mathcal{U}\neq\emptyset$.
+
+A *$z$-ultrafilter* is an ultrafilter on the lattice of zero-sets of $X$ (see {{wikipedia:Ultrafilter}} for the general definition of an ultrafilter on a poset). A *real $z$-ultrafilter* is a $z$-ultrafilter with countable intersection property, that is, for any countable $\mathcal{F}\subseteq \mathcal{U}$ we have $\bigcap\mathcal{F}\neq \emptyset$.
 
 See also section 3.11 in {{zb:0684.54001}}.

properties/P000164.md

@@ -1,14 +1,28 @@
 ---
 uid: P000164
-name: Non-measurable cardinality
+name: Cardinality less than every measurable cardinal
+aliases:
+  - Non-measurable cardinality
 refs:
 - wikipedia: Measurable_cardinal
   name: Measurable cardinal on Wikipedia
 - doi: 10.1007/978-1-4615-7819-2
   name: Rings of Continuous Functions (Gillman and Jerison)
+- doi: 10.1007/3-540-44761-X
+  name: Set Theory (Jech)
 ---
 
-The cardinality of the space is not a measurable cardinal ({{wikipedia:Measurable_cardinal}}).
+The cardinality of the space is smaller than every measurable cardinal, if one exists.
 
-(Note that the existence of a measurable cardinal cannot be proven in ZFC, so no space should ever have this
-property marked as false. See {T383}.)
+A cardinal $\kappa$ is called *measurable* if $\kappa$ is uncountable and there exists a measure $\mu:2^\kappa\to \{0, 1\}$ such that 
+
+1. $\mu$ is *$\kappa$-additive*: $\mu(\bigcup_{i\in I} A_i) = \sum_{i\in I} \mu(A_i)$ for any family $(A_i)_{i\in I}\subseteq 2^\kappa$ of pairwise disjoint sets with $|I| < \kappa$,
+
+2. $\mu$ is non-trivial: $\mu(\kappa) = 1$ and $\mu(\{x\}) = 0$ for all $x\in \kappa$. 
+
+Equivalently, $\kappa$ is uncountable and there exists a free ultrafilter $\mathcal{U}$ on $\kappa$ such that $\mathcal{U}$ is *$\kappa$-complete*, i.e., if $\mathcal{F}\subseteq \mathcal{U}$ and $|\mathcal{F}| < \kappa$ then $\bigcap\mathcal{F}\in \mathcal{U}$. (See {{wikipedia:Measurable_cardinal}} for more details.)
+
+Note: Some authors, for example {{doi:10.1007/978-1-4615-7819-2}}, refer to measurable cardinals as those cardinals $\kappa$ for which there exists a $\sigma$-additive measure $\mu:2^\kappa\to \{0, 1\}$ which is non-trivial. If $\kappa$ is the smallest such cardinal, then a non-trivial $\sigma$-additive measure $\mu:2^\kappa\to \{0, 1\}$ is $\kappa$-additive (see lemma 10.2 of {{doi:10.1007/3-540-44761-X}} and comments preceeding it), so $\kappa$ is also measurable by the above definition.
+
+(The existence of a measurable cardinal cannot be proven in ZFC.
+So spaces whose construction does not depend on set-theoretic axioms beyond ZFC should never have this property marked as false.)

spaces/S000017/README.md

@@ -11,7 +11,7 @@ refs:
   - wikipedia: Cocountable_topology
     name: Cocountable topology on Wikipedia
 ---
-Let $X=\mathbb R$ be the set of real numbers.  Define the open sets on $X$ by a letting a set $U \subset X$ be open iff its complement is countable.  Taking the collection of all such sets, $U$, together with both the $\emptyset$ and $X$ yields a topology on $X$.
+Let $X=\mathbb R$ be the set of real numbers.  Define the topology on $X$ by letting a set in $X$ be open iff it is empty or its complement is countable. 
 
 Defined as counterexample #20 ("Countable Complement Topology")
 in {{doi:10.1007/978-1-4612-6290-9}}.

spaces/S000017/properties/P000071.md

@@ -0,0 +1,9 @@
+---
+space: S000017
+property: P000071
+value: false
+---
+
+Suppose $A\subseteq X$ is infinite and let $(x_n)\subseteq A$ be a sequence of distinct elements of $A$. If $U_n = X\setminus \{x_m : m\geq n\}$ then the sequence $(U_n)$ is increasing, consists of open sets, and $\bigcup_n U_n = X$. If $A$ were relatively compact then there would be $n$ for which $A\subseteq U_n$. But $x_n\in A\setminus U_n$ so that's impossible.
+
+Since relatively compact subsets of $X$ are finite, if $X$ were {P71} it would be countable. But {S17|P65}.

spaces/S000017/properties/P000126.md

@@ -0,0 +1,7 @@
+---
+space: S000017
+property: P000126
+value: false
+---
+
+$[0, \infty)$ is neither closed nor open in $X$.

spaces/S000017/properties/P000172.md

@@ -0,0 +1,7 @@
+---
+space: S000017
+property: P000172
+value: true
+---
+
+If $p\in\overline{A}$ and $A$ is countable, then $p\in A$ so one can take a constant sequence. If $A$ is uncountable, pick any sequence $(x_\lambda)_{\lambda < \omega_1}$ of distinct points such that $x_\lambda\in A$. If $U$ is any neighourhood of $p$, then $x_\lambda\notin U$ for at most countably many $\lambda$. If $\beta$ is supremum of those $\lambda$, then $\beta < \omega_1$ since $\text{cf}(\omega_1) = \omega_1$, and so $x_\lambda\in U$ for $\lambda > \beta$. It follows that $(x_\lambda)_{\lambda < \omega_1}$ converges to $p$.

spaces/S000017/properties/P000189.md

@@ -0,0 +1,7 @@
+---
+space: S000017
+property: P000189
+value: true
+---
+
+Since closed subsets properly contained in $X$ are countable, if $X$ were not $\sigma$-connected then it would be a countable union of countable sets, and so countable. But {S17|P65}.

spaces/S000018/properties/P000071.md

@@ -0,0 +1,7 @@
+---
+space: S000018
+property: P000071
+value: false
+---
+
+The Kolmogorov quotient of {S18} is {S17} and {S17|P71}.

spaces/S000018/properties/P000147.md

@@ -0,0 +1,7 @@
+---
+space: S000018
+property: P000147
+value: true
+---
+
+The Kolmogorov quotient of {S18} is {S17} and {S17|P147}.

spaces/S000018/properties/P000189.md

@@ -0,0 +1,7 @@
+---
+space: S000018
+property: P000189
+value: true
+---
+
+The Kolmogorov quotient of {S18} is {S17} and {S17|P189}.

spaces/S000018/properties/P000206.md

@@ -0,0 +1,7 @@
+---
+space: S000018
+property: P000206
+value: true
+---
+
+The Kolmogorov quotient of {S18} is {S17} and {S17|P206}.

spaces/S000048/properties/P000001.md

@@ -2,11 +2,6 @@
 space: S000048
 property: P000001
 value: true
-refs:
-- doi: 10.1007/978-1-4612-6290-9_6
-  name: Counterexamples in Topology
 ---
 
-If $P,Q \in X$ are distinct, in if $P = (p)$ and $Q = (q)$. Then, without loss of generality, there is some integer $r$ dividing $p$ but not $q$. Thus $P \notin V_r$ but $Q \in V_r$.
-
-See item #3 for space #56 in {{doi:10.1007/978-1-4612-6290-9_6}}.
+By inspection.

spaces/S000048/properties/P000013.md

@@ -2,11 +2,6 @@
 space: S000048
 property: P000013
 value: false
-refs:
-- doi: 10.1007/978-1-4612-6290-9_6
-  name: Counterexamples in Topology
 ---
 
-The ideals $(2)$ and $(3)$ are disjoint closed sets but there are no disjoint open sets in $X$.
-
-See item #3 for space #56 in {{doi:10.1007/978-1-4612-6290-9_6}}.
+$\{0\}$ and $\{1\}$ are disjoint closed sets but there are no nonempty disjoint open sets in $X$.

spaces/S000048/properties/P000027.md

@@ -2,11 +2,6 @@
 space: S000048
 property: P000027
 value: true
-refs:
-- doi: 10.1007/978-1-4612-6290-9_6
-  name: Counterexamples in Topology
 ---
 
-The sets $V_x$ for $x \in \mathbb{Z}^+$ form a countable basis for $X$.
-
-See item #4 for space #56 in {{doi:10.1007/978-1-4612-6290-9_6}}.
+By construction, it is easy to see that there are only countably many open sets.

spaces/S000048/properties/P000042.md

@@ -2,9 +2,7 @@
 space: S000048
 property: P000042
 value: true
-refs:
-- doi: 10.1007/978-1-4612-6290-9_6
-  name: Counterexamples in Topology
 ---
 
-See item #5 for space #56 in {{doi:10.1007/978-1-4612-6290-9_6}}.
+Every nonempty open subset of $X$ has a generic point, namely $p$. And {P201} implies
+{P37} [(Explore)](https://topology.pi-base.org/spaces?q=Has+a+generic+point+%2B+not+Path+connected).

spaces/S000048/properties/P000044.md

@@ -2,10 +2,6 @@
 space: S000048
 property: P000044
 value: false
-refs:
-- doi: 10.1007/978-1-4612-6290-9_6
-  name: Counterexamples in Topology
 ---
 
-Asserted in the General Reference Chart for space #56 in
-{{doi:10.1007/978-1-4612-6290-9_6}}.
+$\{0,p\}$ and $\omega\setminus\{0\}$ form a partition of $X$ into two connected subspaces, each with at least two points.

spaces/S000048/properties/P000138.md

@@ -0,0 +1,7 @@
+---
+space: S000048
+property: P000138
+value: false
+---
+
+Any $f: X \to X$ s.t. $f(p) = p$ and restricts to a permutation of $\omega$ is easily seen to be continuous. Since there are continuum many permutations of $\omega$, $X$ has at least continuum many self-maps.

spaces/S000048/properties/P000181.md

@@ -0,0 +1,7 @@
+---
+space: S000048
+property: P000181
+value: true
+---
+
+By definition.

spaces/S000048/properties/P000192.md

@@ -0,0 +1,9 @@
+---
+space: S000048
+property: P000192
+value: true
+---
+
+A nonempty closed subset $A \subset X$, by definition, is either the entirety of $X$, which has $p$ as a generic point; or is a
+nonempty finite subset of $\omega$. In the latter case, $A$ has discrete topology, so it is {P39} iff it is a
+singleton, which, of course, has a generic point as well.

spaces/S000050/README.md

@@ -0,0 +1,10 @@
+---
+uid: S000050
+name: Rationals extended by a focal point
+aliases:
+  - Non-Hausdorff cone over the rationals
+  - Open extension of the rationals
+---
+
+Let $X = \mathbb{Q} \cup \{\infty\}$, where $\mathbb{Q}$ has the usual topology on {S27} and is an open subset
+of $X$, and where the only neighborhood of $\infty$ is $X$. See {P202}.

spaces/S000050/properties/P000010.md

@@ -0,0 +1,8 @@
+---
+space: S000050
+property: P000010
+value: true
+---
+
+It is easy to check any regular open $U \subsetneq \mathbb{Q}$ is still regular open in $X$. And, of course, $X$ is regular open in itself. The result now easily follows from {S27|P10} and
+{S27|P202}.

spaces/S000050/properties/P000014.md

@@ -0,0 +1,10 @@
+---
+space: S000050
+property: P000014
+value: true
+---
+
+Note that as {S50|P202}, it is
+{P13} [(Explore)](https://topology.pi-base.org/spaces?q=Has+a+point+with+a+unique+neighborhood+%2B+not+Normal).
+Now, any open subset of $X$ is either $X$ itself, which is normal; or an open subset of $\mathbb{Q}$, which is normal
+since {S27|P14}.

spaces/S000050/properties/P000027.md

@@ -0,0 +1,7 @@
+---
+space: S000050
+property: P000027
+value: true
+---
+
+{S27|P27} and any open basis of {S27} together with $X$ forms an open basis of $X$.

spaces/S000050/properties/P000041.md

@@ -0,0 +1,8 @@
+---
+space: S000050
+property: P000041
+value: false
+---
+
+Being {P41} passes to open subspaces. {S27} is an open subspace of $X$ and
+{S27|P41}.

spaces/S000050/properties/P000045.md

@@ -0,0 +1,9 @@
+---
+space: S000050
+property: P000045
+value: true
+---
+
+Since {S50|P202}, it is
+{P36} [(Explore)](https://topology.pi-base.org/spaces?q=Has+a+point+with+a+unique+neighborhood+%2B+not+Connected).
+And, $X\setminus\{\infty\}$ is {S27}. {S27|P47}.

spaces/S000050/properties/P000056.md

@@ -0,0 +1,7 @@
+---
+space: S000050
+property: P000056
+value: true
+---
+
+It is easy to see that $\{q,\infty\}$ is nowhere dense for all $q \in \mathbb{Q}$, and $X$ is the countable union of all such sets.

spaces/S000050/properties/P000130.md

@@ -0,0 +1,8 @@
+---
+space: S000050
+property: P000130
+value: false
+---
+
+Being {P130} passes to open subspaces. {S27} is an open subspace of $X$ and
+{S27|P130}.

spaces/S000050/properties/P000181.md

@@ -0,0 +1,7 @@
+---
+space: S000050
+property: P000181
+value: true
+---
+
+By construction.

spaces/S000050/properties/P000192.md

@@ -0,0 +1,10 @@
+---
+space: S000050
+property: P000192
+value: true
+---
+
+Any nonempty closed subset $A$ of $X$ has to contain $\infty$. If $A = \{\infty\}$, then $A$ clearly has a generic point. Otherwise,
+assume $\{\infty\} \subsetneq A$ and $A$ is {P39}. We see that $A$ cannot contain more than one point of
+$\mathbb{Q}$, as {S27|P3} and $\mathbb{Q}$ is open in $X$. Thus, $A = \{q,\infty\}$ for some
+$q \in \mathbb{Q}$, and $q$ is a generic point of $A$.

spaces/S000050/properties/P000202.md

@@ -0,0 +1,7 @@
+---
+space: S000050
+property: P000202
+value: true
+---
+
+By definition, the only neighborhood of $\infty$ is $X$.