Weakly First countable

properties/P000228.md

@@ -0,0 +1,25 @@
+---
+uid: P000228
+name: Weakly first countable
+aliases:
+  - g-first countable
+refs:
+- zb: "0171.43603"
+  name: Mappings and spaces (Arkhangel’skiĭ)
+- zb: "0285.54022"
+  name: On defining a space by a weak base. (Siwiec)
+---
+
+One can choose for each $x\in X$ a decreasing sequence $\mathcal T_x=(V_n(x))_{n\in\mathbb N}$ of subsets of $X$ containing $x$, such that the family $\{\mathcal T_x:x\in X\}$ satisfies:
+
+$\quad$A subset $U\subseteq X$ is open in $X$ iff for every $x\in U$ there is some $n$ such that $V_n(x)\subseteq U$.
+
+*Note*: The family $\{\mathcal T_x:x\in X\}$ is a special case of a *weak base* for $X$.  See the references for details.
+
+See Definition 2.3 in {{zb:0171.43603}}. This property is called *g-first countable* in {{zb:0285.54022}}.
+
+----
+#### Meta-properties
+
+- $X$ satisfies this property if its Kolmogorov quotient $\text{Kol}(X)$ does.
+- This property is preserved by arbitrary disjoint unions.

theorems/T000309.md

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+---
+uid: T000309
+if:
+  P000028: true
+then:
+  P000228: true
+---
+
+For every $x \in X$, let $(C_n(x))_{n \in \mathbb{N}}$ be a countable neighborhood basis induced by {P28}.
+Then $V_n(x) := \bigcap_{k \leq n}C_k(x)$ works.

theorems/T000626.md

@@ -3,7 +3,8 @@ uid: T000626
 if:
   P000104: true
 then:
-  P000079: true
+  P000228: true
 ---
 
-Let $A \subseteq X$ be sequentially closed and $x \in X \setminus A$. It suffices to show $d(x,A)>0$. Assume to the contrary that $d(x,A)=0$. Then there exists a sequence of elements $a_n$ of $A$ s.t. $d(x,a_n)\to 0$. It is easy to see that this implies $a_n \to x$, so as $A$ is sequentially closed, we have $x \in A$, a contradiction.
+For every $x \in X, n \in \mathbb{N}$ set $V_n(x) := B_d(x, \frac{1}{n}) = \{y\in X:d(x,y)<\frac{1}{n}\}$
+with $d$ being the symmetric in the definition of {P104}.

theorems/T000840.md

@@ -0,0 +1,15 @@
+---
+uid: T000840
+if:
+  P000228: true
+then:
+  P000079: true
+---
+
+Suppose $A\subseteq X$ is not a closed set.
+Then $A^c=X\setminus A$ is not open and by the {P228} property
+there is some $x\in A^c$ such that $A^c$ does not contain any $V_n(x)$, that is, every $V_n(x)$ meets $A$.
+Choose $x_n\in V_n(x)\cap A$ for each $n$.
+The sequence $x_1,x_2,\dots$ converges to $x$,
+since every neighborhood of $x$ contains all $V_n(x)$ for $n$ sufficiently large.
+Thus $A$ is not sequentially closed.