Talk:Deleted Integer Topology is Weakly Countably Compact

Now you've brought this up I believe that ths space may not be w.c.c after all.

Take the set $T = \left\{{n + \dfrac 1 2: n \in \Z_{\ge 0}}\right\}$. Then $S \supseteq T$ and $T$ is infinite, but any open set $\left({n .. n+1}\right) \subseteq S$ contains only one point $n + \frac 1 2 \in T$ and so $x$ is not a limit point of $T$ after all. So by definition $S$ is not w.c.c.

Please check my reasoning and we can then rewrite and so rename this page to d.i.t is not w.c.c.

My fault, I was seduced by S&S who said "[The deleted integer topology] has most of the properties of the odd-even topology". In the glossary at the back they have omitted to specify any properties for this set. --prime mover 00:34, 3 December 2011 (CST)

The reasoning seems to be correct (well, I am almost sure it is). I will, however, first occupy myself with the implementation of $\R_{\ge 0}$ and the like. --Lord_Farin 02:26, 3 December 2011 (CST)

No worries, I can get onto it myself (later, the machine's battery needs recharging). --prime mover 03:12, 3 December 2011 (CST)

The definition of w.c.c. that I read here only needs a limit point in $S$ not in $T$. Your example has a lot of limit points in $S\setminus T$. For example $\dfrac{1}{3}\not\in T$, but any open neighbourhood of $\dfrac{1}{3}$ contains $(0 .. 1)$ and thus $\dfrac{1}{2}\in T$; and by definition of limit point, $\dfrac{1}{3}$ is a limit of $T$ in $S$. --Dan232 07:36, 3 December 2011 (CST)

!!! Dead right. I must have been more tired this morning than I thought. I was certain it said "limit point in $T$". Okay, so the d.i.t. may well be w.c.c. after all. --prime mover 07:52, 3 December 2011 (CST)