2
$\begingroup$

I think I have a proof, but it's a little convoluted.

Since $\tau$ is strictly finer than than $\tau_{st}$, there is an open set $U \in \tau$ such that $U \notin \tau_{st}$. For every $x \in \mathbb{R}$, the sets $U \cup (-\infty, x+1) $ and $U \cup (x, +\infty)$ are both open in $\tau$. However, at most one of the two can be in $\tau_{st}$. For the sake of contradiction, assume both are in $\tau_{st}$, then so is their union, which is $U$, which cannot be true. Thus for every $x \in \mathbb{R}$ we have constructed at least one set open in $\tau$ which is not in $\tau_{st}$, proving their difference is not countable.

Is this correct? Can it be done easier?

$\endgroup$
1
  • $\begingroup$ The argument is incomplete. You have to consider the possibility that $U \cup (x,\infty) =U \cup (y,\infty)$ may hold for lots of pairs $x,y$. $\endgroup$ Jan 24, 2019 at 23:33

1 Answer 1

4
$\begingroup$

Instead of the sets you are considering look at $U\setminus \{x\}$ for $x \in \mathbb R$. You can see that $U\setminus \{x\}\neq U\setminus \{y\}$ whenever $x,y \in U$ and $x \neq y$ ; also $U\setminus \{x\} \in \tau_{st} $ for at most one $x$ (because $U =U\setminus \{x\})\cup U\setminus \{y\}))$ for $x \neq y$ and $U\setminus \{x\} \in \tau $ for all $x$. When $U$ is countable consideration of the intervals $U \cup (x,x+1)$ can be used. I leave the details to you.

$\endgroup$
7
  • $\begingroup$ Won't $U \cup \{x\}^c$ equal $U \cup \{y\}^c$ whenever $x$ and $y$ are either both in $U$ or both not in $U$? $\endgroup$
    – Kasper
    Jan 24, 2019 at 23:45
  • $\begingroup$ @kmm You are right. I have corrected the proof. $\endgroup$ Jan 24, 2019 at 23:49
  • $\begingroup$ I am sorry if I am missing something obvious, but if $x \in U^c$, won't $U \cup \{x\}^c$ just be $\mathbb{R} \setminus \{x\}$, which is in $\tau_{st}$, so it's not in $\tau \setminus \tau_{st}$ $\endgroup$
    – Kasper
    Jan 24, 2019 at 23:53
  • $\begingroup$ @kmm Please check if the proof is OK now. $\endgroup$ Jan 25, 2019 at 0:19
  • $\begingroup$ It looks good to me, thank you very much. $\endgroup$
    – Kasper
    Jan 25, 2019 at 0:24

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .