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?

  • $\begingroup$ 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 $x,y$. $\endgroup$ – Kabo Murphy Jan 24 at 23:33

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.

  • $\begingroup$ 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 at 23:45
  • $\begingroup$ @kmm You are right. I have corrected the proof. $\endgroup$ – Kabo Murphy Jan 24 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 at 23:53
  • $\begingroup$ @kmm Please check if the proof is OK now. $\endgroup$ – Kabo Murphy Jan 25 at 0:19
  • $\begingroup$ It looks good to me, thank you very much. $\endgroup$ – Kasper Jan 25 at 0:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.