# If $\tau$ is strictly finer than the standard topology $\tau_{st}$ on $\mathbb{R}$, prove their difference $\tau \setminus \tau_{st}$ is uncountable

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?

• 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 $x,y$. Jan 24, 2019 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.
• Won't $U \cup \{x\}^c$ equal $U \cup \{y\}^c$ whenever $x$ and $y$ are either both in $U$ or both not in $U$? Jan 24, 2019 at 23:45
• 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 $\tau \setminus \tau_{st}$ Jan 24, 2019 at 23:53