# Is the distinguish topology finer than the usual topology?

Let x$$_0$$ be any real number, the distinguished point topology on $$\mathbb{R}$$ is given by T$$_{x_0}$$ = {B $$\subset$$ $$\mathbb{R}$$: x$$_0$$ $$\in$$ B or B = $$\emptyset$$ }

Let U be the usual topology on $$\mathbb{R}$$ such that U = {V $$\subset$$ $$\mathbb{R}$$ $$\vert$$ if x $$\in$$ V, then there exists a,b $$\in$$ $$\mathbb{R}$$ such that x $$\in$$ (a, b) $$\subset$$ V}

Is the distinguish point topology strictly finer than the usual topology?

I can tell you that the distinguish point is not coarser than the usual because, say x$$_0$$ = 1, {1} is x$$_0$$-open, but certainly any singletons in the usual are not open.

However, I'm trying to think of U-open sets and considering if they're distinguish point open. Say (0,1), it contains all numbers between 0 and 1, so I think the distinguish point of any points in between would be true, but (0,1) isn't a distinguish point for 1 or 0. Could someone clarify to me if the distinguish point is finer or non-comparable to the usual?

• You answered Your question by yourself, $(0,1)$ is open in the usual topology but not in $T_1$ in the same way it is easy to find open sets that are not open in a given distinguish topology $T_{x_0}$. So an arbitrary distinguish topology is neither coarser nor finer than the usual topology – Peter Melech Sep 28 '18 at 10:21
• oh thank you, I was just a bit hesitant because if x $\in$ (0,1), then x $\in$ \{x\} which is a subset to (0,1), but if I can use a specific distinguish point topology like x$_0$=1, then I understand. – Ren Sep 28 '18 at 10:24

No, it's not strictly finer. Let's take $x_0 = 0$ for definiteness. Then $(1,2) \in \mathcal{U}$ but $(1,2) \notin \mathcal{T}_0$. Also, $\{0\} \in \mathcal{T}_0$ but $\{0\} \notin \mathcal{U}$. So the topologies are not comparable.
$\mathcal{T}_0$ does have way more members (namely $2^\mathfrak{c}$) than $\mathcal{U}$ that has "only" $|\mathbb{R}|= \mathfrak{c}$.
• +1. Note for the OP that in particular, more open sets $\not=$ finer if by "more" we simply mean "greater cardinality." – Noah Schweber Sep 28 '18 at 18:14
• $$[x_0-b,x_0+b]\in\mathcal{T}_{x_0}\setminus\mathcal U$$;
• $$(x_0,x_0+1)\in\mathcal{U}\setminus\mathcal{T}_{x_0}$$.