Is the distinguish topology finer than the usual topology?

Question

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?

Answer 1

Neither is finer nor coerser than the other one. For instance,

• $[x_0-b,x_0+b]\in\mathcal{T}_{x_0}\setminus\mathcal U$;
• $(x_0,x_0+1)\in\mathcal{U}\setminus\mathcal{T}_{x_0}$.

Answer 2

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" $|\mathbb{R}|= \mathfrak{c}$.