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?

  • $\begingroup$ 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 $\endgroup$ – Peter Melech Sep 28 '18 at 10:21
  • $\begingroup$ 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. $\endgroup$ – 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" $|\mathbb{R}|= \mathfrak{c}$.

  • $\begingroup$ +1. Note for the OP that in particular, more open sets $\not=$ finer if by "more" we simply mean "greater cardinality." $\endgroup$ – Noah Schweber Sep 28 '18 at 18:14

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

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

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.