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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?

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

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  • $\begingroup$ +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
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Neither is finer nor coerser than the other one. For instance,

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