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Let $\tau$ be the coarsest topology on $\mathbb{R}$ that contains the euclidean topology and all subsets of $\mathbb{Q}$. We already know that $\tau_{Eucl} \cup P(\mathbb{Q}$ is a base for $\tau$. We also have a metric $d$ on $\mathbb{R}$ defined as the following. Let $(r_k)_{k\in \mathbb{Z}_{>0}}$ be a sequence of different rational such that $\mathbb{Q}=\{r_k\mid k\in \mathbb{Z}_{>0}\}$.
Now let $d(x,y)$ for a $x,y \in \mathbb{R}$. $$ d(x,y) = \left\{ \begin{array}{lll} 0 & \mbox{if} \: x=y,\\ \sup\{1/k|x\leq r_k \leq y\} & \mbox{if} \:x<y, \\ \sup\{1/k|y\leq r_k \leq x\} & \mbox{if} \:y<x. \end{array} \right. $$ Now we're asked to prove that the topology defined by this metric is exactly $\tau$.
My attempt:

To prove this statement we need to prove that this topology is both finer and coarser than $\tau$. Since the topology defined by this metric, is generated by unions of open balls. We need to prove on one hand that a random open ball is open in $\tau$. On the other hand we need to prove that a random open in $\tau$ can be described as a union of open balls. But that's how far I get. I don't really see how to characterize these open balls. Does anybody maybe have a tip to get started? We also know that for the metric topology, being open is characterized as follows. A subset $U$ is open iff for every $x \in U$ there exists an $\epsilon >0$ such that $B(x,\epsilon)\subseteq U$. so that if we can prove that this holds for every open in $\tau$ we know that $\tau$ is coarser then the metric topology. I am also thinking I should use the fact that $\tau$ is the coarsest topology that contains the euclidean topology and all subsets of $\mathbb{Q}$.

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Let start with this lemma:

Let $r_h< r_k \in \mathbb{Q}$, then $[r_h,r_k]$ is open in the metric topology.

To prove this, consider $A=\mathbb{R} / [r_h,r_k] $ and let $\{x_n\}$ a sequence contained in $A$. Suppose there exists $\ell \in [r_h,r_k]$ such that, for each $\varepsilon >0$ there is $N$ such that, for each $n>N$ $$ d(x_n,\ell)< \varepsilon $$ but we know that $d(x_n,\ell) \ge \min\left(\frac{1}{h},\frac{1}k\right)$, for each $n$. so $A$ must contains all its limit points, i.e $A$ is closed. So the claim follow

Let $(a,b)$ an Euclidean open interval. For each $x \in (a,b)$ there exist $r_h<r_k \in \mathbb{Q}$ such that $$ a<r_h<x<r_k<b $$ so $(a,b)$ is open in the metric topology.

Let $r_{h_1} < r_{h_2}< r_{h_3}$ rational numbers. We have

$$ \{r_{h_2}\} = [r_{h_1}, r_{h_2}] \cap [r_{h_2},r_{h_3}] $$ so the singleton of $\mathbb{Q}$ are open in the metric topology.

So the metric topology is finer than $\tau_{\text{Eucl}} \cup \mathcal{P}(\mathbb{Q})$

Let now $B_{1/N}(z)$ an open ball of the metric topology with radius $\frac{1}N$ centered at $z \in \mathbb{R}$.

We need the following lemma:

Let $x,y \in B_{1/N}(z)$ such that $d(x,y)>0$ than there exists $r_{h_1},r_{h_2} \in B_{1/N}(z)$ such that $y \in [r_{h_1},r_{h_2}]$

Indeed we have $d(x,y)=\frac{1}{K}$ for some $K \ge N$. Assume $y>x$. By the definition of distance it follows that $x \le r_K \le y$ and that, for any $r_h \in [x,y]$, $h \ge K \ge N$.

Now consider $r_{h_1}\ge y$ with $h_1 \ge N$. If $r_{h_1} \in B_{1/N}(z)$ we have finished, otherwise there is $r_{h_2} \in [y, r_{h_1})$ such that $h_2 <N$( by the definition of the distance) i.e. $h_2 \in \{ 1, \cdots N -1 \}$.

Let now $r_\tilde{h} \in [y, r_{h_2})$ with $\tilde{h} \ge N$ (it exists by the density of the rationals) . If $r_{\tilde{h}} \in B_{1/N}(z)$ we have finished, otherwise there is $r_{h_3} \in [y, r_{\tilde{h}})$ such that $h_3 \le N$, so $h_3 \in \{1,\cdots N-1 \} / \{h_2\}$.

By iterating this procedure at most $N-1$ times we will obtain $r_{h_i} \ge y$ such that $r_{h_i} \in B_{1/N}(z) $

The proof is identical if $y<x$

So suppose $B_{1/N}(z)$ is not a singleton, by the lemma above, any $ y \in B_{1/N}(z)$ is contained in $[r_{h_1},r_{h_2}]$ that is open in $\tau_{\text{Eucl}} \cup \mathcal{P}(\mathbb{Q})$ as such is an open set in that topology.

If $B_{1/N}(z)$ is a singleton then $z$ is rational and so it is open

Consider a general open ball $B_{\lambda}(z)$ for some $\lambda<1$. Let $N$ the smallest positive integer such that $\frac{1}{N}< \lambda$. We easily see that $B_\lambda(z)=B_{1/N}(z)$ and so this prove that open ball in the metric topology are open in $\tau_{\text{Eucl}} \cup \mathcal{P}(\mathbb{Q})$ so the topology are equivalent.

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  • $\begingroup$ thanks, I personally used for the singletons of $\mathbb{Q}$ that for a rational number $r_k$ the open ball $B(r_k,1/k)$ is just the singleton $\{r_k\}$. And for the interval $(a,b)$ that a random $x\in (a,b)$ $a<x<b$. So that we have $d(x,a)=\epsilon_1$ en $d(x,b)=\epsilon_2$. Then the open ball $B(x,\min(\epsilon_1,\epsilon_2))\subseteq (a,b)$. Is this correct as well? $\endgroup$ Commented Mar 25, 2024 at 17:11
  • $\begingroup$ thanks for the second part. I don't fully understand how it proves that the open ball in the metric topology is open in $\tau_{Eucl} \cup P(\mathbb{Q}$ though? $\endgroup$ Commented Mar 25, 2024 at 19:14
  • $\begingroup$ @riescharlison $[r_h,r_k]$ is open in $\tau_{Eucl} \cup P(\mathbb{Q})$ so if $y \in [r_h,r_k] \subseteq B_{1/N}$, for any $y \in B_{1/N}$ than there is an open neightbourhood of $y$ contained in $B_{1/N}$, for each $y \in B_{1/N}$ and so i$B_{1/N}$ is open in $\tau_{Eucl} \cup P(\mathbb{Q})$. $\quad$ If $B_{1/N}$ is a rational singleton than it is open (as it belongs to $P(\mathbb{Q})$) so any open ball is open in $\tau_{Eucl} \cup P(\mathbb{Q})$, as all the open balls of the metric topology satisfy one of these conditions $\endgroup$ Commented Mar 25, 2024 at 19:33
  • $\begingroup$ I still don't fully understand why $y\in [r_h,r_k]$ is always true when it's not a singleton. Can't it be the te case that $y \notin [r_h,r_k]$ but still $y\in B_{1/N}$. For example $y>r_k$ but $y<r_n$ for any $n$ in $0<n \leq N$? $\endgroup$ Commented Mar 25, 2024 at 20:28
  • $\begingroup$ As well as the sequence you described. It's not necessarily the case that $\{r_n : n \ge N\} \cap [x, \infty) $ is a subset of $B_{1/N}(x)$ right? $\endgroup$ Commented Mar 25, 2024 at 20:37

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