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Based on this setup and definitions, I need to solve this.

How are the three topologies on X related to each other (in terms of coarser and finer)?

Definition of coarser and finer topology: Let $X$ be a set and $T_1$ and $T_2$ be topologies on $X$. If $T_1$ $\subset T_2$, we say that $T_1$ is coarser than $T_2$ and that $T_2$ is finer than $T_1$. If $T_1$ $\subset$ $T_2$ and $T_1$ $\neq$ $T_2$, we say that $T_1$ is strictly coarser than $T_2$ and that $T_2$ is strictly finer than $T_1$.

Taking a look at the definitions where we have 3 topologies, we can see that $T_s$ is the subspace topology that is inherited from the product topology,$T_r$ is the topology by railroad metric and $T_c$ is the coherent topology. I think it should be $T_s \subset T_c$ but not sure whether $T_s \subset T_r$. The other direction should be similar to one direction.

Need help in this. Appreciate your help & support.


Yes, $T_s\subseteq T_c$. If $U\in T_s$, there is an ordinary open set $V$ in the plane such that $V\cap X=U$; clearly $U\cap S_n=V\cap S_n$ is open in each $S_n$, so $U\in T_c$. We still have to determine, however, whether the two topologies are equal. It’s not too hard to verify that they are not; for instance $\{0\}\times(0,1)\in T_c\setminus T_s$. (Why?) Thus, $T_s\subsetneqq T_c$.

It’s also the case that $T_s\subseteq T_r$. Let $U\in T_s$, and suppose that $x\in U$. If $x\in S_n\setminus\{v\}$ for some $n$, choose $\epsilon>0$ small enough that $B_d(x,\epsilon)\subseteq U\setminus\{v\}$, where $d$ is the Euclidean metric. Then for each $k$ we have $B_d(x,\epsilon)\cap S_k\in T_r$, because it is either empty or an open interval of $S_k\setminus\{v\}$, and therefore $B_d(x,\epsilon)=\bigcup_k(B_d(x,\epsilon)\cap S_k)\in T_r$. If $x=v$, just choose $\epsilon>0$ small enough so that $B_d(x,\epsilon)\subseteq U$ and check that $B_d(x,\epsilon)=B_r(x,\epsilon)$. Thus, every point of $U$ has a $T_r$-open nbhd contained in $U$, so $U\in T_r$.

Are the two topologies equal? No, and again we can use the set $\{0\}\times(0,1)$: I leave it to you to check that it’s in $T_r\setminus T_s$, so that $T_s\subsetneqq T_r$.

We still have to compare $T_c$ with $T_r$. You shouldn’t have too much trouble checking that $T_r\subseteq T_c$. To see that in fact $T_r\subsetneqq T_c$, for $n\ge 1$ let

$$U_n=\left\{x\in S_n:r(v,x)=d(v,x)<\frac1n\right\}\;,$$

let $U_0=S_0$, let $U=\bigcup_{n\ge 0}U_n$, and show that $U\in T_c\setminus T_r$.

We conclude that $T_s\subsetneqq T_r\subsetneqq T_c$.

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  • $\begingroup$ I figured out few things bt I am sorry to say I am unclear on few things so trying to make sure that I am correct there: {0}×(0,1)∈Tc∖Ts ?? b/c {$0$} X $(0,1)$ is an open interval ($0 X 0, 0 X 1$) which is open in $T_c$ due to condition 3(b) of definition 0.2 but not in $T_s$ (product topology or equivalently metric topology on $\mathbb{R^{2}}$, right ? If not pls let me know. $\endgroup$ – Math_Is_Fun Apr 25 at 22:15
  • $\begingroup$ check that Bd(x,ϵ)=Br(x,ϵ)....So, $B_d(x, \epsilon)$ = {$ y \in X, s.t. d(x,y) < \epsilon$} and $B_r(x, \epsilon)$ = {$ y \in X, s.t. r(x,y) < \epsilon$}. Therefore, we get $d(x,y) < \epsilon$ and $r(x,y) < \epsilon$, and from definition 0.2, 2(a), we know that $r(x,y) = d(x,y)$, therefore we can say that $B_d$ and $B_r$ are equal. right? $\endgroup$ – Math_Is_Fun Apr 25 at 22:19
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    $\begingroup$ @Math_Is_Fun: Let $U=\{0\}\times(0,1)$; then $U$ is open in $S_0$, and for $n\ge 1$ we have $U\cap S_n=\varnothing$, which is open in $S_n$, so $U$ is open in $T_c$. It is not open in $T_s$, because any open set $V$ in $\Bbb R^2$ such that $V\cap S_0=U$ must include points of some $S_n$ for $n\ge 1$ (why?). $\endgroup$ – Brian M. Scott Apr 25 at 22:27
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    $\begingroup$ @Math_Is_Fun: In the previous comment I sketched the argument that $U\notin T_s$; to show that $U\in T_r$ just verify that $$U=B_r\left(\left\langle 0,\frac12\right\rangle,\frac12\right)\;.$$ $\endgroup$ – Brian M. Scott Apr 25 at 22:30
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    $\begingroup$ @Math_Is_Fun: $U\cap S_n=(V\cap X)\cap S_n=V\cap(X\cap S_n)=V\cap S_n$, since $S_n\subseteq X$. And $V\cap S_n$ is open in $S_n$ by the definition of the subspace topology: there’s nothing to prove here. $\endgroup$ – Brian M. Scott Apr 25 at 22:54

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