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

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 $$T_s\subsetneqq T_r\subsetneqq T_c$$.

• 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. – Math_Is_Fun Apr 25 at 22:15
• 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? – Math_Is_Fun Apr 25 at 22:19
• @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?). – Brian M. Scott Apr 25 at 22:27
• @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 $$U=B_r\left(\left\langle 0,\frac12\right\rangle,\frac12\right)\;.$$ – Brian M. Scott Apr 25 at 22:30
• @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. – Brian M. Scott Apr 25 at 22:54