In which of the three topologies is X connected?

Question

In which of the three topologies is X connected?

Below is how I did it but I am not sure, hence it will be great if someone helps me in this.

1. $T_s$ is the Subspace topology obtained from $\mathbb{R^2}$ since each $S_n$ is a Connected Subspace in $\mathbb{R^2}$ and $S_n \cap S_{n+1}$ $\neq$ $\emptyset$ $\forall n$. So, the unions of $S_n$ is connected Subspace of $\mathbb{R^2}$.

2. Since $T_s$ is finer than $T_r$ and $(X,T_s)$ is connected which implies $(X,T_r)$ is Connected.

3.In $T_c$ topology, each $S_n$ is open subset and $S_0$ \ {$v$} is also open as {$v$} is a Closed subset in $X$. Hence, $X = U_n$ = $S_n \cup S_o$ \ {$v$}. So $X$ is not connected in this topology.

I know that I have the right idea somewhere but I am not sure how to explain it better by showing all the calculations. Will be great if someone helps me on this. Appreciate your support & help.

Answer 1

I'm pretty sure $X$ is actually connected in all topologies. The point is that in all topologies, the subspace topology on each $S_n$ will be Euclidean. If you have two separating open sets $U$ and $V$, then either $v \in U$ or $v \in V$. This means that we can choose $S_n$ so that $U \cap S_n$ and $V \cap S_n$ will be nonempty, which contradicts the fact that $S_n$ is connected in the Euclidean topology.

$\mathbf{Remark} \, 1$ The Euclidean topology is not finer than the railroad topology. To see this, choose $x \in S_0 -\{v\}$. Then there exists an open neighborhood of $x$ in $X$ in the railroad topology that is completely contained in $S_0$. This is not true in the Euclidean topology: Any open ball of radius $\epsilon$ about $x$ will contain a point of some $S_n$, $n \neq 0$, for which $1/n < \epsilon$ is small enough.

$\mathbf{Remark} \, 2$ It's worth noting (and I think this might be where the text is going) that $X$ is not locally connected with respect to the Euclidean topology, but that it is with respect to the other two topologies.

Answer 2

In all three topologies, $S_n$ is homeomorphic to a standard interval $[0,1]$, and hence connected, and as all $S_n$ intersect at the top, their union is connected in all of these topologies, by a standard theorem on union of connected subspaces.