Topology on $\Bbb{R}$ where elements are $(a,+\infty)$

Question

Let $\mathcal{T}_1$ the collection of subsets of $\Bbb{R}$ such that elements are $\emptyset,\Bbb{R}$ and all intervals on the form $(a,+\infty).$

I proved that $\mathcal{T}_1$ is a topology where proper open sets are dense and all closed set are empty interior.

1. I would like to find connected sets.

Let $A\subset\Bbb{R}$, $A$ is connected if and only if any continuous application $f:A\to \{0,1\}$ is constant.

As $f$ is continuous, it means that for any open set of $\{0,1\}$, noted $U$ (which are $\emptyset,\{0\},\{1\}\{0,1\})$ we have $f^{-1}(U)$ is open.

I get the result for all the set of the form $\emptyset,\{0\},\{1\}$ I think, but when $U=\{0,1\}$ for example I get that $f^{-1}(\{0,1\})=(a,+\infty)$ which is not constant function.

2. Let $H_0$ be the group of homeomorphisms of $\Bbb{R}$ for the usual topology, and so $H_1$ for the $\mathcal{T}_1$ topology. I would like to prove that $H_1$ a is normal subgroup of index $2$ of $H_0.$

I can prove that $H_1$ is a subgroup of $H_0.$ Now to prove that it's normal I have to prove that $$\forall h_1\in H_1,\forall h_0\in H_0;\quad h_0h_1h_0^{-1}\in H_1.$$

So let $\psi:=h_0h_1h_0^{-1}$ I have to prove that $\psi$ is continuous, bijective, and $\psi^{-1}$ is continuous for the $\mathcal{T}_1$ topology.

Let $U$ be an open set of $\mathcal{T}_1\subset \mathcal{T}_0$, we have $\psi^{-1}(U)=h_0h_1^{-1}h_0^{-1}(U).$ As $h_0$ is a homeomorphism for $\Bbb{R}$ with the usual topology, I don't "see" why $h_0^{-1}(U)$ is an open set for $\mathcal{T}_1$.

I don't don't either how to prove that is of index $2.$

Answer 1

As to connectedness: note that the topology is so-called hyperconnected: for every two open non-empty sets $U$ and $V$ of $X$, $U \cap V \neq \emptyset$, in this case even $U \subset V$ or $V \subset U$. We can say the following about subsets: if $A \subset \mathbb{R}$ has two points or more, say $p,q \in A$, then if $p < q$, then any open set that contains $p$ also contains $q$, but this implies we cannot write $A$ as a disjoint union of relatively open subsets, so $A$ is connected. Conclusion: all subsets of $\mathbb{R}$ are connected in $\mathcal{T}_1$

Answer 2

Concerning the first point I think you could exploit the fact that

$f^{-1}(A\cup B)=f^{-1}(A)\cup f^{-1}(B)$

Hence

$f^{-1}(\{0,1\})=f^{-1}(\{0\}\cup\{1\})=f^{-1}(\{0\})\cup f^{-1}(\{1\})$