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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.

  1. 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 $2.$

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    You can refer [this](http://math.stackexchange.com/questions/331654/imposing-the-topology-of-open-rays-in-bbb-r?rq=1) for getting some idea about connectedness in this space. – 2017-02-11
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    For (1), pretty sure every set subset, $U\subset\mathbb{R}$ is connected (hint: assume not, then there exist disjoint $A$ and $B$ such that $A$ and $B$ are open and their union is an open set as well, that is, they disconnect $A\cup B$, this is not possible: why?) – 2017-02-11
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    As for (2), consider a homeomorphism of $\mathbb{R}$ with standard topology that is a continuous deformation of $\mathbb{R}$ within itself: (consider $f(x)=x^3$ for example, it is a homeomorphism of $\mathbb{R}$, how does it deform $\mathbb{R}$?). The homeomorphism of $\mathbb{R}$ that is not a homeomorphism for $\mathbb{R}$ under the topology $\mathcal{T}_1$ is the inversional map $f(x)=-x$. Every homeomorphism for $\mathcal{T}_1$ corresponds to itself for $\mathcal{T}_0$ as well as itself composed with $f(x)=-x$. – 2017-02-11
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    I don't understand your argument for $(2)$ @JustinBenfield – 2017-02-11
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    @Alex: Every continuous (w.r.t. std top.) bijective map from $\mathbb{R}$ to itself is monotone (thanks to order properties of $\mathbb{R}$), those maps that are increasing also work for $\mathcal{T}_1$ (consider which happens to an open set, $(a,+\infty )$). However, $f(x)=-x$ works for standard topology, but what about the open set $(a,+\infty )$ in $\mathcal{T}_1$? It's image is the set $(-\infty , -a)$ but that isn't open in $\mathcal{T}_1$ but is for std top. The rest of my comment was pointing out this explains why $H_1$ has index $2$ in $H_0$, which btw, implies it is normal. – 2017-02-11
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    @JustinBenfield Really interesting explanation, and yeah if it's of index 2 it's normal, I forgot that! +1 – 2017-02-11

2 Answers 2

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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 $\mathcal{T}_1$

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    Ah! Nice, thanks. – 2017-02-11
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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

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