The question is in the title: 



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> **Q1:** Is there a topological space $X$ containing a copy of the real line and having the property that all the nonempty open subsets of $X$ are homeomorphic?
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Let us say that $X$ is a *homeomorphic open set space*, or a *hoss* for short, if all the nonempty open subsets of $X$ are homeomorphic. 
Such spaces were asked about [here](https://mathoverflow.net/questions/300253/t-2-spaces-where-all-non-empty-open-sets-are-homeomorphic), and N. de Rancourt's answer shows that if $D$ is infinite and discrete then $D^\omega$ is a hoss. It follows that every ``ultrametrizable'' space embeds in a hoss. (A space is called *ultrametrizable* if it is homeomorphic to an [ultrametric space](https://en.wikipedia.org/wiki/Ultrametric_space). Spaces of the form $D^\omega$ are themselves ultrametrizable, and every other ultrametrizable space embeds in one of this form.) This is just about all I know about hosses and spaces that embed in them -- any other information or insight is welcome.


Familiar examples of hosses include the space $\mathbb Q$ of rational numbers and the space $\mathbb R \setminus \mathbb Q$ of irrational numbers.


**EDIT:**


User bof has answered my question by finding a $T\_1$ space $X$ containing the real line, and with the property that all nonempty open subsets of $X$ are homeomorphic. However, I still wonder if there is a Hausdorff space with this property:



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> **Q2:** Is there a Hausdorff space $X$ containing a copy of the real line and having the property that all the nonempty open subsets of $X$ are homeomorphic?
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Can the real line be embedded in a space $X$ such that all the nonempty open subsets of $X$ are homeomorphic?
A metrizable example can be constructed as follows. In the plane consider the subset $$\Xi:=\big\{(x,\tfrac{2k+1}{2^n}):k,n\in\mathbb Z,\;x\in\mathbb R\setminus \tfrac1{2^n}\mathbb Z\big\}.$$


It is clear that $\Xi$ contain (countably many) topological copies of the real line.


There are at least two ways of proving that any non-empty open subset of $\Xi$ is homeomorphic to $\Xi$. One is more geometric and is due to Volodymyr Mykhaylyuk (from Chernivtsi). He observed that for any open set $U\subset \Xi$ and any connected component $C$ of $U$ the interval $C$ has a base of clopen neighborhoods homeomorphic to the strip $\Xi\cap(\mathbb R\times(-\sqrt{2},\sqrt{2}))$ in $\Xi$. Then $U$ can be decomposed into countably many pairwise homeomorphic clopen sets and the same can be done with the space $\Xi$.


Another way is more global. Just to prove a characterization theorem for the space $\Xi$:


**Theorem.** A topological space $X$ is homeomorphic to the space $\Xi$ if and only if 


1) $X$ is a metrizable space;


2) for any point $x\in X$ the connected component $C\_x$ containing $x$ is homeomorphic to the real line;


3) the family $\mathcal C=\{C\_x\}\_{x\in X}$ of connected components of $X$ is countable;


4) for any point $x\in X$ and a neighborhood $O\_x\subset X$ of $x$ there exists a non-empty set $V=\bigcup\{C\in \mathcal C:C\cap V\ne\emptyset\}$ in $O\_x\setminus C\_x$ such that $V$ is clopen in $X\setminus C\_x$, $C\_x\cup V$ is a neighborhood of $x$.


The proof of this characterization uses the back-and-forth argument: We enumerate the connected components and at the $n$-th step construct clopen neighborhoods of $n$-th intervals and establish the combinatorial correspondence between these neighborhoods. I will write down the details in a preprint (with many authors with whom I discussed this problem being on a Summer School in Carpathian mountains).


The characterization theorem implies the following corollary:


**Corollary.** Let $(C\_n)\_{n\in\omega}$ be a family of parwise disjoint curves in $\mathbb R^d$ such that each $C\_n$ is homeomorphic to $\mathbb R$, is closed and nowhere dense in the union $X:=\bigcup\_{n\in\omega}C\_n$ and $\sum\_{n=1}^\infty lenth(C\_n)<\infty$. Then the space $X$ is homeomorphic to $\Xi$.


