[This question](https://mathoverflow.net/questions/300253/t-2-spaces-where-all-non-empty-open-sets-are-homeomorphic) was about spaces in which all non-empty open sets "look alike".


Now I am interested in the opposite: Is there a $T\_2$-space $(X,\tau)$ with $|X|>1$ such that whenever $U\neq V$ are open subsets of $X$, they are not homeomorphic when endowed with the subspace topology?



$T_2$-spaces in which no two open sets are homeomorphic
There is an example in this paper, [A method for constructing ordered continua](https://fa.ewi.tudelft.nl/%7Ehart/37/publications/the_papers/published/ordered-continua.pdf), of an ordered continuum in which no two intervals are homeomorphic.


Because an open set is a union of a disjoint family of intervals a homeomorphism between two open sets it has to map constituents intervals to constituent intervals; as that can only be done by the identity we see that homeomorphic open sets are equal.


And to answer @bof's question: yes there is also a subset of $\mathbb{R}$ with this property. To see that enumerate the set of triples $\langle f,A,B\rangle$, where $A$ and $B$ are disjoint uncountable $G\_\delta$-sets and $f:A\to B$ is a homeomorphism as $\bigl<\langle f\_\alpha,A\_\alpha,B\_\alpha\rangle:\alpha<\mathfrak{c}\bigr>$. Recursively, using that the sets $A\_\alpha$ and $B\_\alpha$ have cardinality $\mathfrak{c}$, choose $x\_\alpha\in A\_\alpha$ such that $x\_\alpha$ and $f\_\alpha(x\_\alpha)$ are both not in the union $\{x\_\beta:\beta<\alpha\}\cup\{f\_\beta(x\_\beta):\beta<\alpha\}$.
In the end let $X=\{x\_\alpha:\alpha<\mathfrak{c}\}$. Claim: $X$ is as required. Assume $f:U\to V$ is a homeomorphism between open subsets and assume $f(x)\neq x$ for some $x$. Pick an open interval $I$ around $x$ such that $f[I\cap X]\cap I=\emptyset$. Apply Lavrentieff's theorem to find $G\_\delta$-sets $A$ around $I\cap X$ and $B$ around $f[I\cap X]$ and a homeomorphism $\tilde f:A\to B$ that extends $f$. We can assume $A\subseteq I$, so $A\cap B=\emptyset$. Then $\langle f,A,B\rangle = \langle f\_\alpha,A\_\alpha,B\_\alpha\rangle$ for some $\alpha$ and we find that $x\_\alpha\in X$ but $f(x\_\alpha)=f\_\alpha(x\_\alpha)\notin X$. This contradiction shows that $f$ must be the identity and hence that $U=V$.


