# $T_2$-spaces in which no two open sets are homeomorphic

This question 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?

• Re: close, is there an easy example or an easy argument for a negative answer that I missed? – Dominic van der Zypen Nov 1 '18 at 10:27
• I seem to recall that this is a problem which appeared at the Vojtěch Jarník International Mathematical Competition held in Ostrava something like 10+ years ago. – Tomek Kania Nov 1 '18 at 10:51
• I wonder if there is a subspace of $\mathbb R$ with that property. – bof Nov 1 '18 at 10:59

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 $$U=V$$.