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My Real Analysis professor and I have been trying to construct a particular example in the irrationals but to no avail. The criteria are as follows:

Let $\mathbb{J}$ be the set of irrationals and let $\mathbb{J}$ be given the topology inherited from the usual topology of the Reals. Find an uncountable proper dense subset $A$ of $\mathbb{J}$ such that $A$ is homeomorphic to $\mathbb{J}$.

I have been trying to create the subset by some embedding of $\mathbb{J}$ into itself but when I force the function to miss irrationals I create problems in the domain as well.

Russle
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1 Answers1

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How about $\{x^{1/3}\mid x\in\mathbb J\}$?

This is homeomorphic to $\mathbb J$ because it is its image under a homeomorphism $\mathbb R\to\mathbb R$. It is a subset of $\mathbb J$ because no irrational has a rational cube root. And it's a proper subset because it doesn't contain $\sqrt[3]2$.

More generally, if $X$ and $Y$ are any two countable dense subsets of $\mathbb R$, then there is a homeomorphism $f:\mathbb R\to \mathbb R$ that maps $X$ to $Y$. (Any countable dense subset of $\mathbb R$ is order-isomorphic to $\mathbb Q$, and the order isomorphism $X\to Y$ extends to a homeomorphism on $\mathbb R$ by continuity).

So we could also take $X=\mathbb Q$ and $Y=\mathbb Q\cup\{\pi\}$, and conclude that you could take $A=\mathbb J\setminus\{\pi\}$ -- or in general $A$ can be $\mathbb J$ minus any nonempty finite or countable set of irrationals.

hmakholm left over Monica
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