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Let

\begin{align} S&=\{(x,y,z)\in\mathbb R^3:\text{exactly one of }x,y,z\text{ is rational}\}\\ &=\mathbb Q^c\times\mathbb Q^c\times\mathbb Q\cup \mathbb Q^c\times\mathbb Q\times\mathbb Q^c\cup \mathbb Q\times\mathbb Q^c\times\mathbb Q^c\\ &= S_1\cup S_2\cup S_3. \end{align}

Is $S$ connected? Is it path-connected? The analogous problem one dimension lower (connectedness of the points in the plane with exactly one rational coordinate) can be answered by noticing that the line $\{(x,x)\}$ does not intersect your set, so it separates it into two connected components.

I suspect $S$ is connected but not path-connected because if we had an open set $A\subset \mathbb R^3$ which is not empty and not the entire space, we would want $\partial A\subset S^c$ and $S^c$ is the union of $\mathbb Q^c\times \mathbb Q^c\times \mathbb Q^c$ and a set of (Hausdorff, topological) dimension $1$. The first is totally disconnected, and the second is really small so I think constructing such a set $A$ is impossible. On the other hand, constructing a nontrivial path $\gamma:[0,1]\to S$ seems really hard because $S_1\cup S_2=\mathbb Q^c\times(\mathbb Q^c\times\mathbb Q\cup \mathbb Q^c\times\mathbb Q)$ so if you don't visit $S_3$ you can't change your $x$ coordinate, and the same for the other permutations.

Martin Sleziak
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Derivative
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    Engelking, General topology, 7.2.11 proves rather easily that $S$ is even 0-dimensional. The proof is essentially based on the countable sum theorem for dim. – Ulli Aug 04 '25 at 08:04
  • @Ulli what does this imply wrt connectedness? – Derivative Aug 04 '25 at 16:16
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    0-dimensional, $T_1$ spaces are totally disconnected, i.e. connected subsets have at most 1 point. Hence, $S$ is disconnected. But this is much, much stronger! – Ulli Aug 04 '25 at 17:01
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    BTW: the same question was raised today at MO. – Ulli Aug 04 '25 at 17:33

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