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It seems too good to be possible, but:

Is there a uniformly continuous injective image of $(0,1)\setminus\Bbb Q$ in the Cantor space?

Here, the Cantor space $\{0,1\}^{\Bbb N}$ is equipped with the standard metric $$d(a,b):=\frac{1}{2^{\min\{n: a(n)\neq b(n)\}}}.$$

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    $\begingroup$ Doesn't a uniformly continuous map from $(0,1)\setminus\mathbb Q$ to a compact metric space have a continuous extension to $[0,1]$? But a continuous map from $[0,1]$ to the Cantor space is constant. What am I missing? $\endgroup$
    – bof
    Commented Feb 5 at 12:36
  • $\begingroup$ @bof You are right. Thanks! For interested readers, here is one proof of your statement: math.stackexchange.com/questions/245237/… $\endgroup$
    – Boaz Tsaban
    Commented Feb 5 at 13:01

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Indeed it isn't possible. Consider any uniformly continuous image of those irrationals in the Cantor set. The inverse image $X$ of a clopen set, i.e. a finite union of the smaller "copies" of the Cantor set in itself, is going to be "uniformly open" (there exists a $\delta>0$ such that every ball of radius $\delta$ with a center in $X$ is entirely contained in $X$) and closed.

Such sets in $[0,1]\setminus\mathbb Q$ are either empty or the full set, by the usual proof that $[0,1]$ is connected. In particular, $y\neq z$ in the Cantor set can never both be values for the map, since the preimages of two disjoint clopen sets cannot both be full. Hence the map is constant.

Edit. This answer is basically an unfolding of the argument of bof in the comments (existence of an extension to $[0,1]$ corresponds to the fact that the preimage is "uniformly open").

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