How to construct an order-preserving injection $f:Q\rightarrow R$ , such that the image of $f$ is discrete subspace of $R$ (set of reals).


1 Answer 1


Enumerate $Q = \{q_n : n \in \omega \}$. We are going to define $f(q_n)$ by induction on $n$. This is done by defining an auxiliary sequence $(I_n)_{n \in \omega}$ of disjoint open intervals which do not overlap and have distinct endpoints so that $q_n \in I_n$ for every $n$.

Start by mapping $q_0$ anywhere you want and take $I_0$ to be any bounded open interval around $f(q_0)$.

Having defined $(f(q_i))_{i<n}$, map $q_n$ aproprietly so that $f(q_n) \notin \bigcup_{i<n} I_i$. Now we can still find an open interval $I_n$ around $f(q_n)$ which is disjoint from $\bigcup_{i<n} I_i$ and does not share any endpoints.

I leave the details to you.

  • $\begingroup$ How is such $f$ order preserving? And do I have to specify these disjoint open intervals, say from Cantors middle third set? $\endgroup$
    – ersh
    May 7, 2019 at 18:54
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    $\begingroup$ This is what I meant with "map $q_n$ approprietly". $q_n$ has a certain position within $\{q_i : i \leq n \}$, say for example $q_j < q_n < q_i$ with $j,i <n$ where $q_j$ is chosen largest with this property and $q_i$ smallest. Then we would map $q_n$ to something in the intervall $(f(q_j), f(q_i))$. Let $a$ be the right endpoint of $I_j$ and $b$ the left endpoint of $I_i$, then $q_j < a < b < q_i$. So we can actually chose $f(q_n)$ between $a$ and $b$. Then let $I_n = (c,d)$ so that $a<c<d<b$ and $f(q_n) \in (c,d)$. $\endgroup$ May 7, 2019 at 19:17
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    $\begingroup$ Any choice of $(c,d)$ is ok, e.g. $c = f(q_n) - \frac{f(q_n) -a}{2}$, $d = f(q_n) + \frac{b -f(q_n)}{2}$. $\endgroup$ May 7, 2019 at 19:17
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    $\begingroup$ You can also see that the set $\{ \sum_{i < \vert s \vert } (-1)^{1-s(i)}3^{-i} : s \in 2^{< \mathbb{N}} \}$ is countable dense in itself and discrete. Maybe that's the example you were looking for (or asked for), since you are talking about the Cantor set. $\endgroup$ May 7, 2019 at 19:30
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    $\begingroup$ @ersh Yes the construction works for any countable linear order. $\endgroup$ May 7, 2019 at 21:22

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