5 added 4 characters in body

Strong version: no. Consider $[0,1]$ with distance $d(x,y)=|x-y|^{1/3}$. There is no even a triple of points with rational distances - otherwise there would be a nonzero rational solution of $x^3+y^3=z^3$.

Weak version: yes. Let $(X,d)$ be the space in question. Construct sets $S_1\subset S_2\subset\dots$ such that each $S_k$ is a maximal $(2^{-k})$-separated net in $X$. Let $S$ be the union of these nets; then $S$ is countable and dense in $X$.

Now construct the following metric graph on $S$. For every $k$, connect every pair of points $x,y\in S_k$ by an edge whose length is $(1-10^{-k})d(x,y)$ rounded down to a multiple of $10^{-2k}$. The new distance $d'$ on $S$ is the induced length distance in this graph. It is easy to see that the edges outside $S_k$ do not affect the distances in $S_k$, hence all these distances are rational (multiples of $10^{-2k}$). The new metric $d'$ on $S$ satisfies $\frac12d\le d'\le d$, hence the completion of $(S,d')$ is the same set $X$ with an equivalent metric.

UPDATE. Here is a more detailed description without the term "metric graph".

For each $k$, define a function $f_k:\mathbb R_+\to\mathbb R_+$ by $$f_k(t) = 10^{-2k}\left\lfloor 10^{2k}(1-10^{-k})t \right\rfloor .$$ The actual form of $f_k$ does not matter, we only need the following properties:

• $f_k$ takes only rational values with bounded denominators (by $10^{-k}$).

• Let $a_k$ and $b_k$ denote the infimum and the supremum of $f_k(t)/t$ over the set $\{t\ge 2^{-k}\}$. Then $\frac12\le a_k\le b_k\le a_{k+1}\le 1$ for all $k$. (Indeed, we have $1-2\cdot10^k\le a_k\le b_k\le 1-10^k$.)

For every $x,y\in S_k$, define $\ell(x,y)=f_k(d(x,y))$ where $k=k(x,y)$ is the minimum number such that $x,y\in S_k$. Note that $$a_k d(x,y) \le \ell(x,y) \le b_k d(x,y)$$ for all such pairs $x,y\in S_k$, x,y$, since$S_k$is a$(2^{-k})$-separated set. For a finite sequence$x_0,x_1,\dots,x_n\in S$define $$\ell(x_0,x_1,\dots,x_n) = \sum_{i=1}^n \ell(x_{i-1},x_i) .$$ I will refer to this expression as the$\ell$-length of the sequence$x_0,\dots,x_n$. Define $$d'(x,y) = \inf\{ \ell(x_0,x_1,\dots,x_n) \}$$ where the infimum is taken over all finite sequences$x_0,x_1,\dots,x_n$in$S$such that$x_0=x$and$x_n=y$. Clearly$d'$is a metric and$\frac12d\le d'\le d$. It remains to show that$d'$takes only rational values. Lemma: If$x,y\in S_k$, then$d'(x,y)$equals the infimum of$\ell$-lengths of sequences contained in$S_k$. Proof: Consider any sequence$x_0,\dots,x_n$in$S$such that$x_0=x$and$y_0=y$. Remove all points that do not belong to$S_k$from this sequence. I claim that the$\ell$-length became shorter. Indeed, it suffices to prove that $$\ell(x_r,x_s) \le \ell(x_r,x_{r+1},\dots,x_{s-1},x_s)$$ if$x_r$and$x_s$are in$S_k$and the intermediate points are not. By the second property of the functions$f_k$, the left-hand side is bounded above by$b_k d(x_r,x_s)$and every term$\ell(x_i,x_{i+1})$in the right-hand side is bounded below by$b_k d(x_i,x_{i+1})$. So it suffices to prove that $$b_k d(x_r,x_s) \le b_k\sum_{i=r}^{s-1} d(x_i,x_{i+1}),$$ and this is a triangle inequality multiplied by$b_k$. Q.E.D. All$\ell$-lengths of sequences in$S_k$are multiples of some fixed rational number (namely$10^{-2k}$). Hence$d'(x,y)$is a multiple of the same number if$x,y\in S_k$. Thus all values of$d'$are rational. 4 added more details Let me answer Question 2. Strong version: no. Consider$[0,1]$with distance$d(x,y)=|x-y|^{1/3}$. There is no even a triple of points with rational distances - otherwise there would be a nonzero rational solution of$x^3+y^3=z^3$. Weak version: yes. Let$(X,d)$be the space in question. Construct sets$S_1\subset S_2\subset\dots$such that each$S_k$is a maximal$(2^{-k})$-separated net in$X$. Let$S$be the union of these nets; then$S$is countable and dense in$X$. Now construct the following metric graph on$S$. For every$k$, connect every pair of points$x,y\in S_k$by an edge whose length is$(1-10^{-k})d(x,y)$rounded down to a multiple of$10^{-2k}$. The new distance$d'$on$S$is the induced length distance in this graph. It is easy to see that the edges outside$S_k$do not affect the distances in$S_k$, hence all these distances are rational (multiples of$10^{-2k}$). The new metric$d'$on$S$satisfies$\frac12d\le d'\le d$, hence the completion of$(S,d')$is the same set$X$with an equivalent metric. UPDATE. Here is a more detailed description without the term "metric graph". For each$k$, define a function$f_k:\mathbb R_+\to\mathbb R_+$by $$f_k(t) = 10^{-2k}\left\lfloor 10^{2k}(1-10^{-k})t \right\rfloor .$$ The actual form of$f_k$does not matter, we only need the following properties: •$f_k$takes only rational values with bounded denominators (by$10^{-k}$). • Let$a_k$and$b_k$denote the infimum and the supremum of$f_k(t)/t$over the set $\{t\ge 2^{-k}\}$. Then$\frac12\le a_k\le b_k\le a_{k+1}\le 1$for all$k$. (Indeed, we have$1-2\cdot10^k\le a_k\le b_k\le 1-10^k$.) For every$x,y\in S_k$, define$\ell(x,y)=f_k(d(x,y))$where$k=k(x,y)$is the minimum number such that$x,y\in S_k$. Note that $$a_k d(x,y) \le \ell(x,y) \le b_k d(x,y)$$ for all$x,y\in S_k$, since$S_k$is a$(2^{-k})$-separated set. For a finite sequence$x_0,x_1,\dots,x_n\in S$define $$\ell(x_0,x_1,\dots,x_n) = \sum_{i=1}^n \ell(x_{i-1},x_i) .$$ I will refer to this expression as the$\ell$-length of the sequence$x_0,\dots,x_n$. Define $$d'(x,y) = \inf\{ \ell(x_0,x_1,\dots,x_n) \}$$ where the infimum is taken over all finite sequences$x_0,x_1,\dots,x_n$in$S$such that$x_0=x$and$x_n=y$. Clearly$d'$is a metric and$\frac12d\le d'\le d$. It remains to show that$d'$takes only rational values. Lemma: If$x,y\in S_k$, then$d'(x,y)$equals the infimum of$\ell$-lengths of sequences contained in$S_k$. Proof: Consider any sequence$x_0,\dots,x_n$in$S$such that$x_0=x$and$y_0=y$. Remove all points that do not belong to$S_k$from this sequence. I claim that the$\ell$-length became shorter. Indeed, it suffices to prove that $$\ell(x_r,x_s) \le \ell(x_r,x_{r+1},\dots,x_{s-1},x_s)$$ if$x_r$and$x_s$are in$S_k$and the intermediate points are not. By the second property of the functions$f_k$, the left-hand side is bounded above by$b_k d(x_r,x_s)$and every term$\ell(x_i,x_{i+1})$in the right-hand side is bounded below by$b_k d(x_i,x_{i+1})$. So it suffices to prove that $$b_k d(x_r,x_s) \le b_k\sum_{i=r}^{s-1} d(x_i,x_{i+1}),$$ and this is a triangle inequality multiplied by$b_k$. Q.E.D. All$\ell$-lengths of sequences in$S_k$are multiples of some fixed rational number (namely$10^{-2k}$). Hence$d'(x,y)$is a multiple of the same number if$x,y\in S_k$. Thus all values of$d'$are rational. 3 added details Weak version: yes. Let$(X,d)$be the space is in question. Construct sets$S_1\subset S_2\subset\dots$such that each$S_k$is a maximal$(2^{-k})$-separated net in$X$. Let$S$be the union of these nets; then$S$is countable and dense in$X$. Now construct the following metric graph on$S$. For every$k$, connect every pair of points$x,y\in S_k$by an edge whose length is$(1-10^{-k})d(x,y)$rounded down to a multiple of$10^{-2k}$. The new distance$d'$on$S$is the induced length distance in this graph. It is easy to see that the edges outside$S_k$do not affect the distances in$S_k$, hence all these distances are rational (multiples of$10^{-2k}$). The new metric$d'$on$S$satisfies$\frac12d\le d'\le d$, hence the completion of$(S,d')$is the same set$X$with an equivalent metric. UPDATE.Here is a more detailed description without the term "metric graph". For each$k$, define a function$f_k:\mathbb R_+\to\mathbb R_+$by f_k(t) = 10^{-2k}\left\lfloor 10^{2k}(1-10^{-k})t \right\rfloor .The actual form of$f_k$does not matter, we only need the following properties: •$f_k$takes only rational values with bounded denominators (by$10^{-k}$). • Let$a_k$and$b_k$denote the infimum and the supremum of$f_k(t)/t$over the set $\{t\ge 2^{-k}\}$. Then$\frac12\le a_k\le b_k\le a_{k+1}\le 1$for all$k$. (Indeed, we have$1-2\cdot10^k\le a_k\le b_k\le 1-10^k$.) • For every$x,y\in S_k$, define$\ell(x,y)=f_k(d(x,y))$where$k=k(x,y)$is the minimum number such that$x,y\in S_k$. For a finite sequence$x_0,x_1,\dots,x_n\in S$define \ell(x_0,x_1,\dots,x_n) = \sum_{i=1}^n \ell(x_{i-1},x_i) .I will refer to this expression as the$\ell$-length of the sequence$x_0,\dots,x_n$. Define$$d'(x,y) = \inf\{ \ell(x_0,x_1,\dots,x_n) \}$$where the infimum is taken over all finite sequences$x_0,x_1,\dots,x_n$in$S$such that$x_0=x$and$x_n=y$. Clearly$d'$is a metric and$\frac12d\le d'\le d$. It remains to show that$d'$takes only rational values. Lemma: If$x,y\in S_k$, then$d'(x,y)$equals the infimum of$\ell$-lengths of sequences contained in$S_k$. Proof: Consider any sequence$x_0,\dots,x_n$in$S$such that$x_0=x$and$y_0=y$. Remove all points that do not belong to$S_k$from this sequence. I claim that the$\ell$-length became shorter. Indeed, it suffices to prove that \ell(x_r,x_s) \le \ell(x_r,x_{r+1},\dots,x_{s-1},x_s)if$x_r$and$x_s$are in$S_k$and the intermediate points are not. By the second property of the functions$f_k$, the left-hand side is bounded above by$b_k d(x_r,x_s)$and every term$\ell(x_i,x_{i+1})$in the right-hand side is bounded below by$b_k d(x_i,x_{i+1})$. So it suffices to prove that b_k d(x_r,x_s) \le b_k\sum_{i=r}^{s-1} d(x_i,x_{i+1}),and this is a triangle inequality multiplied by$b_k$. Q.E.D. All$\ell$-lengths of sequences in$S_k$are multiples of some fixed rational number (namely$10^{-2k}$). Hence$d'(x,y)$is a multiple of the same number if$x,y\in S_k$. Thus all values of$d'\$ are rational.

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