Let's start by generalizing the concept of a metric space. An $S$-metric space is a set $X$ with a function $d : X \times X \to S$ such that
- $d(x,y) = 0 \iff x = y$
- $d(x,y) = d(y,x)$
- $d(x,z) \leq d(x,y) + d(y,z)$
This is just a metric space which need not necessarily map into $\mathbb{R}$. So my question is:
Are all $\mathbb{R}$-metrizable spaces also $\mathbb{Q}$-metrizable spaces?
I suspect the answer is "No", but I have yet to come up with a counter example. I have shown that a few metrizable spaces are $\mathbb{Q}$-metrizable.
For example discrete spaces are $\mathbb{Q}$-metrizable since the usual metric has a range of $\{0,1\}$. Additionally $\mathbb{R}^n$ is $\mathbb{Q}$-metrizable. If we take $d$ to be the normal Euclidean metric then we can define $d'$ such that:
$
d'(x,y)=\left\lceil d(x,y)\right\rceil
$
The first two conditions follow trivially from the fact that $d$ is a metric and the third is true by virtue of the fact that $\lceil x+y\rceil \leq \lceil x\rceil + \lceil y\rceil$.
Since $\mathbb{R}^n$ is generated by unit balls, this metric generates the usual topology on $\mathbb{R}^n$.
This gives us a whole lot more spaces which are homeomorphic to a subspace of $\mathbb{R}^n$ as well, but I don't see a way to adjust this more generally.
Is there an example of a space which is $\mathbb{R}$-metrizable but not $\mathbb{Q}$-metrizable by the above definition?
Best Answer
${\bf R}$ is not ${\bf Q}$-metrizable. In fact no connected set $X$ is $\bf Q$-metrizable. Let $$\phi(x,y) = d(x,y) \in {\bf Q}$$ This is a continuous function. If $X$ is connected, so is $X\times X$ and so is its image $\phi(X\times X)$. The only connected subset of $\bf Q$ containing $0$ is $\{0\}$ hence $\phi$ is zero everywhere.
Here I am assuming that the topology given by the $\bf Q$-distance is the euclidean one. If we do not make this assumption, then we can put a $\bf Q$-distance on all spaces that are in bijection with $\bf R$ because $\bf R$ itself is in bijection with $\{0,1\}^{\bf N}$ and $\{0,1\}^{\bf N}$ possesses the $\bf Q$-distance $$ d(\{x_n\}_{n \in {\bf N}},\{y_n\}_{n \in {\bf N}}) = 2^{-\min\{n \in {\bf N} \mid x_n \neq y_n\}}. $$