show/hide this revision's text 3 edited body

My gut feeling is that no list of axioms could simultaneously cover Lebesgue dimension, vector space dimension, Krull dimension, fractal dimension,...

It is not clear to me, for example, how axioms would decide whether $\mathbb C$ has dimension $0$, as required by Krull, dimension $1$ as wished by complex geometers or dimension $2$, the topologists' choice.
(And I haven't even begun to examine the logicians' claim that it has dimension $2^{\aleph_0}$ over $\mathbb Q$)

But this is subjective , so let me say something indisputable: your axiom $A\subset B\Rightarrow dim A \leq dim B$ is already false does not hold for Krull dimension .
Indeed, if $A$ is any domain of Krull dimension $n\gt 0$ and if $K$ is its field of fractions, we have $dim K=0$ and the inequality $dim A=n \leq dim K=0$ is falsenot true.
show/hide this revision's text 2 added example of possible ambiguity; deleted 43 characters in body

My gut feeling is that no list of axioms could simultaneously cover Lebesgue dimension, vector space dimension, Krull dimension, fractal dimension,...

It is not clear to me, for example, how axioms would decide whether $\mathbb C$ has dimension $0$, as required by Krull, dimension $1$ as wished by complex geometers or dimension $2$, the topologists' choice.
(And I haven't even begun to examine the logicians' claim that it has dimension $2^{\aleph_0}$ over $\mathbb Q$)

But this is subjective , so let me say something indisputable: your axiom $A\subset B\Rightarrow dim A \leq dim B$ is already false for Krull dimension :.
Indeed, if $A$ is any domain of Krull dimension $n\gt 0$ and if $K$ is its field of fractions, we have $dim K=0$ and the inequality $dim A=n \leq dim K=0$ is false.
show/hide this revision's text 1

My gut feeling is that no list of axioms could simultaneously cover Lebesgue dimension, vector space dimension, Krull dimension, fractal dimension,...

But this is subjective, so let me say something indisputable: your axiom $A\subset B\Rightarrow dim A \leq dim B$ is already false for Krull dimension :
Indeed, if $A$ is any domain of Krull dimension $n\gt 0$ and if $K$ is its field of fractions, we have $dim K=0$ and the inequality $dim A=n \leq dim K=0$ is false.