Continuous injections preserve Hausdorff dimension

Nonnegativity, symmetry and separation for $\rho$ follow from those for $d$. For the triangle inequality, $\sqrt{a+b}\le \sqrt{a}+\sqrt{b}$ for $a,b\ge 0$, because squaring the right side gives $a+b+2\sqrt{ab}\ge a+b$. Apply this to the triangle inequality for $d$. Also $B_{\rho }(x,r)=B_d(x,r^2)$ for every $r>0$, so both metrics have identical open sets and the identity is a homeomorphism. The usual compact interval is therefore compact in both metrics.

For every subset $U\subseteq I$, ${\mathrm{diam}}_{\rho }U=({\mathrm{diam}}_{d}U{)}^{1\mathrm{/}2}$; for nonempty sets this follows from monotonicity and continuity of the square root applied to the supremum of distances, and for the empty set both sides are zero. Thus for $t\ge 0$ and $\delta >0$, the same cover families give ${\mathcal{H}}_{\rho ,\delta }^t(I)={\mathcal{H}}_{d,{\delta }^2}^{t\mathrm{/}2}(I)$. Nonempty singleton costs match also when $t=0$. Passing to the small-scale suprema yields ${\mathcal{H}}_{\rho }^t(I)={\mathcal{H}}_d^{t\mathrm{/}2}(I)$. Covers in the line may be intersected with $I$ without increasing their costs, and covers in $I$ are line covers, so the usual ambient and subspace values agree.

The unit interval has Lebesgue length one. At $t=2$ the preceding identity gives ${\mathcal{H}}_{\rho }^2(I)={\mathcal{H}}_d^1(I)=1$. The finite-positive criterion gives dimensions two and one in the two metrics. The identity is bijective and hence injective, so this is a counterexample to the asserted invariance.