Hausdorff dimension is monotone and countably stable

If $A\subseteq B$, every exponent with ${\mathcal{H}}^s(B)=0$ also has ${\mathcal{H}}^s(A)=0$. Infimising zero exponents gives ${\dim }_{H}A\le {\dim }_{H}B$, including empty zero-exponent sets. Thus the dimension of the union is at least $D={\sup }_k{\dim }_H{A}_k$.

If $D=\mathrm{\infty }$ that lower bound is equality. If $D<\mathrm{\infty }$, for every $t>D$ all ${\mathcal{H}}^t(A_k)$ vanish, so countable subadditivity makes their union null. Its dimension is at most every $t>D$, hence at most $D$. This includes $D=0$.

An at most countable set has a cover by its singletons, whose total cost is zero for each $t>0$. The empty set has the empty cover. Thus their dimensions are zero; one point and every finite set are included.