The question has been answered in the comments, but just for the record: The homeomorphic image of a disc (even embedded in $\mathbb{R}^n$ for $n>2$, or indeed in any metric space) cannot have Hausdorff dimension less than two, since the topological dimension is a lower bound for the Hausdorff dimension. The diffeomorphic image of a disc will always have Hausdorff dimension two.

The quantity in question is related to what is sometimes called Hausdorff content; it involves not having the diameter of the covering sets tending to zero. Compare e.g. "Hausdorff content and Hausdorff measure" on math.stackexchange.

The point is that Hausdorff content is always finite for sets with bounded diameter, while Hausdorff measure need not be. On the other hand, both have the same sets of zero measure (since getting a small sum clearly requires you to have the diameters of the sets being small).

The question has been answered in the comments, but just for the record: The homeomorphic image of a disc (even embedded in $\mathbb{R}^n$ for $n>2$, or indeed in any metric space) cannot have Hausdorff dimension less than two, since the topological dimension is a lower bound for the Hausdorff dimension. The diffeomorphic image of a disc will always have Hausdorff dimension two.

The quantity in question is related to what is sometimes called Hausdorff content; it involves not having the diameter of the covering sets tending to zero. Compare e.g. "Hausdorff content and Hausdorff measure" on math.stackexchange.

The point is that Hausdorff content is always finite for sets with bounded diameter, while Hausdorff measure need not be. On the other hand, both have the same sets of zero measure (since getting a small sum clearly requires you to have the diameters of the sets being small).

The question has been answered in the comments, but just for the record: The homeomorphic image of a disc cannot have Hausdorff dimension less than two, since the topological dimension is a lower bound for the Hausdorff dimension. The diffeomorphic image of a disc will always have Hausdorff dimension two.

The quantity in question is related to what is sometimes called Hausdorff content; it involves not having the diameter of the covering sets tending to zero. Compare e.g. "Hausdorff content and Hausdorff measure" on math.stackexchange.

The point is that Hausdorff content is always finite for sets with bounded diameter, while Hausdorff measure need not be. On the other hand, both have the same sets of zero measure (since getting a small sum clearly requires you to have the diameters of the sets being small).

The question has been answered in the comments, but just for the record: The homeomorphic image of a disc (even embedded in $\mathbb{R}^n$ for $n>2$, or indeed in any metric space) cannot have Hausdorff dimension less than two, since the topological dimension is a lower bound for the Hausdorff dimension. The diffeomorphic image of a disc will always have Hausdorff dimension two.

The quantity in question is related to what is sometimes called Hausdorff content; it involves not having the diameter of the covering sets tending to zero. Compare e.g. "Hausdorff content and Hausdorff measure" on math.stackexchange.

The point is that Hausdorff content is always finite for sets with bounded diameter, while Hausdorff measure need not be. On the other hand, both have the same sets of zero measure (since getting a small sum clearly requires you to have the diameters of the sets being small).