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A Euclidean similarity is a map $S:{\mathbb{R}}^n\to {\mathbb{R}}^n$ satisfying $|S(x)-S(y)|=c|x-y|$ for a fixed $c>0$. It has the form $S(x)=cRx+b$ with $R$ orthogonal: after subtracting $S(0)$ and dividing by $c$, the map fixes zero and preserves distances, hence inner products by polarization, so it is a linear orthogonal map. It is a contraction if $c<1$. Contracting Euclidean similarities define self-similar compact sets; the open set condition is a useful separation hypothesis for computing their Hausdorff dimension.

For finitely many contracting Euclidean similarities, the map $\mathcal{T}$ on nonempty compact subsets of ${\mathbb{R}}^n$ is a contraction in the Hausdorff metric, with contraction constant ${\max }_i{c}_i<1$. The contraction mapping theorem gives a unique nonempty compact invariant set. If the open set condition holds, its Hausdorff dimension is the unique $s$ satisfying ${\sum }_i{c}_i^s=1$.

The middle-third Cantor set has Hausdorff dimensionIts Lebesgue measure is zero, yet $0<{\mathcal{H}}^d(C)<\infty$. These assertions measure different aspects of size: interval length is too coarse to distinguish this set from a null set, while the critical fractional measuring rule captures its remaining structure.

At level $n$, the construction covers $C$ by $2^n$ intervals of diameter $3^{-n}$. Their total $s$-cost isFor $s>d$ it tends to zero, proving the upper dimension bound. At $s=d$ it equals one, giving ${\mathcal{H}}^d(C)\le 1$. For $s<d$ this particular cover's cost grows, but that alone would not prove a lower bound on Hausdorff measure.

For the matching lower bound, use the Cantor distribution, which assigns mass $2^{-n}$ to each level-$n$ Cantor cylinder. An interval of length $r$ with $3^{-(n+1)}<r\le 3^{-n}$ meets at most two level-$n$ construction intervals, since their gaps are at least $3^{-n}$. Its mass is therefore at mostEvery covering set of diameter $r$ in the line lies in an interval of length $r$, so any small cover must have total $d$-cost at least $1/4$. This is the mass distribution principle in its interval form: $1/4\le {\mathcal{H}}^d(C)\le 1$, sufficient to prove the claimed dimension. The constants are convenient bounds, rather than an assertion of the exact critical Hausdorff measure. In the diagram, the right panel is a finite-stage approximation to the Cantor function, showing how probability is shared while the middle third receives none.

The reusable pattern is to obtain an upper bound by constructing covers and a lower bound by distributing mass. The same strategy underlies Cartesian powers of the Cantor set and the dimension formula for a finite similarity iterated function system satisfying the open set condition.

The Minkowski sausage replaces a segment by eight directed unit segments whose endpoint displacement is four units. A middle straight run of length two represents two collinear unit segments. After normalizing the displacement to one, the eight contracting Euclidean similarities all have ratio $1/4$. The open diamond $\{|x-1/2|+|y|<1/2\}$ is an open set condition witness: its eight image diamonds have disjoint interiors and lie within it. Hence $8(1/4{)}^s=1$, giving Hausdorff dimension $3/2$. Treating the middle run as a single similarity of ratio $1/2$ instead defines a different iterated scheme.

For $s\ge 0$ and $\delta >0$, define the scale-restricted Hausdorff content at scale $\delta$ byusing countable covers. The Hausdorff measure is ${\mathcal{H}}^s(F)={\lim }_{\delta \downarrow 0}{\mathcal{H}}_{\delta }^s(F)$; the limit exists by monotonicity. A conventional dimension-dependent normalizing constant does not change the Hausdorff dimensionAt $s=0$, a nonempty singleton has contribution one, so the measure counts points.

Let $\mathcal{T}(K)={\bigcup }_{i=1}^k{S}_i(K)$ and $c={\max }_i{c}_i<1$. For compact nonempty $K,L$, matching a point of $S_i(K)$ with the image of a nearby point of $L$ gives $d_H(\mathcal{T}K,\mathcal{T}L)\le c{d}_H(K,L)$. Choose a sufficiently large closed ball $B$ with every $S_i(B)\subset B$. The nonempty compact sets ${\mathcal{T}}^m(B)$ are nested, so their intersection $I$ is nonempty and compact. The finite union of similarity images commutes with this decreasing compact intersection: a point lying in all unions has some repeated branch index, and compactness supplies a preimage in all the nested sets. Consequently $\mathcal{T}(I)=I$. If $J$ is another invariant compact set, the contraction estimate gives $d_H(I,J)\le c{d}_H(I,J)$, hence $I=J$. This proves existence and uniqueness of the invariant set of a finite similarity iterated function system.

Under the open set condition, there is a nonempty bounded open set $O$ such that the $S_i(O)$ lie inside $O$ and are pairwise disjoint. The dimension formula isOverlaps can invalidate this formula if that assumption is omitted.

For the printed template, subdivide the central straight run of length two into its two collinear unit steps. The eight directed steps, before dividing all coordinates by four, have successive verticesEach step defines a similarity with ratio $1/4$, with rotation matching its direction. The open diamond $O=\{|x-1/2|+|y|<1/2\}$ is an explicit witness for the open set condition. Its eight image diamonds have radius $1/8$ in the ${\ell }^1$ metric, lie in $O$, and their centres have pairwise ${\ell }^1$ distance at least $1/4$; their interiors are therefore disjoint. The Minkowski sausage dimension is consequentlyThis makes the subdivision convention explicit. If the length-two run were instead treated as one unsplit edge replaced at scale $1/2$, there would be six maps of ratio $1/4$ and one of ratio $1/2$. The same diamond separation works, but the equation becomes $64^{-s}+2^{-s}=1$, giving $s={\log }_{2}3$, a different fractal. The requested $3/2$ corresponds to the eight-unit-step interpretation.

Under the open set condition there is a nonempty bounded open set $O$ such that the sets $S_i(O)$ are pairwise disjoint and all contained in $O$. The Hausdorff dimension of the nonempty compact invariant set is the unique $s\ge 0$ satisfyingFor the carpet here the 24 retained sub-squares give Euclidean similarities $S_{ij}(x,y)=((x+i)/5,(y+j)/5)$ for $0\le i,j\le 4$, $(i,j)\ne (2,2)$. Their images of $O=(0,1{)}^2$ are disjoint and lie in $O$, so the condition holds even though some closed images share boundary points. All contraction factors are $1/5$. Thus