ax = 1 i

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In the particular case where all distances dist(Ai , Ai+1 ) are the same, the ratios ai are equal to a value a such that N a > 1 (necessary condition for the continuity of ) and N a2 < 1 (necessary condition for the simplicity of ) Clearly, ( ) = ( ) = log N/|log a|

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Figure 11 Von Koch curve, the attractor of a system of four similarities with common ratio

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Function scales The previous de nitions all involve ratios of logarithms This is an immediate consequence of the fact that a dimension is de ned as an order of growth related to the scale of functions {t , > 0} In general, a scale of functions F in the neighborhood of 0 is a family of functions which are all comparable in the Hardy sense, that is, for any f and g in F, the ratio f (x)/g(x) tends to a limit (possibly + or ) when x tends to 0 Function scales are de ned in a similar way in the neighborhood of + Scales other than {t } will yield other types of dimensions A dimension must be considered as a Dedekind cut in a given scale of functions The following expressions will make this clearer: (E) = inf{ such that N (E) 0} (E) = sup{ such that N (E) + } these are equivalent to equations (11) and (12) (see [TRI 99]) Complementary intervals on the line In the particular case where the compact E lies in an interval J of the line, the complementary set of E in J is a union of disjoint open intervals, whose lengths will be denoted by cn Let |E| be the Lebesgue measure of E (which means, for an interval, its length) The dimension of E may be written as: (E) = lim sup 1

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(14) (15)

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log E( ) log

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If |E| = 0, the sum of the cn is equal to the length of J The dimension is then equal to the convergence exponent of the series cn : (E) = inf such that

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c < + n

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(16)

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Proof This result may be obtained by calculating an approximation of the length of Minkowski sausage E( ) Let us assume that the complementary intervals are ranked in decreasing lengths: c1 If |E| = 0 and if cn c2 cn

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> cn+1 , then: |E( )| n +

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thus 1 L(E( ))

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n + 1

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It may be shown that both values such that 1

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inf{ such that n < + } and

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ci < +

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are equal to the convergence exponent It is therefore equal to (E) EXERCISE 11 Verify formula (16) for the perfect symmetric sets of Example 11 If |E| = 0, then the convergence exponent of cn still makes sense It characterizes a degree of proximity of the exterior with the set E More precisely, we obtain inf such that

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c < + = lim sup 1 n

log|E( ) E| log

(17)

where the set E( ) E refers to the Minkowski sausage of E deprived of the points of E How can we generalize the study of the complementary set in Rn with n 2 The open intervals must be replaced with an appropriate paving The results connecting the elements of this paving to the dimension depend both on the geometry of the tiles and on their respective positions The topology of the complementary set must be investigated more deeply [TRI 87] The index that generalizes (17) (replacing the 1 of the space dimension by n) is the fact fractal exponent, studied in [GRE 85, TRI 86b] In the case of a zero area curve in R2 , this also leads to the notion of lateral dimension Note that the dimensions corresponding to each side of the curve are not necessarily equal [TRI 99]

122 Packing dimension The packing dimension is, to some extent, a regularization of the box dimension [TRI 82] Indeed, is not -stable, but we may derive a -stable dimension from any index thanks to the operation described below PROPOSITION 11 Let B be the family of all bounded sets of Rn and : B R+ Then, the function de ned for any subsets of Rn as: (E) = inf{sup (Ei )/E = Ei , Ei B} is monotonous and -stable