Nested family of sets with given Hausdorff dimensionExistence of a set with given Hausdorff...
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Existence of subset with given Hausdorff dimension
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Nested family of sets with given Hausdorff dimension
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$begingroup$
Just before (namely here Existence of a set with given Hausdorff dimension) I asked whether one can find for any real number $alpha>0$ a set $A_alpha$ such that $A_alpha$ has Hausdorff dimension $alpha$. The answer was constructive and used fat Cantor sets and taking the cartesian product with some unit intervall. This made me wonder whether one could push this a bit further, namely
Can we find a family of sets $(A_alpha)_{alpha>0}$ such that $A_alpha$ has Hausdorff dimension $alpha$, $A_n $ is homotopic to $[0,1]^n$ for $nin mathbb{N}$ and $A_alphasubset A_beta$ for $alpha < beta$?
measure-theory geometric-measure-theory
asked Mar 10 at 21:36
Severin SchravenSeverin Schraven
6,4401935
$endgroup$
|
show 4 more comments
$begingroup$
Just before (namely here Existence of a set with given Hausdorff dimension) I asked whether one can find for any real number $alpha>0$ a set $A_alpha$ such that $A_alpha$ has Hausdorff dimension $alpha$. The answer was constructive and used fat Cantor sets and taking the cartesian product with some unit intervall. This made me wonder whether one could push this a bit further, namely
Can we find a family of sets $(A_alpha)_{alpha>0}$ such that $A_alpha$ has Hausdorff dimension $alpha$, $A_n $ is homotopic to $[0,1]^n$ for $nin mathbb{N}$ and $A_alphasubset A_beta$ for $alpha < beta$?
measure-theory geometric-measure-theory
asked Mar 10 at 21:36
Severin SchravenSeverin Schraven
6,4401935
$endgroup$
$begingroup$
Interesting question ! Unlike in the link you mentioned, we cannot really consider the case $0<alpha<1$ and then generalize with induction, because we would not be sure that $A_{alpha} subset A_{beta}$ when $alpha to n^-$ and $beta to n^+$. Thinking about it
$endgroup$
– Charles Madeline
Mar 10 at 22:23
$begingroup$
@CharlesMadeline I think the inclusion you describe are not problematic, as we can modify our family such that $A_{gamma+1}:= A_gamma times [0;1] cup [0;1]^{lfloor gamma rfloor}times {0}$. The thing I do not see is how we get inclusions for $0<gamma<1$.
$endgroup$
– Severin Schraven
2 days ago
$begingroup$
ah yes, that is right indeed. The Cantor sets are a priori not really nested ; what would suffice is a lemma like: if $A subset B$ and $mbox{dim}_H(A) < mbox{dim}_H(B)$, then there exists $C$ with $A subset C subset B$ and $mbox{dim}_H(A)<mbox{dim}_H(C) < mbox{dim}_H(C)$
$endgroup$
– Charles Madeline
2 days ago
$begingroup$
@CharlesMadeline I am not quite sure whether that would suffice (as we are dealing with an uncountable family).
$endgroup$
– Severin Schraven
2 days ago
$begingroup$
I think that would do. There is a classic exercise that says: if $f$ is a continuous function on $[0,1]$ such that for all $x<y$, there is $lambda in ]0,1[$ s.t. $f(lambda x +(1-lambda)y)le $lambda f(x)+(1-lambda)f(y)$. Then $f$ is convex. In our case, you have a dense set of correct values $alpha$. Take unions and conclude
$endgroup$
– Charles Madeline
2 days ago
|
show 4 more comments
$begingroup$
Just before (namely here Existence of a set with given Hausdorff dimension) I asked whether one can find for any real number $alpha>0$ a set $A_alpha$ such that $A_alpha$ has Hausdorff dimension $alpha$. The answer was constructive and used fat Cantor sets and taking the cartesian product with some unit intervall. This made me wonder whether one could push this a bit further, namely
Can we find a family of sets $(A_alpha)_{alpha>0}$ such that $A_alpha$ has Hausdorff dimension $alpha$, $A_n $ is homotopic to $[0,1]^n$ for $nin mathbb{N}$ and $A_alphasubset A_beta$ for $alpha < beta$?
measure-theory geometric-measure-theory
asked Mar 10 at 21:36
Severin SchravenSeverin Schraven
6,4401935
$endgroup$
Just before (namely here Existence of a set with given Hausdorff dimension) I asked whether one can find for any real number $alpha>0$ a set $A_alpha$ such that $A_alpha$ has Hausdorff dimension $alpha$. The answer was constructive and used fat Cantor sets and taking the cartesian product with some unit intervall. This made me wonder whether one could push this a bit further, namely
Can we find a family of sets $(A_alpha)_{alpha>0}$ such that $A_alpha$ has Hausdorff dimension $alpha$, $A_n $ is homotopic to $[0,1]^n$ for $nin mathbb{N}$ and $A_alphasubset A_beta$ for $alpha < beta$?
measure-theory geometric-measure-theory
measure-theory geometric-measure-theory
asked Mar 10 at 21:36
Severin SchravenSeverin Schraven
6,4401935
asked Mar 10 at 21:36
Severin SchravenSeverin Schraven
6,4401935
edited 2 hours ago
Severin Schraven
asked Mar 10 at 21:36
Severin SchravenSeverin Schraven
6,4401935
asked Mar 10 at 21:36
Severin SchravenSeverin Schraven
6,4401935
asked Mar 10 at 21:36
Severin SchravenSeverin Schraven
6,4401935
6,4401935
$begingroup$
Interesting question ! Unlike in the link you mentioned, we cannot really consider the case $0<alpha<1$ and then generalize with induction, because we would not be sure that $A_{alpha} subset A_{beta}$ when $alpha to n^-$ and $beta to n^+$. Thinking about it
$endgroup$
– Charles Madeline
Mar 10 at 22:23
$begingroup$
@CharlesMadeline I think the inclusion you describe are not problematic, as we can modify our family such that $A_{gamma+1}:= A_gamma times [0;1] cup [0;1]^{lfloor gamma rfloor}times {0}$. The thing I do not see is how we get inclusions for $0<gamma<1$.
$endgroup$
– Severin Schraven
2 days ago
$begingroup$
ah yes, that is right indeed. The Cantor sets are a priori not really nested ; what would suffice is a lemma like: if $A subset B$ and $mbox{dim}_H(A) < mbox{dim}_H(B)$, then there exists $C$ with $A subset C subset B$ and $mbox{dim}_H(A)<mbox{dim}_H(C) < mbox{dim}_H(C)$
$endgroup$
– Charles Madeline
2 days ago
$begingroup$
@CharlesMadeline I am not quite sure whether that would suffice (as we are dealing with an uncountable family).
$endgroup$
– Severin Schraven
2 days ago
$begingroup$
I think that would do. There is a classic exercise that says: if $f$ is a continuous function on $[0,1]$ such that for all $x<y$, there is $lambda in ]0,1[$ s.t. $f(lambda x +(1-lambda)y)le $lambda f(x)+(1-lambda)f(y)$. Then $f$ is convex. In our case, you have a dense set of correct values $alpha$. Take unions and conclude
$endgroup$
– Charles Madeline
2 days ago
|
show 4 more comments
$begingroup$
Interesting question ! Unlike in the link you mentioned, we cannot really consider the case $0<alpha<1$ and then generalize with induction, because we would not be sure that $A_{alpha} subset A_{beta}$ when $alpha to n^-$ and $beta to n^+$. Thinking about it
$endgroup$
– Charles Madeline
Mar 10 at 22:23
$begingroup$
@CharlesMadeline I think the inclusion you describe are not problematic, as we can modify our family such that $A_{gamma+1}:= A_gamma times [0;1] cup [0;1]^{lfloor gamma rfloor}times {0}$. The thing I do not see is how we get inclusions for $0<gamma<1$.
$endgroup$
– Severin Schraven
2 days ago
$begingroup$
ah yes, that is right indeed. The Cantor sets are a priori not really nested ; what would suffice is a lemma like: if $A subset B$ and $mbox{dim}_H(A) < mbox{dim}_H(B)$, then there exists $C$ with $A subset C subset B$ and $mbox{dim}_H(A)<mbox{dim}_H(C) < mbox{dim}_H(C)$
$endgroup$
– Charles Madeline
2 days ago
$begingroup$
@CharlesMadeline I am not quite sure whether that would suffice (as we are dealing with an uncountable family).
$endgroup$
– Severin Schraven
2 days ago
$begingroup$
I think that would do. There is a classic exercise that says: if $f$ is a continuous function on $[0,1]$ such that for all $x<y$, there is $lambda in ]0,1[$ s.t. $f(lambda x +(1-lambda)y)le $lambda f(x)+(1-lambda)f(y)$. Then $f$ is convex. In our case, you have a dense set of correct values $alpha$. Take unions and conclude
$endgroup$
– Charles Madeline
2 days ago
$begingroup$
Interesting question ! Unlike in the link you mentioned, we cannot really consider the case $0<alpha<1$ and then generalize with induction, because we would not be sure that $A_{alpha} subset A_{beta}$ when $alpha to n^-$ and $beta to n^+$. Thinking about it
$endgroup$
– Charles Madeline
Mar 10 at 22:23
$begingroup$
Interesting question ! Unlike in the link you mentioned, we cannot really consider the case $0<alpha<1$ and then generalize with induction, because we would not be sure that $A_{alpha} subset A_{beta}$ when $alpha to n^-$ and $beta to n^+$. Thinking about it
$endgroup$
– Charles Madeline
Mar 10 at 22:23
$begingroup$
@CharlesMadeline I think the inclusion you describe are not problematic, as we can modify our family such that $A_{gamma+1}:= A_gamma times [0;1] cup [0;1]^{lfloor gamma rfloor}times {0}$. The thing I do not see is how we get inclusions for $0<gamma<1$.
$endgroup$
– Severin Schraven
2 days ago
$begingroup$
@CharlesMadeline I think the inclusion you describe are not problematic, as we can modify our family such that $A_{gamma+1}:= A_gamma times [0;1] cup [0;1]^{lfloor gamma rfloor}times {0}$. The thing I do not see is how we get inclusions for $0<gamma<1$.
$endgroup$
– Severin Schraven
2 days ago
$begingroup$
ah yes, that is right indeed. The Cantor sets are a priori not really nested ; what would suffice is a lemma like: if $A subset B$ and $mbox{dim}_H(A) < mbox{dim}_H(B)$, then there exists $C$ with $A subset C subset B$ and $mbox{dim}_H(A)<mbox{dim}_H(C) < mbox{dim}_H(C)$
$endgroup$
– Charles Madeline
2 days ago
$begingroup$
ah yes, that is right indeed. The Cantor sets are a priori not really nested ; what would suffice is a lemma like: if $A subset B$ and $mbox{dim}_H(A) < mbox{dim}_H(B)$, then there exists $C$ with $A subset C subset B$ and $mbox{dim}_H(A)<mbox{dim}_H(C) < mbox{dim}_H(C)$
$endgroup$
– Charles Madeline
2 days ago
$begingroup$
@CharlesMadeline I am not quite sure whether that would suffice (as we are dealing with an uncountable family).
$endgroup$
– Severin Schraven
2 days ago
$begingroup$
@CharlesMadeline I am not quite sure whether that would suffice (as we are dealing with an uncountable family).
$endgroup$
– Severin Schraven
2 days ago
$begingroup$
I think that would do. There is a classic exercise that says: if $f$ is a continuous function on $[0,1]$ such that for all $x<y$, there is $lambda in ]0,1[$ s.t. $f(lambda x +(1-lambda)y)le $lambda f(x)+(1-lambda)f(y)$. Then $f$ is convex. In our case, you have a dense set of correct values $alpha$. Take unions and conclude
$endgroup$
– Charles Madeline
2 days ago
$begingroup$
I think that would do. There is a classic exercise that says: if $f$ is a continuous function on $[0,1]$ such that for all $x<y$, there is $lambda in ]0,1[$ s.t. $f(lambda x +(1-lambda)y)le $lambda f(x)+(1-lambda)f(y)$. Then $f$ is convex. In our case, you have a dense set of correct values $alpha$. Take unions and conclude
$endgroup$
– Charles Madeline
2 days ago
|
show 4 more comments
1 Answer
1
active
oldest
votes
$begingroup$
Is a wrap up of the comments by Charles Madeline and the reference given by Skeeve in this mathoverflow question https://mathoverflow.net/questions/325532/existence-of-subset-with-given-hausdorff-dimension.
In Theorem 5.6 from The Geometry of Fractal Sets by Falconer stating that for any Souslin space $Asubseteq mathbb{R}^n$ and any $0<alpha < text{dim}_H(A)$ there exists a compact set $Ksubset A$ such that $alpha= text{dim}_H(K)$.
From this we get that for all Souslin spaces $Asubset B subseteq [0;1]$ and any $gamma in (text{dim}_H(A); text{dim}_H(B)) $ there exists $Asubseteq C subseteq B$ such that $text{dim}_H(C)= gamma$.
Indeed, this follows from the fact that $Bsetminus A$ is a Souslin space again and that $dim_H(Bsetminus A)= dim_H(B)$ as $dim_H(B) = dim_H(Bsetminus A cup A)= max{ dim_H(Bsetminus A), dim_H(A) }$. This allows us to construct (by induction) a family $(A_alpha)_{alphain [0;1]cap mathbb{Q}}$ sucht that $A_alpha subset A_beta$ for $alpha<beta$ and $dim_H(A_alpha)= alpha$. Furthermore, we may choose $A_1 = [0;1]$.
Finally we define for $gamma in [0;1]$
$$ A_gamma = bigcup_{alpha in [0;gamma]cap mathbb{Q}} A_alpha $$
then we get
$$ dim_H(A_gamma) = dim_Hleft( bigcup_{alpha in [0;gamma]cap mathbb{Q}} A_alpha right) = sup_{alpha in [0;gamma]cap mathbb{Q}} dim_H(A_alpha) = gamma. $$
This family satisfies all the conditions we want. Identifying $[0;1]$ with $[0;1]times {0 }$ we play the same game for $gamma in [1;2]$, respectively with similar identification for all $gamma in [0,infty)$ (we could also use the construction suggested by Charles Madeline in the comments taking the cartesian product with the unit interval to pass from $[n; n+1]$ to $[n+1;n+2]$).
answered 2 hours ago
Severin SchravenSeverin Schraven
6,4401935
$endgroup$
add a comment |
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$begingroup$
Is a wrap up of the comments by Charles Madeline and the reference given by Skeeve in this mathoverflow question https://mathoverflow.net/questions/325532/existence-of-subset-with-given-hausdorff-dimension.
In Theorem 5.6 from The Geometry of Fractal Sets by Falconer stating that for any Souslin space $Asubseteq mathbb{R}^n$ and any $0<alpha < text{dim}_H(A)$ there exists a compact set $Ksubset A$ such that $alpha= text{dim}_H(K)$.
From this we get that for all Souslin spaces $Asubset B subseteq [0;1]$ and any $gamma in (text{dim}_H(A); text{dim}_H(B)) $ there exists $Asubseteq C subseteq B$ such that $text{dim}_H(C)= gamma$.
Indeed, this follows from the fact that $Bsetminus A$ is a Souslin space again and that $dim_H(Bsetminus A)= dim_H(B)$ as $dim_H(B) = dim_H(Bsetminus A cup A)= max{ dim_H(Bsetminus A), dim_H(A) }$. This allows us to construct (by induction) a family $(A_alpha)_{alphain [0;1]cap mathbb{Q}}$ sucht that $A_alpha subset A_beta$ for $alpha<beta$ and $dim_H(A_alpha)= alpha$. Furthermore, we may choose $A_1 = [0;1]$.
Finally we define for $gamma in [0;1]$
$$ A_gamma = bigcup_{alpha in [0;gamma]cap mathbb{Q}} A_alpha $$
then we get
$$ dim_H(A_gamma) = dim_Hleft( bigcup_{alpha in [0;gamma]cap mathbb{Q}} A_alpha right) = sup_{alpha in [0;gamma]cap mathbb{Q}} dim_H(A_alpha) = gamma. $$
This family satisfies all the conditions we want. Identifying $[0;1]$ with $[0;1]times {0 }$ we play the same game for $gamma in [1;2]$, respectively with similar identification for all $gamma in [0,infty)$ (we could also use the construction suggested by Charles Madeline in the comments taking the cartesian product with the unit interval to pass from $[n; n+1]$ to $[n+1;n+2]$).
answered 2 hours ago
Severin SchravenSeverin Schraven
6,4401935
$endgroup$
add a comment |
$begingroup$
Is a wrap up of the comments by Charles Madeline and the reference given by Skeeve in this mathoverflow question https://mathoverflow.net/questions/325532/existence-of-subset-with-given-hausdorff-dimension.
In Theorem 5.6 from The Geometry of Fractal Sets by Falconer stating that for any Souslin space $Asubseteq mathbb{R}^n$ and any $0<alpha < text{dim}_H(A)$ there exists a compact set $Ksubset A$ such that $alpha= text{dim}_H(K)$.
From this we get that for all Souslin spaces $Asubset B subseteq [0;1]$ and any $gamma in (text{dim}_H(A); text{dim}_H(B)) $ there exists $Asubseteq C subseteq B$ such that $text{dim}_H(C)= gamma$.
Indeed, this follows from the fact that $Bsetminus A$ is a Souslin space again and that $dim_H(Bsetminus A)= dim_H(B)$ as $dim_H(B) = dim_H(Bsetminus A cup A)= max{ dim_H(Bsetminus A), dim_H(A) }$. This allows us to construct (by induction) a family $(A_alpha)_{alphain [0;1]cap mathbb{Q}}$ sucht that $A_alpha subset A_beta$ for $alpha<beta$ and $dim_H(A_alpha)= alpha$. Furthermore, we may choose $A_1 = [0;1]$.
Finally we define for $gamma in [0;1]$
$$ A_gamma = bigcup_{alpha in [0;gamma]cap mathbb{Q}} A_alpha $$
then we get
$$ dim_H(A_gamma) = dim_Hleft( bigcup_{alpha in [0;gamma]cap mathbb{Q}} A_alpha right) = sup_{alpha in [0;gamma]cap mathbb{Q}} dim_H(A_alpha) = gamma. $$
This family satisfies all the conditions we want. Identifying $[0;1]$ with $[0;1]times {0 }$ we play the same game for $gamma in [1;2]$, respectively with similar identification for all $gamma in [0,infty)$ (we could also use the construction suggested by Charles Madeline in the comments taking the cartesian product with the unit interval to pass from $[n; n+1]$ to $[n+1;n+2]$).
answered 2 hours ago
Severin SchravenSeverin Schraven
6,4401935
$endgroup$
add a comment |
$begingroup$
Is a wrap up of the comments by Charles Madeline and the reference given by Skeeve in this mathoverflow question https://mathoverflow.net/questions/325532/existence-of-subset-with-given-hausdorff-dimension.
In Theorem 5.6 from The Geometry of Fractal Sets by Falconer stating that for any Souslin space $Asubseteq mathbb{R}^n$ and any $0<alpha < text{dim}_H(A)$ there exists a compact set $Ksubset A$ such that $alpha= text{dim}_H(K)$.
From this we get that for all Souslin spaces $Asubset B subseteq [0;1]$ and any $gamma in (text{dim}_H(A); text{dim}_H(B)) $ there exists $Asubseteq C subseteq B$ such that $text{dim}_H(C)= gamma$.
Indeed, this follows from the fact that $Bsetminus A$ is a Souslin space again and that $dim_H(Bsetminus A)= dim_H(B)$ as $dim_H(B) = dim_H(Bsetminus A cup A)= max{ dim_H(Bsetminus A), dim_H(A) }$. This allows us to construct (by induction) a family $(A_alpha)_{alphain [0;1]cap mathbb{Q}}$ sucht that $A_alpha subset A_beta$ for $alpha<beta$ and $dim_H(A_alpha)= alpha$. Furthermore, we may choose $A_1 = [0;1]$.
Finally we define for $gamma in [0;1]$
$$ A_gamma = bigcup_{alpha in [0;gamma]cap mathbb{Q}} A_alpha $$
then we get
$$ dim_H(A_gamma) = dim_Hleft( bigcup_{alpha in [0;gamma]cap mathbb{Q}} A_alpha right) = sup_{alpha in [0;gamma]cap mathbb{Q}} dim_H(A_alpha) = gamma. $$
This family satisfies all the conditions we want. Identifying $[0;1]$ with $[0;1]times {0 }$ we play the same game for $gamma in [1;2]$, respectively with similar identification for all $gamma in [0,infty)$ (we could also use the construction suggested by Charles Madeline in the comments taking the cartesian product with the unit interval to pass from $[n; n+1]$ to $[n+1;n+2]$).
answered 2 hours ago
Severin SchravenSeverin Schraven
6,4401935
$endgroup$
Is a wrap up of the comments by Charles Madeline and the reference given by Skeeve in this mathoverflow question https://mathoverflow.net/questions/325532/existence-of-subset-with-given-hausdorff-dimension.
In Theorem 5.6 from The Geometry of Fractal Sets by Falconer stating that for any Souslin space $Asubseteq mathbb{R}^n$ and any $0<alpha < text{dim}_H(A)$ there exists a compact set $Ksubset A$ such that $alpha= text{dim}_H(K)$.
From this we get that for all Souslin spaces $Asubset B subseteq [0;1]$ and any $gamma in (text{dim}_H(A); text{dim}_H(B)) $ there exists $Asubseteq C subseteq B$ such that $text{dim}_H(C)= gamma$.
Indeed, this follows from the fact that $Bsetminus A$ is a Souslin space again and that $dim_H(Bsetminus A)= dim_H(B)$ as $dim_H(B) = dim_H(Bsetminus A cup A)= max{ dim_H(Bsetminus A), dim_H(A) }$. This allows us to construct (by induction) a family $(A_alpha)_{alphain [0;1]cap mathbb{Q}}$ sucht that $A_alpha subset A_beta$ for $alpha<beta$ and $dim_H(A_alpha)= alpha$. Furthermore, we may choose $A_1 = [0;1]$.
Finally we define for $gamma in [0;1]$
$$ A_gamma = bigcup_{alpha in [0;gamma]cap mathbb{Q}} A_alpha $$
then we get
$$ dim_H(A_gamma) = dim_Hleft( bigcup_{alpha in [0;gamma]cap mathbb{Q}} A_alpha right) = sup_{alpha in [0;gamma]cap mathbb{Q}} dim_H(A_alpha) = gamma. $$
This family satisfies all the conditions we want. Identifying $[0;1]$ with $[0;1]times {0 }$ we play the same game for $gamma in [1;2]$, respectively with similar identification for all $gamma in [0,infty)$ (we could also use the construction suggested by Charles Madeline in the comments taking the cartesian product with the unit interval to pass from $[n; n+1]$ to $[n+1;n+2]$).
answered 2 hours ago
Severin SchravenSeverin Schraven
6,4401935
answered 2 hours ago
Severin SchravenSeverin Schraven
6,4401935
answered 2 hours ago
Severin SchravenSeverin Schraven
6,4401935
answered 2 hours ago
Severin SchravenSeverin Schraven
6,4401935
6,4401935
add a comment |
add a comment |
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$begingroup$
Interesting question ! Unlike in the link you mentioned, we cannot really consider the case $0<alpha<1$ and then generalize with induction, because we would not be sure that $A_{alpha} subset A_{beta}$ when $alpha to n^-$ and $beta to n^+$. Thinking about it
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– Charles Madeline
Mar 10 at 22:23
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@CharlesMadeline I think the inclusion you describe are not problematic, as we can modify our family such that $A_{gamma+1}:= A_gamma times [0;1] cup [0;1]^{lfloor gamma rfloor}times {0}$. The thing I do not see is how we get inclusions for $0<gamma<1$.
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– Severin Schraven
2 days ago
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ah yes, that is right indeed. The Cantor sets are a priori not really nested ; what would suffice is a lemma like: if $A subset B$ and $mbox{dim}_H(A) < mbox{dim}_H(B)$, then there exists $C$ with $A subset C subset B$ and $mbox{dim}_H(A)<mbox{dim}_H(C) < mbox{dim}_H(C)$
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– Charles Madeline
2 days ago
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@CharlesMadeline I am not quite sure whether that would suffice (as we are dealing with an uncountable family).
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– Severin Schraven
2 days ago
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I think that would do. There is a classic exercise that says: if $f$ is a continuous function on $[0,1]$ such that for all $x<y$, there is $lambda in ]0,1[$ s.t. $f(lambda x +(1-lambda)y)le $lambda f(x)+(1-lambda)f(y)$. Then $f$ is convex. In our case, you have a dense set of correct values $alpha$. Take unions and conclude
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– Charles Madeline
2 days ago