How do max and union commute in Hausdorff measure?












0












$begingroup$


Recall that $d_H(A,B) = max{max_{a in A} min_{b in B} d(a,b),max_{b in B} min_{a in A} d(a,b)}$




Theorem: Let $A,B,C in H(X)$ (where $H(X)$ is the set of non-empty compact subsets of $X$). Then $d_H(A cup B,C) = max {d_H(A,C),d_H(B,C)}$




This claim appears in the book on fractals and splines by Peter Massopust. However, I'm not able to connect the hint below with the statement of the theorem:




Proof: Note that
begin{align}
d(Acup B, C) &= max_{alphain Acup B} d(alpha, C) = maxleft{ max_{alphain A} d(alpha,C), max_{alphain B} d(alpha,C)right} \
&= max{ d(A,C), d(B,C)}
end{align}
which implies that claim. (The reader is encouraged to verify this).




It seems that the definition of $d$ is (see the book Fractals everywhere) taken to be $d(A,B) = max{d(a,B):a in A}$. So the above hint makes sense.



So how can I use this hint to prove my theorem?



My try:



$d_H(A cup B,C) = max {d(A cup B,C) , d(C, A cup B) }$



by definition. Then, by the hint:



$d_H(A cup B,C) = max {max{d(A cup B,C), d(C,A cup B)} , d(C, A cup B) }$



then, I take the other member:



$max{d_H(A,C),d_H(B,C)} = max {max{d(A,C), d(C,A)} , {max{d(B,C), d(C,B)} }$



again by definition. Two terms are in the two sides, but I would need an equality of the sort $d(C,A cup B) = max {d(C,A),d(C,B)}$. This reduces to show:



$max_{c in C}{min_{a in A cup B} d(c,a)} = max{max_{c in C}{min_{a in A} d(c,a),max_{c in C}{min_{a in B} d(c,a)}}$






functional-analysis analysis metric-spaces fractals







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$endgroup$












  • $begingroup$
    Where, specifically, are you stuck? There are three equalities in the quoted text, as well as, perhaps, some ambiguity in the last sentence. What, specifically, are you having trouble understanding?
    $endgroup$
    – Xander Henderson
    Nov 30 '18 at 20:55










  • $begingroup$
    @XanderHenderson I would appreciate your feedback with respect to the answer below
    $endgroup$
    – Javier
    Dec 3 '18 at 15:00
















0












$begingroup$


Recall that $d_H(A,B) = max{max_{a in A} min_{b in B} d(a,b),max_{b in B} min_{a in A} d(a,b)}$




Theorem: Let $A,B,C in H(X)$ (where $H(X)$ is the set of non-empty compact subsets of $X$). Then $d_H(A cup B,C) = max {d_H(A,C),d_H(B,C)}$




This claim appears in the book on fractals and splines by Peter Massopust. However, I'm not able to connect the hint below with the statement of the theorem:




Proof: Note that
begin{align}
d(Acup B, C) &= max_{alphain Acup B} d(alpha, C) = maxleft{ max_{alphain A} d(alpha,C), max_{alphain B} d(alpha,C)right} \
&= max{ d(A,C), d(B,C)}
end{align}
which implies that claim. (The reader is encouraged to verify this).




It seems that the definition of $d$ is (see the book Fractals everywhere) taken to be $d(A,B) = max{d(a,B):a in A}$. So the above hint makes sense.



So how can I use this hint to prove my theorem?



My try:



$d_H(A cup B,C) = max {d(A cup B,C) , d(C, A cup B) }$



by definition. Then, by the hint:



$d_H(A cup B,C) = max {max{d(A cup B,C), d(C,A cup B)} , d(C, A cup B) }$



then, I take the other member:



$max{d_H(A,C),d_H(B,C)} = max {max{d(A,C), d(C,A)} , {max{d(B,C), d(C,B)} }$



again by definition. Two terms are in the two sides, but I would need an equality of the sort $d(C,A cup B) = max {d(C,A),d(C,B)}$. This reduces to show:



$max_{c in C}{min_{a in A cup B} d(c,a)} = max{max_{c in C}{min_{a in A} d(c,a),max_{c in C}{min_{a in B} d(c,a)}}$






functional-analysis analysis metric-spaces fractals







share|cite|improve this question











$endgroup$












  • $begingroup$
    Where, specifically, are you stuck? There are three equalities in the quoted text, as well as, perhaps, some ambiguity in the last sentence. What, specifically, are you having trouble understanding?
    $endgroup$
    – Xander Henderson
    Nov 30 '18 at 20:55










  • $begingroup$
    @XanderHenderson I would appreciate your feedback with respect to the answer below
    $endgroup$
    – Javier
    Dec 3 '18 at 15:00














0












0








0





$begingroup$


Recall that $d_H(A,B) = max{max_{a in A} min_{b in B} d(a,b),max_{b in B} min_{a in A} d(a,b)}$




Theorem: Let $A,B,C in H(X)$ (where $H(X)$ is the set of non-empty compact subsets of $X$). Then $d_H(A cup B,C) = max {d_H(A,C),d_H(B,C)}$




This claim appears in the book on fractals and splines by Peter Massopust. However, I'm not able to connect the hint below with the statement of the theorem:




Proof: Note that
begin{align}
d(Acup B, C) &= max_{alphain Acup B} d(alpha, C) = maxleft{ max_{alphain A} d(alpha,C), max_{alphain B} d(alpha,C)right} \
&= max{ d(A,C), d(B,C)}
end{align}
which implies that claim. (The reader is encouraged to verify this).




It seems that the definition of $d$ is (see the book Fractals everywhere) taken to be $d(A,B) = max{d(a,B):a in A}$. So the above hint makes sense.



So how can I use this hint to prove my theorem?



My try:



$d_H(A cup B,C) = max {d(A cup B,C) , d(C, A cup B) }$



by definition. Then, by the hint:



$d_H(A cup B,C) = max {max{d(A cup B,C), d(C,A cup B)} , d(C, A cup B) }$



then, I take the other member:



$max{d_H(A,C),d_H(B,C)} = max {max{d(A,C), d(C,A)} , {max{d(B,C), d(C,B)} }$



again by definition. Two terms are in the two sides, but I would need an equality of the sort $d(C,A cup B) = max {d(C,A),d(C,B)}$. This reduces to show:



$max_{c in C}{min_{a in A cup B} d(c,a)} = max{max_{c in C}{min_{a in A} d(c,a),max_{c in C}{min_{a in B} d(c,a)}}$






functional-analysis analysis metric-spaces fractals







share|cite|improve this question











$endgroup$




Recall that $d_H(A,B) = max{max_{a in A} min_{b in B} d(a,b),max_{b in B} min_{a in A} d(a,b)}$




Theorem: Let $A,B,C in H(X)$ (where $H(X)$ is the set of non-empty compact subsets of $X$). Then $d_H(A cup B,C) = max {d_H(A,C),d_H(B,C)}$




This claim appears in the book on fractals and splines by Peter Massopust. However, I'm not able to connect the hint below with the statement of the theorem:




Proof: Note that
begin{align}
d(Acup B, C) &= max_{alphain Acup B} d(alpha, C) = maxleft{ max_{alphain A} d(alpha,C), max_{alphain B} d(alpha,C)right} \
&= max{ d(A,C), d(B,C)}
end{align}
which implies that claim. (The reader is encouraged to verify this).




It seems that the definition of $d$ is (see the book Fractals everywhere) taken to be $d(A,B) = max{d(a,B):a in A}$. So the above hint makes sense.



So how can I use this hint to prove my theorem?



My try:



$d_H(A cup B,C) = max {d(A cup B,C) , d(C, A cup B) }$



by definition. Then, by the hint:



$d_H(A cup B,C) = max {max{d(A cup B,C), d(C,A cup B)} , d(C, A cup B) }$



then, I take the other member:



$max{d_H(A,C),d_H(B,C)} = max {max{d(A,C), d(C,A)} , {max{d(B,C), d(C,B)} }$



again by definition. Two terms are in the two sides, but I would need an equality of the sort $d(C,A cup B) = max {d(C,A),d(C,B)}$. This reduces to show:



$max_{c in C}{min_{a in A cup B} d(c,a)} = max{max_{c in C}{min_{a in A} d(c,a),max_{c in C}{min_{a in B} d(c,a)}}$






functional-analysis analysis metric-spaces fractals




functional-analysis analysis metric-spaces fractals






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 30 '18 at 23:49







Javier

















asked Nov 30 '18 at 18:19









JavierJavier

2,01621133




2,01621133












  • $begingroup$
    Where, specifically, are you stuck? There are three equalities in the quoted text, as well as, perhaps, some ambiguity in the last sentence. What, specifically, are you having trouble understanding?
    $endgroup$
    – Xander Henderson
    Nov 30 '18 at 20:55










  • $begingroup$
    @XanderHenderson I would appreciate your feedback with respect to the answer below
    $endgroup$
    – Javier
    Dec 3 '18 at 15:00


















  • $begingroup$
    Where, specifically, are you stuck? There are three equalities in the quoted text, as well as, perhaps, some ambiguity in the last sentence. What, specifically, are you having trouble understanding?
    $endgroup$
    – Xander Henderson
    Nov 30 '18 at 20:55










  • $begingroup$
    @XanderHenderson I would appreciate your feedback with respect to the answer below
    $endgroup$
    – Javier
    Dec 3 '18 at 15:00
















$begingroup$
Where, specifically, are you stuck? There are three equalities in the quoted text, as well as, perhaps, some ambiguity in the last sentence. What, specifically, are you having trouble understanding?
$endgroup$
– Xander Henderson
Nov 30 '18 at 20:55




$begingroup$
Where, specifically, are you stuck? There are three equalities in the quoted text, as well as, perhaps, some ambiguity in the last sentence. What, specifically, are you having trouble understanding?
$endgroup$
– Xander Henderson
Nov 30 '18 at 20:55












$begingroup$
@XanderHenderson I would appreciate your feedback with respect to the answer below
$endgroup$
– Javier
Dec 3 '18 at 15:00




$begingroup$
@XanderHenderson I would appreciate your feedback with respect to the answer below
$endgroup$
– Javier
Dec 3 '18 at 15:00










1 Answer
1






active

oldest

votes


















0












$begingroup$

Up to now I have an equality, this comes from Barnsley's "Superfractals", 2006:



The keypoint is the following non-trivial inequality:



$d(C, A cup B) = max_{c in C} min_{x in A cup B} d(c,x) = max_{c in C} min { min_{a in A} d(c,a), min_{b in B} d(c,b)} le min{ max_{c in C} min_{a in A} d(c,a), max_{c in C} min_{b in B} d(c,b) } = min { d(C,A),d(C,B) }$



This inequality gives us that $d(C,A cup B) le d(C,A),d(C,B)$ and $d(C,A cup B) le max{d(C,A),d(C,B)}$.



On the other hand, we also know that $d(B cup C, A) = max{d(B,A),d(C,A)}$.



Finally,



$d_H(A cup B,C) = max {d(C,A cup B),d(A cup B,C) } le max { d(C,A),d(A,C),d(C,B),d(B,C)} = max{d_H(A,C),d_H(B,C)}$



The other inequality is immediate, since:



$d_H(A,C),d_H(B,C) le d_H(A cup B,C)$



so that $max{d_H(A,C),d_H(B,C)} le d_H(A cup B,C)$.






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    0












    $begingroup$

    Up to now I have an equality, this comes from Barnsley's "Superfractals", 2006:



    The keypoint is the following non-trivial inequality:



    $d(C, A cup B) = max_{c in C} min_{x in A cup B} d(c,x) = max_{c in C} min { min_{a in A} d(c,a), min_{b in B} d(c,b)} le min{ max_{c in C} min_{a in A} d(c,a), max_{c in C} min_{b in B} d(c,b) } = min { d(C,A),d(C,B) }$



    This inequality gives us that $d(C,A cup B) le d(C,A),d(C,B)$ and $d(C,A cup B) le max{d(C,A),d(C,B)}$.



    On the other hand, we also know that $d(B cup C, A) = max{d(B,A),d(C,A)}$.



    Finally,



    $d_H(A cup B,C) = max {d(C,A cup B),d(A cup B,C) } le max { d(C,A),d(A,C),d(C,B),d(B,C)} = max{d_H(A,C),d_H(B,C)}$



    The other inequality is immediate, since:



    $d_H(A,C),d_H(B,C) le d_H(A cup B,C)$



    so that $max{d_H(A,C),d_H(B,C)} le d_H(A cup B,C)$.






    share|cite|improve this answer











    $endgroup$


















      0












      $begingroup$

      Up to now I have an equality, this comes from Barnsley's "Superfractals", 2006:



      The keypoint is the following non-trivial inequality:



      $d(C, A cup B) = max_{c in C} min_{x in A cup B} d(c,x) = max_{c in C} min { min_{a in A} d(c,a), min_{b in B} d(c,b)} le min{ max_{c in C} min_{a in A} d(c,a), max_{c in C} min_{b in B} d(c,b) } = min { d(C,A),d(C,B) }$



      This inequality gives us that $d(C,A cup B) le d(C,A),d(C,B)$ and $d(C,A cup B) le max{d(C,A),d(C,B)}$.



      On the other hand, we also know that $d(B cup C, A) = max{d(B,A),d(C,A)}$.



      Finally,



      $d_H(A cup B,C) = max {d(C,A cup B),d(A cup B,C) } le max { d(C,A),d(A,C),d(C,B),d(B,C)} = max{d_H(A,C),d_H(B,C)}$



      The other inequality is immediate, since:



      $d_H(A,C),d_H(B,C) le d_H(A cup B,C)$



      so that $max{d_H(A,C),d_H(B,C)} le d_H(A cup B,C)$.






      share|cite|improve this answer











      $endgroup$
















        0












        0








        0





        $begingroup$

        Up to now I have an equality, this comes from Barnsley's "Superfractals", 2006:



        The keypoint is the following non-trivial inequality:



        $d(C, A cup B) = max_{c in C} min_{x in A cup B} d(c,x) = max_{c in C} min { min_{a in A} d(c,a), min_{b in B} d(c,b)} le min{ max_{c in C} min_{a in A} d(c,a), max_{c in C} min_{b in B} d(c,b) } = min { d(C,A),d(C,B) }$



        This inequality gives us that $d(C,A cup B) le d(C,A),d(C,B)$ and $d(C,A cup B) le max{d(C,A),d(C,B)}$.



        On the other hand, we also know that $d(B cup C, A) = max{d(B,A),d(C,A)}$.



        Finally,



        $d_H(A cup B,C) = max {d(C,A cup B),d(A cup B,C) } le max { d(C,A),d(A,C),d(C,B),d(B,C)} = max{d_H(A,C),d_H(B,C)}$



        The other inequality is immediate, since:



        $d_H(A,C),d_H(B,C) le d_H(A cup B,C)$



        so that $max{d_H(A,C),d_H(B,C)} le d_H(A cup B,C)$.






        share|cite|improve this answer











        $endgroup$



        Up to now I have an equality, this comes from Barnsley's "Superfractals", 2006:



        The keypoint is the following non-trivial inequality:



        $d(C, A cup B) = max_{c in C} min_{x in A cup B} d(c,x) = max_{c in C} min { min_{a in A} d(c,a), min_{b in B} d(c,b)} le min{ max_{c in C} min_{a in A} d(c,a), max_{c in C} min_{b in B} d(c,b) } = min { d(C,A),d(C,B) }$



        This inequality gives us that $d(C,A cup B) le d(C,A),d(C,B)$ and $d(C,A cup B) le max{d(C,A),d(C,B)}$.



        On the other hand, we also know that $d(B cup C, A) = max{d(B,A),d(C,A)}$.



        Finally,



        $d_H(A cup B,C) = max {d(C,A cup B),d(A cup B,C) } le max { d(C,A),d(A,C),d(C,B),d(B,C)} = max{d_H(A,C),d_H(B,C)}$



        The other inequality is immediate, since:



        $d_H(A,C),d_H(B,C) le d_H(A cup B,C)$



        so that $max{d_H(A,C),d_H(B,C)} le d_H(A cup B,C)$.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 3 '18 at 14:57

























        answered Dec 3 '18 at 14:45









        JavierJavier

        2,01621133




        2,01621133






























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