How do max and union commute in Hausdorff measure?
$begingroup$
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
asked Nov 30 '18 at 18:19
JavierJavier
2,01621133
$endgroup$
add a comment |
$begingroup$
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
asked Nov 30 '18 at 18:19
JavierJavier
2,01621133
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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
asked Nov 30 '18 at 18:19
JavierJavier
2,01621133
$endgroup$
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
asked Nov 30 '18 at 18:19
JavierJavier
2,01621133
asked Nov 30 '18 at 18:19
JavierJavier
2,01621133
edited Nov 30 '18 at 23:49
Javier
asked Nov 30 '18 at 18:19
JavierJavier
2,01621133
asked Nov 30 '18 at 18:19
JavierJavier
2,01621133
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
add a comment |
$begingroup$
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
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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 $max{d_H(A,C),d_H(B,C)} le d_H(A cup B,C)$.
answered Dec 3 '18 at 14:45
JavierJavier
2,01621133
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3020435%2fhow-do-max-and-union-commute-in-hausdorff-measure%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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 $max{d_H(A,C),d_H(B,C)} le d_H(A cup B,C)$.
answered Dec 3 '18 at 14:45
JavierJavier
2,01621133
$endgroup$
add a comment |
$begingroup$
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 $max{d_H(A,C),d_H(B,C)} le d_H(A cup B,C)$.
answered Dec 3 '18 at 14:45
JavierJavier
2,01621133
$endgroup$
add a comment |
$begingroup$
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 $max{d_H(A,C),d_H(B,C)} le d_H(A cup B,C)$.
answered Dec 3 '18 at 14:45
JavierJavier
2,01621133
$endgroup$
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 $max{d_H(A,C),d_H(B,C)} le d_H(A cup B,C)$.
answered Dec 3 '18 at 14:45
JavierJavier
2,01621133
edited Dec 3 '18 at 14:57
answered Dec 3 '18 at 14:45
JavierJavier
2,01621133
answered Dec 3 '18 at 14:45
JavierJavier
2,01621133
answered Dec 3 '18 at 14:45
JavierJavier
2,01621133
2,01621133
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3020435%2fhow-do-max-and-union-commute-in-hausdorff-measure%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
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