Mixing
The Hausdorff dimension (implication)
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Let: $s geq 0, s in mathbb{R}$ and $E subset mathbb{R^n}$.
Suppose that for every $x in E$ exists an open subset $U subset mathbb{R^n}$ which contains $x$
and dim$_mathcal{H}(E cap U) leq s Rightarrow$ dim$_mathcal{H}(E) leq s$.
How to prove that implication?
I tried to use that:
The Hausdorff dimension is given by $s_0=$inf$lbrace s in [0, infty): mathcal{H^s}(E)=0 rbrace$, there is a $s=m in mathbb{N}$ such that $s_0(E) leq m$.
$U subset mathbb{R^n}$ is open $Rightarrow s_0(E) geq m$
So I got another result than instead of dim$_mathcal{H}(E) leq s$
How to conclude this?
measure-theory
asked 1 hour ago
Olsgur
302
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0
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favorite
Let: $s geq 0, s in mathbb{R}$ and $E subset mathbb{R^n}$.
Suppose that for every $x in E$ exists an open subset $U subset mathbb{R^n}$ which contains $x$
and dim$_mathcal{H}(E cap U) leq s Rightarrow$ dim$_mathcal{H}(E) leq s$.
How to prove that implication?
I tried to use that:
The Hausdorff dimension is given by $s_0=$inf$lbrace s in [0, infty): mathcal{H^s}(E)=0 rbrace$, there is a $s=m in mathbb{N}$ such that $s_0(E) leq m$.
$U subset mathbb{R^n}$ is open $Rightarrow s_0(E) geq m$
So I got another result than instead of dim$_mathcal{H}(E) leq s$
How to conclude this?
measure-theory
asked 1 hour ago
Olsgur
302
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let: $s geq 0, s in mathbb{R}$ and $E subset mathbb{R^n}$.
Suppose that for every $x in E$ exists an open subset $U subset mathbb{R^n}$ which contains $x$
and dim$_mathcal{H}(E cap U) leq s Rightarrow$ dim$_mathcal{H}(E) leq s$.
How to prove that implication?
I tried to use that:
The Hausdorff dimension is given by $s_0=$inf$lbrace s in [0, infty): mathcal{H^s}(E)=0 rbrace$, there is a $s=m in mathbb{N}$ such that $s_0(E) leq m$.
$U subset mathbb{R^n}$ is open $Rightarrow s_0(E) geq m$
So I got another result than instead of dim$_mathcal{H}(E) leq s$
How to conclude this?
measure-theory
asked 1 hour ago
Olsgur
302
Let: $s geq 0, s in mathbb{R}$ and $E subset mathbb{R^n}$.
Suppose that for every $x in E$ exists an open subset $U subset mathbb{R^n}$ which contains $x$
and dim$_mathcal{H}(E cap U) leq s Rightarrow$ dim$_mathcal{H}(E) leq s$.
How to prove that implication?
I tried to use that:
The Hausdorff dimension is given by $s_0=$inf$lbrace s in [0, infty): mathcal{H^s}(E)=0 rbrace$, there is a $s=m in mathbb{N}$ such that $s_0(E) leq m$.
$U subset mathbb{R^n}$ is open $Rightarrow s_0(E) geq m$
So I got another result than instead of dim$_mathcal{H}(E) leq s$
How to conclude this?
measure-theory
measure-theory
asked 1 hour ago
Olsgur
302
asked 1 hour ago
Olsgur
302
asked 1 hour ago
Olsgur
302
asked 1 hour ago
Olsgur
302
asked 1 hour ago
Olsgur
302
302
add a comment |
add a comment |
2 Answers
2
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Pick a $t>s$ to be arbitrary. What you need to show now is that
$$mathcal{H}^{t}(E)=0.$$
Now for every $xin E$ you find a Ball $B_r(x)$, such that
$$mathcal{H}^t(B_r(x)cap E)=0.$$
Now you can choose a countable subcollection of these balls (see e.g. Besicovitch covering theorem) such that
$$Esubset bigcup_{Jin N}B_{r_j}(x_j)cap E$$
and $N$ is countable. Since all these balls satisfy the assumption and $mathcal{H}^t$ is an outer measure you get
$$mathcal{H}^t(E)leq sum_{jin N}mathcal{H}^t(B_{r_j}(x_j)cap E) = 0.$$
Hence $dim_{mathcal{H}}(E)leq t$. Since $t>s$ arbitrary the result follows.
answered 39 mins ago
humanStampedist
2,148213
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The Hausdorff measures satisfies that $mathcal{H}^t(E)=0$ for every $t>d$ if and only if $dim(E)leq d$. So we have to prove that $mathcal{H}^r(E)=0$ for every $r>s$. Let $r$ such a number.
We know that for every $x$ in $E$ there exist an open neighborhood $U_x$ of $x$ such that $mathcal{H}^r(Ecap U_x)=0$. So we have a cover of $E$ by open sets of $mathbb{R}^n$ and we can take a countable subcover ${U_n}_{n in mathbb{N}}$.
By $sigma$-subadditivity we get that $mathcal{H}^r(E)leqsum_{nin mathbb{N}}mathcal{H}^r(Ecap U_n)=0$. And we are done.
answered 22 mins ago
Dante Grevino
363
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Pick a $t>s$ to be arbitrary. What you need to show now is that
$$mathcal{H}^{t}(E)=0.$$
Now for every $xin E$ you find a Ball $B_r(x)$, such that
$$mathcal{H}^t(B_r(x)cap E)=0.$$
Now you can choose a countable subcollection of these balls (see e.g. Besicovitch covering theorem) such that
$$Esubset bigcup_{Jin N}B_{r_j}(x_j)cap E$$
and $N$ is countable. Since all these balls satisfy the assumption and $mathcal{H}^t$ is an outer measure you get
$$mathcal{H}^t(E)leq sum_{jin N}mathcal{H}^t(B_{r_j}(x_j)cap E) = 0.$$
Hence $dim_{mathcal{H}}(E)leq t$. Since $t>s$ arbitrary the result follows.
answered 39 mins ago
humanStampedist
2,148213
add a comment |
up vote
0
down vote
Pick a $t>s$ to be arbitrary. What you need to show now is that
$$mathcal{H}^{t}(E)=0.$$
Now for every $xin E$ you find a Ball $B_r(x)$, such that
$$mathcal{H}^t(B_r(x)cap E)=0.$$
Now you can choose a countable subcollection of these balls (see e.g. Besicovitch covering theorem) such that
$$Esubset bigcup_{Jin N}B_{r_j}(x_j)cap E$$
and $N$ is countable. Since all these balls satisfy the assumption and $mathcal{H}^t$ is an outer measure you get
$$mathcal{H}^t(E)leq sum_{jin N}mathcal{H}^t(B_{r_j}(x_j)cap E) = 0.$$
Hence $dim_{mathcal{H}}(E)leq t$. Since $t>s$ arbitrary the result follows.
answered 39 mins ago
humanStampedist
2,148213
add a comment |
up vote
0
down vote
up vote
0
down vote
Pick a $t>s$ to be arbitrary. What you need to show now is that
$$mathcal{H}^{t}(E)=0.$$
Now for every $xin E$ you find a Ball $B_r(x)$, such that
$$mathcal{H}^t(B_r(x)cap E)=0.$$
Now you can choose a countable subcollection of these balls (see e.g. Besicovitch covering theorem) such that
$$Esubset bigcup_{Jin N}B_{r_j}(x_j)cap E$$
and $N$ is countable. Since all these balls satisfy the assumption and $mathcal{H}^t$ is an outer measure you get
$$mathcal{H}^t(E)leq sum_{jin N}mathcal{H}^t(B_{r_j}(x_j)cap E) = 0.$$
Hence $dim_{mathcal{H}}(E)leq t$. Since $t>s$ arbitrary the result follows.
answered 39 mins ago
humanStampedist
2,148213
Pick a $t>s$ to be arbitrary. What you need to show now is that
$$mathcal{H}^{t}(E)=0.$$
Now for every $xin E$ you find a Ball $B_r(x)$, such that
$$mathcal{H}^t(B_r(x)cap E)=0.$$
Now you can choose a countable subcollection of these balls (see e.g. Besicovitch covering theorem) such that
$$Esubset bigcup_{Jin N}B_{r_j}(x_j)cap E$$
and $N$ is countable. Since all these balls satisfy the assumption and $mathcal{H}^t$ is an outer measure you get
$$mathcal{H}^t(E)leq sum_{jin N}mathcal{H}^t(B_{r_j}(x_j)cap E) = 0.$$
Hence $dim_{mathcal{H}}(E)leq t$. Since $t>s$ arbitrary the result follows.
answered 39 mins ago
humanStampedist
2,148213
answered 39 mins ago
humanStampedist
2,148213
answered 39 mins ago
humanStampedist
2,148213
answered 39 mins ago
humanStampedist
2,148213
2,148213
add a comment |
add a comment |
up vote
0
down vote
The Hausdorff measures satisfies that $mathcal{H}^t(E)=0$ for every $t>d$ if and only if $dim(E)leq d$. So we have to prove that $mathcal{H}^r(E)=0$ for every $r>s$. Let $r$ such a number.
We know that for every $x$ in $E$ there exist an open neighborhood $U_x$ of $x$ such that $mathcal{H}^r(Ecap U_x)=0$. So we have a cover of $E$ by open sets of $mathbb{R}^n$ and we can take a countable subcover ${U_n}_{n in mathbb{N}}$.
By $sigma$-subadditivity we get that $mathcal{H}^r(E)leqsum_{nin mathbb{N}}mathcal{H}^r(Ecap U_n)=0$. And we are done.
answered 22 mins ago
Dante Grevino
363
New contributor
Dante Grevino is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
up vote
0
down vote
The Hausdorff measures satisfies that $mathcal{H}^t(E)=0$ for every $t>d$ if and only if $dim(E)leq d$. So we have to prove that $mathcal{H}^r(E)=0$ for every $r>s$. Let $r$ such a number.
We know that for every $x$ in $E$ there exist an open neighborhood $U_x$ of $x$ such that $mathcal{H}^r(Ecap U_x)=0$. So we have a cover of $E$ by open sets of $mathbb{R}^n$ and we can take a countable subcover ${U_n}_{n in mathbb{N}}$.
By $sigma$-subadditivity we get that $mathcal{H}^r(E)leqsum_{nin mathbb{N}}mathcal{H}^r(Ecap U_n)=0$. And we are done.
answered 22 mins ago
Dante Grevino
363
New contributor
Dante Grevino is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
up vote
0
down vote
up vote
0
down vote
The Hausdorff measures satisfies that $mathcal{H}^t(E)=0$ for every $t>d$ if and only if $dim(E)leq d$. So we have to prove that $mathcal{H}^r(E)=0$ for every $r>s$. Let $r$ such a number.
We know that for every $x$ in $E$ there exist an open neighborhood $U_x$ of $x$ such that $mathcal{H}^r(Ecap U_x)=0$. So we have a cover of $E$ by open sets of $mathbb{R}^n$ and we can take a countable subcover ${U_n}_{n in mathbb{N}}$.
By $sigma$-subadditivity we get that $mathcal{H}^r(E)leqsum_{nin mathbb{N}}mathcal{H}^r(Ecap U_n)=0$. And we are done.
answered 22 mins ago
Dante Grevino
363
New contributor
Dante Grevino is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
The Hausdorff measures satisfies that $mathcal{H}^t(E)=0$ for every $t>d$ if and only if $dim(E)leq d$. So we have to prove that $mathcal{H}^r(E)=0$ for every $r>s$. Let $r$ such a number.
We know that for every $x$ in $E$ there exist an open neighborhood $U_x$ of $x$ such that $mathcal{H}^r(Ecap U_x)=0$. So we have a cover of $E$ by open sets of $mathbb{R}^n$ and we can take a countable subcover ${U_n}_{n in mathbb{N}}$.
By $sigma$-subadditivity we get that $mathcal{H}^r(E)leqsum_{nin mathbb{N}}mathcal{H}^r(Ecap U_n)=0$. And we are done.
answered 22 mins ago
Dante Grevino
363
New contributor
Dante Grevino is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 22 mins ago
Dante Grevino
363
New contributor
Dante Grevino is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 22 mins ago
Dante Grevino
363
answered 22 mins ago
Dante Grevino
363
363
New contributor
Dante Grevino is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Dante Grevino is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Dante Grevino is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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