Mixing
Any set of positive Lebesgue measure in k-dimensional space contains all finite sets in the space to within a similarity transformation
Clash Royale CLAN TAG#URR8PPP
up vote
1
down vote
favorite
In the third page of My Scottish Book 'Problems', Paul Erdös writes that
There is a very simple theorem of Steinhaus which says that the difference set of a set of positive measure (say on the line) contains an interval. ... It follows instantly from the Lebesgue density theorem, and therefore by this method one obtains the following theorem: Any set of positive measure on the line, or, more generally, in k-dimensional space, contains all finite sets in the space to within a similarity transformation. The proof is almost immediate because by the Lebesgue density theorem there is an interval or a sphere in which the density is as close to 1 as one wishes, and therefore it follows that set will contain a set which is similar to any finite set.
I just cannot follow the reasoning here-- why does that theorem follows from Lebesgue density theorem?
Any hint or reference will be appreciated!
real-analysis lebesgue-measure
asked Jul 24 at 1:42
user392347
183
add a comment |Â
up vote
1
down vote
favorite
In the third page of My Scottish Book 'Problems', Paul Erdös writes that
There is a very simple theorem of Steinhaus which says that the difference set of a set of positive measure (say on the line) contains an interval. ... It follows instantly from the Lebesgue density theorem, and therefore by this method one obtains the following theorem: Any set of positive measure on the line, or, more generally, in k-dimensional space, contains all finite sets in the space to within a similarity transformation. The proof is almost immediate because by the Lebesgue density theorem there is an interval or a sphere in which the density is as close to 1 as one wishes, and therefore it follows that set will contain a set which is similar to any finite set.
I just cannot follow the reasoning here-- why does that theorem follows from Lebesgue density theorem?
Any hint or reference will be appreciated!
real-analysis lebesgue-measure
asked Jul 24 at 1:42
user392347
183
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
In the third page of My Scottish Book 'Problems', Paul Erdös writes that
There is a very simple theorem of Steinhaus which says that the difference set of a set of positive measure (say on the line) contains an interval. ... It follows instantly from the Lebesgue density theorem, and therefore by this method one obtains the following theorem: Any set of positive measure on the line, or, more generally, in k-dimensional space, contains all finite sets in the space to within a similarity transformation. The proof is almost immediate because by the Lebesgue density theorem there is an interval or a sphere in which the density is as close to 1 as one wishes, and therefore it follows that set will contain a set which is similar to any finite set.
I just cannot follow the reasoning here-- why does that theorem follows from Lebesgue density theorem?
Any hint or reference will be appreciated!
real-analysis lebesgue-measure
asked Jul 24 at 1:42
user392347
183
In the third page of My Scottish Book 'Problems', Paul Erdös writes that
There is a very simple theorem of Steinhaus which says that the difference set of a set of positive measure (say on the line) contains an interval. ... It follows instantly from the Lebesgue density theorem, and therefore by this method one obtains the following theorem: Any set of positive measure on the line, or, more generally, in k-dimensional space, contains all finite sets in the space to within a similarity transformation. The proof is almost immediate because by the Lebesgue density theorem there is an interval or a sphere in which the density is as close to 1 as one wishes, and therefore it follows that set will contain a set which is similar to any finite set.
I just cannot follow the reasoning here-- why does that theorem follows from Lebesgue density theorem?
Any hint or reference will be appreciated!
real-analysis lebesgue-measure
asked Jul 24 at 1:42
user392347
183
asked Jul 24 at 1:42
user392347
183
asked Jul 24 at 1:42
user392347
183
asked Jul 24 at 1:42
user392347
183
183
add a comment |Â
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
Let $B$ be a set of positive measure, and by translation and scaling we can assume that $0in B$ is a point of density $1$ in $B$, that $x_1=0<x_2<...<x_n=1$ is the finite set, and we can also assume that $1in B$.
Scaling furter we can assume that $Bcap [0,1]$ has measure $>1-epsilon$, that $Bcap[0,x_n-1]$ has measure $>x_n-1-epsilon$, and so on, $Bcap[0,x_2]$ has measure $> x_2-epsilon$.
Assume that for all $rin[0,1]$, there is an $i_rin1,2,...,n$ such that $rx_inotin B$.
The points of the form $rx_nnotin B$ can only be a set of measure $epsilon$. The set of points of the form $rx_n-1notin B$ can only be a set of measure $epsilon$, ... and likewise for all the others. Therefore, the set of values of $r$ for which at least one $rx_inotin B$ can at most have measure $nepsilon$. Choosing $epsilon << 1/n$, we get that there must be some $rin[0,1]$ such that all $rx_1,...,rx_nin B$.
Perfect! Thank you!
– user392347
Jul 24 at 2:36
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Let $B$ be a set of positive measure, and by translation and scaling we can assume that $0in B$ is a point of density $1$ in $B$, that $x_1=0<x_2<...<x_n=1$ is the finite set, and we can also assume that $1in B$.
Scaling furter we can assume that $Bcap [0,1]$ has measure $>1-epsilon$, that $Bcap[0,x_n-1]$ has measure $>x_n-1-epsilon$, and so on, $Bcap[0,x_2]$ has measure $> x_2-epsilon$.
Assume that for all $rin[0,1]$, there is an $i_rin1,2,...,n$ such that $rx_inotin B$.
The points of the form $rx_nnotin B$ can only be a set of measure $epsilon$. The set of points of the form $rx_n-1notin B$ can only be a set of measure $epsilon$, ... and likewise for all the others. Therefore, the set of values of $r$ for which at least one $rx_inotin B$ can at most have measure $nepsilon$. Choosing $epsilon << 1/n$, we get that there must be some $rin[0,1]$ such that all $rx_1,...,rx_nin B$.
Perfect! Thank you!
– user392347
Jul 24 at 2:36
add a comment |Â
up vote
1
down vote
accepted
Let $B$ be a set of positive measure, and by translation and scaling we can assume that $0in B$ is a point of density $1$ in $B$, that $x_1=0<x_2<...<x_n=1$ is the finite set, and we can also assume that $1in B$.
Scaling furter we can assume that $Bcap [0,1]$ has measure $>1-epsilon$, that $Bcap[0,x_n-1]$ has measure $>x_n-1-epsilon$, and so on, $Bcap[0,x_2]$ has measure $> x_2-epsilon$.
Assume that for all $rin[0,1]$, there is an $i_rin1,2,...,n$ such that $rx_inotin B$.
The points of the form $rx_nnotin B$ can only be a set of measure $epsilon$. The set of points of the form $rx_n-1notin B$ can only be a set of measure $epsilon$, ... and likewise for all the others. Therefore, the set of values of $r$ for which at least one $rx_inotin B$ can at most have measure $nepsilon$. Choosing $epsilon << 1/n$, we get that there must be some $rin[0,1]$ such that all $rx_1,...,rx_nin B$.
Perfect! Thank you!
– user392347
Jul 24 at 2:36
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Let $B$ be a set of positive measure, and by translation and scaling we can assume that $0in B$ is a point of density $1$ in $B$, that $x_1=0<x_2<...<x_n=1$ is the finite set, and we can also assume that $1in B$.
Scaling furter we can assume that $Bcap [0,1]$ has measure $>1-epsilon$, that $Bcap[0,x_n-1]$ has measure $>x_n-1-epsilon$, and so on, $Bcap[0,x_2]$ has measure $> x_2-epsilon$.
Assume that for all $rin[0,1]$, there is an $i_rin1,2,...,n$ such that $rx_inotin B$.
The points of the form $rx_nnotin B$ can only be a set of measure $epsilon$. The set of points of the form $rx_n-1notin B$ can only be a set of measure $epsilon$, ... and likewise for all the others. Therefore, the set of values of $r$ for which at least one $rx_inotin B$ can at most have measure $nepsilon$. Choosing $epsilon << 1/n$, we get that there must be some $rin[0,1]$ such that all $rx_1,...,rx_nin B$.
Let $B$ be a set of positive measure, and by translation and scaling we can assume that $0in B$ is a point of density $1$ in $B$, that $x_1=0<x_2<...<x_n=1$ is the finite set, and we can also assume that $1in B$.
Scaling furter we can assume that $Bcap [0,1]$ has measure $>1-epsilon$, that $Bcap[0,x_n-1]$ has measure $>x_n-1-epsilon$, and so on, $Bcap[0,x_2]$ has measure $> x_2-epsilon$.
Assume that for all $rin[0,1]$, there is an $i_rin1,2,...,n$ such that $rx_inotin B$.
The points of the form $rx_nnotin B$ can only be a set of measure $epsilon$. The set of points of the form $rx_n-1notin B$ can only be a set of measure $epsilon$, ... and likewise for all the others. Therefore, the set of values of $r$ for which at least one $rx_inotin B$ can at most have measure $nepsilon$. Choosing $epsilon << 1/n$, we get that there must be some $rin[0,1]$ such that all $rx_1,...,rx_nin B$.
answered Jul 24 at 2:30
user578878
Perfect! Thank you!
– user392347
Jul 24 at 2:36
add a comment |Â
Perfect! Thank you!
– user392347
Jul 24 at 2:36
Perfect! Thank you!
– user392347
Jul 24 at 2:36
Perfect! Thank you!
– user392347
Jul 24 at 2:36
add a comment |Â
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
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2860926%2fany-set-of-positive-lebesgue-measure-in-k-dimensional-space-contains-all-finite%23new-answer', 'question_page');
);
Post as a guest
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
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
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