Mixing
The cantor set $K$ contains no intervals
$begingroup$
Let $K$ the Cantor set. I already proved the following properties
Properties.begin{equation}
begin{split}
(1)&quad|K|=|mathbb{R}|\
(2)&quadlambda(K)=0,text{where};lambda;text{is the Lebesgue measure};\
(3)&quadtext{If};Ein 2^{K}Rightarrow E;text{is Lebesgue measurable;}
end{split}
end{equation}
I want to prove that
$K$ does not contain intervals.
Now, if $I=[a,b]subseteq K$ is an interval, $ane b$, then for $(3)$ it is measurable and $lambda(I)>0$, absurd. Therefore, $K$ does not contain intervals.
Question. How can I show that $K$ contains no intervals using only the method by which it is constructed, ie without using the monotony of the measure?
Thanks!
measure-theory cantor-set
asked Jan 24 at 15:09
Jack J.Jack J.
3492419
$endgroup$
add a comment |
$begingroup$
Let $K$ the Cantor set. I already proved the following properties
Properties.begin{equation}
begin{split}
(1)&quad|K|=|mathbb{R}|\
(2)&quadlambda(K)=0,text{where};lambda;text{is the Lebesgue measure};\
(3)&quadtext{If};Ein 2^{K}Rightarrow E;text{is Lebesgue measurable;}
end{split}
end{equation}
I want to prove that
$K$ does not contain intervals.
Now, if $I=[a,b]subseteq K$ is an interval, $ane b$, then for $(3)$ it is measurable and $lambda(I)>0$, absurd. Therefore, $K$ does not contain intervals.
Question. How can I show that $K$ contains no intervals using only the method by which it is constructed, ie without using the monotony of the measure?
Thanks!
measure-theory cantor-set
asked Jan 24 at 15:09
Jack J.Jack J.
3492419
$endgroup$
$begingroup$
What is the method by which it is constructed?
$endgroup$
– Praphulla Koushik
Jan 24 at 15:10
$begingroup$
math.stackexchange.com/questions/3083396/…
$endgroup$
– Jack J.
Jan 24 at 15:14
$begingroup$
(there are many different methods to construct the Cantor set as evidenced by the different answers, so I believe Praphulla was asking which method /you/ considered to be the method by which it is constructed)
$endgroup$
– Calvin Khor
Jan 24 at 16:28
$begingroup$
In fact I replied to Praphulla, although I did not apologize for the lack of clarity. I take the opportunity to do it now.
$endgroup$
– Jack J.
Jan 24 at 16:31
add a comment |
$begingroup$
Let $K$ the Cantor set. I already proved the following properties
Properties.begin{equation}
begin{split}
(1)&quad|K|=|mathbb{R}|\
(2)&quadlambda(K)=0,text{where};lambda;text{is the Lebesgue measure};\
(3)&quadtext{If};Ein 2^{K}Rightarrow E;text{is Lebesgue measurable;}
end{split}
end{equation}
I want to prove that
$K$ does not contain intervals.
Now, if $I=[a,b]subseteq K$ is an interval, $ane b$, then for $(3)$ it is measurable and $lambda(I)>0$, absurd. Therefore, $K$ does not contain intervals.
Question. How can I show that $K$ contains no intervals using only the method by which it is constructed, ie without using the monotony of the measure?
Thanks!
measure-theory cantor-set
asked Jan 24 at 15:09
Jack J.Jack J.
3492419
$endgroup$
Let $K$ the Cantor set. I already proved the following properties
Properties.begin{equation}
begin{split}
(1)&quad|K|=|mathbb{R}|\
(2)&quadlambda(K)=0,text{where};lambda;text{is the Lebesgue measure};\
(3)&quadtext{If};Ein 2^{K}Rightarrow E;text{is Lebesgue measurable;}
end{split}
end{equation}
I want to prove that
$K$ does not contain intervals.
Now, if $I=[a,b]subseteq K$ is an interval, $ane b$, then for $(3)$ it is measurable and $lambda(I)>0$, absurd. Therefore, $K$ does not contain intervals.
Question. How can I show that $K$ contains no intervals using only the method by which it is constructed, ie without using the monotony of the measure?
Thanks!
measure-theory cantor-set
measure-theory cantor-set
asked Jan 24 at 15:09
Jack J.Jack J.
3492419
asked Jan 24 at 15:09
Jack J.Jack J.
3492419
asked Jan 24 at 15:09
Jack J.Jack J.
3492419
asked Jan 24 at 15:09
Jack J.Jack J.
3492419
asked Jan 24 at 15:09
Jack J.Jack J.
3492419
3492419
$begingroup$
What is the method by which it is constructed?
$endgroup$
– Praphulla Koushik
Jan 24 at 15:10
$begingroup$
math.stackexchange.com/questions/3083396/…
$endgroup$
– Jack J.
Jan 24 at 15:14
$begingroup$
(there are many different methods to construct the Cantor set as evidenced by the different answers, so I believe Praphulla was asking which method /you/ considered to be the method by which it is constructed)
$endgroup$
– Calvin Khor
Jan 24 at 16:28
$begingroup$
In fact I replied to Praphulla, although I did not apologize for the lack of clarity. I take the opportunity to do it now.
$endgroup$
– Jack J.
Jan 24 at 16:31
add a comment |
$begingroup$
What is the method by which it is constructed?
$endgroup$
– Praphulla Koushik
Jan 24 at 15:10
$begingroup$
math.stackexchange.com/questions/3083396/…
$endgroup$
– Jack J.
Jan 24 at 15:14
$begingroup$
(there are many different methods to construct the Cantor set as evidenced by the different answers, so I believe Praphulla was asking which method /you/ considered to be the method by which it is constructed)
$endgroup$
– Calvin Khor
Jan 24 at 16:28
$begingroup$
In fact I replied to Praphulla, although I did not apologize for the lack of clarity. I take the opportunity to do it now.
$endgroup$
– Jack J.
Jan 24 at 16:31
$begingroup$
What is the method by which it is constructed?
$endgroup$
– Praphulla Koushik
Jan 24 at 15:10
$begingroup$
What is the method by which it is constructed?
$endgroup$
– Praphulla Koushik
Jan 24 at 15:10
$begingroup$
math.stackexchange.com/questions/3083396/…
$endgroup$
– Jack J.
Jan 24 at 15:14
$begingroup$
math.stackexchange.com/questions/3083396/…
$endgroup$
– Jack J.
Jan 24 at 15:14
$begingroup$
(there are many different methods to construct the Cantor set as evidenced by the different answers, so I believe Praphulla was asking which method /you/ considered to be the method by which it is constructed)
$endgroup$
– Calvin Khor
Jan 24 at 16:28
$begingroup$
(there are many different methods to construct the Cantor set as evidenced by the different answers, so I believe Praphulla was asking which method /you/ considered to be the method by which it is constructed)
$endgroup$
– Calvin Khor
Jan 24 at 16:28
$begingroup$
In fact I replied to Praphulla, although I did not apologize for the lack of clarity. I take the opportunity to do it now.
$endgroup$
– Jack J.
Jan 24 at 16:31
$begingroup$
In fact I replied to Praphulla, although I did not apologize for the lack of clarity. I take the opportunity to do it now.
$endgroup$
– Jack J.
Jan 24 at 16:31
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Given any interval $[a,b]$ you can prove it is not in $K$ by finding an interval within it that is removed. Find an $n$ such that $3^{-n} lt b-a$ and you will remove a segment of length $3^{-(n+1)}$ from it at stage $n+1$ unless you have removed some sooner. This shows $[a,b]$ is not in the set.
answered Jan 24 at 15:16
Ross MillikanRoss Millikan
299k24200374
$endgroup$
add a comment |
$begingroup$
The Cantor set can be described as the set of numbers in $[0,1]$ that do not require the digit $1$ to be written in base $3$. You can prove this is equivalent to the definition you provided - in the first step you remove all numbers of the form $0.1dots$ (except $0.1$ itself which can also be written as $0.0222cdots$). In the second step you remove all numbers of the form $0.x1cdots$ where $x$ is $0$ or $2$ (except $0.01$ and $0.21$ which can be written as $0.00222cdots$ and $0.20222 cdots$, respectively). And so on.
If $[a,b] subseteq K$ with $a neq b$, then consider the base $3$ expansions of $a$ and $b$: $a=0.a_1a_2a_3dots$, $b=0.b_1b_2b_3dots$. These numbers are distinct, so consider the first digit at which they differ. Say $a_k neq b_k$ but $a_i=b_i$ for all $i <k$. The expansions of $a$ and $b$ don't contain any $1$'s, so all the digits are either $0$ or $2$.
Since $a<b$ and they agree in their first $k-1$ digits, we must have $a_k=0$ and $b_k=2$. Then the number $c=0.a_1a_2cdots a_{k-1}111 cdots$ is between $a$ and $b$, and it requires a $1$ in its ternary expansion, so $c notin K$, a contradiction.
answered Jan 24 at 15:24
kccukccu
10.6k11229
$endgroup$
add a comment |
$begingroup$
$C=bigcap_k C_k $, where each $C_k$ is the union of $2^k$ closed intervals in $[0,1]$ of length $3^{-k}.$ Therefore, as soon as $mathbb N ni N>-log_3(b-a), (a,b)nsubseteq C_N$ and so of course, $(a,b)$ cannot be contained in the intersection of the $C_k$.
answered Jan 24 at 16:06
MatematletaMatematleta
11.5k2920
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Given any interval $[a,b]$ you can prove it is not in $K$ by finding an interval within it that is removed. Find an $n$ such that $3^{-n} lt b-a$ and you will remove a segment of length $3^{-(n+1)}$ from it at stage $n+1$ unless you have removed some sooner. This shows $[a,b]$ is not in the set.
answered Jan 24 at 15:16
Ross MillikanRoss Millikan
299k24200374
$endgroup$
add a comment |
$begingroup$
Given any interval $[a,b]$ you can prove it is not in $K$ by finding an interval within it that is removed. Find an $n$ such that $3^{-n} lt b-a$ and you will remove a segment of length $3^{-(n+1)}$ from it at stage $n+1$ unless you have removed some sooner. This shows $[a,b]$ is not in the set.
answered Jan 24 at 15:16
Ross MillikanRoss Millikan
299k24200374
$endgroup$
add a comment |
$begingroup$
Given any interval $[a,b]$ you can prove it is not in $K$ by finding an interval within it that is removed. Find an $n$ such that $3^{-n} lt b-a$ and you will remove a segment of length $3^{-(n+1)}$ from it at stage $n+1$ unless you have removed some sooner. This shows $[a,b]$ is not in the set.
answered Jan 24 at 15:16
Ross MillikanRoss Millikan
299k24200374
$endgroup$
Given any interval $[a,b]$ you can prove it is not in $K$ by finding an interval within it that is removed. Find an $n$ such that $3^{-n} lt b-a$ and you will remove a segment of length $3^{-(n+1)}$ from it at stage $n+1$ unless you have removed some sooner. This shows $[a,b]$ is not in the set.
answered Jan 24 at 15:16
Ross MillikanRoss Millikan
299k24200374
edited Jan 24 at 15:22
answered Jan 24 at 15:16
Ross MillikanRoss Millikan
299k24200374
answered Jan 24 at 15:16
Ross MillikanRoss Millikan
299k24200374
answered Jan 24 at 15:16
Ross MillikanRoss Millikan
299k24200374
299k24200374
add a comment |
add a comment |
$begingroup$
The Cantor set can be described as the set of numbers in $[0,1]$ that do not require the digit $1$ to be written in base $3$. You can prove this is equivalent to the definition you provided - in the first step you remove all numbers of the form $0.1dots$ (except $0.1$ itself which can also be written as $0.0222cdots$). In the second step you remove all numbers of the form $0.x1cdots$ where $x$ is $0$ or $2$ (except $0.01$ and $0.21$ which can be written as $0.00222cdots$ and $0.20222 cdots$, respectively). And so on.
If $[a,b] subseteq K$ with $a neq b$, then consider the base $3$ expansions of $a$ and $b$: $a=0.a_1a_2a_3dots$, $b=0.b_1b_2b_3dots$. These numbers are distinct, so consider the first digit at which they differ. Say $a_k neq b_k$ but $a_i=b_i$ for all $i <k$. The expansions of $a$ and $b$ don't contain any $1$'s, so all the digits are either $0$ or $2$.
Since $a<b$ and they agree in their first $k-1$ digits, we must have $a_k=0$ and $b_k=2$. Then the number $c=0.a_1a_2cdots a_{k-1}111 cdots$ is between $a$ and $b$, and it requires a $1$ in its ternary expansion, so $c notin K$, a contradiction.
answered Jan 24 at 15:24
kccukccu
10.6k11229
$endgroup$
add a comment |
$begingroup$
The Cantor set can be described as the set of numbers in $[0,1]$ that do not require the digit $1$ to be written in base $3$. You can prove this is equivalent to the definition you provided - in the first step you remove all numbers of the form $0.1dots$ (except $0.1$ itself which can also be written as $0.0222cdots$). In the second step you remove all numbers of the form $0.x1cdots$ where $x$ is $0$ or $2$ (except $0.01$ and $0.21$ which can be written as $0.00222cdots$ and $0.20222 cdots$, respectively). And so on.
If $[a,b] subseteq K$ with $a neq b$, then consider the base $3$ expansions of $a$ and $b$: $a=0.a_1a_2a_3dots$, $b=0.b_1b_2b_3dots$. These numbers are distinct, so consider the first digit at which they differ. Say $a_k neq b_k$ but $a_i=b_i$ for all $i <k$. The expansions of $a$ and $b$ don't contain any $1$'s, so all the digits are either $0$ or $2$.
Since $a<b$ and they agree in their first $k-1$ digits, we must have $a_k=0$ and $b_k=2$. Then the number $c=0.a_1a_2cdots a_{k-1}111 cdots$ is between $a$ and $b$, and it requires a $1$ in its ternary expansion, so $c notin K$, a contradiction.
answered Jan 24 at 15:24
kccukccu
10.6k11229
$endgroup$
add a comment |
$begingroup$
The Cantor set can be described as the set of numbers in $[0,1]$ that do not require the digit $1$ to be written in base $3$. You can prove this is equivalent to the definition you provided - in the first step you remove all numbers of the form $0.1dots$ (except $0.1$ itself which can also be written as $0.0222cdots$). In the second step you remove all numbers of the form $0.x1cdots$ where $x$ is $0$ or $2$ (except $0.01$ and $0.21$ which can be written as $0.00222cdots$ and $0.20222 cdots$, respectively). And so on.
If $[a,b] subseteq K$ with $a neq b$, then consider the base $3$ expansions of $a$ and $b$: $a=0.a_1a_2a_3dots$, $b=0.b_1b_2b_3dots$. These numbers are distinct, so consider the first digit at which they differ. Say $a_k neq b_k$ but $a_i=b_i$ for all $i <k$. The expansions of $a$ and $b$ don't contain any $1$'s, so all the digits are either $0$ or $2$.
Since $a<b$ and they agree in their first $k-1$ digits, we must have $a_k=0$ and $b_k=2$. Then the number $c=0.a_1a_2cdots a_{k-1}111 cdots$ is between $a$ and $b$, and it requires a $1$ in its ternary expansion, so $c notin K$, a contradiction.
answered Jan 24 at 15:24
kccukccu
10.6k11229
$endgroup$
The Cantor set can be described as the set of numbers in $[0,1]$ that do not require the digit $1$ to be written in base $3$. You can prove this is equivalent to the definition you provided - in the first step you remove all numbers of the form $0.1dots$ (except $0.1$ itself which can also be written as $0.0222cdots$). In the second step you remove all numbers of the form $0.x1cdots$ where $x$ is $0$ or $2$ (except $0.01$ and $0.21$ which can be written as $0.00222cdots$ and $0.20222 cdots$, respectively). And so on.
If $[a,b] subseteq K$ with $a neq b$, then consider the base $3$ expansions of $a$ and $b$: $a=0.a_1a_2a_3dots$, $b=0.b_1b_2b_3dots$. These numbers are distinct, so consider the first digit at which they differ. Say $a_k neq b_k$ but $a_i=b_i$ for all $i <k$. The expansions of $a$ and $b$ don't contain any $1$'s, so all the digits are either $0$ or $2$.
Since $a<b$ and they agree in their first $k-1$ digits, we must have $a_k=0$ and $b_k=2$. Then the number $c=0.a_1a_2cdots a_{k-1}111 cdots$ is between $a$ and $b$, and it requires a $1$ in its ternary expansion, so $c notin K$, a contradiction.
answered Jan 24 at 15:24
kccukccu
10.6k11229
answered Jan 24 at 15:24
kccukccu
10.6k11229
answered Jan 24 at 15:24
kccukccu
10.6k11229
answered Jan 24 at 15:24
kccukccu
10.6k11229
10.6k11229
add a comment |
add a comment |
$begingroup$
$C=bigcap_k C_k $, where each $C_k$ is the union of $2^k$ closed intervals in $[0,1]$ of length $3^{-k}.$ Therefore, as soon as $mathbb N ni N>-log_3(b-a), (a,b)nsubseteq C_N$ and so of course, $(a,b)$ cannot be contained in the intersection of the $C_k$.
answered Jan 24 at 16:06
MatematletaMatematleta
11.5k2920
$endgroup$
add a comment |
$begingroup$
$C=bigcap_k C_k $, where each $C_k$ is the union of $2^k$ closed intervals in $[0,1]$ of length $3^{-k}.$ Therefore, as soon as $mathbb N ni N>-log_3(b-a), (a,b)nsubseteq C_N$ and so of course, $(a,b)$ cannot be contained in the intersection of the $C_k$.
answered Jan 24 at 16:06
MatematletaMatematleta
11.5k2920
$endgroup$
add a comment |
$begingroup$
$C=bigcap_k C_k $, where each $C_k$ is the union of $2^k$ closed intervals in $[0,1]$ of length $3^{-k}.$ Therefore, as soon as $mathbb N ni N>-log_3(b-a), (a,b)nsubseteq C_N$ and so of course, $(a,b)$ cannot be contained in the intersection of the $C_k$.
answered Jan 24 at 16:06
MatematletaMatematleta
11.5k2920
$endgroup$
$C=bigcap_k C_k $, where each $C_k$ is the union of $2^k$ closed intervals in $[0,1]$ of length $3^{-k}.$ Therefore, as soon as $mathbb N ni N>-log_3(b-a), (a,b)nsubseteq C_N$ and so of course, $(a,b)$ cannot be contained in the intersection of the $C_k$.
answered Jan 24 at 16:06
MatematletaMatematleta
11.5k2920
edited Jan 24 at 16:25
answered Jan 24 at 16:06
MatematletaMatematleta
11.5k2920
answered Jan 24 at 16:06
MatematletaMatematleta
11.5k2920
answered Jan 24 at 16:06
MatematletaMatematleta
11.5k2920
11.5k2920
add a comment |
add a comment |
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$begingroup$
What is the method by which it is constructed?
$endgroup$
– Praphulla Koushik
Jan 24 at 15:10
$begingroup$
math.stackexchange.com/questions/3083396/…
$endgroup$
– Jack J.
Jan 24 at 15:14
$begingroup$
(there are many different methods to construct the Cantor set as evidenced by the different answers, so I believe Praphulla was asking which method /you/ considered to be the method by which it is constructed)
$endgroup$
– Calvin Khor
Jan 24 at 16:28
$begingroup$
In fact I replied to Praphulla, although I did not apologize for the lack of clarity. I take the opportunity to do it now.
$endgroup$
– Jack J.
Jan 24 at 16:31