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Problem in finding the Euclidean measure of a set in $mathbbR^3$
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Problem. For each $alpha in mathbbR$, let $S_alpha=~ x^2+y^2+z^2=alpha^2$.
Let $E=bigcup_alphain mathbbRsetminus mathbbQS_alpha$. Which of the followings are true?
- The Lebesgue measure of $E$ is infinite?
- E contains an nonempty open set.
- E is path-connected .
- Every open set containing $E^c$ has infinite Lebesgue measure.
My Solution.
True. Since $E^c=bigcup_alphain mathbbQS_alpha$ and $mu(S_alpha)=0$ So by countable additivity of $mu$, $E^c$ has measure zero.
False. Since $E$ and $E^c$ both are Dense in $mathbbR^3$.
False. Since any two sphere of irrational radius always there exists an intermediate sphere of rational radius between them.
True. I think there is only one open set containing $E^c$ namely $mathbbR^3$. And $mu(mathbbR^3)=infty.$
But the answer key indicates the options: 1 is only True.
Then what is wrong with my conclusion about option 4. Please let me know where I made mistake. Thank You..
analysis measure-theory
asked Aug 6 at 2:14
Indrajit Ghosh
602415
add a comment |Â
up vote
1
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favorite
Problem. For each $alpha in mathbbR$, let $S_alpha=~ x^2+y^2+z^2=alpha^2$.
Let $E=bigcup_alphain mathbbRsetminus mathbbQS_alpha$. Which of the followings are true?
- The Lebesgue measure of $E$ is infinite?
- E contains an nonempty open set.
- E is path-connected .
- Every open set containing $E^c$ has infinite Lebesgue measure.
My Solution.
True. Since $E^c=bigcup_alphain mathbbQS_alpha$ and $mu(S_alpha)=0$ So by countable additivity of $mu$, $E^c$ has measure zero.
False. Since $E$ and $E^c$ both are Dense in $mathbbR^3$.
False. Since any two sphere of irrational radius always there exists an intermediate sphere of rational radius between them.
True. I think there is only one open set containing $E^c$ namely $mathbbR^3$. And $mu(mathbbR^3)=infty.$
But the answer key indicates the options: 1 is only True.
Then what is wrong with my conclusion about option 4. Please let me know where I made mistake. Thank You..
analysis measure-theory
asked Aug 6 at 2:14
Indrajit Ghosh
602415
2
The open set only needs to contain a countable collection of the $S_alpha$. Each 'shell' can be contained in an open set of arbitrarily small measure. Sum them up. Basically the same way you would show that the rationals have measure zero in the reals.
– copper.hat
Aug 6 at 2:23
Oho ... u r right....!!!!
– Indrajit Ghosh
Aug 6 at 2:28
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Problem. For each $alpha in mathbbR$, let $S_alpha=~ x^2+y^2+z^2=alpha^2$.
Let $E=bigcup_alphain mathbbRsetminus mathbbQS_alpha$. Which of the followings are true?
- The Lebesgue measure of $E$ is infinite?
- E contains an nonempty open set.
- E is path-connected .
- Every open set containing $E^c$ has infinite Lebesgue measure.
My Solution.
True. Since $E^c=bigcup_alphain mathbbQS_alpha$ and $mu(S_alpha)=0$ So by countable additivity of $mu$, $E^c$ has measure zero.
False. Since $E$ and $E^c$ both are Dense in $mathbbR^3$.
False. Since any two sphere of irrational radius always there exists an intermediate sphere of rational radius between them.
True. I think there is only one open set containing $E^c$ namely $mathbbR^3$. And $mu(mathbbR^3)=infty.$
But the answer key indicates the options: 1 is only True.
Then what is wrong with my conclusion about option 4. Please let me know where I made mistake. Thank You..
analysis measure-theory
asked Aug 6 at 2:14
Indrajit Ghosh
602415
Problem. For each $alpha in mathbbR$, let $S_alpha=~ x^2+y^2+z^2=alpha^2$.
Let $E=bigcup_alphain mathbbRsetminus mathbbQS_alpha$. Which of the followings are true?
- The Lebesgue measure of $E$ is infinite?
- E contains an nonempty open set.
- E is path-connected .
- Every open set containing $E^c$ has infinite Lebesgue measure.
My Solution.
True. Since $E^c=bigcup_alphain mathbbQS_alpha$ and $mu(S_alpha)=0$ So by countable additivity of $mu$, $E^c$ has measure zero.
False. Since $E$ and $E^c$ both are Dense in $mathbbR^3$.
False. Since any two sphere of irrational radius always there exists an intermediate sphere of rational radius between them.
True. I think there is only one open set containing $E^c$ namely $mathbbR^3$. And $mu(mathbbR^3)=infty.$
But the answer key indicates the options: 1 is only True.
Then what is wrong with my conclusion about option 4. Please let me know where I made mistake. Thank You..
analysis measure-theory
asked Aug 6 at 2:14
Indrajit Ghosh
602415
asked Aug 6 at 2:14
Indrajit Ghosh
602415
asked Aug 6 at 2:14
Indrajit Ghosh
602415
asked Aug 6 at 2:14
Indrajit Ghosh
602415
602415
2
The open set only needs to contain a countable collection of the $S_alpha$. Each 'shell' can be contained in an open set of arbitrarily small measure. Sum them up. Basically the same way you would show that the rationals have measure zero in the reals.
– copper.hat
Aug 6 at 2:23
Oho ... u r right....!!!!
– Indrajit Ghosh
Aug 6 at 2:28
add a comment |Â
2
The open set only needs to contain a countable collection of the $S_alpha$. Each 'shell' can be contained in an open set of arbitrarily small measure. Sum them up. Basically the same way you would show that the rationals have measure zero in the reals.
– copper.hat
Aug 6 at 2:23
Oho ... u r right....!!!!
– Indrajit Ghosh
Aug 6 at 2:28
2
2
The open set only needs to contain a countable collection of the $S_alpha$. Each 'shell' can be contained in an open set of arbitrarily small measure. Sum them up. Basically the same way you would show that the rationals have measure zero in the reals.
– copper.hat
Aug 6 at 2:23
The open set only needs to contain a countable collection of the $S_alpha$. Each 'shell' can be contained in an open set of arbitrarily small measure. Sum them up. Basically the same way you would show that the rationals have measure zero in the reals.
– copper.hat
Aug 6 at 2:23
Oho ... u r right....!!!!
– Indrajit Ghosh
Aug 6 at 2:28
Oho ... u r right....!!!!
– Indrajit Ghosh
Aug 6 at 2:28
add a comment |Â
1 Answer
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As you pointed out, we know $E^c = bigcup_alpha in mathbbQ S_alpha$. Let $(a_n)_n$ be an enumeration of the rationals. Then for each $n in mathbbN$, define:
$$
r_n = min left frac12^n, frac1a_n^4 2^n right
$$
where $r_n = 1/2^n$ if $a_n = 0$, and
$$
R_n = (x,y,z) in mathbbR^3 ,
$$
Remark that each $R_n$ is open, and that the size of $R_n$ is given by the difference of two spheres, i.e.:
beginalign*
mu(R_n) &= mu(B(0,a_n^2+r_n)) - mu(B(a_n^2-r_n))
\ &= frac43 pi (a_n^2 + r_n)^3 - frac43 pi (a_n^2 - r_n)^3
\ &= frac43 pi left( 6 a_n^4 r_n + 2 r_n^3 right)
\ &leq frac43 pi left( frac62^n + frac22^3n right)
\ &leq frac48 pi3 cdot frac12^n
endalign*
Since each $R_n$ is open, their union is open. It is also clear that the union of the $R_n$ contains $E^c$, and finally we have:
$$
mu left( bigcup_n in mathbbN R_n right)
leq
sum limits_n=1^infty frac48 pi3 cdot frac12^n
= frac48 pi3
$$
This gives an example of an open set that contains $E^c$ which has finite measure.
answered Aug 6 at 16:24
Sambo
1,2651427
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
As you pointed out, we know $E^c = bigcup_alpha in mathbbQ S_alpha$. Let $(a_n)_n$ be an enumeration of the rationals. Then for each $n in mathbbN$, define:
$$
r_n = min left frac12^n, frac1a_n^4 2^n right
$$
where $r_n = 1/2^n$ if $a_n = 0$, and
$$
R_n = (x,y,z) in mathbbR^3 ,
$$
Remark that each $R_n$ is open, and that the size of $R_n$ is given by the difference of two spheres, i.e.:
beginalign*
mu(R_n) &= mu(B(0,a_n^2+r_n)) - mu(B(a_n^2-r_n))
\ &= frac43 pi (a_n^2 + r_n)^3 - frac43 pi (a_n^2 - r_n)^3
\ &= frac43 pi left( 6 a_n^4 r_n + 2 r_n^3 right)
\ &leq frac43 pi left( frac62^n + frac22^3n right)
\ &leq frac48 pi3 cdot frac12^n
endalign*
Since each $R_n$ is open, their union is open. It is also clear that the union of the $R_n$ contains $E^c$, and finally we have:
$$
mu left( bigcup_n in mathbbN R_n right)
leq
sum limits_n=1^infty frac48 pi3 cdot frac12^n
= frac48 pi3
$$
This gives an example of an open set that contains $E^c$ which has finite measure.
answered Aug 6 at 16:24
Sambo
1,2651427
add a comment |Â
up vote
1
down vote
As you pointed out, we know $E^c = bigcup_alpha in mathbbQ S_alpha$. Let $(a_n)_n$ be an enumeration of the rationals. Then for each $n in mathbbN$, define:
$$
r_n = min left frac12^n, frac1a_n^4 2^n right
$$
where $r_n = 1/2^n$ if $a_n = 0$, and
$$
R_n = (x,y,z) in mathbbR^3 ,
$$
Remark that each $R_n$ is open, and that the size of $R_n$ is given by the difference of two spheres, i.e.:
beginalign*
mu(R_n) &= mu(B(0,a_n^2+r_n)) - mu(B(a_n^2-r_n))
\ &= frac43 pi (a_n^2 + r_n)^3 - frac43 pi (a_n^2 - r_n)^3
\ &= frac43 pi left( 6 a_n^4 r_n + 2 r_n^3 right)
\ &leq frac43 pi left( frac62^n + frac22^3n right)
\ &leq frac48 pi3 cdot frac12^n
endalign*
Since each $R_n$ is open, their union is open. It is also clear that the union of the $R_n$ contains $E^c$, and finally we have:
$$
mu left( bigcup_n in mathbbN R_n right)
leq
sum limits_n=1^infty frac48 pi3 cdot frac12^n
= frac48 pi3
$$
This gives an example of an open set that contains $E^c$ which has finite measure.
answered Aug 6 at 16:24
Sambo
1,2651427
add a comment |Â
up vote
1
down vote
up vote
1
down vote
As you pointed out, we know $E^c = bigcup_alpha in mathbbQ S_alpha$. Let $(a_n)_n$ be an enumeration of the rationals. Then for each $n in mathbbN$, define:
$$
r_n = min left frac12^n, frac1a_n^4 2^n right
$$
where $r_n = 1/2^n$ if $a_n = 0$, and
$$
R_n = (x,y,z) in mathbbR^3 ,
$$
Remark that each $R_n$ is open, and that the size of $R_n$ is given by the difference of two spheres, i.e.:
beginalign*
mu(R_n) &= mu(B(0,a_n^2+r_n)) - mu(B(a_n^2-r_n))
\ &= frac43 pi (a_n^2 + r_n)^3 - frac43 pi (a_n^2 - r_n)^3
\ &= frac43 pi left( 6 a_n^4 r_n + 2 r_n^3 right)
\ &leq frac43 pi left( frac62^n + frac22^3n right)
\ &leq frac48 pi3 cdot frac12^n
endalign*
Since each $R_n$ is open, their union is open. It is also clear that the union of the $R_n$ contains $E^c$, and finally we have:
$$
mu left( bigcup_n in mathbbN R_n right)
leq
sum limits_n=1^infty frac48 pi3 cdot frac12^n
= frac48 pi3
$$
This gives an example of an open set that contains $E^c$ which has finite measure.
answered Aug 6 at 16:24
Sambo
1,2651427
As you pointed out, we know $E^c = bigcup_alpha in mathbbQ S_alpha$. Let $(a_n)_n$ be an enumeration of the rationals. Then for each $n in mathbbN$, define:
$$
r_n = min left frac12^n, frac1a_n^4 2^n right
$$
where $r_n = 1/2^n$ if $a_n = 0$, and
$$
R_n = (x,y,z) in mathbbR^3 ,
$$
Remark that each $R_n$ is open, and that the size of $R_n$ is given by the difference of two spheres, i.e.:
beginalign*
mu(R_n) &= mu(B(0,a_n^2+r_n)) - mu(B(a_n^2-r_n))
\ &= frac43 pi (a_n^2 + r_n)^3 - frac43 pi (a_n^2 - r_n)^3
\ &= frac43 pi left( 6 a_n^4 r_n + 2 r_n^3 right)
\ &leq frac43 pi left( frac62^n + frac22^3n right)
\ &leq frac48 pi3 cdot frac12^n
endalign*
Since each $R_n$ is open, their union is open. It is also clear that the union of the $R_n$ contains $E^c$, and finally we have:
$$
mu left( bigcup_n in mathbbN R_n right)
leq
sum limits_n=1^infty frac48 pi3 cdot frac12^n
= frac48 pi3
$$
This gives an example of an open set that contains $E^c$ which has finite measure.
answered Aug 6 at 16:24
Sambo
1,2651427
answered Aug 6 at 16:24
Sambo
1,2651427
answered Aug 6 at 16:24
Sambo
1,2651427
answered Aug 6 at 16:24
Sambo
1,2651427
1,2651427
add a comment |Â
add a comment |Â
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2
The open set only needs to contain a countable collection of the $S_alpha$. Each 'shell' can be contained in an open set of arbitrarily small measure. Sum them up. Basically the same way you would show that the rationals have measure zero in the reals.
– copper.hat
Aug 6 at 2:23
Oho ... u r right....!!!!
– Indrajit Ghosh
Aug 6 at 2:28