“Fat” Cantor Set
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So the standard Cantor set has an outer measure equal to 0, but how can you construct a "fat" Cantor set with a positive outer measure? I was told that it is even possible to produce one with an outer measure of 1! I don't see how changing the size of the "chuck" taken out will change the value of the outer measure. Regardless of the size, I feel like it will inevitably reach a value of 0 as well...
Are there other constraints that need to be made in order to accomplish this?
real-analysis cantor-set
edited Jul 5 '15 at 8:56
Martin Sleziak
44.5k7115268
asked Jan 27 '13 at 4:03
Frank
4112
|
show 1 more comment
up vote
8
down vote
favorite
So the standard Cantor set has an outer measure equal to 0, but how can you construct a "fat" Cantor set with a positive outer measure? I was told that it is even possible to produce one with an outer measure of 1! I don't see how changing the size of the "chuck" taken out will change the value of the outer measure. Regardless of the size, I feel like it will inevitably reach a value of 0 as well...
Are there other constraints that need to be made in order to accomplish this?
real-analysis cantor-set
edited Jul 5 '15 at 8:56
Martin Sleziak
44.5k7115268
asked Jan 27 '13 at 4:03
Frank
4112
10
The very first Google hit for "fat Cantor set" has the answer. (If I could, I would vote to close as general-reference.)
– Rahul
Jan 27 '13 at 4:06
If the bits you remove at each stage have total length less than $1$, then what's left has positive measure.
– Gerry Myerson
Jan 27 '13 at 4:06
You can't have outer measure 1. Otherwise the set would be dense in [0,1], contradicting its compactness.
– user53153
Jan 27 '13 at 4:09
You can, however, have outer measure arbitrarily close to one.
– Brian M. Scott
Jan 27 '13 at 4:10
1
Here are some details of what's contained on the Wikipedia page.
– Martin
Jan 27 '13 at 4:13
|
show 1 more comment
up vote
8
down vote
favorite
up vote
8
down vote
favorite
So the standard Cantor set has an outer measure equal to 0, but how can you construct a "fat" Cantor set with a positive outer measure? I was told that it is even possible to produce one with an outer measure of 1! I don't see how changing the size of the "chuck" taken out will change the value of the outer measure. Regardless of the size, I feel like it will inevitably reach a value of 0 as well...
Are there other constraints that need to be made in order to accomplish this?
real-analysis cantor-set
edited Jul 5 '15 at 8:56
Martin Sleziak
44.5k7115268
asked Jan 27 '13 at 4:03
Frank
4112
So the standard Cantor set has an outer measure equal to 0, but how can you construct a "fat" Cantor set with a positive outer measure? I was told that it is even possible to produce one with an outer measure of 1! I don't see how changing the size of the "chuck" taken out will change the value of the outer measure. Regardless of the size, I feel like it will inevitably reach a value of 0 as well...
Are there other constraints that need to be made in order to accomplish this?
real-analysis cantor-set
real-analysis cantor-set
edited Jul 5 '15 at 8:56
Martin Sleziak
44.5k7115268
asked Jan 27 '13 at 4:03
Frank
4112
edited Jul 5 '15 at 8:56
Martin Sleziak
44.5k7115268
asked Jan 27 '13 at 4:03
Frank
4112
edited Jul 5 '15 at 8:56
Martin Sleziak
44.5k7115268
edited Jul 5 '15 at 8:56
Martin Sleziak
44.5k7115268
edited Jul 5 '15 at 8:56
Martin Sleziak
44.5k7115268
44.5k7115268
asked Jan 27 '13 at 4:03
Frank
4112
asked Jan 27 '13 at 4:03
Frank
4112
asked Jan 27 '13 at 4:03
Frank
4112
4112
10
The very first Google hit for "fat Cantor set" has the answer. (If I could, I would vote to close as general-reference.)
– Rahul
Jan 27 '13 at 4:06
If the bits you remove at each stage have total length less than $1$, then what's left has positive measure.
– Gerry Myerson
Jan 27 '13 at 4:06
You can't have outer measure 1. Otherwise the set would be dense in [0,1], contradicting its compactness.
– user53153
Jan 27 '13 at 4:09
You can, however, have outer measure arbitrarily close to one.
– Brian M. Scott
Jan 27 '13 at 4:10
1
Here are some details of what's contained on the Wikipedia page.
– Martin
Jan 27 '13 at 4:13
|
show 1 more comment
10
The very first Google hit for "fat Cantor set" has the answer. (If I could, I would vote to close as general-reference.)
– Rahul
Jan 27 '13 at 4:06
If the bits you remove at each stage have total length less than $1$, then what's left has positive measure.
– Gerry Myerson
Jan 27 '13 at 4:06
You can't have outer measure 1. Otherwise the set would be dense in [0,1], contradicting its compactness.
– user53153
Jan 27 '13 at 4:09
You can, however, have outer measure arbitrarily close to one.
– Brian M. Scott
Jan 27 '13 at 4:10
1
Here are some details of what's contained on the Wikipedia page.
– Martin
Jan 27 '13 at 4:13
10
10
The very first Google hit for "fat Cantor set" has the answer. (If I could, I would vote to close as general-reference.)
– Rahul
Jan 27 '13 at 4:06
The very first Google hit for "fat Cantor set" has the answer. (If I could, I would vote to close as general-reference.)
– Rahul
Jan 27 '13 at 4:06
If the bits you remove at each stage have total length less than $1$, then what's left has positive measure.
– Gerry Myerson
Jan 27 '13 at 4:06
If the bits you remove at each stage have total length less than $1$, then what's left has positive measure.
– Gerry Myerson
Jan 27 '13 at 4:06
You can't have outer measure 1. Otherwise the set would be dense in [0,1], contradicting its compactness.
– user53153
Jan 27 '13 at 4:09
You can't have outer measure 1. Otherwise the set would be dense in [0,1], contradicting its compactness.
– user53153
Jan 27 '13 at 4:09
You can, however, have outer measure arbitrarily close to one.
– Brian M. Scott
Jan 27 '13 at 4:10
You can, however, have outer measure arbitrarily close to one.
– Brian M. Scott
Jan 27 '13 at 4:10
1
1
Here are some details of what's contained on the Wikipedia page.
– Martin
Jan 27 '13 at 4:13
Here are some details of what's contained on the Wikipedia page.
– Martin
Jan 27 '13 at 4:13
|
show 1 more comment
1 Answer
1
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up vote
11
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Say you delete the middle third.
Then delete two intervals the sum of whose lengths is $1/6$ from the two remaining intervals.
Then delete intervals the sum of whose lengths is $1/12$, one from each of the four remaining intervals.
And so on. The amount you delete is $displaystylefrac13+frac16+frac{1}{12}+cdots= frac23.$ That is less than the whole measure of the interval $[0,1]$ from which you're deleting things.
answered Jan 27 '13 at 4:23
Michael Hardy
1
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
11
down vote
Say you delete the middle third.
Then delete two intervals the sum of whose lengths is $1/6$ from the two remaining intervals.
Then delete intervals the sum of whose lengths is $1/12$, one from each of the four remaining intervals.
And so on. The amount you delete is $displaystylefrac13+frac16+frac{1}{12}+cdots= frac23.$ That is less than the whole measure of the interval $[0,1]$ from which you're deleting things.
answered Jan 27 '13 at 4:23
Michael Hardy
1
add a comment |
up vote
11
down vote
Say you delete the middle third.
Then delete two intervals the sum of whose lengths is $1/6$ from the two remaining intervals.
Then delete intervals the sum of whose lengths is $1/12$, one from each of the four remaining intervals.
And so on. The amount you delete is $displaystylefrac13+frac16+frac{1}{12}+cdots= frac23.$ That is less than the whole measure of the interval $[0,1]$ from which you're deleting things.
answered Jan 27 '13 at 4:23
Michael Hardy
1
add a comment |
up vote
11
down vote
up vote
11
down vote
Say you delete the middle third.
Then delete two intervals the sum of whose lengths is $1/6$ from the two remaining intervals.
Then delete intervals the sum of whose lengths is $1/12$, one from each of the four remaining intervals.
And so on. The amount you delete is $displaystylefrac13+frac16+frac{1}{12}+cdots= frac23.$ That is less than the whole measure of the interval $[0,1]$ from which you're deleting things.
answered Jan 27 '13 at 4:23
Michael Hardy
1
Say you delete the middle third.
Then delete two intervals the sum of whose lengths is $1/6$ from the two remaining intervals.
Then delete intervals the sum of whose lengths is $1/12$, one from each of the four remaining intervals.
And so on. The amount you delete is $displaystylefrac13+frac16+frac{1}{12}+cdots= frac23.$ That is less than the whole measure of the interval $[0,1]$ from which you're deleting things.
answered Jan 27 '13 at 4:23
Michael Hardy
1
edited Jan 9 '14 at 3:56
answered Jan 27 '13 at 4:23
Michael Hardy
1
answered Jan 27 '13 at 4:23
Michael Hardy
1
answered Jan 27 '13 at 4:23
Michael Hardy
1
1
add a comment |
add a comment |
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10
The very first Google hit for "fat Cantor set" has the answer. (If I could, I would vote to close as general-reference.)
– Rahul
Jan 27 '13 at 4:06
If the bits you remove at each stage have total length less than $1$, then what's left has positive measure.
– Gerry Myerson
Jan 27 '13 at 4:06
You can't have outer measure 1. Otherwise the set would be dense in [0,1], contradicting its compactness.
– user53153
Jan 27 '13 at 4:09
You can, however, have outer measure arbitrarily close to one.
– Brian M. Scott
Jan 27 '13 at 4:10
1
Here are some details of what's contained on the Wikipedia page.
– Martin
Jan 27 '13 at 4:13