Mixing
Meaning of “absolute zero” in set theory
$begingroup$
With great pleasure I am reading "Believing the axioms" by Penelope Maddy after someone linked to it here on MSE. (https://www.jstor.org/stable/2274520). However on page 495 there is a sentence I don't understand:
For example, a subset of the unit interval is called "absolute zero" if it can be covered by any countable collection of intervals.
(The 'for example' refers to 'strong smallness properties', of which being absolute zero is an example)
I don't understand this definition and due to the overwhelming number of articles on the more famous "absolute zero" from physics it is impossible to google. I think my problem is with the interpretation of the word 'any'.
It seems to me that for every point $x$ in the unit interval it is easy to construct a countable collection of intervals not covering $x$, hence showing that no non-empty set can be absolute zero. But clearly that is not what is meant. But then, what is?
The next sentence from the article does not make it any clearer:
If covering is only required when the intervals are of equal length, then the set would have Lebesgue measure zero, but would not necessarily be absolute zero.
I was hoping I could think of some criterion that sounds vaguely like 'can be covered by any countable collection of intervals of the same length' that would imply being measure zero but I failed. The next sentence from the article elaborates on the distinction between measure zero and absolute zero:
Thus Cantor's discontinuum has Lebesgue measure zero, but is not absolute zero, because it cannot be covered by countable manby intervals of length $(1/3)^n$.
'Ok', I thought, 'this gives some hint of what might be meant'. Perhaps they meant 'A set $Z$ is absolute zero if for each sequence $a_1, a_2 ,ldots$ of numbers in $(0, 1]$ there is a collection of intervals $I_1, I_2, ldots$ such that $I_n$ has length $a_n$ and together the intervals cover $Z$.' At least this sounds like a well defined property. EDITED IN: Due to Asaf Karagila' comment below I now know that this property actually has name and a Wikipedia page, increasing the likelihood that this indeed is what is meant. END OF EDIT
But then I thought what the analogue with 'intervals of equal length' would be and everything collapses: obviously the property 'for each number $a in (0, 1]$ there is a countable collection of intervals, each of length $a$, that together cover $X$' is true for every subset of $[0,1]$, not just those of measure zero.
So I am back to square one. Can anyone tell me what the definition of 'absolute zero' is and what property implying 'ordinary' measure zero is intended in the second sentence?
UPDATE: after reading the wikipedia page on 'strong measure zero' I actually believe that indeed (as Asaf wrote) strong measure zero and absolute zero are the same thing. The relations of both properties to the continuum hypothesis and conjectures by Borel discussed in both articles are just too similar for them not to be the same property.
But I am still mystified what property of a given set involving collections of intervals of the same length and implying the set to have measure zero is alluded to in the second of the quoted sentences. Does anyone know?
measure-theory set-theory
asked Jan 11 at 11:57
VincentVincent
3,16111229
$endgroup$
add a comment |
$begingroup$
With great pleasure I am reading "Believing the axioms" by Penelope Maddy after someone linked to it here on MSE. (https://www.jstor.org/stable/2274520). However on page 495 there is a sentence I don't understand:
For example, a subset of the unit interval is called "absolute zero" if it can be covered by any countable collection of intervals.
(The 'for example' refers to 'strong smallness properties', of which being absolute zero is an example)
I don't understand this definition and due to the overwhelming number of articles on the more famous "absolute zero" from physics it is impossible to google. I think my problem is with the interpretation of the word 'any'.
It seems to me that for every point $x$ in the unit interval it is easy to construct a countable collection of intervals not covering $x$, hence showing that no non-empty set can be absolute zero. But clearly that is not what is meant. But then, what is?
The next sentence from the article does not make it any clearer:
If covering is only required when the intervals are of equal length, then the set would have Lebesgue measure zero, but would not necessarily be absolute zero.
I was hoping I could think of some criterion that sounds vaguely like 'can be covered by any countable collection of intervals of the same length' that would imply being measure zero but I failed. The next sentence from the article elaborates on the distinction between measure zero and absolute zero:
Thus Cantor's discontinuum has Lebesgue measure zero, but is not absolute zero, because it cannot be covered by countable manby intervals of length $(1/3)^n$.
'Ok', I thought, 'this gives some hint of what might be meant'. Perhaps they meant 'A set $Z$ is absolute zero if for each sequence $a_1, a_2 ,ldots$ of numbers in $(0, 1]$ there is a collection of intervals $I_1, I_2, ldots$ such that $I_n$ has length $a_n$ and together the intervals cover $Z$.' At least this sounds like a well defined property. EDITED IN: Due to Asaf Karagila' comment below I now know that this property actually has name and a Wikipedia page, increasing the likelihood that this indeed is what is meant. END OF EDIT
But then I thought what the analogue with 'intervals of equal length' would be and everything collapses: obviously the property 'for each number $a in (0, 1]$ there is a countable collection of intervals, each of length $a$, that together cover $X$' is true for every subset of $[0,1]$, not just those of measure zero.
So I am back to square one. Can anyone tell me what the definition of 'absolute zero' is and what property implying 'ordinary' measure zero is intended in the second sentence?
UPDATE: after reading the wikipedia page on 'strong measure zero' I actually believe that indeed (as Asaf wrote) strong measure zero and absolute zero are the same thing. The relations of both properties to the continuum hypothesis and conjectures by Borel discussed in both articles are just too similar for them not to be the same property.
But I am still mystified what property of a given set involving collections of intervals of the same length and implying the set to have measure zero is alluded to in the second of the quoted sentences. Does anyone know?
measure-theory set-theory
asked Jan 11 at 11:57
VincentVincent
3,16111229
$endgroup$
3
$begingroup$
I think she means strong measure zero. But saying a set is covered by any countable collection of intervals means that the set is empty, since for any $xin[0,1]$ we can find a collection of intervals which do not include $x$ itself. So that can't be right...
$endgroup$
– Asaf Karagila♦
Jan 11 at 12:00
$begingroup$
Wasn' specidied which kind of intervals can be used ? Otherwise, the unit interval can trivially be covered by finite many intervals.
$endgroup$
– Peter
Jan 11 at 12:03
$begingroup$
I think it just means that take the definition of a strong measure zero set and instead of requiring it to hold for all sequences ${varepsilon_n }_{n lt omega}$ of positive real numbers, require it to hold for all constant sequences of positive real numbers.
$endgroup$
– Shervin Sorouri
Jan 11 at 12:29
$begingroup$
@ShervinSorouri but I wrote about that in the original post. For EVERY subset of the interval (not just measure zero) there is a sequence of intervals, each of length smaller than a given fixed $epsilon$ that covers the set, so that seems wrong.
$endgroup$
– Vincent
Jan 11 at 14:57
$begingroup$
It may also be that "absolute zero" is meant to be "universal measure zero", i.e. "absolute null", a related notion to "strongly measure zero". This does not help explain the characterization in the paper, though.
$endgroup$
– Carl Mummert
Jan 11 at 16:58
add a comment |
$begingroup$
With great pleasure I am reading "Believing the axioms" by Penelope Maddy after someone linked to it here on MSE. (https://www.jstor.org/stable/2274520). However on page 495 there is a sentence I don't understand:
For example, a subset of the unit interval is called "absolute zero" if it can be covered by any countable collection of intervals.
(The 'for example' refers to 'strong smallness properties', of which being absolute zero is an example)
I don't understand this definition and due to the overwhelming number of articles on the more famous "absolute zero" from physics it is impossible to google. I think my problem is with the interpretation of the word 'any'.
It seems to me that for every point $x$ in the unit interval it is easy to construct a countable collection of intervals not covering $x$, hence showing that no non-empty set can be absolute zero. But clearly that is not what is meant. But then, what is?
The next sentence from the article does not make it any clearer:
If covering is only required when the intervals are of equal length, then the set would have Lebesgue measure zero, but would not necessarily be absolute zero.
I was hoping I could think of some criterion that sounds vaguely like 'can be covered by any countable collection of intervals of the same length' that would imply being measure zero but I failed. The next sentence from the article elaborates on the distinction between measure zero and absolute zero:
Thus Cantor's discontinuum has Lebesgue measure zero, but is not absolute zero, because it cannot be covered by countable manby intervals of length $(1/3)^n$.
'Ok', I thought, 'this gives some hint of what might be meant'. Perhaps they meant 'A set $Z$ is absolute zero if for each sequence $a_1, a_2 ,ldots$ of numbers in $(0, 1]$ there is a collection of intervals $I_1, I_2, ldots$ such that $I_n$ has length $a_n$ and together the intervals cover $Z$.' At least this sounds like a well defined property. EDITED IN: Due to Asaf Karagila' comment below I now know that this property actually has name and a Wikipedia page, increasing the likelihood that this indeed is what is meant. END OF EDIT
But then I thought what the analogue with 'intervals of equal length' would be and everything collapses: obviously the property 'for each number $a in (0, 1]$ there is a countable collection of intervals, each of length $a$, that together cover $X$' is true for every subset of $[0,1]$, not just those of measure zero.
So I am back to square one. Can anyone tell me what the definition of 'absolute zero' is and what property implying 'ordinary' measure zero is intended in the second sentence?
UPDATE: after reading the wikipedia page on 'strong measure zero' I actually believe that indeed (as Asaf wrote) strong measure zero and absolute zero are the same thing. The relations of both properties to the continuum hypothesis and conjectures by Borel discussed in both articles are just too similar for them not to be the same property.
But I am still mystified what property of a given set involving collections of intervals of the same length and implying the set to have measure zero is alluded to in the second of the quoted sentences. Does anyone know?
measure-theory set-theory
asked Jan 11 at 11:57
VincentVincent
3,16111229
$endgroup$
With great pleasure I am reading "Believing the axioms" by Penelope Maddy after someone linked to it here on MSE. (https://www.jstor.org/stable/2274520). However on page 495 there is a sentence I don't understand:
For example, a subset of the unit interval is called "absolute zero" if it can be covered by any countable collection of intervals.
(The 'for example' refers to 'strong smallness properties', of which being absolute zero is an example)
I don't understand this definition and due to the overwhelming number of articles on the more famous "absolute zero" from physics it is impossible to google. I think my problem is with the interpretation of the word 'any'.
It seems to me that for every point $x$ in the unit interval it is easy to construct a countable collection of intervals not covering $x$, hence showing that no non-empty set can be absolute zero. But clearly that is not what is meant. But then, what is?
The next sentence from the article does not make it any clearer:
If covering is only required when the intervals are of equal length, then the set would have Lebesgue measure zero, but would not necessarily be absolute zero.
I was hoping I could think of some criterion that sounds vaguely like 'can be covered by any countable collection of intervals of the same length' that would imply being measure zero but I failed. The next sentence from the article elaborates on the distinction between measure zero and absolute zero:
Thus Cantor's discontinuum has Lebesgue measure zero, but is not absolute zero, because it cannot be covered by countable manby intervals of length $(1/3)^n$.
'Ok', I thought, 'this gives some hint of what might be meant'. Perhaps they meant 'A set $Z$ is absolute zero if for each sequence $a_1, a_2 ,ldots$ of numbers in $(0, 1]$ there is a collection of intervals $I_1, I_2, ldots$ such that $I_n$ has length $a_n$ and together the intervals cover $Z$.' At least this sounds like a well defined property. EDITED IN: Due to Asaf Karagila' comment below I now know that this property actually has name and a Wikipedia page, increasing the likelihood that this indeed is what is meant. END OF EDIT
But then I thought what the analogue with 'intervals of equal length' would be and everything collapses: obviously the property 'for each number $a in (0, 1]$ there is a countable collection of intervals, each of length $a$, that together cover $X$' is true for every subset of $[0,1]$, not just those of measure zero.
So I am back to square one. Can anyone tell me what the definition of 'absolute zero' is and what property implying 'ordinary' measure zero is intended in the second sentence?
UPDATE: after reading the wikipedia page on 'strong measure zero' I actually believe that indeed (as Asaf wrote) strong measure zero and absolute zero are the same thing. The relations of both properties to the continuum hypothesis and conjectures by Borel discussed in both articles are just too similar for them not to be the same property.
But I am still mystified what property of a given set involving collections of intervals of the same length and implying the set to have measure zero is alluded to in the second of the quoted sentences. Does anyone know?
measure-theory set-theory
measure-theory set-theory
asked Jan 11 at 11:57
VincentVincent
3,16111229
asked Jan 11 at 11:57
VincentVincent
3,16111229
edited Jan 11 at 12:12
Vincent
asked Jan 11 at 11:57
VincentVincent
3,16111229
asked Jan 11 at 11:57
VincentVincent
3,16111229
asked Jan 11 at 11:57
VincentVincent
3,16111229
3,16111229
3
$begingroup$
I think she means strong measure zero. But saying a set is covered by any countable collection of intervals means that the set is empty, since for any $xin[0,1]$ we can find a collection of intervals which do not include $x$ itself. So that can't be right...
$endgroup$
– Asaf Karagila♦
Jan 11 at 12:00
$begingroup$
Wasn' specidied which kind of intervals can be used ? Otherwise, the unit interval can trivially be covered by finite many intervals.
$endgroup$
– Peter
Jan 11 at 12:03
$begingroup$
I think it just means that take the definition of a strong measure zero set and instead of requiring it to hold for all sequences ${varepsilon_n }_{n lt omega}$ of positive real numbers, require it to hold for all constant sequences of positive real numbers.
$endgroup$
– Shervin Sorouri
Jan 11 at 12:29
$begingroup$
@ShervinSorouri but I wrote about that in the original post. For EVERY subset of the interval (not just measure zero) there is a sequence of intervals, each of length smaller than a given fixed $epsilon$ that covers the set, so that seems wrong.
$endgroup$
– Vincent
Jan 11 at 14:57
$begingroup$
It may also be that "absolute zero" is meant to be "universal measure zero", i.e. "absolute null", a related notion to "strongly measure zero". This does not help explain the characterization in the paper, though.
$endgroup$
– Carl Mummert
Jan 11 at 16:58
add a comment |
3
$begingroup$
I think she means strong measure zero. But saying a set is covered by any countable collection of intervals means that the set is empty, since for any $xin[0,1]$ we can find a collection of intervals which do not include $x$ itself. So that can't be right...
$endgroup$
– Asaf Karagila♦
Jan 11 at 12:00
$begingroup$
Wasn' specidied which kind of intervals can be used ? Otherwise, the unit interval can trivially be covered by finite many intervals.
$endgroup$
– Peter
Jan 11 at 12:03
$begingroup$
I think it just means that take the definition of a strong measure zero set and instead of requiring it to hold for all sequences ${varepsilon_n }_{n lt omega}$ of positive real numbers, require it to hold for all constant sequences of positive real numbers.
$endgroup$
– Shervin Sorouri
Jan 11 at 12:29
$begingroup$
@ShervinSorouri but I wrote about that in the original post. For EVERY subset of the interval (not just measure zero) there is a sequence of intervals, each of length smaller than a given fixed $epsilon$ that covers the set, so that seems wrong.
$endgroup$
– Vincent
Jan 11 at 14:57
$begingroup$
It may also be that "absolute zero" is meant to be "universal measure zero", i.e. "absolute null", a related notion to "strongly measure zero". This does not help explain the characterization in the paper, though.
$endgroup$
– Carl Mummert
Jan 11 at 16:58
3
3
$begingroup$
I think she means strong measure zero. But saying a set is covered by any countable collection of intervals means that the set is empty, since for any $xin[0,1]$ we can find a collection of intervals which do not include $x$ itself. So that can't be right...
$endgroup$
– Asaf Karagila♦
Jan 11 at 12:00
$begingroup$
I think she means strong measure zero. But saying a set is covered by any countable collection of intervals means that the set is empty, since for any $xin[0,1]$ we can find a collection of intervals which do not include $x$ itself. So that can't be right...
$endgroup$
– Asaf Karagila♦
Jan 11 at 12:00
$begingroup$
Wasn' specidied which kind of intervals can be used ? Otherwise, the unit interval can trivially be covered by finite many intervals.
$endgroup$
– Peter
Jan 11 at 12:03
$begingroup$
Wasn' specidied which kind of intervals can be used ? Otherwise, the unit interval can trivially be covered by finite many intervals.
$endgroup$
– Peter
Jan 11 at 12:03
$begingroup$
I think it just means that take the definition of a strong measure zero set and instead of requiring it to hold for all sequences ${varepsilon_n }_{n lt omega}$ of positive real numbers, require it to hold for all constant sequences of positive real numbers.
$endgroup$
– Shervin Sorouri
Jan 11 at 12:29
$begingroup$
I think it just means that take the definition of a strong measure zero set and instead of requiring it to hold for all sequences ${varepsilon_n }_{n lt omega}$ of positive real numbers, require it to hold for all constant sequences of positive real numbers.
$endgroup$
– Shervin Sorouri
Jan 11 at 12:29
$begingroup$
@ShervinSorouri but I wrote about that in the original post. For EVERY subset of the interval (not just measure zero) there is a sequence of intervals, each of length smaller than a given fixed $epsilon$ that covers the set, so that seems wrong.
$endgroup$
– Vincent
Jan 11 at 14:57
$begingroup$
@ShervinSorouri but I wrote about that in the original post. For EVERY subset of the interval (not just measure zero) there is a sequence of intervals, each of length smaller than a given fixed $epsilon$ that covers the set, so that seems wrong.
$endgroup$
– Vincent
Jan 11 at 14:57
$begingroup$
It may also be that "absolute zero" is meant to be "universal measure zero", i.e. "absolute null", a related notion to "strongly measure zero". This does not help explain the characterization in the paper, though.
$endgroup$
– Carl Mummert
Jan 11 at 16:58
$begingroup$
It may also be that "absolute zero" is meant to be "universal measure zero", i.e. "absolute null", a related notion to "strongly measure zero". This does not help explain the characterization in the paper, though.
$endgroup$
– Carl Mummert
Jan 11 at 16:58
add a comment |
1 Answer
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oldest
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$begingroup$
Unfortunately, Maddy is being imprecise in her use of terminology and the surrounding explanation. The mention of Borel in page 496 is a good hint that the notion she is discussing is that of being strong measure zero, as suggested in the comments.
A set of reals is (or has) measure zero if and only if for any $epsilon>0$ it can be covered by countably many open intervals whose lengths add up to less than $epsilon$. Cantor's set is a good example.
A set is strong measure zero if and only if, for any sequence of numbers $epsilon_n>0$, it can be covered by a sequence of open intervals $I_n$ with the length of $I_n$ being at most $epsilon_n$. The notion is due to Borel, in 1919.
Clearly any set satisfying this latter notion is measure 0, as for any $epsilon>0$ we can take $epsilon_n=epsilon/2^{n+1}$. What the new notion brings to the table is that we can control very strongly how the set is covered by intervals, essentially ensuring that it cannot have many accumulation points. Easily, any countable set is strong measure 0, but one can show that no perfect set is (Maddy also mentions this).
One can prove this as follows: First, given any perfect set $Psubseteq mathbb R$, there are continuous maps $f:mathbb Rtomathbb R$ that send $P$ onto the unit interval: when $P$ is the Cantor set, a famous example of such a function $f$ is the Cantor function; in general, any perfect set contains a set homeomorphic to the Cantor set, and from this the claim easily follows. Second, if $Xsubseteq mathbb R$ is strong measure 0, then its image under any continuous map $f:mathbb Rtomathbb R$ is again strong measure 0, this is an easy consequence of continuity. It follows that no perfect set is strong measure 0.
Borel conjectured that every strong measure 0 set is countable. This fails under CH and it is a famous result of Laver from 1976 that it is consistent (in fact, later work by Laver, Shelah and Woodin shows that it is consistent with $mathbb R$ being arbitrarily large).
[Note that, as pointed out in the comments, only the empty set can be covered by any countable collection of intervals. Also, we do not recover the notion of measure zero if instead we require that the set can be covered by a countable collection of intervals of equal length, even if we insist that this length be less than any given $epsilon>0$. Indeed, any set can be covered by such a collection. I suspect what Maddy meant to say was that the set can be covered by a countable collection of intervals of any given length, meaning that the whole collection has the given length, rather than the individual intervals.]
answered Jan 11 at 21:06
Andrés E. CaicedoAndrés E. Caicedo
65.4k8158248
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add a comment |
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$begingroup$
Unfortunately, Maddy is being imprecise in her use of terminology and the surrounding explanation. The mention of Borel in page 496 is a good hint that the notion she is discussing is that of being strong measure zero, as suggested in the comments.
A set of reals is (or has) measure zero if and only if for any $epsilon>0$ it can be covered by countably many open intervals whose lengths add up to less than $epsilon$. Cantor's set is a good example.
A set is strong measure zero if and only if, for any sequence of numbers $epsilon_n>0$, it can be covered by a sequence of open intervals $I_n$ with the length of $I_n$ being at most $epsilon_n$. The notion is due to Borel, in 1919.
Clearly any set satisfying this latter notion is measure 0, as for any $epsilon>0$ we can take $epsilon_n=epsilon/2^{n+1}$. What the new notion brings to the table is that we can control very strongly how the set is covered by intervals, essentially ensuring that it cannot have many accumulation points. Easily, any countable set is strong measure 0, but one can show that no perfect set is (Maddy also mentions this).
One can prove this as follows: First, given any perfect set $Psubseteq mathbb R$, there are continuous maps $f:mathbb Rtomathbb R$ that send $P$ onto the unit interval: when $P$ is the Cantor set, a famous example of such a function $f$ is the Cantor function; in general, any perfect set contains a set homeomorphic to the Cantor set, and from this the claim easily follows. Second, if $Xsubseteq mathbb R$ is strong measure 0, then its image under any continuous map $f:mathbb Rtomathbb R$ is again strong measure 0, this is an easy consequence of continuity. It follows that no perfect set is strong measure 0.
Borel conjectured that every strong measure 0 set is countable. This fails under CH and it is a famous result of Laver from 1976 that it is consistent (in fact, later work by Laver, Shelah and Woodin shows that it is consistent with $mathbb R$ being arbitrarily large).
[Note that, as pointed out in the comments, only the empty set can be covered by any countable collection of intervals. Also, we do not recover the notion of measure zero if instead we require that the set can be covered by a countable collection of intervals of equal length, even if we insist that this length be less than any given $epsilon>0$. Indeed, any set can be covered by such a collection. I suspect what Maddy meant to say was that the set can be covered by a countable collection of intervals of any given length, meaning that the whole collection has the given length, rather than the individual intervals.]
answered Jan 11 at 21:06
Andrés E. CaicedoAndrés E. Caicedo
65.4k8158248
$endgroup$
add a comment |
$begingroup$
Unfortunately, Maddy is being imprecise in her use of terminology and the surrounding explanation. The mention of Borel in page 496 is a good hint that the notion she is discussing is that of being strong measure zero, as suggested in the comments.
A set of reals is (or has) measure zero if and only if for any $epsilon>0$ it can be covered by countably many open intervals whose lengths add up to less than $epsilon$. Cantor's set is a good example.
A set is strong measure zero if and only if, for any sequence of numbers $epsilon_n>0$, it can be covered by a sequence of open intervals $I_n$ with the length of $I_n$ being at most $epsilon_n$. The notion is due to Borel, in 1919.
Clearly any set satisfying this latter notion is measure 0, as for any $epsilon>0$ we can take $epsilon_n=epsilon/2^{n+1}$. What the new notion brings to the table is that we can control very strongly how the set is covered by intervals, essentially ensuring that it cannot have many accumulation points. Easily, any countable set is strong measure 0, but one can show that no perfect set is (Maddy also mentions this).
One can prove this as follows: First, given any perfect set $Psubseteq mathbb R$, there are continuous maps $f:mathbb Rtomathbb R$ that send $P$ onto the unit interval: when $P$ is the Cantor set, a famous example of such a function $f$ is the Cantor function; in general, any perfect set contains a set homeomorphic to the Cantor set, and from this the claim easily follows. Second, if $Xsubseteq mathbb R$ is strong measure 0, then its image under any continuous map $f:mathbb Rtomathbb R$ is again strong measure 0, this is an easy consequence of continuity. It follows that no perfect set is strong measure 0.
Borel conjectured that every strong measure 0 set is countable. This fails under CH and it is a famous result of Laver from 1976 that it is consistent (in fact, later work by Laver, Shelah and Woodin shows that it is consistent with $mathbb R$ being arbitrarily large).
[Note that, as pointed out in the comments, only the empty set can be covered by any countable collection of intervals. Also, we do not recover the notion of measure zero if instead we require that the set can be covered by a countable collection of intervals of equal length, even if we insist that this length be less than any given $epsilon>0$. Indeed, any set can be covered by such a collection. I suspect what Maddy meant to say was that the set can be covered by a countable collection of intervals of any given length, meaning that the whole collection has the given length, rather than the individual intervals.]
answered Jan 11 at 21:06
Andrés E. CaicedoAndrés E. Caicedo
65.4k8158248
$endgroup$
add a comment |
$begingroup$
Unfortunately, Maddy is being imprecise in her use of terminology and the surrounding explanation. The mention of Borel in page 496 is a good hint that the notion she is discussing is that of being strong measure zero, as suggested in the comments.
A set of reals is (or has) measure zero if and only if for any $epsilon>0$ it can be covered by countably many open intervals whose lengths add up to less than $epsilon$. Cantor's set is a good example.
A set is strong measure zero if and only if, for any sequence of numbers $epsilon_n>0$, it can be covered by a sequence of open intervals $I_n$ with the length of $I_n$ being at most $epsilon_n$. The notion is due to Borel, in 1919.
Clearly any set satisfying this latter notion is measure 0, as for any $epsilon>0$ we can take $epsilon_n=epsilon/2^{n+1}$. What the new notion brings to the table is that we can control very strongly how the set is covered by intervals, essentially ensuring that it cannot have many accumulation points. Easily, any countable set is strong measure 0, but one can show that no perfect set is (Maddy also mentions this).
One can prove this as follows: First, given any perfect set $Psubseteq mathbb R$, there are continuous maps $f:mathbb Rtomathbb R$ that send $P$ onto the unit interval: when $P$ is the Cantor set, a famous example of such a function $f$ is the Cantor function; in general, any perfect set contains a set homeomorphic to the Cantor set, and from this the claim easily follows. Second, if $Xsubseteq mathbb R$ is strong measure 0, then its image under any continuous map $f:mathbb Rtomathbb R$ is again strong measure 0, this is an easy consequence of continuity. It follows that no perfect set is strong measure 0.
Borel conjectured that every strong measure 0 set is countable. This fails under CH and it is a famous result of Laver from 1976 that it is consistent (in fact, later work by Laver, Shelah and Woodin shows that it is consistent with $mathbb R$ being arbitrarily large).
[Note that, as pointed out in the comments, only the empty set can be covered by any countable collection of intervals. Also, we do not recover the notion of measure zero if instead we require that the set can be covered by a countable collection of intervals of equal length, even if we insist that this length be less than any given $epsilon>0$. Indeed, any set can be covered by such a collection. I suspect what Maddy meant to say was that the set can be covered by a countable collection of intervals of any given length, meaning that the whole collection has the given length, rather than the individual intervals.]
answered Jan 11 at 21:06
Andrés E. CaicedoAndrés E. Caicedo
65.4k8158248
$endgroup$
Unfortunately, Maddy is being imprecise in her use of terminology and the surrounding explanation. The mention of Borel in page 496 is a good hint that the notion she is discussing is that of being strong measure zero, as suggested in the comments.
A set of reals is (or has) measure zero if and only if for any $epsilon>0$ it can be covered by countably many open intervals whose lengths add up to less than $epsilon$. Cantor's set is a good example.
A set is strong measure zero if and only if, for any sequence of numbers $epsilon_n>0$, it can be covered by a sequence of open intervals $I_n$ with the length of $I_n$ being at most $epsilon_n$. The notion is due to Borel, in 1919.
Clearly any set satisfying this latter notion is measure 0, as for any $epsilon>0$ we can take $epsilon_n=epsilon/2^{n+1}$. What the new notion brings to the table is that we can control very strongly how the set is covered by intervals, essentially ensuring that it cannot have many accumulation points. Easily, any countable set is strong measure 0, but one can show that no perfect set is (Maddy also mentions this).
One can prove this as follows: First, given any perfect set $Psubseteq mathbb R$, there are continuous maps $f:mathbb Rtomathbb R$ that send $P$ onto the unit interval: when $P$ is the Cantor set, a famous example of such a function $f$ is the Cantor function; in general, any perfect set contains a set homeomorphic to the Cantor set, and from this the claim easily follows. Second, if $Xsubseteq mathbb R$ is strong measure 0, then its image under any continuous map $f:mathbb Rtomathbb R$ is again strong measure 0, this is an easy consequence of continuity. It follows that no perfect set is strong measure 0.
Borel conjectured that every strong measure 0 set is countable. This fails under CH and it is a famous result of Laver from 1976 that it is consistent (in fact, later work by Laver, Shelah and Woodin shows that it is consistent with $mathbb R$ being arbitrarily large).
[Note that, as pointed out in the comments, only the empty set can be covered by any countable collection of intervals. Also, we do not recover the notion of measure zero if instead we require that the set can be covered by a countable collection of intervals of equal length, even if we insist that this length be less than any given $epsilon>0$. Indeed, any set can be covered by such a collection. I suspect what Maddy meant to say was that the set can be covered by a countable collection of intervals of any given length, meaning that the whole collection has the given length, rather than the individual intervals.]
answered Jan 11 at 21:06
Andrés E. CaicedoAndrés E. Caicedo
65.4k8158248
answered Jan 11 at 21:06
Andrés E. CaicedoAndrés E. Caicedo
65.4k8158248
answered Jan 11 at 21:06
Andrés E. CaicedoAndrés E. Caicedo
65.4k8158248
answered Jan 11 at 21:06
Andrés E. CaicedoAndrés E. Caicedo
65.4k8158248
65.4k8158248
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I think she means strong measure zero. But saying a set is covered by any countable collection of intervals means that the set is empty, since for any $xin[0,1]$ we can find a collection of intervals which do not include $x$ itself. So that can't be right...
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– Asaf Karagila♦
Jan 11 at 12:00
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Wasn' specidied which kind of intervals can be used ? Otherwise, the unit interval can trivially be covered by finite many intervals.
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– Peter
Jan 11 at 12:03
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I think it just means that take the definition of a strong measure zero set and instead of requiring it to hold for all sequences ${varepsilon_n }_{n lt omega}$ of positive real numbers, require it to hold for all constant sequences of positive real numbers.
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– Shervin Sorouri
Jan 11 at 12:29
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@ShervinSorouri but I wrote about that in the original post. For EVERY subset of the interval (not just measure zero) there is a sequence of intervals, each of length smaller than a given fixed $epsilon$ that covers the set, so that seems wrong.
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– Vincent
Jan 11 at 14:57
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It may also be that "absolute zero" is meant to be "universal measure zero", i.e. "absolute null", a related notion to "strongly measure zero". This does not help explain the characterization in the paper, though.
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– Carl Mummert
Jan 11 at 16:58