Why is the Cantor set not defined by a limit? The 2019 Stack Overflow Developer Survey Results Are In Unicorn Meta Zoo #1: Why another podcast? Announcing the arrival of Valued Associate #679: Cesar Manaralim sup and lim inf of sequence of sets.measure preserving transformations on the ternary Cantor setAbout Cantor set: Cantor set + Cantor setHow to construct binary sequences associated to points of the Cantor set?Function on Cantor setWondering if something is an algebra. If it is, question about closure under complements.Are countably infinite unions limits?Transition from countable to uncountableFormal representation of the numbers of the Cantor set.Alternate definition of middle-$alpha$ Cantor set using affine transformations
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Why is the Cantor set not defined by a limit?
The 2019 Stack Overflow Developer Survey Results Are In
Unicorn Meta Zoo #1: Why another podcast?
Announcing the arrival of Valued Associate #679: Cesar Manaralim sup and lim inf of sequence of sets.measure preserving transformations on the ternary Cantor setAbout Cantor set: Cantor set + Cantor setHow to construct binary sequences associated to points of the Cantor set?Function on Cantor setWondering if something is an algebra. If it is, question about closure under complements.Are countably infinite unions limits?Transition from countable to uncountableFormal representation of the numbers of the Cantor set.Alternate definition of middle-$alpha$ Cantor set using affine transformations
$begingroup$
The Cantor set, as showed in books and Wikipedia, is defined in terms of $C_k$, the finite Cantor set of level $k$:
$$ mathcalC = bigcap_k=1^infty C_k $$
But after intersection only the "last" remains, so, why not to define it by a limit?
$$mathcalC = lim_k to infty C_k$$
It is perhaps a naive intuition, but I not see a good justification.
(adding here a note after first answer, only for comment the comments)
NOTE: if it is not only a question of choice of notation, but also about context and semantics.
Can I tell an engineer that the intersection is a kind of specification, something like a project to explain "what I need", and the limit is a "what I get", the end result? ...Or perhaps the inverse, as suggested by @HansLundmark (thanks the comment! also thnaks @SangchulLee!). I am supposing that $C_k$ is a "decreasing sequence", $C_1 supseteq C_2 supseteq C_3 supseteq dotsb$,
so using your anser we can say
"it's natural to define the limit as the intersection: $C_n to mathcalC$ as $ntoinfty$",
where $mathcalC$ is defined by intersection.
About comment of @LordShark: its is possible to use "limits" notation in the context of set sequences without "develop a theory" for it? @HansLundmark's link is a satisfactory answer for it?
cantor-set
edited Mar 25 at 12:24
Matt Samuel
39.2k63870
asked Mar 24 at 18:48
Peter KraussPeter Krauss
1065
$endgroup$
add a comment |
$begingroup$
The Cantor set, as showed in books and Wikipedia, is defined in terms of $C_k$, the finite Cantor set of level $k$:
$$ mathcalC = bigcap_k=1^infty C_k $$
But after intersection only the "last" remains, so, why not to define it by a limit?
$$mathcalC = lim_k to infty C_k$$
It is perhaps a naive intuition, but I not see a good justification.
(adding here a note after first answer, only for comment the comments)
NOTE: if it is not only a question of choice of notation, but also about context and semantics.
Can I tell an engineer that the intersection is a kind of specification, something like a project to explain "what I need", and the limit is a "what I get", the end result? ...Or perhaps the inverse, as suggested by @HansLundmark (thanks the comment! also thnaks @SangchulLee!). I am supposing that $C_k$ is a "decreasing sequence", $C_1 supseteq C_2 supseteq C_3 supseteq dotsb$,
so using your anser we can say
"it's natural to define the limit as the intersection: $C_n to mathcalC$ as $ntoinfty$",
where $mathcalC$ is defined by intersection.
About comment of @LordShark: its is possible to use "limits" notation in the context of set sequences without "develop a theory" for it? @HansLundmark's link is a satisfactory answer for it?
cantor-set
edited Mar 25 at 12:24
Matt Samuel
39.2k63870
asked Mar 24 at 18:48
Peter KraussPeter Krauss
1065
$endgroup$
2
$begingroup$
Why develop a theory of "limits" of sets to define $cal C$ when it's simply an intersection?
$endgroup$
– Lord Shark the Unknown
Mar 24 at 18:53
1
$begingroup$
The intersection is the limit in this case (when you have a decreasing sequence of sets). See here: math.stackexchange.com/questions/107931/…
$endgroup$
– Hans Lundmark
Mar 24 at 18:54
1
$begingroup$
Of course we can perfectly make sense of $mathcalC = lim_ktoinfty C_k$. On the other hand, we need some works to put that notion to a rigorous mathematical framework. And that is sort of asking too much for this very specific case, as pointed out by others.
$endgroup$
– Sangchul Lee
Mar 24 at 18:56
$begingroup$
Hi @SangchulLee, I edited a note, can you check if it is ok?
$endgroup$
– Peter Krauss
Mar 24 at 20:48
1
$begingroup$
The notion of limit explained in HansLundmark's answer is sufficient for this case, which is an instance of limit notion in lattice theory. It definitely works for your sequence. Since $mathcalC$ can be realized as a bona-fide limit, you can interpret $mathcalC$ as the ideal target and $C_n$ as approximations of $mathcalC$ whose accuracy improves progressively in $n$, just as for limits in $mathbbR$.
$endgroup$
– Sangchul Lee
Mar 24 at 21:43
add a comment |
$begingroup$
The Cantor set, as showed in books and Wikipedia, is defined in terms of $C_k$, the finite Cantor set of level $k$:
$$ mathcalC = bigcap_k=1^infty C_k $$
But after intersection only the "last" remains, so, why not to define it by a limit?
$$mathcalC = lim_k to infty C_k$$
It is perhaps a naive intuition, but I not see a good justification.
(adding here a note after first answer, only for comment the comments)
NOTE: if it is not only a question of choice of notation, but also about context and semantics.
Can I tell an engineer that the intersection is a kind of specification, something like a project to explain "what I need", and the limit is a "what I get", the end result? ...Or perhaps the inverse, as suggested by @HansLundmark (thanks the comment! also thnaks @SangchulLee!). I am supposing that $C_k$ is a "decreasing sequence", $C_1 supseteq C_2 supseteq C_3 supseteq dotsb$,
so using your anser we can say
"it's natural to define the limit as the intersection: $C_n to mathcalC$ as $ntoinfty$",
where $mathcalC$ is defined by intersection.
About comment of @LordShark: its is possible to use "limits" notation in the context of set sequences without "develop a theory" for it? @HansLundmark's link is a satisfactory answer for it?
cantor-set
edited Mar 25 at 12:24
Matt Samuel
39.2k63870
asked Mar 24 at 18:48
Peter KraussPeter Krauss
1065
$endgroup$
The Cantor set, as showed in books and Wikipedia, is defined in terms of $C_k$, the finite Cantor set of level $k$:
$$ mathcalC = bigcap_k=1^infty C_k $$
But after intersection only the "last" remains, so, why not to define it by a limit?
$$mathcalC = lim_k to infty C_k$$
It is perhaps a naive intuition, but I not see a good justification.
(adding here a note after first answer, only for comment the comments)
NOTE: if it is not only a question of choice of notation, but also about context and semantics.
Can I tell an engineer that the intersection is a kind of specification, something like a project to explain "what I need", and the limit is a "what I get", the end result? ...Or perhaps the inverse, as suggested by @HansLundmark (thanks the comment! also thnaks @SangchulLee!). I am supposing that $C_k$ is a "decreasing sequence", $C_1 supseteq C_2 supseteq C_3 supseteq dotsb$,
so using your anser we can say
"it's natural to define the limit as the intersection: $C_n to mathcalC$ as $ntoinfty$",
where $mathcalC$ is defined by intersection.
About comment of @LordShark: its is possible to use "limits" notation in the context of set sequences without "develop a theory" for it? @HansLundmark's link is a satisfactory answer for it?
cantor-set
cantor-set
edited Mar 25 at 12:24
Matt Samuel
39.2k63870
asked Mar 24 at 18:48
Peter KraussPeter Krauss
1065
edited Mar 25 at 12:24
Matt Samuel
39.2k63870
asked Mar 24 at 18:48
Peter KraussPeter Krauss
1065
edited Mar 25 at 12:24
Matt Samuel
39.2k63870
edited Mar 25 at 12:24
Matt Samuel
39.2k63870
edited Mar 25 at 12:24
Matt Samuel
39.2k63870
39.2k63870
asked Mar 24 at 18:48
Peter KraussPeter Krauss
1065
asked Mar 24 at 18:48
Peter KraussPeter Krauss
1065
asked Mar 24 at 18:48
Peter KraussPeter Krauss
1065
1065
2
$begingroup$
Why develop a theory of "limits" of sets to define $cal C$ when it's simply an intersection?
$endgroup$
– Lord Shark the Unknown
Mar 24 at 18:53
1
$begingroup$
The intersection is the limit in this case (when you have a decreasing sequence of sets). See here: math.stackexchange.com/questions/107931/…
$endgroup$
– Hans Lundmark
Mar 24 at 18:54
1
$begingroup$
Of course we can perfectly make sense of $mathcalC = lim_ktoinfty C_k$. On the other hand, we need some works to put that notion to a rigorous mathematical framework. And that is sort of asking too much for this very specific case, as pointed out by others.
$endgroup$
– Sangchul Lee
Mar 24 at 18:56
$begingroup$
Hi @SangchulLee, I edited a note, can you check if it is ok?
$endgroup$
– Peter Krauss
Mar 24 at 20:48
1
$begingroup$
The notion of limit explained in HansLundmark's answer is sufficient for this case, which is an instance of limit notion in lattice theory. It definitely works for your sequence. Since $mathcalC$ can be realized as a bona-fide limit, you can interpret $mathcalC$ as the ideal target and $C_n$ as approximations of $mathcalC$ whose accuracy improves progressively in $n$, just as for limits in $mathbbR$.
$endgroup$
– Sangchul Lee
Mar 24 at 21:43
add a comment |
2
$begingroup$
Why develop a theory of "limits" of sets to define $cal C$ when it's simply an intersection?
$endgroup$
– Lord Shark the Unknown
Mar 24 at 18:53
1
$begingroup$
The intersection is the limit in this case (when you have a decreasing sequence of sets). See here: math.stackexchange.com/questions/107931/…
$endgroup$
– Hans Lundmark
Mar 24 at 18:54
1
$begingroup$
Of course we can perfectly make sense of $mathcalC = lim_ktoinfty C_k$. On the other hand, we need some works to put that notion to a rigorous mathematical framework. And that is sort of asking too much for this very specific case, as pointed out by others.
$endgroup$
– Sangchul Lee
Mar 24 at 18:56
$begingroup$
Hi @SangchulLee, I edited a note, can you check if it is ok?
$endgroup$
– Peter Krauss
Mar 24 at 20:48
1
$begingroup$
The notion of limit explained in HansLundmark's answer is sufficient for this case, which is an instance of limit notion in lattice theory. It definitely works for your sequence. Since $mathcalC$ can be realized as a bona-fide limit, you can interpret $mathcalC$ as the ideal target and $C_n$ as approximations of $mathcalC$ whose accuracy improves progressively in $n$, just as for limits in $mathbbR$.
$endgroup$
– Sangchul Lee
Mar 24 at 21:43
2
2
$begingroup$
Why develop a theory of "limits" of sets to define $cal C$ when it's simply an intersection?
$endgroup$
– Lord Shark the Unknown
Mar 24 at 18:53
$begingroup$
Why develop a theory of "limits" of sets to define $cal C$ when it's simply an intersection?
$endgroup$
– Lord Shark the Unknown
Mar 24 at 18:53
1
1
$begingroup$
The intersection is the limit in this case (when you have a decreasing sequence of sets). See here: math.stackexchange.com/questions/107931/…
$endgroup$
– Hans Lundmark
Mar 24 at 18:54
$begingroup$
The intersection is the limit in this case (when you have a decreasing sequence of sets). See here: math.stackexchange.com/questions/107931/…
$endgroup$
– Hans Lundmark
Mar 24 at 18:54
1
1
$begingroup$
Of course we can perfectly make sense of $mathcalC = lim_ktoinfty C_k$. On the other hand, we need some works to put that notion to a rigorous mathematical framework. And that is sort of asking too much for this very specific case, as pointed out by others.
$endgroup$
– Sangchul Lee
Mar 24 at 18:56
$begingroup$
Of course we can perfectly make sense of $mathcalC = lim_ktoinfty C_k$. On the other hand, we need some works to put that notion to a rigorous mathematical framework. And that is sort of asking too much for this very specific case, as pointed out by others.
$endgroup$
– Sangchul Lee
Mar 24 at 18:56
$begingroup$
Hi @SangchulLee, I edited a note, can you check if it is ok?
$endgroup$
– Peter Krauss
Mar 24 at 20:48
$begingroup$
Hi @SangchulLee, I edited a note, can you check if it is ok?
$endgroup$
– Peter Krauss
Mar 24 at 20:48
1
1
$begingroup$
The notion of limit explained in HansLundmark's answer is sufficient for this case, which is an instance of limit notion in lattice theory. It definitely works for your sequence. Since $mathcalC$ can be realized as a bona-fide limit, you can interpret $mathcalC$ as the ideal target and $C_n$ as approximations of $mathcalC$ whose accuracy improves progressively in $n$, just as for limits in $mathbbR$.
$endgroup$
– Sangchul Lee
Mar 24 at 21:43
$begingroup$
The notion of limit explained in HansLundmark's answer is sufficient for this case, which is an instance of limit notion in lattice theory. It definitely works for your sequence. Since $mathcalC$ can be realized as a bona-fide limit, you can interpret $mathcalC$ as the ideal target and $C_n$ as approximations of $mathcalC$ whose accuracy improves progressively in $n$, just as for limits in $mathbbR$.
$endgroup$
– Sangchul Lee
Mar 24 at 21:43
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Essentially, the intersection is the limit. There is no final set, and we don't want to appeal to any concept of convergence of sets or categorical limits when this is usually introduced in an undergraduate course because this would not be understood by the students. But you're allowed to take arbitrary intersections, and this is an elementary way to get the limit of a nested decreasing sequence of sets.
answered Mar 24 at 18:53
Matt SamuelMatt Samuel
39.2k63870
$endgroup$
$begingroup$
Can I tell an engineer that the intersection is a specification, something like a project to explain "what I need", and the limit is a "what I get", the end result?
$endgroup$
– Peter Krauss
Mar 24 at 19:17
1
$begingroup$
@PeterKrauss: No, that is totally wrong. The intersection is equal to the limit (in the sense as discussed in the comments), and they are both the end result.
$endgroup$
– Eric Wofsey
Mar 24 at 23:34
$begingroup$
Thanks Matt and @EricWofsey, I am clicking here solved. Eric, I'm not sure, but I tend to agree. I improved my comment by editing the question, adding a note, there is a suggestion from LordShark, that is also in this direction of "they are both the end result".
$endgroup$
– Peter Krauss
Mar 24 at 23:54
add a comment |
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1 Answer
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$begingroup$
Essentially, the intersection is the limit. There is no final set, and we don't want to appeal to any concept of convergence of sets or categorical limits when this is usually introduced in an undergraduate course because this would not be understood by the students. But you're allowed to take arbitrary intersections, and this is an elementary way to get the limit of a nested decreasing sequence of sets.
answered Mar 24 at 18:53
Matt SamuelMatt Samuel
39.2k63870
$endgroup$
$begingroup$
Can I tell an engineer that the intersection is a specification, something like a project to explain "what I need", and the limit is a "what I get", the end result?
$endgroup$
– Peter Krauss
Mar 24 at 19:17
1
$begingroup$
@PeterKrauss: No, that is totally wrong. The intersection is equal to the limit (in the sense as discussed in the comments), and they are both the end result.
$endgroup$
– Eric Wofsey
Mar 24 at 23:34
$begingroup$
Thanks Matt and @EricWofsey, I am clicking here solved. Eric, I'm not sure, but I tend to agree. I improved my comment by editing the question, adding a note, there is a suggestion from LordShark, that is also in this direction of "they are both the end result".
$endgroup$
– Peter Krauss
Mar 24 at 23:54
add a comment |
$begingroup$
Essentially, the intersection is the limit. There is no final set, and we don't want to appeal to any concept of convergence of sets or categorical limits when this is usually introduced in an undergraduate course because this would not be understood by the students. But you're allowed to take arbitrary intersections, and this is an elementary way to get the limit of a nested decreasing sequence of sets.
answered Mar 24 at 18:53
Matt SamuelMatt Samuel
39.2k63870
$endgroup$
$begingroup$
Can I tell an engineer that the intersection is a specification, something like a project to explain "what I need", and the limit is a "what I get", the end result?
$endgroup$
– Peter Krauss
Mar 24 at 19:17
1
$begingroup$
@PeterKrauss: No, that is totally wrong. The intersection is equal to the limit (in the sense as discussed in the comments), and they are both the end result.
$endgroup$
– Eric Wofsey
Mar 24 at 23:34
$begingroup$
Thanks Matt and @EricWofsey, I am clicking here solved. Eric, I'm not sure, but I tend to agree. I improved my comment by editing the question, adding a note, there is a suggestion from LordShark, that is also in this direction of "they are both the end result".
$endgroup$
– Peter Krauss
Mar 24 at 23:54
add a comment |
$begingroup$
Essentially, the intersection is the limit. There is no final set, and we don't want to appeal to any concept of convergence of sets or categorical limits when this is usually introduced in an undergraduate course because this would not be understood by the students. But you're allowed to take arbitrary intersections, and this is an elementary way to get the limit of a nested decreasing sequence of sets.
answered Mar 24 at 18:53
Matt SamuelMatt Samuel
39.2k63870
$endgroup$
Essentially, the intersection is the limit. There is no final set, and we don't want to appeal to any concept of convergence of sets or categorical limits when this is usually introduced in an undergraduate course because this would not be understood by the students. But you're allowed to take arbitrary intersections, and this is an elementary way to get the limit of a nested decreasing sequence of sets.
answered Mar 24 at 18:53
Matt SamuelMatt Samuel
39.2k63870
answered Mar 24 at 18:53
Matt SamuelMatt Samuel
39.2k63870
answered Mar 24 at 18:53
Matt SamuelMatt Samuel
39.2k63870
answered Mar 24 at 18:53
Matt SamuelMatt Samuel
39.2k63870
39.2k63870
$begingroup$
Can I tell an engineer that the intersection is a specification, something like a project to explain "what I need", and the limit is a "what I get", the end result?
$endgroup$
– Peter Krauss
Mar 24 at 19:17
1
$begingroup$
@PeterKrauss: No, that is totally wrong. The intersection is equal to the limit (in the sense as discussed in the comments), and they are both the end result.
$endgroup$
– Eric Wofsey
Mar 24 at 23:34
$begingroup$
Thanks Matt and @EricWofsey, I am clicking here solved. Eric, I'm not sure, but I tend to agree. I improved my comment by editing the question, adding a note, there is a suggestion from LordShark, that is also in this direction of "they are both the end result".
$endgroup$
– Peter Krauss
Mar 24 at 23:54
add a comment |
$begingroup$
Can I tell an engineer that the intersection is a specification, something like a project to explain "what I need", and the limit is a "what I get", the end result?
$endgroup$
– Peter Krauss
Mar 24 at 19:17
1
$begingroup$
@PeterKrauss: No, that is totally wrong. The intersection is equal to the limit (in the sense as discussed in the comments), and they are both the end result.
$endgroup$
– Eric Wofsey
Mar 24 at 23:34
$begingroup$
Thanks Matt and @EricWofsey, I am clicking here solved. Eric, I'm not sure, but I tend to agree. I improved my comment by editing the question, adding a note, there is a suggestion from LordShark, that is also in this direction of "they are both the end result".
$endgroup$
– Peter Krauss
Mar 24 at 23:54
$begingroup$
Can I tell an engineer that the intersection is a specification, something like a project to explain "what I need", and the limit is a "what I get", the end result?
$endgroup$
– Peter Krauss
Mar 24 at 19:17
$begingroup$
Can I tell an engineer that the intersection is a specification, something like a project to explain "what I need", and the limit is a "what I get", the end result?
$endgroup$
– Peter Krauss
Mar 24 at 19:17
1
1
$begingroup$
@PeterKrauss: No, that is totally wrong. The intersection is equal to the limit (in the sense as discussed in the comments), and they are both the end result.
$endgroup$
– Eric Wofsey
Mar 24 at 23:34
$begingroup$
@PeterKrauss: No, that is totally wrong. The intersection is equal to the limit (in the sense as discussed in the comments), and they are both the end result.
$endgroup$
– Eric Wofsey
Mar 24 at 23:34
$begingroup$
Thanks Matt and @EricWofsey, I am clicking here solved. Eric, I'm not sure, but I tend to agree. I improved my comment by editing the question, adding a note, there is a suggestion from LordShark, that is also in this direction of "they are both the end result".
$endgroup$
– Peter Krauss
Mar 24 at 23:54
$begingroup$
Thanks Matt and @EricWofsey, I am clicking here solved. Eric, I'm not sure, but I tend to agree. I improved my comment by editing the question, adding a note, there is a suggestion from LordShark, that is also in this direction of "they are both the end result".
$endgroup$
– Peter Krauss
Mar 24 at 23:54
add a comment |
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jd4gvEG6jyx9c vPZU53DHL78RDdPKJDXrnajOdLNnX,2rB7cYkIwYK,YYb4WRvs3jNLzkhZu9XNLO5
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Why develop a theory of "limits" of sets to define $cal C$ when it's simply an intersection?
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– Lord Shark the Unknown
Mar 24 at 18:53
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The intersection is the limit in this case (when you have a decreasing sequence of sets). See here: math.stackexchange.com/questions/107931/…
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– Hans Lundmark
Mar 24 at 18:54
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Of course we can perfectly make sense of $mathcalC = lim_ktoinfty C_k$. On the other hand, we need some works to put that notion to a rigorous mathematical framework. And that is sort of asking too much for this very specific case, as pointed out by others.
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– Sangchul Lee
Mar 24 at 18:56
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Hi @SangchulLee, I edited a note, can you check if it is ok?
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– Peter Krauss
Mar 24 at 20:48
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The notion of limit explained in HansLundmark's answer is sufficient for this case, which is an instance of limit notion in lattice theory. It definitely works for your sequence. Since $mathcalC$ can be realized as a bona-fide limit, you can interpret $mathcalC$ as the ideal target and $C_n$ as approximations of $mathcalC$ whose accuracy improves progressively in $n$, just as for limits in $mathbbR$.
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– Sangchul Lee
Mar 24 at 21:43