Natural Definition of Continuity [on hold]
The argueably simplest definition for continuous function $f$ is: both ${x|f(x)>a}$ and ${x|f(x)<a}$ are open sets $forall ainmathbb R$.
How to generalize this set-theoretical definition to Lipshitz continuity? I don't want to use any addition or substraction operation between $f(x)$. Plus and minus between $x$ is fine.
(Sorry the definition was wrong. Edited to the standard one.)
real-analysis functions continuity
asked Dec 26 at 8:37
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put on hold as off-topic by Lord_Farin, Cesareo, user91500, mrtaurho, amWhy Dec 26 at 12:55
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The argueably simplest definition for continuous function $f$ is: both ${x|f(x)>a}$ and ${x|f(x)<a}$ are open sets $forall ainmathbb R$.
How to generalize this set-theoretical definition to Lipshitz continuity? I don't want to use any addition or substraction operation between $f(x)$. Plus and minus between $x$ is fine.
(Sorry the definition was wrong. Edited to the standard one.)
real-analysis functions continuity
asked Dec 26 at 8:37
High GPA
894419
put on hold as off-topic by Lord_Farin, Cesareo, user91500, mrtaurho, amWhy Dec 26 at 12:55
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Lord_Farin, Cesareo, user91500, mrtaurho, amWhy
If this question can be reworded to fit the rules in the help center, please edit the question.
4
is that even a valid definition?
– mathworker21
Dec 26 at 8:42
Please edit your question to clarify what you mean with "absolute continuity".
– Lord_Farin
Dec 26 at 9:43
add a comment |
The argueably simplest definition for continuous function $f$ is: both ${x|f(x)>a}$ and ${x|f(x)<a}$ are open sets $forall ainmathbb R$.
How to generalize this set-theoretical definition to Lipshitz continuity? I don't want to use any addition or substraction operation between $f(x)$. Plus and minus between $x$ is fine.
(Sorry the definition was wrong. Edited to the standard one.)
real-analysis functions continuity
asked Dec 26 at 8:37
High GPA
894419
The argueably simplest definition for continuous function $f$ is: both ${x|f(x)>a}$ and ${x|f(x)<a}$ are open sets $forall ainmathbb R$.
How to generalize this set-theoretical definition to Lipshitz continuity? I don't want to use any addition or substraction operation between $f(x)$. Plus and minus between $x$ is fine.
(Sorry the definition was wrong. Edited to the standard one.)
real-analysis functions continuity
real-analysis functions continuity
asked Dec 26 at 8:37
High GPA
894419
asked Dec 26 at 8:37
High GPA
894419
edited Dec 26 at 9:56
asked Dec 26 at 8:37
High GPA
894419
asked Dec 26 at 8:37
High GPA
894419
asked Dec 26 at 8:37
High GPA
894419
894419
put on hold as off-topic by Lord_Farin, Cesareo, user91500, mrtaurho, amWhy Dec 26 at 12:55
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Lord_Farin, Cesareo, user91500, mrtaurho, amWhy
If this question can be reworded to fit the rules in the help center, please edit the question.
put on hold as off-topic by Lord_Farin, Cesareo, user91500, mrtaurho, amWhy Dec 26 at 12:55
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Lord_Farin, Cesareo, user91500, mrtaurho, amWhy
If this question can be reworded to fit the rules in the help center, please edit the question.
4
is that even a valid definition?
– mathworker21
Dec 26 at 8:42
Please edit your question to clarify what you mean with "absolute continuity".
– Lord_Farin
Dec 26 at 9:43
add a comment |
4
is that even a valid definition?
– mathworker21
Dec 26 at 8:42
Please edit your question to clarify what you mean with "absolute continuity".
– Lord_Farin
Dec 26 at 9:43
4
4
is that even a valid definition?
– mathworker21
Dec 26 at 8:42
is that even a valid definition?
– mathworker21
Dec 26 at 8:42
Please edit your question to clarify what you mean with "absolute continuity".
– Lord_Farin
Dec 26 at 9:43
Please edit your question to clarify what you mean with "absolute continuity".
– Lord_Farin
Dec 26 at 9:43
add a comment |
1 Answer
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Your definition is invalid. Consider $f(x) := -x^2$ for $x not = 0$ and $f(0) = 1$. Then ${x : f(x) > f(y)}$ is open for each $y$, but $f$ is not continuous (since, e.g., ${x : f(x) > .5}$ is not open)
Added: to answer your updated question, I don't think there is a set theoretic definition of absolute continuity, since the definition has to do with sums of lengths of intervals, which cannot be easily encoded just in terms of sets without the use of quantifiers.
answered Dec 26 at 8:47
mathworker21
8,6371928
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
Your definition is invalid. Consider $f(x) := -x^2$ for $x not = 0$ and $f(0) = 1$. Then ${x : f(x) > f(y)}$ is open for each $y$, but $f$ is not continuous (since, e.g., ${x : f(x) > .5}$ is not open)
Added: to answer your updated question, I don't think there is a set theoretic definition of absolute continuity, since the definition has to do with sums of lengths of intervals, which cannot be easily encoded just in terms of sets without the use of quantifiers.
answered Dec 26 at 8:47
mathworker21
8,6371928
add a comment |
Your definition is invalid. Consider $f(x) := -x^2$ for $x not = 0$ and $f(0) = 1$. Then ${x : f(x) > f(y)}$ is open for each $y$, but $f$ is not continuous (since, e.g., ${x : f(x) > .5}$ is not open)
Added: to answer your updated question, I don't think there is a set theoretic definition of absolute continuity, since the definition has to do with sums of lengths of intervals, which cannot be easily encoded just in terms of sets without the use of quantifiers.
answered Dec 26 at 8:47
mathworker21
8,6371928
add a comment |
Your definition is invalid. Consider $f(x) := -x^2$ for $x not = 0$ and $f(0) = 1$. Then ${x : f(x) > f(y)}$ is open for each $y$, but $f$ is not continuous (since, e.g., ${x : f(x) > .5}$ is not open)
Added: to answer your updated question, I don't think there is a set theoretic definition of absolute continuity, since the definition has to do with sums of lengths of intervals, which cannot be easily encoded just in terms of sets without the use of quantifiers.
answered Dec 26 at 8:47
mathworker21
8,6371928
Your definition is invalid. Consider $f(x) := -x^2$ for $x not = 0$ and $f(0) = 1$. Then ${x : f(x) > f(y)}$ is open for each $y$, but $f$ is not continuous (since, e.g., ${x : f(x) > .5}$ is not open)
Added: to answer your updated question, I don't think there is a set theoretic definition of absolute continuity, since the definition has to do with sums of lengths of intervals, which cannot be easily encoded just in terms of sets without the use of quantifiers.
answered Dec 26 at 8:47
mathworker21
8,6371928
edited Dec 26 at 9:34
answered Dec 26 at 8:47
mathworker21
8,6371928
answered Dec 26 at 8:47
mathworker21
8,6371928
answered Dec 26 at 8:47
mathworker21
8,6371928
8,6371928
add a comment |
add a comment |
4
is that even a valid definition?
– mathworker21
Dec 26 at 8:42
Please edit your question to clarify what you mean with "absolute continuity".
– Lord_Farin
Dec 26 at 9:43