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Is the empty function differentiable?
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Given the empty function $varnothing: emptyset to X$ where $X$ is a set, is $varnothing$ a differentiable function? If so, what is its derivative?
Also, is the empty set a differentiable manifold? If so, what is it's dimension?
calculus real-analysis functions differential-topology
asked 5 hours ago
Mobius
528
add a comment |Â
up vote
2
down vote
favorite
Given the empty function $varnothing: emptyset to X$ where $X$ is a set, is $varnothing$ a differentiable function? If so, what is its derivative?
Also, is the empty set a differentiable manifold? If so, what is it's dimension?
calculus real-analysis functions differential-topology
asked 5 hours ago
Mobius
528
3
In order for a function to be differentiable you have to be able to define "differentialbe", which requires a fair amount of structure on the domain and codomain that's not there in your question.
– Ethan Bolker
5 hours ago
4
Long ago at MIT, this sort of question was repeatedly asked by a student in a topology class. The prof eventually got fed up, and declared all such questions SO important that they really needed a single-person oracle to know all the answers, and named that particular student "the keeper of the empty set." Several of my grad school classmates were part of that class, and I've remembered the story over the (many) intervening years. Take from it what you will.
– John Hughes
5 hours ago
@EthanBolker Can the empty set not be considered a manifold? I agree that X needs more structure before differentiability is defined, but if X is a manifold, I would argue there is a notion of differentiability for the map.
– MightyTyGuy
5 hours ago
2
The first paragraph of the definition of a manifold at en.wikipedia.org/wiki/Manifold seems to say that the empty set is a manifold of every dimension. I suggest you take this question up with the oracle @JohnHughes tells us about.
– Ethan Bolker
5 hours ago
To answer: Assuming in your question that $X$ is a manifold, and that your particular keeper of the empty set declares it, too, to be a manifold, then the derivative would be the empty map from the tangent bundle of the empty set (also empty) to the tangent bundle of $X$.
– John Hughes
2 hours ago
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Given the empty function $varnothing: emptyset to X$ where $X$ is a set, is $varnothing$ a differentiable function? If so, what is its derivative?
Also, is the empty set a differentiable manifold? If so, what is it's dimension?
calculus real-analysis functions differential-topology
asked 5 hours ago
Mobius
528
Given the empty function $varnothing: emptyset to X$ where $X$ is a set, is $varnothing$ a differentiable function? If so, what is its derivative?
Also, is the empty set a differentiable manifold? If so, what is it's dimension?
calculus real-analysis functions differential-topology
calculus real-analysis functions differential-topology
asked 5 hours ago
Mobius
528
asked 5 hours ago
Mobius
528
asked 5 hours ago
Mobius
528
asked 5 hours ago
Mobius
528
asked 5 hours ago
Mobius
528
528
3
In order for a function to be differentiable you have to be able to define "differentialbe", which requires a fair amount of structure on the domain and codomain that's not there in your question.
– Ethan Bolker
5 hours ago
4
Long ago at MIT, this sort of question was repeatedly asked by a student in a topology class. The prof eventually got fed up, and declared all such questions SO important that they really needed a single-person oracle to know all the answers, and named that particular student "the keeper of the empty set." Several of my grad school classmates were part of that class, and I've remembered the story over the (many) intervening years. Take from it what you will.
– John Hughes
5 hours ago
@EthanBolker Can the empty set not be considered a manifold? I agree that X needs more structure before differentiability is defined, but if X is a manifold, I would argue there is a notion of differentiability for the map.
– MightyTyGuy
5 hours ago
2
The first paragraph of the definition of a manifold at en.wikipedia.org/wiki/Manifold seems to say that the empty set is a manifold of every dimension. I suggest you take this question up with the oracle @JohnHughes tells us about.
– Ethan Bolker
5 hours ago
To answer: Assuming in your question that $X$ is a manifold, and that your particular keeper of the empty set declares it, too, to be a manifold, then the derivative would be the empty map from the tangent bundle of the empty set (also empty) to the tangent bundle of $X$.
– John Hughes
2 hours ago
add a comment |Â
3
In order for a function to be differentiable you have to be able to define "differentialbe", which requires a fair amount of structure on the domain and codomain that's not there in your question.
– Ethan Bolker
5 hours ago
4
Long ago at MIT, this sort of question was repeatedly asked by a student in a topology class. The prof eventually got fed up, and declared all such questions SO important that they really needed a single-person oracle to know all the answers, and named that particular student "the keeper of the empty set." Several of my grad school classmates were part of that class, and I've remembered the story over the (many) intervening years. Take from it what you will.
– John Hughes
5 hours ago
@EthanBolker Can the empty set not be considered a manifold? I agree that X needs more structure before differentiability is defined, but if X is a manifold, I would argue there is a notion of differentiability for the map.
– MightyTyGuy
5 hours ago
2
The first paragraph of the definition of a manifold at en.wikipedia.org/wiki/Manifold seems to say that the empty set is a manifold of every dimension. I suggest you take this question up with the oracle @JohnHughes tells us about.
– Ethan Bolker
5 hours ago
To answer: Assuming in your question that $X$ is a manifold, and that your particular keeper of the empty set declares it, too, to be a manifold, then the derivative would be the empty map from the tangent bundle of the empty set (also empty) to the tangent bundle of $X$.
– John Hughes
2 hours ago
3
3
In order for a function to be differentiable you have to be able to define "differentialbe", which requires a fair amount of structure on the domain and codomain that's not there in your question.
– Ethan Bolker
5 hours ago
In order for a function to be differentiable you have to be able to define "differentialbe", which requires a fair amount of structure on the domain and codomain that's not there in your question.
– Ethan Bolker
5 hours ago
4
4
Long ago at MIT, this sort of question was repeatedly asked by a student in a topology class. The prof eventually got fed up, and declared all such questions SO important that they really needed a single-person oracle to know all the answers, and named that particular student "the keeper of the empty set." Several of my grad school classmates were part of that class, and I've remembered the story over the (many) intervening years. Take from it what you will.
– John Hughes
5 hours ago
Long ago at MIT, this sort of question was repeatedly asked by a student in a topology class. The prof eventually got fed up, and declared all such questions SO important that they really needed a single-person oracle to know all the answers, and named that particular student "the keeper of the empty set." Several of my grad school classmates were part of that class, and I've remembered the story over the (many) intervening years. Take from it what you will.
– John Hughes
5 hours ago
@EthanBolker Can the empty set not be considered a manifold? I agree that X needs more structure before differentiability is defined, but if X is a manifold, I would argue there is a notion of differentiability for the map.
– MightyTyGuy
5 hours ago
@EthanBolker Can the empty set not be considered a manifold? I agree that X needs more structure before differentiability is defined, but if X is a manifold, I would argue there is a notion of differentiability for the map.
– MightyTyGuy
5 hours ago
2
2
The first paragraph of the definition of a manifold at en.wikipedia.org/wiki/Manifold seems to say that the empty set is a manifold of every dimension. I suggest you take this question up with the oracle @JohnHughes tells us about.
– Ethan Bolker
5 hours ago
The first paragraph of the definition of a manifold at en.wikipedia.org/wiki/Manifold seems to say that the empty set is a manifold of every dimension. I suggest you take this question up with the oracle @JohnHughes tells us about.
– Ethan Bolker
5 hours ago
To answer: Assuming in your question that $X$ is a manifold, and that your particular keeper of the empty set declares it, too, to be a manifold, then the derivative would be the empty map from the tangent bundle of the empty set (also empty) to the tangent bundle of $X$.
– John Hughes
2 hours ago
To answer: Assuming in your question that $X$ is a manifold, and that your particular keeper of the empty set declares it, too, to be a manifold, then the derivative would be the empty map from the tangent bundle of the empty set (also empty) to the tangent bundle of $X$.
– John Hughes
2 hours ago
add a comment |Â
2 Answers
2
active
oldest
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up vote
4
down vote
Yes, the empty set is a smooth manifold (it is covered by the empty collection of coordinate charts!). It has every dimension. (That is, for any $n$, it is true that $emptyset$ is a manifold of dimension $n$. Note that there is not really a unified definition of "manifold" but rather a separate definition of "manifold of dimension $n$" for each $n$, so there is nothing wrong with a single object satisfying the definition for multiple different values of $n$.)
For any smooth manifold $X$, the empty function $emptysetto X$ is smooth. After all, this just means that it gives a smooth function in every pair coordinate charts of a pair of atlases on the domain and codomain, which is vacuously true since the empty set is an atlas for $emptyset$. (Or, in the context of just open subset of Euclidean space, if we consider $emptyset$ as an open subset of $mathbbR^m$, then the empty function $emptysettomathbbR^n$ is smooth because it is vacuously infinitely differentiable at every point in the domain. Its derivative is then the empty function $emptysetto mathbbR^ntimes m$.)
answered 1 hour ago
Eric Wofsey
171k12198317
I would say that since there is no point in the domain of the empty function where it fails to be differentiable, you can say that it is differentiable. And the domain of a differentiable $f$ is equal to the domain of $f.$ So $emptyset'=emptyset.$
– DanielWainfleet
26 mins ago
add a comment |Â
up vote
1
down vote
Yes, the empty function is differentiable because at
every point in the empty set, it is differentiable.
answered 2 hours ago
William Elliot
6,1382517
Yes. Because there are no points in its domain where it fails to have a derivative.
– DanielWainfleet
22 mins ago
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
Yes, the empty set is a smooth manifold (it is covered by the empty collection of coordinate charts!). It has every dimension. (That is, for any $n$, it is true that $emptyset$ is a manifold of dimension $n$. Note that there is not really a unified definition of "manifold" but rather a separate definition of "manifold of dimension $n$" for each $n$, so there is nothing wrong with a single object satisfying the definition for multiple different values of $n$.)
For any smooth manifold $X$, the empty function $emptysetto X$ is smooth. After all, this just means that it gives a smooth function in every pair coordinate charts of a pair of atlases on the domain and codomain, which is vacuously true since the empty set is an atlas for $emptyset$. (Or, in the context of just open subset of Euclidean space, if we consider $emptyset$ as an open subset of $mathbbR^m$, then the empty function $emptysettomathbbR^n$ is smooth because it is vacuously infinitely differentiable at every point in the domain. Its derivative is then the empty function $emptysetto mathbbR^ntimes m$.)
answered 1 hour ago
Eric Wofsey
171k12198317
I would say that since there is no point in the domain of the empty function where it fails to be differentiable, you can say that it is differentiable. And the domain of a differentiable $f$ is equal to the domain of $f.$ So $emptyset'=emptyset.$
– DanielWainfleet
26 mins ago
add a comment |Â
up vote
4
down vote
Yes, the empty set is a smooth manifold (it is covered by the empty collection of coordinate charts!). It has every dimension. (That is, for any $n$, it is true that $emptyset$ is a manifold of dimension $n$. Note that there is not really a unified definition of "manifold" but rather a separate definition of "manifold of dimension $n$" for each $n$, so there is nothing wrong with a single object satisfying the definition for multiple different values of $n$.)
For any smooth manifold $X$, the empty function $emptysetto X$ is smooth. After all, this just means that it gives a smooth function in every pair coordinate charts of a pair of atlases on the domain and codomain, which is vacuously true since the empty set is an atlas for $emptyset$. (Or, in the context of just open subset of Euclidean space, if we consider $emptyset$ as an open subset of $mathbbR^m$, then the empty function $emptysettomathbbR^n$ is smooth because it is vacuously infinitely differentiable at every point in the domain. Its derivative is then the empty function $emptysetto mathbbR^ntimes m$.)
answered 1 hour ago
Eric Wofsey
171k12198317
I would say that since there is no point in the domain of the empty function where it fails to be differentiable, you can say that it is differentiable. And the domain of a differentiable $f$ is equal to the domain of $f.$ So $emptyset'=emptyset.$
– DanielWainfleet
26 mins ago
add a comment |Â
up vote
4
down vote
up vote
4
down vote
Yes, the empty set is a smooth manifold (it is covered by the empty collection of coordinate charts!). It has every dimension. (That is, for any $n$, it is true that $emptyset$ is a manifold of dimension $n$. Note that there is not really a unified definition of "manifold" but rather a separate definition of "manifold of dimension $n$" for each $n$, so there is nothing wrong with a single object satisfying the definition for multiple different values of $n$.)
For any smooth manifold $X$, the empty function $emptysetto X$ is smooth. After all, this just means that it gives a smooth function in every pair coordinate charts of a pair of atlases on the domain and codomain, which is vacuously true since the empty set is an atlas for $emptyset$. (Or, in the context of just open subset of Euclidean space, if we consider $emptyset$ as an open subset of $mathbbR^m$, then the empty function $emptysettomathbbR^n$ is smooth because it is vacuously infinitely differentiable at every point in the domain. Its derivative is then the empty function $emptysetto mathbbR^ntimes m$.)
answered 1 hour ago
Eric Wofsey
171k12198317
Yes, the empty set is a smooth manifold (it is covered by the empty collection of coordinate charts!). It has every dimension. (That is, for any $n$, it is true that $emptyset$ is a manifold of dimension $n$. Note that there is not really a unified definition of "manifold" but rather a separate definition of "manifold of dimension $n$" for each $n$, so there is nothing wrong with a single object satisfying the definition for multiple different values of $n$.)
For any smooth manifold $X$, the empty function $emptysetto X$ is smooth. After all, this just means that it gives a smooth function in every pair coordinate charts of a pair of atlases on the domain and codomain, which is vacuously true since the empty set is an atlas for $emptyset$. (Or, in the context of just open subset of Euclidean space, if we consider $emptyset$ as an open subset of $mathbbR^m$, then the empty function $emptysettomathbbR^n$ is smooth because it is vacuously infinitely differentiable at every point in the domain. Its derivative is then the empty function $emptysetto mathbbR^ntimes m$.)
answered 1 hour ago
Eric Wofsey
171k12198317
edited 1 hour ago
answered 1 hour ago
Eric Wofsey
171k12198317
answered 1 hour ago
Eric Wofsey
171k12198317
answered 1 hour ago
Eric Wofsey
171k12198317
171k12198317
I would say that since there is no point in the domain of the empty function where it fails to be differentiable, you can say that it is differentiable. And the domain of a differentiable $f$ is equal to the domain of $f.$ So $emptyset'=emptyset.$
– DanielWainfleet
26 mins ago
add a comment |Â
I would say that since there is no point in the domain of the empty function where it fails to be differentiable, you can say that it is differentiable. And the domain of a differentiable $f$ is equal to the domain of $f.$ So $emptyset'=emptyset.$
– DanielWainfleet
26 mins ago
I would say that since there is no point in the domain of the empty function where it fails to be differentiable, you can say that it is differentiable. And the domain of a differentiable $f$ is equal to the domain of $f.$ So $emptyset'=emptyset.$
– DanielWainfleet
26 mins ago
I would say that since there is no point in the domain of the empty function where it fails to be differentiable, you can say that it is differentiable. And the domain of a differentiable $f$ is equal to the domain of $f.$ So $emptyset'=emptyset.$
– DanielWainfleet
26 mins ago
add a comment |Â
up vote
1
down vote
Yes, the empty function is differentiable because at
every point in the empty set, it is differentiable.
answered 2 hours ago
William Elliot
6,1382517
Yes. Because there are no points in its domain where it fails to have a derivative.
– DanielWainfleet
22 mins ago
add a comment |Â
up vote
1
down vote
Yes, the empty function is differentiable because at
every point in the empty set, it is differentiable.
answered 2 hours ago
William Elliot
6,1382517
Yes. Because there are no points in its domain where it fails to have a derivative.
– DanielWainfleet
22 mins ago
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Yes, the empty function is differentiable because at
every point in the empty set, it is differentiable.
answered 2 hours ago
William Elliot
6,1382517
Yes, the empty function is differentiable because at
every point in the empty set, it is differentiable.
answered 2 hours ago
William Elliot
6,1382517
answered 2 hours ago
William Elliot
6,1382517
answered 2 hours ago
William Elliot
6,1382517
answered 2 hours ago
William Elliot
6,1382517
6,1382517
Yes. Because there are no points in its domain where it fails to have a derivative.
– DanielWainfleet
22 mins ago
add a comment |Â
Yes. Because there are no points in its domain where it fails to have a derivative.
– DanielWainfleet
22 mins ago
Yes. Because there are no points in its domain where it fails to have a derivative.
– DanielWainfleet
22 mins ago
Yes. Because there are no points in its domain where it fails to have a derivative.
– DanielWainfleet
22 mins ago
add a comment |Â
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3
In order for a function to be differentiable you have to be able to define "differentialbe", which requires a fair amount of structure on the domain and codomain that's not there in your question.
– Ethan Bolker
5 hours ago
4
Long ago at MIT, this sort of question was repeatedly asked by a student in a topology class. The prof eventually got fed up, and declared all such questions SO important that they really needed a single-person oracle to know all the answers, and named that particular student "the keeper of the empty set." Several of my grad school classmates were part of that class, and I've remembered the story over the (many) intervening years. Take from it what you will.
– John Hughes
5 hours ago
@EthanBolker Can the empty set not be considered a manifold? I agree that X needs more structure before differentiability is defined, but if X is a manifold, I would argue there is a notion of differentiability for the map.
– MightyTyGuy
5 hours ago
2
The first paragraph of the definition of a manifold at en.wikipedia.org/wiki/Manifold seems to say that the empty set is a manifold of every dimension. I suggest you take this question up with the oracle @JohnHughes tells us about.
– Ethan Bolker
5 hours ago
To answer: Assuming in your question that $X$ is a manifold, and that your particular keeper of the empty set declares it, too, to be a manifold, then the derivative would be the empty map from the tangent bundle of the empty set (also empty) to the tangent bundle of $X$.
– John Hughes
2 hours ago