Is there a “topology” of functions?
$begingroup$
A pair of real variables could be the surface coordinates for a 2 dimensional surface such as a sphere or a torus or a genus n surface for example.
Going from the space of real numbers to the space of functions is there an equivalent "topology" possible? One might imagine a functional of a pair of functions with certain properties defining a "topological surface" paramaterised by pairs of functions.
I.e. do function spaces also have a notion of topology that is a higher order version of topology?
general-topology functional-analysis
asked Jan 1 at 17:01
zoobyzooby
1,052716
$endgroup$
add a comment |
$begingroup$
A pair of real variables could be the surface coordinates for a 2 dimensional surface such as a sphere or a torus or a genus n surface for example.
Going from the space of real numbers to the space of functions is there an equivalent "topology" possible? One might imagine a functional of a pair of functions with certain properties defining a "topological surface" paramaterised by pairs of functions.
I.e. do function spaces also have a notion of topology that is a higher order version of topology?
general-topology functional-analysis
asked Jan 1 at 17:01
zoobyzooby
1,052716
$endgroup$
$begingroup$
Are you looking for a topology on a collection of functions, or something more specifically like a 2 dimensional surface (say, of a manifold). You can always take the product of two topological spaces (e.g. the product of two circles is a torus), so I'm trying to see if the "pair of functions" idea is crucial or incidental to what you're asking.
$endgroup$
– Mark S.
Jan 1 at 19:07
$begingroup$
@Mark For every point on a 2D surface one can associate a pair of real numbers such that near points have near numbers. I wondered if there was a functional version where points on the "surface" are associated with tuples of functions. This would have to be an infinite dimensional surface in normal language.
$endgroup$
– zooby
Jan 1 at 23:31
add a comment |
$begingroup$
A pair of real variables could be the surface coordinates for a 2 dimensional surface such as a sphere or a torus or a genus n surface for example.
Going from the space of real numbers to the space of functions is there an equivalent "topology" possible? One might imagine a functional of a pair of functions with certain properties defining a "topological surface" paramaterised by pairs of functions.
I.e. do function spaces also have a notion of topology that is a higher order version of topology?
general-topology functional-analysis
asked Jan 1 at 17:01
zoobyzooby
1,052716
$endgroup$
A pair of real variables could be the surface coordinates for a 2 dimensional surface such as a sphere or a torus or a genus n surface for example.
Going from the space of real numbers to the space of functions is there an equivalent "topology" possible? One might imagine a functional of a pair of functions with certain properties defining a "topological surface" paramaterised by pairs of functions.
I.e. do function spaces also have a notion of topology that is a higher order version of topology?
general-topology functional-analysis
general-topology functional-analysis
asked Jan 1 at 17:01
zoobyzooby
1,052716
asked Jan 1 at 17:01
zoobyzooby
1,052716
asked Jan 1 at 17:01
zoobyzooby
1,052716
asked Jan 1 at 17:01
zoobyzooby
1,052716
asked Jan 1 at 17:01
zoobyzooby
1,052716
1,052716
$begingroup$
Are you looking for a topology on a collection of functions, or something more specifically like a 2 dimensional surface (say, of a manifold). You can always take the product of two topological spaces (e.g. the product of two circles is a torus), so I'm trying to see if the "pair of functions" idea is crucial or incidental to what you're asking.
$endgroup$
– Mark S.
Jan 1 at 19:07
$begingroup$
@Mark For every point on a 2D surface one can associate a pair of real numbers such that near points have near numbers. I wondered if there was a functional version where points on the "surface" are associated with tuples of functions. This would have to be an infinite dimensional surface in normal language.
$endgroup$
– zooby
Jan 1 at 23:31
add a comment |
$begingroup$
Are you looking for a topology on a collection of functions, or something more specifically like a 2 dimensional surface (say, of a manifold). You can always take the product of two topological spaces (e.g. the product of two circles is a torus), so I'm trying to see if the "pair of functions" idea is crucial or incidental to what you're asking.
$endgroup$
– Mark S.
Jan 1 at 19:07
$begingroup$
@Mark For every point on a 2D surface one can associate a pair of real numbers such that near points have near numbers. I wondered if there was a functional version where points on the "surface" are associated with tuples of functions. This would have to be an infinite dimensional surface in normal language.
$endgroup$
– zooby
Jan 1 at 23:31
$begingroup$
Are you looking for a topology on a collection of functions, or something more specifically like a 2 dimensional surface (say, of a manifold). You can always take the product of two topological spaces (e.g. the product of two circles is a torus), so I'm trying to see if the "pair of functions" idea is crucial or incidental to what you're asking.
$endgroup$
– Mark S.
Jan 1 at 19:07
$begingroup$
Are you looking for a topology on a collection of functions, or something more specifically like a 2 dimensional surface (say, of a manifold). You can always take the product of two topological spaces (e.g. the product of two circles is a torus), so I'm trying to see if the "pair of functions" idea is crucial or incidental to what you're asking.
$endgroup$
– Mark S.
Jan 1 at 19:07
$begingroup$
@Mark For every point on a 2D surface one can associate a pair of real numbers such that near points have near numbers. I wondered if there was a functional version where points on the "surface" are associated with tuples of functions. This would have to be an infinite dimensional surface in normal language.
$endgroup$
– zooby
Jan 1 at 23:31
$begingroup$
@Mark For every point on a 2D surface one can associate a pair of real numbers such that near points have near numbers. I wondered if there was a functional version where points on the "surface" are associated with tuples of functions. This would have to be an infinite dimensional surface in normal language.
$endgroup$
– zooby
Jan 1 at 23:31
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
I suppose a Banach manifold could be considered what you're looking for, when the underlying Banach space is a space of functions. Here is a list of mathoverflow questions that have the tag "Banach manifold".
answered Jan 1 at 17:09
Dave L. RenfroDave L. Renfro
25.4k34082
$endgroup$
$begingroup$
Could be. I guess a function can be expressed as an infinite series. So they would be infinite dimensional manifolds in some sense. I have heard the phrase before but I think it is beyond my understanding. From Wikipedia it seems that the classification of Banach manifolds is not so interesting compared to finite dimensional manifolds.
$endgroup$
– zooby
Jan 1 at 23:32
add a comment |
$begingroup$
I don't know if this answer will be satisfying to you, but there is some general topology context that might change how you look at this question.
Firstly, if you want to handle pairs of anything-with-topologies, you can use the product topology. This applies equally well to anything you have topologies on, and often gives nice answers. For example, if $mathbb R$ has the standard topology and the circle $S^1$ has its standard topology (induced from its embedding in the plane, say), then:
- The product topology on $mathbb Rtimesmathbb R$ is indeed the standard topology on $mathbb R^2$
- The product topology on $S^1timesmathbb R$ corresponds exactly to the standard topology on an infinite cylinder living in $mathbb R^3$.
- The product topology on $S^1times S^1$ corresponds exactly to the standard topology on a torus living in $mathbb R^3$.
So if we had a reasonable topology on a collection of functions, then we could get a reasonable topology on pairs of functions via this same product-topology construction.
Now the question is: what is a reasonable topology on a bunch of functions $f:Xto Y$? Well, suppose the functions are (some of) the continuous maps between two (possibly identical) topological spaces (maybe nice ones like metric spaces or something).
One candidate would be to treat the space of these functions just like (infinite if $X$ is infinite) products of the space $Y$, but this is ignoring the topology on $X$. Another common one which takes that into account is called the compact-open topology. For a number of reasons (one is mentioned at this MSE answer), the compact-open topology is usually nicer/more appropriate.
Since you used the tag functional analysis, if we're looking at the continuous (aka bounded) linear maps on a Hilbert space $H$, then there are lots of topologies in use. And this MO answer mentions that on norm-bounded subsets, the "strong operator topology" agrees with the compact-open topology I mentioned above.
answered Jan 3 at 2:21
Mark S.Mark S.
12.5k22772
$endgroup$
$begingroup$
The product topologies are maybe not the most interesting topologies. The genus-2 surface for example is not a product. I wonder if in function space there are these non-product topologies? Maybe not. Perhaps just like polytopes in n-dimensions they get a bit boring after 4 dimensions. The operator topologies are a bit over my head I think!
$endgroup$
– zooby
Jan 3 at 5:09
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
I suppose a Banach manifold could be considered what you're looking for, when the underlying Banach space is a space of functions. Here is a list of mathoverflow questions that have the tag "Banach manifold".
answered Jan 1 at 17:09
Dave L. RenfroDave L. Renfro
25.4k34082
$endgroup$
$begingroup$
Could be. I guess a function can be expressed as an infinite series. So they would be infinite dimensional manifolds in some sense. I have heard the phrase before but I think it is beyond my understanding. From Wikipedia it seems that the classification of Banach manifolds is not so interesting compared to finite dimensional manifolds.
$endgroup$
– zooby
Jan 1 at 23:32
add a comment |
$begingroup$
I suppose a Banach manifold could be considered what you're looking for, when the underlying Banach space is a space of functions. Here is a list of mathoverflow questions that have the tag "Banach manifold".
answered Jan 1 at 17:09
Dave L. RenfroDave L. Renfro
25.4k34082
$endgroup$
$begingroup$
Could be. I guess a function can be expressed as an infinite series. So they would be infinite dimensional manifolds in some sense. I have heard the phrase before but I think it is beyond my understanding. From Wikipedia it seems that the classification of Banach manifolds is not so interesting compared to finite dimensional manifolds.
$endgroup$
– zooby
Jan 1 at 23:32
add a comment |
$begingroup$
I suppose a Banach manifold could be considered what you're looking for, when the underlying Banach space is a space of functions. Here is a list of mathoverflow questions that have the tag "Banach manifold".
answered Jan 1 at 17:09
Dave L. RenfroDave L. Renfro
25.4k34082
$endgroup$
I suppose a Banach manifold could be considered what you're looking for, when the underlying Banach space is a space of functions. Here is a list of mathoverflow questions that have the tag "Banach manifold".
answered Jan 1 at 17:09
Dave L. RenfroDave L. Renfro
25.4k34082
answered Jan 1 at 17:09
Dave L. RenfroDave L. Renfro
25.4k34082
answered Jan 1 at 17:09
Dave L. RenfroDave L. Renfro
25.4k34082
answered Jan 1 at 17:09
Dave L. RenfroDave L. Renfro
25.4k34082
25.4k34082
$begingroup$
Could be. I guess a function can be expressed as an infinite series. So they would be infinite dimensional manifolds in some sense. I have heard the phrase before but I think it is beyond my understanding. From Wikipedia it seems that the classification of Banach manifolds is not so interesting compared to finite dimensional manifolds.
$endgroup$
– zooby
Jan 1 at 23:32
add a comment |
$begingroup$
Could be. I guess a function can be expressed as an infinite series. So they would be infinite dimensional manifolds in some sense. I have heard the phrase before but I think it is beyond my understanding. From Wikipedia it seems that the classification of Banach manifolds is not so interesting compared to finite dimensional manifolds.
$endgroup$
– zooby
Jan 1 at 23:32
$begingroup$
Could be. I guess a function can be expressed as an infinite series. So they would be infinite dimensional manifolds in some sense. I have heard the phrase before but I think it is beyond my understanding. From Wikipedia it seems that the classification of Banach manifolds is not so interesting compared to finite dimensional manifolds.
$endgroup$
– zooby
Jan 1 at 23:32
$begingroup$
Could be. I guess a function can be expressed as an infinite series. So they would be infinite dimensional manifolds in some sense. I have heard the phrase before but I think it is beyond my understanding. From Wikipedia it seems that the classification of Banach manifolds is not so interesting compared to finite dimensional manifolds.
$endgroup$
– zooby
Jan 1 at 23:32
add a comment |
$begingroup$
I don't know if this answer will be satisfying to you, but there is some general topology context that might change how you look at this question.
Firstly, if you want to handle pairs of anything-with-topologies, you can use the product topology. This applies equally well to anything you have topologies on, and often gives nice answers. For example, if $mathbb R$ has the standard topology and the circle $S^1$ has its standard topology (induced from its embedding in the plane, say), then:
- The product topology on $mathbb Rtimesmathbb R$ is indeed the standard topology on $mathbb R^2$
- The product topology on $S^1timesmathbb R$ corresponds exactly to the standard topology on an infinite cylinder living in $mathbb R^3$.
- The product topology on $S^1times S^1$ corresponds exactly to the standard topology on a torus living in $mathbb R^3$.
So if we had a reasonable topology on a collection of functions, then we could get a reasonable topology on pairs of functions via this same product-topology construction.
Now the question is: what is a reasonable topology on a bunch of functions $f:Xto Y$? Well, suppose the functions are (some of) the continuous maps between two (possibly identical) topological spaces (maybe nice ones like metric spaces or something).
One candidate would be to treat the space of these functions just like (infinite if $X$ is infinite) products of the space $Y$, but this is ignoring the topology on $X$. Another common one which takes that into account is called the compact-open topology. For a number of reasons (one is mentioned at this MSE answer), the compact-open topology is usually nicer/more appropriate.
Since you used the tag functional analysis, if we're looking at the continuous (aka bounded) linear maps on a Hilbert space $H$, then there are lots of topologies in use. And this MO answer mentions that on norm-bounded subsets, the "strong operator topology" agrees with the compact-open topology I mentioned above.
answered Jan 3 at 2:21
Mark S.Mark S.
12.5k22772
$endgroup$
$begingroup$
The product topologies are maybe not the most interesting topologies. The genus-2 surface for example is not a product. I wonder if in function space there are these non-product topologies? Maybe not. Perhaps just like polytopes in n-dimensions they get a bit boring after 4 dimensions. The operator topologies are a bit over my head I think!
$endgroup$
– zooby
Jan 3 at 5:09
add a comment |
$begingroup$
I don't know if this answer will be satisfying to you, but there is some general topology context that might change how you look at this question.
Firstly, if you want to handle pairs of anything-with-topologies, you can use the product topology. This applies equally well to anything you have topologies on, and often gives nice answers. For example, if $mathbb R$ has the standard topology and the circle $S^1$ has its standard topology (induced from its embedding in the plane, say), then:
- The product topology on $mathbb Rtimesmathbb R$ is indeed the standard topology on $mathbb R^2$
- The product topology on $S^1timesmathbb R$ corresponds exactly to the standard topology on an infinite cylinder living in $mathbb R^3$.
- The product topology on $S^1times S^1$ corresponds exactly to the standard topology on a torus living in $mathbb R^3$.
So if we had a reasonable topology on a collection of functions, then we could get a reasonable topology on pairs of functions via this same product-topology construction.
Now the question is: what is a reasonable topology on a bunch of functions $f:Xto Y$? Well, suppose the functions are (some of) the continuous maps between two (possibly identical) topological spaces (maybe nice ones like metric spaces or something).
One candidate would be to treat the space of these functions just like (infinite if $X$ is infinite) products of the space $Y$, but this is ignoring the topology on $X$. Another common one which takes that into account is called the compact-open topology. For a number of reasons (one is mentioned at this MSE answer), the compact-open topology is usually nicer/more appropriate.
Since you used the tag functional analysis, if we're looking at the continuous (aka bounded) linear maps on a Hilbert space $H$, then there are lots of topologies in use. And this MO answer mentions that on norm-bounded subsets, the "strong operator topology" agrees with the compact-open topology I mentioned above.
answered Jan 3 at 2:21
Mark S.Mark S.
12.5k22772
$endgroup$
$begingroup$
The product topologies are maybe not the most interesting topologies. The genus-2 surface for example is not a product. I wonder if in function space there are these non-product topologies? Maybe not. Perhaps just like polytopes in n-dimensions they get a bit boring after 4 dimensions. The operator topologies are a bit over my head I think!
$endgroup$
– zooby
Jan 3 at 5:09
add a comment |
$begingroup$
I don't know if this answer will be satisfying to you, but there is some general topology context that might change how you look at this question.
Firstly, if you want to handle pairs of anything-with-topologies, you can use the product topology. This applies equally well to anything you have topologies on, and often gives nice answers. For example, if $mathbb R$ has the standard topology and the circle $S^1$ has its standard topology (induced from its embedding in the plane, say), then:
- The product topology on $mathbb Rtimesmathbb R$ is indeed the standard topology on $mathbb R^2$
- The product topology on $S^1timesmathbb R$ corresponds exactly to the standard topology on an infinite cylinder living in $mathbb R^3$.
- The product topology on $S^1times S^1$ corresponds exactly to the standard topology on a torus living in $mathbb R^3$.
So if we had a reasonable topology on a collection of functions, then we could get a reasonable topology on pairs of functions via this same product-topology construction.
Now the question is: what is a reasonable topology on a bunch of functions $f:Xto Y$? Well, suppose the functions are (some of) the continuous maps between two (possibly identical) topological spaces (maybe nice ones like metric spaces or something).
One candidate would be to treat the space of these functions just like (infinite if $X$ is infinite) products of the space $Y$, but this is ignoring the topology on $X$. Another common one which takes that into account is called the compact-open topology. For a number of reasons (one is mentioned at this MSE answer), the compact-open topology is usually nicer/more appropriate.
Since you used the tag functional analysis, if we're looking at the continuous (aka bounded) linear maps on a Hilbert space $H$, then there are lots of topologies in use. And this MO answer mentions that on norm-bounded subsets, the "strong operator topology" agrees with the compact-open topology I mentioned above.
answered Jan 3 at 2:21
Mark S.Mark S.
12.5k22772
$endgroup$
I don't know if this answer will be satisfying to you, but there is some general topology context that might change how you look at this question.
Firstly, if you want to handle pairs of anything-with-topologies, you can use the product topology. This applies equally well to anything you have topologies on, and often gives nice answers. For example, if $mathbb R$ has the standard topology and the circle $S^1$ has its standard topology (induced from its embedding in the plane, say), then:
- The product topology on $mathbb Rtimesmathbb R$ is indeed the standard topology on $mathbb R^2$
- The product topology on $S^1timesmathbb R$ corresponds exactly to the standard topology on an infinite cylinder living in $mathbb R^3$.
- The product topology on $S^1times S^1$ corresponds exactly to the standard topology on a torus living in $mathbb R^3$.
So if we had a reasonable topology on a collection of functions, then we could get a reasonable topology on pairs of functions via this same product-topology construction.
Now the question is: what is a reasonable topology on a bunch of functions $f:Xto Y$? Well, suppose the functions are (some of) the continuous maps between two (possibly identical) topological spaces (maybe nice ones like metric spaces or something).
One candidate would be to treat the space of these functions just like (infinite if $X$ is infinite) products of the space $Y$, but this is ignoring the topology on $X$. Another common one which takes that into account is called the compact-open topology. For a number of reasons (one is mentioned at this MSE answer), the compact-open topology is usually nicer/more appropriate.
Since you used the tag functional analysis, if we're looking at the continuous (aka bounded) linear maps on a Hilbert space $H$, then there are lots of topologies in use. And this MO answer mentions that on norm-bounded subsets, the "strong operator topology" agrees with the compact-open topology I mentioned above.
answered Jan 3 at 2:21
Mark S.Mark S.
12.5k22772
answered Jan 3 at 2:21
Mark S.Mark S.
12.5k22772
answered Jan 3 at 2:21
Mark S.Mark S.
12.5k22772
answered Jan 3 at 2:21
Mark S.Mark S.
12.5k22772
12.5k22772
$begingroup$
The product topologies are maybe not the most interesting topologies. The genus-2 surface for example is not a product. I wonder if in function space there are these non-product topologies? Maybe not. Perhaps just like polytopes in n-dimensions they get a bit boring after 4 dimensions. The operator topologies are a bit over my head I think!
$endgroup$
– zooby
Jan 3 at 5:09
add a comment |
$begingroup$
The product topologies are maybe not the most interesting topologies. The genus-2 surface for example is not a product. I wonder if in function space there are these non-product topologies? Maybe not. Perhaps just like polytopes in n-dimensions they get a bit boring after 4 dimensions. The operator topologies are a bit over my head I think!
$endgroup$
– zooby
Jan 3 at 5:09
$begingroup$
The product topologies are maybe not the most interesting topologies. The genus-2 surface for example is not a product. I wonder if in function space there are these non-product topologies? Maybe not. Perhaps just like polytopes in n-dimensions they get a bit boring after 4 dimensions. The operator topologies are a bit over my head I think!
$endgroup$
– zooby
Jan 3 at 5:09
$begingroup$
The product topologies are maybe not the most interesting topologies. The genus-2 surface for example is not a product. I wonder if in function space there are these non-product topologies? Maybe not. Perhaps just like polytopes in n-dimensions they get a bit boring after 4 dimensions. The operator topologies are a bit over my head I think!
$endgroup$
– zooby
Jan 3 at 5:09
add a comment |
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$begingroup$
Are you looking for a topology on a collection of functions, or something more specifically like a 2 dimensional surface (say, of a manifold). You can always take the product of two topological spaces (e.g. the product of two circles is a torus), so I'm trying to see if the "pair of functions" idea is crucial or incidental to what you're asking.
$endgroup$
– Mark S.
Jan 1 at 19:07
$begingroup$
@Mark For every point on a 2D surface one can associate a pair of real numbers such that near points have near numbers. I wondered if there was a functional version where points on the "surface" are associated with tuples of functions. This would have to be an infinite dimensional surface in normal language.
$endgroup$
– zooby
Jan 1 at 23:31