Set of continous functions writable as cartesian product
$begingroup$
let $(X,d_X)$ and $(mathbb{R}^n,d_2)$ be metric spaces with $X$ compact.
Then the set of continuous functions is defined by
$$
C_n(X):={f:Xrightarrow mathbb{R}^n ;|;f text{ continuous }}
$$
$left(C_n(X),d_{sup}right)$ is a complete and separable metric space
First question
(i): Is it necessary, that $X$ must be a metric space or is it also fine if $X$ is a compact topological space?
(ii) We can write $C_n(X)$ as the cartesian product $C_n(X)=C_1(X)times C_{n-1}(X)$. Is this really obvious or do we have to proof this? I know, if we have $C_1$ and $C_{n-1}$ than it is not obvious that we can write $C_1(X)times C_{n-1}(X)=C_n(X)$ because it depends on the properties of $C_1$ and $C_{n-1}$.
If it is clear, that we can write $C_n(X)$ as cartesian product, will $C_1$ and $C_{n-1}(X)$ be also complete and separable and why?
thanks
general-topology functional-analysis product-space
edited Dec 1 '18 at 11:08
Davide Giraudo
126k16150261
asked Dec 1 '18 at 9:55
GeoRieGeoRie
102
$endgroup$
add a comment |
$begingroup$
let $(X,d_X)$ and $(mathbb{R}^n,d_2)$ be metric spaces with $X$ compact.
Then the set of continuous functions is defined by
$$
C_n(X):={f:Xrightarrow mathbb{R}^n ;|;f text{ continuous }}
$$
$left(C_n(X),d_{sup}right)$ is a complete and separable metric space
First question
(i): Is it necessary, that $X$ must be a metric space or is it also fine if $X$ is a compact topological space?
(ii) We can write $C_n(X)$ as the cartesian product $C_n(X)=C_1(X)times C_{n-1}(X)$. Is this really obvious or do we have to proof this? I know, if we have $C_1$ and $C_{n-1}$ than it is not obvious that we can write $C_1(X)times C_{n-1}(X)=C_n(X)$ because it depends on the properties of $C_1$ and $C_{n-1}$.
If it is clear, that we can write $C_n(X)$ as cartesian product, will $C_1$ and $C_{n-1}(X)$ be also complete and separable and why?
thanks
general-topology functional-analysis product-space
edited Dec 1 '18 at 11:08
Davide Giraudo
126k16150261
asked Dec 1 '18 at 9:55
GeoRieGeoRie
102
$endgroup$
add a comment |
$begingroup$
let $(X,d_X)$ and $(mathbb{R}^n,d_2)$ be metric spaces with $X$ compact.
Then the set of continuous functions is defined by
$$
C_n(X):={f:Xrightarrow mathbb{R}^n ;|;f text{ continuous }}
$$
$left(C_n(X),d_{sup}right)$ is a complete and separable metric space
First question
(i): Is it necessary, that $X$ must be a metric space or is it also fine if $X$ is a compact topological space?
(ii) We can write $C_n(X)$ as the cartesian product $C_n(X)=C_1(X)times C_{n-1}(X)$. Is this really obvious or do we have to proof this? I know, if we have $C_1$ and $C_{n-1}$ than it is not obvious that we can write $C_1(X)times C_{n-1}(X)=C_n(X)$ because it depends on the properties of $C_1$ and $C_{n-1}$.
If it is clear, that we can write $C_n(X)$ as cartesian product, will $C_1$ and $C_{n-1}(X)$ be also complete and separable and why?
thanks
general-topology functional-analysis product-space
edited Dec 1 '18 at 11:08
Davide Giraudo
126k16150261
asked Dec 1 '18 at 9:55
GeoRieGeoRie
102
$endgroup$
let $(X,d_X)$ and $(mathbb{R}^n,d_2)$ be metric spaces with $X$ compact.
Then the set of continuous functions is defined by
$$
C_n(X):={f:Xrightarrow mathbb{R}^n ;|;f text{ continuous }}
$$
$left(C_n(X),d_{sup}right)$ is a complete and separable metric space
First question
(i): Is it necessary, that $X$ must be a metric space or is it also fine if $X$ is a compact topological space?
(ii) We can write $C_n(X)$ as the cartesian product $C_n(X)=C_1(X)times C_{n-1}(X)$. Is this really obvious or do we have to proof this? I know, if we have $C_1$ and $C_{n-1}$ than it is not obvious that we can write $C_1(X)times C_{n-1}(X)=C_n(X)$ because it depends on the properties of $C_1$ and $C_{n-1}$.
If it is clear, that we can write $C_n(X)$ as cartesian product, will $C_1$ and $C_{n-1}(X)$ be also complete and separable and why?
thanks
general-topology functional-analysis product-space
general-topology functional-analysis product-space
edited Dec 1 '18 at 11:08
Davide Giraudo
126k16150261
asked Dec 1 '18 at 9:55
GeoRieGeoRie
102
edited Dec 1 '18 at 11:08
Davide Giraudo
126k16150261
asked Dec 1 '18 at 9:55
GeoRieGeoRie
102
edited Dec 1 '18 at 11:08
Davide Giraudo
126k16150261
edited Dec 1 '18 at 11:08
Davide Giraudo
126k16150261
edited Dec 1 '18 at 11:08
Davide Giraudo
126k16150261
126k16150261
asked Dec 1 '18 at 9:55
GeoRieGeoRie
102
asked Dec 1 '18 at 9:55
GeoRieGeoRie
102
asked Dec 1 '18 at 9:55
GeoRieGeoRie
102
102
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
(i) For $C_n(X)$ to be separable metric, $X$ has to be compact metrisable (e.g. see the references in this post. The result is true for both functions to the reals and finite-dimensional Euclidean space.
(ii) A function $f: X to mathbb{R}^n$ is continuous iff all
$pi_i circ f: X to mathbb{R}$ are continuous. So mapping $f in C_n(X)$ to $(pi_1circ f,ldots, pi_n circ f) in C_1(X)^n$ is a linear isomomorphism between the spaces, and an isometry (I think) if we give $mathbb{R}^n$ and the product $C_1(X)^n$ the max-metric (and base the $d_{text{sup}}$ on the max-metric, which is topologically the same).
It's not a long fact to prove, but it does require a small argument such as I gave above, IMHO.
A finite product of completely metrisable separable spaces is still of that type.
answered Dec 1 '18 at 10:43
Henno BrandsmaHenno Brandsma
106k347114
$endgroup$
$begingroup$
@GeoRie And $C_n(X) = C_1(X)^n$ and $C_1(X)$ is completely metrisable complete, hence so is the power. The proofs for the product are standard: the standard product metric is complete when all factors are complete and separability is obvious: a product of dense sets is dense.
$endgroup$
– Henno Brandsma
Dec 1 '18 at 15:00
$begingroup$
@GeoRie $C_n(X) simeq C_1(X)^n$ is always true. From that: $C_n(X)$ is completely metrisable separable (Polish) iff $C_1(X)$ is iff $X$ is compact metrisable (as in the reference I linked to).
$endgroup$
– Henno Brandsma
Dec 1 '18 at 15:07
$begingroup$
Thank you for answering. (i) is only for continuous functions of that kind ${f:Xrightarrow mathbb{R}|; f ; continuous}$. So you say, that a finite product of completely metrisable separable spaces is still of that type. So we can write $C_n(X)$ as cartesian product $C_n(X)=C_1(X)^n$ isn't it? Do you know a book where I can find the proof, that the finite product of completely metrisable separable spaces is still of that type?
$endgroup$
– GeoRie
Dec 1 '18 at 15:07
$begingroup$
,do you know a book, where I can read that $C_n(X)$ is isomorph to $C_1(X)^n$? Thank you
$endgroup$
– GeoRie
Dec 1 '18 at 15:12
$begingroup$
@GeoRie I gave the argument in the posting...
$endgroup$
– Henno Brandsma
Dec 1 '18 at 15:14
add a comment |
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$begingroup$
(i) For $C_n(X)$ to be separable metric, $X$ has to be compact metrisable (e.g. see the references in this post. The result is true for both functions to the reals and finite-dimensional Euclidean space.
(ii) A function $f: X to mathbb{R}^n$ is continuous iff all
$pi_i circ f: X to mathbb{R}$ are continuous. So mapping $f in C_n(X)$ to $(pi_1circ f,ldots, pi_n circ f) in C_1(X)^n$ is a linear isomomorphism between the spaces, and an isometry (I think) if we give $mathbb{R}^n$ and the product $C_1(X)^n$ the max-metric (and base the $d_{text{sup}}$ on the max-metric, which is topologically the same).
It's not a long fact to prove, but it does require a small argument such as I gave above, IMHO.
A finite product of completely metrisable separable spaces is still of that type.
answered Dec 1 '18 at 10:43
Henno BrandsmaHenno Brandsma
106k347114
$endgroup$
$begingroup$
@GeoRie And $C_n(X) = C_1(X)^n$ and $C_1(X)$ is completely metrisable complete, hence so is the power. The proofs for the product are standard: the standard product metric is complete when all factors are complete and separability is obvious: a product of dense sets is dense.
$endgroup$
– Henno Brandsma
Dec 1 '18 at 15:00
$begingroup$
@GeoRie $C_n(X) simeq C_1(X)^n$ is always true. From that: $C_n(X)$ is completely metrisable separable (Polish) iff $C_1(X)$ is iff $X$ is compact metrisable (as in the reference I linked to).
$endgroup$
– Henno Brandsma
Dec 1 '18 at 15:07
$begingroup$
Thank you for answering. (i) is only for continuous functions of that kind ${f:Xrightarrow mathbb{R}|; f ; continuous}$. So you say, that a finite product of completely metrisable separable spaces is still of that type. So we can write $C_n(X)$ as cartesian product $C_n(X)=C_1(X)^n$ isn't it? Do you know a book where I can find the proof, that the finite product of completely metrisable separable spaces is still of that type?
$endgroup$
– GeoRie
Dec 1 '18 at 15:07
$begingroup$
,do you know a book, where I can read that $C_n(X)$ is isomorph to $C_1(X)^n$? Thank you
$endgroup$
– GeoRie
Dec 1 '18 at 15:12
$begingroup$
@GeoRie I gave the argument in the posting...
$endgroup$
– Henno Brandsma
Dec 1 '18 at 15:14
add a comment |
$begingroup$
(i) For $C_n(X)$ to be separable metric, $X$ has to be compact metrisable (e.g. see the references in this post. The result is true for both functions to the reals and finite-dimensional Euclidean space.
(ii) A function $f: X to mathbb{R}^n$ is continuous iff all
$pi_i circ f: X to mathbb{R}$ are continuous. So mapping $f in C_n(X)$ to $(pi_1circ f,ldots, pi_n circ f) in C_1(X)^n$ is a linear isomomorphism between the spaces, and an isometry (I think) if we give $mathbb{R}^n$ and the product $C_1(X)^n$ the max-metric (and base the $d_{text{sup}}$ on the max-metric, which is topologically the same).
It's not a long fact to prove, but it does require a small argument such as I gave above, IMHO.
A finite product of completely metrisable separable spaces is still of that type.
answered Dec 1 '18 at 10:43
Henno BrandsmaHenno Brandsma
106k347114
$endgroup$
$begingroup$
@GeoRie And $C_n(X) = C_1(X)^n$ and $C_1(X)$ is completely metrisable complete, hence so is the power. The proofs for the product are standard: the standard product metric is complete when all factors are complete and separability is obvious: a product of dense sets is dense.
$endgroup$
– Henno Brandsma
Dec 1 '18 at 15:00
$begingroup$
@GeoRie $C_n(X) simeq C_1(X)^n$ is always true. From that: $C_n(X)$ is completely metrisable separable (Polish) iff $C_1(X)$ is iff $X$ is compact metrisable (as in the reference I linked to).
$endgroup$
– Henno Brandsma
Dec 1 '18 at 15:07
$begingroup$
Thank you for answering. (i) is only for continuous functions of that kind ${f:Xrightarrow mathbb{R}|; f ; continuous}$. So you say, that a finite product of completely metrisable separable spaces is still of that type. So we can write $C_n(X)$ as cartesian product $C_n(X)=C_1(X)^n$ isn't it? Do you know a book where I can find the proof, that the finite product of completely metrisable separable spaces is still of that type?
$endgroup$
– GeoRie
Dec 1 '18 at 15:07
$begingroup$
,do you know a book, where I can read that $C_n(X)$ is isomorph to $C_1(X)^n$? Thank you
$endgroup$
– GeoRie
Dec 1 '18 at 15:12
$begingroup$
@GeoRie I gave the argument in the posting...
$endgroup$
– Henno Brandsma
Dec 1 '18 at 15:14
add a comment |
$begingroup$
(i) For $C_n(X)$ to be separable metric, $X$ has to be compact metrisable (e.g. see the references in this post. The result is true for both functions to the reals and finite-dimensional Euclidean space.
(ii) A function $f: X to mathbb{R}^n$ is continuous iff all
$pi_i circ f: X to mathbb{R}$ are continuous. So mapping $f in C_n(X)$ to $(pi_1circ f,ldots, pi_n circ f) in C_1(X)^n$ is a linear isomomorphism between the spaces, and an isometry (I think) if we give $mathbb{R}^n$ and the product $C_1(X)^n$ the max-metric (and base the $d_{text{sup}}$ on the max-metric, which is topologically the same).
It's not a long fact to prove, but it does require a small argument such as I gave above, IMHO.
A finite product of completely metrisable separable spaces is still of that type.
answered Dec 1 '18 at 10:43
Henno BrandsmaHenno Brandsma
106k347114
$endgroup$
(i) For $C_n(X)$ to be separable metric, $X$ has to be compact metrisable (e.g. see the references in this post. The result is true for both functions to the reals and finite-dimensional Euclidean space.
(ii) A function $f: X to mathbb{R}^n$ is continuous iff all
$pi_i circ f: X to mathbb{R}$ are continuous. So mapping $f in C_n(X)$ to $(pi_1circ f,ldots, pi_n circ f) in C_1(X)^n$ is a linear isomomorphism between the spaces, and an isometry (I think) if we give $mathbb{R}^n$ and the product $C_1(X)^n$ the max-metric (and base the $d_{text{sup}}$ on the max-metric, which is topologically the same).
It's not a long fact to prove, but it does require a small argument such as I gave above, IMHO.
A finite product of completely metrisable separable spaces is still of that type.
answered Dec 1 '18 at 10:43
Henno BrandsmaHenno Brandsma
106k347114
edited Dec 1 '18 at 12:25
answered Dec 1 '18 at 10:43
Henno BrandsmaHenno Brandsma
106k347114
answered Dec 1 '18 at 10:43
Henno BrandsmaHenno Brandsma
106k347114
answered Dec 1 '18 at 10:43
Henno BrandsmaHenno Brandsma
106k347114
106k347114
$begingroup$
@GeoRie And $C_n(X) = C_1(X)^n$ and $C_1(X)$ is completely metrisable complete, hence so is the power. The proofs for the product are standard: the standard product metric is complete when all factors are complete and separability is obvious: a product of dense sets is dense.
$endgroup$
– Henno Brandsma
Dec 1 '18 at 15:00
$begingroup$
@GeoRie $C_n(X) simeq C_1(X)^n$ is always true. From that: $C_n(X)$ is completely metrisable separable (Polish) iff $C_1(X)$ is iff $X$ is compact metrisable (as in the reference I linked to).
$endgroup$
– Henno Brandsma
Dec 1 '18 at 15:07
$begingroup$
Thank you for answering. (i) is only for continuous functions of that kind ${f:Xrightarrow mathbb{R}|; f ; continuous}$. So you say, that a finite product of completely metrisable separable spaces is still of that type. So we can write $C_n(X)$ as cartesian product $C_n(X)=C_1(X)^n$ isn't it? Do you know a book where I can find the proof, that the finite product of completely metrisable separable spaces is still of that type?
$endgroup$
– GeoRie
Dec 1 '18 at 15:07
$begingroup$
,do you know a book, where I can read that $C_n(X)$ is isomorph to $C_1(X)^n$? Thank you
$endgroup$
– GeoRie
Dec 1 '18 at 15:12
$begingroup$
@GeoRie I gave the argument in the posting...
$endgroup$
– Henno Brandsma
Dec 1 '18 at 15:14
add a comment |
$begingroup$
@GeoRie And $C_n(X) = C_1(X)^n$ and $C_1(X)$ is completely metrisable complete, hence so is the power. The proofs for the product are standard: the standard product metric is complete when all factors are complete and separability is obvious: a product of dense sets is dense.
$endgroup$
– Henno Brandsma
Dec 1 '18 at 15:00
$begingroup$
@GeoRie $C_n(X) simeq C_1(X)^n$ is always true. From that: $C_n(X)$ is completely metrisable separable (Polish) iff $C_1(X)$ is iff $X$ is compact metrisable (as in the reference I linked to).
$endgroup$
– Henno Brandsma
Dec 1 '18 at 15:07
$begingroup$
Thank you for answering. (i) is only for continuous functions of that kind ${f:Xrightarrow mathbb{R}|; f ; continuous}$. So you say, that a finite product of completely metrisable separable spaces is still of that type. So we can write $C_n(X)$ as cartesian product $C_n(X)=C_1(X)^n$ isn't it? Do you know a book where I can find the proof, that the finite product of completely metrisable separable spaces is still of that type?
$endgroup$
– GeoRie
Dec 1 '18 at 15:07
$begingroup$
,do you know a book, where I can read that $C_n(X)$ is isomorph to $C_1(X)^n$? Thank you
$endgroup$
– GeoRie
Dec 1 '18 at 15:12
$begingroup$
@GeoRie I gave the argument in the posting...
$endgroup$
– Henno Brandsma
Dec 1 '18 at 15:14
$begingroup$
@GeoRie And $C_n(X) = C_1(X)^n$ and $C_1(X)$ is completely metrisable complete, hence so is the power. The proofs for the product are standard: the standard product metric is complete when all factors are complete and separability is obvious: a product of dense sets is dense.
$endgroup$
– Henno Brandsma
Dec 1 '18 at 15:00
$begingroup$
@GeoRie And $C_n(X) = C_1(X)^n$ and $C_1(X)$ is completely metrisable complete, hence so is the power. The proofs for the product are standard: the standard product metric is complete when all factors are complete and separability is obvious: a product of dense sets is dense.
$endgroup$
– Henno Brandsma
Dec 1 '18 at 15:00
$begingroup$
@GeoRie $C_n(X) simeq C_1(X)^n$ is always true. From that: $C_n(X)$ is completely metrisable separable (Polish) iff $C_1(X)$ is iff $X$ is compact metrisable (as in the reference I linked to).
$endgroup$
– Henno Brandsma
Dec 1 '18 at 15:07
$begingroup$
@GeoRie $C_n(X) simeq C_1(X)^n$ is always true. From that: $C_n(X)$ is completely metrisable separable (Polish) iff $C_1(X)$ is iff $X$ is compact metrisable (as in the reference I linked to).
$endgroup$
– Henno Brandsma
Dec 1 '18 at 15:07
$begingroup$
Thank you for answering. (i) is only for continuous functions of that kind ${f:Xrightarrow mathbb{R}|; f ; continuous}$. So you say, that a finite product of completely metrisable separable spaces is still of that type. So we can write $C_n(X)$ as cartesian product $C_n(X)=C_1(X)^n$ isn't it? Do you know a book where I can find the proof, that the finite product of completely metrisable separable spaces is still of that type?
$endgroup$
– GeoRie
Dec 1 '18 at 15:07
$begingroup$
Thank you for answering. (i) is only for continuous functions of that kind ${f:Xrightarrow mathbb{R}|; f ; continuous}$. So you say, that a finite product of completely metrisable separable spaces is still of that type. So we can write $C_n(X)$ as cartesian product $C_n(X)=C_1(X)^n$ isn't it? Do you know a book where I can find the proof, that the finite product of completely metrisable separable spaces is still of that type?
$endgroup$
– GeoRie
Dec 1 '18 at 15:07
$begingroup$
,do you know a book, where I can read that $C_n(X)$ is isomorph to $C_1(X)^n$? Thank you
$endgroup$
– GeoRie
Dec 1 '18 at 15:12
$begingroup$
,do you know a book, where I can read that $C_n(X)$ is isomorph to $C_1(X)^n$? Thank you
$endgroup$
– GeoRie
Dec 1 '18 at 15:12
$begingroup$
@GeoRie I gave the argument in the posting...
$endgroup$
– Henno Brandsma
Dec 1 '18 at 15:14
$begingroup$
@GeoRie I gave the argument in the posting...
$endgroup$
– Henno Brandsma
Dec 1 '18 at 15:14
add a comment |
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