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Separability of $C(K,K)$
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Assume that $K$ is a compact metric space. I know that using the Stone-Weierstrass theorem it can be proven that $C(K)$, the Banach space of all continuous functions defined on $K$, is separable.
However, I don't know how to prove that the space $space C(K,K):=f:K to K; space f text is continuous $ endowed with the supremum metric is separable. Since $K$ doesn't have a linear structure, it doesn't even make sense to talk about density of polynomials or anything like that.
I will be grateful for any help.
general-topology functional-analysis metric-spaces
asked Jul 22 at 9:00
Jenda358
1077
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up vote
2
down vote
favorite
Assume that $K$ is a compact metric space. I know that using the Stone-Weierstrass theorem it can be proven that $C(K)$, the Banach space of all continuous functions defined on $K$, is separable.
However, I don't know how to prove that the space $space C(K,K):=f:K to K; space f text is continuous $ endowed with the supremum metric is separable. Since $K$ doesn't have a linear structure, it doesn't even make sense to talk about density of polynomials or anything like that.
I will be grateful for any help.
general-topology functional-analysis metric-spaces
asked Jul 22 at 9:00
Jenda358
1077
1
The norm on $C(K, K)$ you suggest also doesn't make sense, you cannot take $sup_x in K|f(x)|$.
– mechanodroid
Jul 22 at 11:59
I agree that such a norm wouldn't make sense, but note that I mentioned just a supremum metric, i.e. the metric given by $ d(f,g)=mathrmsup rho(f(x),g(x)) ; space x in K $ where $rho$ is the metric on $K$. $C(K,K)$ is just a metric space for me.
– Jenda358
Jul 22 at 14:48
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Assume that $K$ is a compact metric space. I know that using the Stone-Weierstrass theorem it can be proven that $C(K)$, the Banach space of all continuous functions defined on $K$, is separable.
However, I don't know how to prove that the space $space C(K,K):=f:K to K; space f text is continuous $ endowed with the supremum metric is separable. Since $K$ doesn't have a linear structure, it doesn't even make sense to talk about density of polynomials or anything like that.
I will be grateful for any help.
general-topology functional-analysis metric-spaces
asked Jul 22 at 9:00
Jenda358
1077
Assume that $K$ is a compact metric space. I know that using the Stone-Weierstrass theorem it can be proven that $C(K)$, the Banach space of all continuous functions defined on $K$, is separable.
However, I don't know how to prove that the space $space C(K,K):=f:K to K; space f text is continuous $ endowed with the supremum metric is separable. Since $K$ doesn't have a linear structure, it doesn't even make sense to talk about density of polynomials or anything like that.
I will be grateful for any help.
general-topology functional-analysis metric-spaces
asked Jul 22 at 9:00
Jenda358
1077
asked Jul 22 at 9:00
Jenda358
1077
asked Jul 22 at 9:00
Jenda358
1077
asked Jul 22 at 9:00
Jenda358
1077
1077
1
The norm on $C(K, K)$ you suggest also doesn't make sense, you cannot take $sup_x in K|f(x)|$.
– mechanodroid
Jul 22 at 11:59
I agree that such a norm wouldn't make sense, but note that I mentioned just a supremum metric, i.e. the metric given by $ d(f,g)=mathrmsup rho(f(x),g(x)) ; space x in K $ where $rho$ is the metric on $K$. $C(K,K)$ is just a metric space for me.
– Jenda358
Jul 22 at 14:48
add a comment |Â
1
The norm on $C(K, K)$ you suggest also doesn't make sense, you cannot take $sup_x in K|f(x)|$.
– mechanodroid
Jul 22 at 11:59
I agree that such a norm wouldn't make sense, but note that I mentioned just a supremum metric, i.e. the metric given by $ d(f,g)=mathrmsup rho(f(x),g(x)) ; space x in K $ where $rho$ is the metric on $K$. $C(K,K)$ is just a metric space for me.
– Jenda358
Jul 22 at 14:48
1
1
The norm on $C(K, K)$ you suggest also doesn't make sense, you cannot take $sup_x in K|f(x)|$.
– mechanodroid
Jul 22 at 11:59
The norm on $C(K, K)$ you suggest also doesn't make sense, you cannot take $sup_x in K|f(x)|$.
– mechanodroid
Jul 22 at 11:59
I agree that such a norm wouldn't make sense, but note that I mentioned just a supremum metric, i.e. the metric given by $ d(f,g)=mathrmsup rho(f(x),g(x)) ; space x in K $ where $rho$ is the metric on $K$. $C(K,K)$ is just a metric space for me.
– Jenda358
Jul 22 at 14:48
I agree that such a norm wouldn't make sense, but note that I mentioned just a supremum metric, i.e. the metric given by $ d(f,g)=mathrmsup rho(f(x),g(x)) ; space x in K $ where $rho$ is the metric on $K$. $C(K,K)$ is just a metric space for me.
– Jenda358
Jul 22 at 14:48
add a comment |Â
1 Answer
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For any space $X$ and any metric space $(Y,d)$ we obtain a metric $d^+$ on the set $C(X,Y;d)$ of continuous $d$-bounded functions $f : X to Y$ by defining
$$d^+(f,g) = sup d(f(x),g(x)) mid x in X .$$
If $X$ is compact, then as a set $C(X,Y;d) = Y^X$ = set of all continuous functions $f : X to Y$ and it is well-known that the topology generated by $d^+$ is nothing else than the compact open topology on $Y^X$.
Now let $X,Y$ be compact metric spaces. We claim that $Y^X$ is separable.
It is well-known that $Y$ can be embedded into the space $s = Pi_i=1^infty mathbbR_i$ where each $mathbbR_i$ is a copy of the real line. Let $e : X to s$ be such an embedding. Then $e_ast : Y^X to s^X, e_ast(f) = e circ f$ is an embedding. It is well-known that the spaces $s^X = (Pi_i=1^infty mathbbR_i)^X$ and $Pi_i=1^infty (mathbbR_i^X)$ are homeomorphic. But each $mathbbR_i^X approx mathbbR^X = C(X)$ is separable, hence the countable infinite product $s^X$ is separable and the subspace $e_ast(Y^X)$ is separable.
answered Jul 22 at 16:09
Paul Frost
3,658420
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
For any space $X$ and any metric space $(Y,d)$ we obtain a metric $d^+$ on the set $C(X,Y;d)$ of continuous $d$-bounded functions $f : X to Y$ by defining
$$d^+(f,g) = sup d(f(x),g(x)) mid x in X .$$
If $X$ is compact, then as a set $C(X,Y;d) = Y^X$ = set of all continuous functions $f : X to Y$ and it is well-known that the topology generated by $d^+$ is nothing else than the compact open topology on $Y^X$.
Now let $X,Y$ be compact metric spaces. We claim that $Y^X$ is separable.
It is well-known that $Y$ can be embedded into the space $s = Pi_i=1^infty mathbbR_i$ where each $mathbbR_i$ is a copy of the real line. Let $e : X to s$ be such an embedding. Then $e_ast : Y^X to s^X, e_ast(f) = e circ f$ is an embedding. It is well-known that the spaces $s^X = (Pi_i=1^infty mathbbR_i)^X$ and $Pi_i=1^infty (mathbbR_i^X)$ are homeomorphic. But each $mathbbR_i^X approx mathbbR^X = C(X)$ is separable, hence the countable infinite product $s^X$ is separable and the subspace $e_ast(Y^X)$ is separable.
answered Jul 22 at 16:09
Paul Frost
3,658420
add a comment |Â
up vote
1
down vote
accepted
For any space $X$ and any metric space $(Y,d)$ we obtain a metric $d^+$ on the set $C(X,Y;d)$ of continuous $d$-bounded functions $f : X to Y$ by defining
$$d^+(f,g) = sup d(f(x),g(x)) mid x in X .$$
If $X$ is compact, then as a set $C(X,Y;d) = Y^X$ = set of all continuous functions $f : X to Y$ and it is well-known that the topology generated by $d^+$ is nothing else than the compact open topology on $Y^X$.
Now let $X,Y$ be compact metric spaces. We claim that $Y^X$ is separable.
It is well-known that $Y$ can be embedded into the space $s = Pi_i=1^infty mathbbR_i$ where each $mathbbR_i$ is a copy of the real line. Let $e : X to s$ be such an embedding. Then $e_ast : Y^X to s^X, e_ast(f) = e circ f$ is an embedding. It is well-known that the spaces $s^X = (Pi_i=1^infty mathbbR_i)^X$ and $Pi_i=1^infty (mathbbR_i^X)$ are homeomorphic. But each $mathbbR_i^X approx mathbbR^X = C(X)$ is separable, hence the countable infinite product $s^X$ is separable and the subspace $e_ast(Y^X)$ is separable.
answered Jul 22 at 16:09
Paul Frost
3,658420
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
For any space $X$ and any metric space $(Y,d)$ we obtain a metric $d^+$ on the set $C(X,Y;d)$ of continuous $d$-bounded functions $f : X to Y$ by defining
$$d^+(f,g) = sup d(f(x),g(x)) mid x in X .$$
If $X$ is compact, then as a set $C(X,Y;d) = Y^X$ = set of all continuous functions $f : X to Y$ and it is well-known that the topology generated by $d^+$ is nothing else than the compact open topology on $Y^X$.
Now let $X,Y$ be compact metric spaces. We claim that $Y^X$ is separable.
It is well-known that $Y$ can be embedded into the space $s = Pi_i=1^infty mathbbR_i$ where each $mathbbR_i$ is a copy of the real line. Let $e : X to s$ be such an embedding. Then $e_ast : Y^X to s^X, e_ast(f) = e circ f$ is an embedding. It is well-known that the spaces $s^X = (Pi_i=1^infty mathbbR_i)^X$ and $Pi_i=1^infty (mathbbR_i^X)$ are homeomorphic. But each $mathbbR_i^X approx mathbbR^X = C(X)$ is separable, hence the countable infinite product $s^X$ is separable and the subspace $e_ast(Y^X)$ is separable.
answered Jul 22 at 16:09
Paul Frost
3,658420
For any space $X$ and any metric space $(Y,d)$ we obtain a metric $d^+$ on the set $C(X,Y;d)$ of continuous $d$-bounded functions $f : X to Y$ by defining
$$d^+(f,g) = sup d(f(x),g(x)) mid x in X .$$
If $X$ is compact, then as a set $C(X,Y;d) = Y^X$ = set of all continuous functions $f : X to Y$ and it is well-known that the topology generated by $d^+$ is nothing else than the compact open topology on $Y^X$.
Now let $X,Y$ be compact metric spaces. We claim that $Y^X$ is separable.
It is well-known that $Y$ can be embedded into the space $s = Pi_i=1^infty mathbbR_i$ where each $mathbbR_i$ is a copy of the real line. Let $e : X to s$ be such an embedding. Then $e_ast : Y^X to s^X, e_ast(f) = e circ f$ is an embedding. It is well-known that the spaces $s^X = (Pi_i=1^infty mathbbR_i)^X$ and $Pi_i=1^infty (mathbbR_i^X)$ are homeomorphic. But each $mathbbR_i^X approx mathbbR^X = C(X)$ is separable, hence the countable infinite product $s^X$ is separable and the subspace $e_ast(Y^X)$ is separable.
answered Jul 22 at 16:09
Paul Frost
3,658420
answered Jul 22 at 16:09
Paul Frost
3,658420
answered Jul 22 at 16:09
Paul Frost
3,658420
answered Jul 22 at 16:09
Paul Frost
3,658420
3,658420
add a comment |Â
add a comment |Â
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1
The norm on $C(K, K)$ you suggest also doesn't make sense, you cannot take $sup_x in K|f(x)|$.
– mechanodroid
Jul 22 at 11:59
I agree that such a norm wouldn't make sense, but note that I mentioned just a supremum metric, i.e. the metric given by $ d(f,g)=mathrmsup rho(f(x),g(x)) ; space x in K $ where $rho$ is the metric on $K$. $C(K,K)$ is just a metric space for me.
– Jenda358
Jul 22 at 14:48