$C(X)$ is separable when $X$ is compact?
Clash Royale CLAN TAG#URR8PPP
up vote
2
down vote
favorite
$X$ is a compact metric space, $C(X)$ is separable when $X$ is compact
where $C(X)$ denotes the space of continuous functions on $X$.
How to prove it?
And if $X$ is just a compact Hausdorff space, then $C(X)$ is still separable?
Or if $X$ is just a compact (not necessarily Hausdorff) space, then $C(X)$ is still separable?
Please help me. Thanks in advance.
functional-analysis
edited Jun 19 '15 at 10:34
egreg
165k1180187
asked Jun 19 '15 at 10:30
David Chan
772410
add a comment |Â
up vote
2
down vote
favorite
$X$ is a compact metric space, $C(X)$ is separable when $X$ is compact
where $C(X)$ denotes the space of continuous functions on $X$.
How to prove it?
And if $X$ is just a compact Hausdorff space, then $C(X)$ is still separable?
Or if $X$ is just a compact (not necessarily Hausdorff) space, then $C(X)$ is still separable?
Please help me. Thanks in advance.
functional-analysis
edited Jun 19 '15 at 10:34
egreg
165k1180187
asked Jun 19 '15 at 10:30
David Chan
772410
1
this result is not trivial: If X is a compact $T_2$ space $X$, then $C(X)$ is separable iff there is a metric $Xtimes Xrightarrow R$ that induces the topology of $X$. You need to use the Stone-Weierstrass Thm, Urysohn Lemma and the Urysohn Metrization Thm.
– Driver 8
Jun 19 '15 at 13:21
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
$X$ is a compact metric space, $C(X)$ is separable when $X$ is compact
where $C(X)$ denotes the space of continuous functions on $X$.
How to prove it?
And if $X$ is just a compact Hausdorff space, then $C(X)$ is still separable?
Or if $X$ is just a compact (not necessarily Hausdorff) space, then $C(X)$ is still separable?
Please help me. Thanks in advance.
functional-analysis
edited Jun 19 '15 at 10:34
egreg
165k1180187
asked Jun 19 '15 at 10:30
David Chan
772410
$X$ is a compact metric space, $C(X)$ is separable when $X$ is compact
where $C(X)$ denotes the space of continuous functions on $X$.
How to prove it?
And if $X$ is just a compact Hausdorff space, then $C(X)$ is still separable?
Or if $X$ is just a compact (not necessarily Hausdorff) space, then $C(X)$ is still separable?
Please help me. Thanks in advance.
functional-analysis
edited Jun 19 '15 at 10:34
egreg
165k1180187
asked Jun 19 '15 at 10:30
David Chan
772410
edited Jun 19 '15 at 10:34
egreg
165k1180187
edited Jun 19 '15 at 10:34
egreg
165k1180187
edited Jun 19 '15 at 10:34
egreg
165k1180187
165k1180187
asked Jun 19 '15 at 10:30
David Chan
772410
asked Jun 19 '15 at 10:30
David Chan
772410
asked Jun 19 '15 at 10:30
David Chan
772410
772410
1
this result is not trivial: If X is a compact $T_2$ space $X$, then $C(X)$ is separable iff there is a metric $Xtimes Xrightarrow R$ that induces the topology of $X$. You need to use the Stone-Weierstrass Thm, Urysohn Lemma and the Urysohn Metrization Thm.
– Driver 8
Jun 19 '15 at 13:21
add a comment |Â
1
this result is not trivial: If X is a compact $T_2$ space $X$, then $C(X)$ is separable iff there is a metric $Xtimes Xrightarrow R$ that induces the topology of $X$. You need to use the Stone-Weierstrass Thm, Urysohn Lemma and the Urysohn Metrization Thm.
– Driver 8
Jun 19 '15 at 13:21
1
1
this result is not trivial: If X is a compact $T_2$ space $X$, then $C(X)$ is separable iff there is a metric $Xtimes Xrightarrow R$ that induces the topology of $X$. You need to use the Stone-Weierstrass Thm, Urysohn Lemma and the Urysohn Metrization Thm.
– Driver 8
Jun 19 '15 at 13:21
this result is not trivial: If X is a compact $T_2$ space $X$, then $C(X)$ is separable iff there is a metric $Xtimes Xrightarrow R$ that induces the topology of $X$. You need to use the Stone-Weierstrass Thm, Urysohn Lemma and the Urysohn Metrization Thm.
– Driver 8
Jun 19 '15 at 13:21
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
5
down vote
Theorem. If $X$ is compact Hausdorff then $C(X)$ is separable iff $X$ is metrizable.
There is a natural embedding $xin Xto delta _xin mathcalM(X)$ (more precisely in the unit ball of $mathcalM(X)$). This is an homeomorphism for the weak*-topology of $mathcalM(X)$. If $C(X)$ is separable then $(mathcalM(X), w*)$ have a compact metrizable unit ball. So $X$ is compact.
For the converse, assume $X$ is a metrizable compact Hausdorff space. Let $d$ be a metric inducing the topology and $(x_n)$ a dense countable subset of $X$. Define $d_n: xin X to d_n(x):=d(x,x_n)$. It is a continuous function. It is easy to check that the algebra generated by $1$ and $(d_n)_n$ separate the points in $X$ so by Stone Weierstrass theorem this subalgebra is dense in $C(X)$. By considering linear combination with rational coefficient of element of this subalgebra it is easy to see that $C(X)$ is separable.
edited Jan 29 at 16:45
user152715
3,60423553
answered Jun 19 '15 at 11:14
Patissot
1,105615
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
Theorem. If $X$ is compact Hausdorff then $C(X)$ is separable iff $X$ is metrizable.
There is a natural embedding $xin Xto delta _xin mathcalM(X)$ (more precisely in the unit ball of $mathcalM(X)$). This is an homeomorphism for the weak*-topology of $mathcalM(X)$. If $C(X)$ is separable then $(mathcalM(X), w*)$ have a compact metrizable unit ball. So $X$ is compact.
For the converse, assume $X$ is a metrizable compact Hausdorff space. Let $d$ be a metric inducing the topology and $(x_n)$ a dense countable subset of $X$. Define $d_n: xin X to d_n(x):=d(x,x_n)$. It is a continuous function. It is easy to check that the algebra generated by $1$ and $(d_n)_n$ separate the points in $X$ so by Stone Weierstrass theorem this subalgebra is dense in $C(X)$. By considering linear combination with rational coefficient of element of this subalgebra it is easy to see that $C(X)$ is separable.
edited Jan 29 at 16:45
user152715
3,60423553
answered Jun 19 '15 at 11:14
Patissot
1,105615
add a comment |Â
up vote
5
down vote
Theorem. If $X$ is compact Hausdorff then $C(X)$ is separable iff $X$ is metrizable.
There is a natural embedding $xin Xto delta _xin mathcalM(X)$ (more precisely in the unit ball of $mathcalM(X)$). This is an homeomorphism for the weak*-topology of $mathcalM(X)$. If $C(X)$ is separable then $(mathcalM(X), w*)$ have a compact metrizable unit ball. So $X$ is compact.
For the converse, assume $X$ is a metrizable compact Hausdorff space. Let $d$ be a metric inducing the topology and $(x_n)$ a dense countable subset of $X$. Define $d_n: xin X to d_n(x):=d(x,x_n)$. It is a continuous function. It is easy to check that the algebra generated by $1$ and $(d_n)_n$ separate the points in $X$ so by Stone Weierstrass theorem this subalgebra is dense in $C(X)$. By considering linear combination with rational coefficient of element of this subalgebra it is easy to see that $C(X)$ is separable.
edited Jan 29 at 16:45
user152715
3,60423553
answered Jun 19 '15 at 11:14
Patissot
1,105615
add a comment |Â
up vote
5
down vote
up vote
5
down vote
Theorem. If $X$ is compact Hausdorff then $C(X)$ is separable iff $X$ is metrizable.
There is a natural embedding $xin Xto delta _xin mathcalM(X)$ (more precisely in the unit ball of $mathcalM(X)$). This is an homeomorphism for the weak*-topology of $mathcalM(X)$. If $C(X)$ is separable then $(mathcalM(X), w*)$ have a compact metrizable unit ball. So $X$ is compact.
For the converse, assume $X$ is a metrizable compact Hausdorff space. Let $d$ be a metric inducing the topology and $(x_n)$ a dense countable subset of $X$. Define $d_n: xin X to d_n(x):=d(x,x_n)$. It is a continuous function. It is easy to check that the algebra generated by $1$ and $(d_n)_n$ separate the points in $X$ so by Stone Weierstrass theorem this subalgebra is dense in $C(X)$. By considering linear combination with rational coefficient of element of this subalgebra it is easy to see that $C(X)$ is separable.
edited Jan 29 at 16:45
user152715
3,60423553
answered Jun 19 '15 at 11:14
Patissot
1,105615
Theorem. If $X$ is compact Hausdorff then $C(X)$ is separable iff $X$ is metrizable.
There is a natural embedding $xin Xto delta _xin mathcalM(X)$ (more precisely in the unit ball of $mathcalM(X)$). This is an homeomorphism for the weak*-topology of $mathcalM(X)$. If $C(X)$ is separable then $(mathcalM(X), w*)$ have a compact metrizable unit ball. So $X$ is compact.
For the converse, assume $X$ is a metrizable compact Hausdorff space. Let $d$ be a metric inducing the topology and $(x_n)$ a dense countable subset of $X$. Define $d_n: xin X to d_n(x):=d(x,x_n)$. It is a continuous function. It is easy to check that the algebra generated by $1$ and $(d_n)_n$ separate the points in $X$ so by Stone Weierstrass theorem this subalgebra is dense in $C(X)$. By considering linear combination with rational coefficient of element of this subalgebra it is easy to see that $C(X)$ is separable.
edited Jan 29 at 16:45
user152715
3,60423553
answered Jun 19 '15 at 11:14
Patissot
1,105615
edited Jan 29 at 16:45
user152715
3,60423553
edited Jan 29 at 16:45
user152715
3,60423553
edited Jan 29 at 16:45
user152715
3,60423553
3,60423553
answered Jun 19 '15 at 11:14
Patissot
1,105615
answered Jun 19 '15 at 11:14
Patissot
1,105615
answered Jun 19 '15 at 11:14
Patissot
1,105615
1,105615
add a comment |Â
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f1331321%2fcx-is-separable-when-x-is-compact%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
1
this result is not trivial: If X is a compact $T_2$ space $X$, then $C(X)$ is separable iff there is a metric $Xtimes Xrightarrow R$ that induces the topology of $X$. You need to use the Stone-Weierstrass Thm, Urysohn Lemma and the Urysohn Metrization Thm.
– Driver 8
Jun 19 '15 at 13:21