Is every metrizable topology induced by a Heine-Borel metric?
$begingroup$
A metric space $(X,d)$ has the Heine-Borel property if for any subset $A$ of $X$, $A$ is compact if and only it is closed and bounded. ($mathbb{R}^n$ is the classic example.) My question is, for any metrizable topological space $X$, does there exist a metric on $X$ which induces the topology on $X$ and which has the Heine-Borel property?
If not, what is an example of a topological space such that all the metrics which induce the topology fail to have the Heine-Borel property?
general-topology metric-spaces compactness examples-counterexamples
asked Dec 19 '18 at 6:03
Keshav SrinivasanKeshav Srinivasan
2,39121446
$endgroup$
add a comment |
$begingroup$
A metric space $(X,d)$ has the Heine-Borel property if for any subset $A$ of $X$, $A$ is compact if and only it is closed and bounded. ($mathbb{R}^n$ is the classic example.) My question is, for any metrizable topological space $X$, does there exist a metric on $X$ which induces the topology on $X$ and which has the Heine-Borel property?
If not, what is an example of a topological space such that all the metrics which induce the topology fail to have the Heine-Borel property?
general-topology metric-spaces compactness examples-counterexamples
asked Dec 19 '18 at 6:03
Keshav SrinivasanKeshav Srinivasan
2,39121446
$endgroup$
add a comment |
$begingroup$
A metric space $(X,d)$ has the Heine-Borel property if for any subset $A$ of $X$, $A$ is compact if and only it is closed and bounded. ($mathbb{R}^n$ is the classic example.) My question is, for any metrizable topological space $X$, does there exist a metric on $X$ which induces the topology on $X$ and which has the Heine-Borel property?
If not, what is an example of a topological space such that all the metrics which induce the topology fail to have the Heine-Borel property?
general-topology metric-spaces compactness examples-counterexamples
asked Dec 19 '18 at 6:03
Keshav SrinivasanKeshav Srinivasan
2,39121446
$endgroup$
A metric space $(X,d)$ has the Heine-Borel property if for any subset $A$ of $X$, $A$ is compact if and only it is closed and bounded. ($mathbb{R}^n$ is the classic example.) My question is, for any metrizable topological space $X$, does there exist a metric on $X$ which induces the topology on $X$ and which has the Heine-Borel property?
If not, what is an example of a topological space such that all the metrics which induce the topology fail to have the Heine-Borel property?
general-topology metric-spaces compactness examples-counterexamples
general-topology metric-spaces compactness examples-counterexamples
asked Dec 19 '18 at 6:03
Keshav SrinivasanKeshav Srinivasan
2,39121446
asked Dec 19 '18 at 6:03
Keshav SrinivasanKeshav Srinivasan
2,39121446
asked Dec 19 '18 at 6:03
Keshav SrinivasanKeshav Srinivasan
2,39121446
asked Dec 19 '18 at 6:03
Keshav SrinivasanKeshav Srinivasan
2,39121446
asked Dec 19 '18 at 6:03
Keshav SrinivasanKeshav Srinivasan
2,39121446
2,39121446
add a comment |
add a comment |
1 Answer
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Any metric space with Heine Borel property is sigma compact. An infinite dimensional Banach space is never sigma compact.
answered Dec 19 '18 at 6:13
Kavi Rama MurthyKavi Rama Murthy
71.5k53170
$endgroup$
$begingroup$
Or any uncountable set with the discrete topology, for the same reason.
$endgroup$
– A. Pongrácz
Dec 19 '18 at 6:16
$begingroup$
Why does the Heine-Borel property imply sigma-compactness?
$endgroup$
– Keshav Srinivasan
Dec 19 '18 at 6:19
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$X=cup_{n=1}^{infty} overline {B} (x,n)$ for any fixed point $x$. @KeshavSrinivasan
$endgroup$
– Kavi Rama Murthy
Dec 19 '18 at 6:21
$begingroup$
OK, that makes sense. By the way, does every sigma compact metrizable space have a Heine-Borel metric which induces its topology?
$endgroup$
– Keshav Srinivasan
Dec 19 '18 at 6:29
$begingroup$
It must also be locally compact. $mathbb Q$ is $sigma$-compact but not locally compact. Maybe a locally compact separable metrizable space has a Heine–Borel metric?
$endgroup$
– bof
Dec 19 '18 at 9:17
|
show 4 more comments
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Any metric space with Heine Borel property is sigma compact. An infinite dimensional Banach space is never sigma compact.
answered Dec 19 '18 at 6:13
Kavi Rama MurthyKavi Rama Murthy
71.5k53170
$endgroup$
$begingroup$
Or any uncountable set with the discrete topology, for the same reason.
$endgroup$
– A. Pongrácz
Dec 19 '18 at 6:16
$begingroup$
Why does the Heine-Borel property imply sigma-compactness?
$endgroup$
– Keshav Srinivasan
Dec 19 '18 at 6:19
$begingroup$
$X=cup_{n=1}^{infty} overline {B} (x,n)$ for any fixed point $x$. @KeshavSrinivasan
$endgroup$
– Kavi Rama Murthy
Dec 19 '18 at 6:21
$begingroup$
OK, that makes sense. By the way, does every sigma compact metrizable space have a Heine-Borel metric which induces its topology?
$endgroup$
– Keshav Srinivasan
Dec 19 '18 at 6:29
$begingroup$
It must also be locally compact. $mathbb Q$ is $sigma$-compact but not locally compact. Maybe a locally compact separable metrizable space has a Heine–Borel metric?
$endgroup$
– bof
Dec 19 '18 at 9:17
|
show 4 more comments
$begingroup$
Any metric space with Heine Borel property is sigma compact. An infinite dimensional Banach space is never sigma compact.
answered Dec 19 '18 at 6:13
Kavi Rama MurthyKavi Rama Murthy
71.5k53170
$endgroup$
$begingroup$
Or any uncountable set with the discrete topology, for the same reason.
$endgroup$
– A. Pongrácz
Dec 19 '18 at 6:16
$begingroup$
Why does the Heine-Borel property imply sigma-compactness?
$endgroup$
– Keshav Srinivasan
Dec 19 '18 at 6:19
$begingroup$
$X=cup_{n=1}^{infty} overline {B} (x,n)$ for any fixed point $x$. @KeshavSrinivasan
$endgroup$
– Kavi Rama Murthy
Dec 19 '18 at 6:21
$begingroup$
OK, that makes sense. By the way, does every sigma compact metrizable space have a Heine-Borel metric which induces its topology?
$endgroup$
– Keshav Srinivasan
Dec 19 '18 at 6:29
$begingroup$
It must also be locally compact. $mathbb Q$ is $sigma$-compact but not locally compact. Maybe a locally compact separable metrizable space has a Heine–Borel metric?
$endgroup$
– bof
Dec 19 '18 at 9:17
|
show 4 more comments
$begingroup$
Any metric space with Heine Borel property is sigma compact. An infinite dimensional Banach space is never sigma compact.
answered Dec 19 '18 at 6:13
Kavi Rama MurthyKavi Rama Murthy
71.5k53170
$endgroup$
Any metric space with Heine Borel property is sigma compact. An infinite dimensional Banach space is never sigma compact.
answered Dec 19 '18 at 6:13
Kavi Rama MurthyKavi Rama Murthy
71.5k53170
answered Dec 19 '18 at 6:13
Kavi Rama MurthyKavi Rama Murthy
71.5k53170
answered Dec 19 '18 at 6:13
Kavi Rama MurthyKavi Rama Murthy
71.5k53170
answered Dec 19 '18 at 6:13
Kavi Rama MurthyKavi Rama Murthy
71.5k53170
71.5k53170
$begingroup$
Or any uncountable set with the discrete topology, for the same reason.
$endgroup$
– A. Pongrácz
Dec 19 '18 at 6:16
$begingroup$
Why does the Heine-Borel property imply sigma-compactness?
$endgroup$
– Keshav Srinivasan
Dec 19 '18 at 6:19
$begingroup$
$X=cup_{n=1}^{infty} overline {B} (x,n)$ for any fixed point $x$. @KeshavSrinivasan
$endgroup$
– Kavi Rama Murthy
Dec 19 '18 at 6:21
$begingroup$
OK, that makes sense. By the way, does every sigma compact metrizable space have a Heine-Borel metric which induces its topology?
$endgroup$
– Keshav Srinivasan
Dec 19 '18 at 6:29
$begingroup$
It must also be locally compact. $mathbb Q$ is $sigma$-compact but not locally compact. Maybe a locally compact separable metrizable space has a Heine–Borel metric?
$endgroup$
– bof
Dec 19 '18 at 9:17
|
show 4 more comments
$begingroup$
Or any uncountable set with the discrete topology, for the same reason.
$endgroup$
– A. Pongrácz
Dec 19 '18 at 6:16
$begingroup$
Why does the Heine-Borel property imply sigma-compactness?
$endgroup$
– Keshav Srinivasan
Dec 19 '18 at 6:19
$begingroup$
$X=cup_{n=1}^{infty} overline {B} (x,n)$ for any fixed point $x$. @KeshavSrinivasan
$endgroup$
– Kavi Rama Murthy
Dec 19 '18 at 6:21
$begingroup$
OK, that makes sense. By the way, does every sigma compact metrizable space have a Heine-Borel metric which induces its topology?
$endgroup$
– Keshav Srinivasan
Dec 19 '18 at 6:29
$begingroup$
It must also be locally compact. $mathbb Q$ is $sigma$-compact but not locally compact. Maybe a locally compact separable metrizable space has a Heine–Borel metric?
$endgroup$
– bof
Dec 19 '18 at 9:17
$begingroup$
Or any uncountable set with the discrete topology, for the same reason.
$endgroup$
– A. Pongrácz
Dec 19 '18 at 6:16
$begingroup$
Or any uncountable set with the discrete topology, for the same reason.
$endgroup$
– A. Pongrácz
Dec 19 '18 at 6:16
$begingroup$
Why does the Heine-Borel property imply sigma-compactness?
$endgroup$
– Keshav Srinivasan
Dec 19 '18 at 6:19
$begingroup$
Why does the Heine-Borel property imply sigma-compactness?
$endgroup$
– Keshav Srinivasan
Dec 19 '18 at 6:19
$begingroup$
$X=cup_{n=1}^{infty} overline {B} (x,n)$ for any fixed point $x$. @KeshavSrinivasan
$endgroup$
– Kavi Rama Murthy
Dec 19 '18 at 6:21
$begingroup$
$X=cup_{n=1}^{infty} overline {B} (x,n)$ for any fixed point $x$. @KeshavSrinivasan
$endgroup$
– Kavi Rama Murthy
Dec 19 '18 at 6:21
$begingroup$
OK, that makes sense. By the way, does every sigma compact metrizable space have a Heine-Borel metric which induces its topology?
$endgroup$
– Keshav Srinivasan
Dec 19 '18 at 6:29
$begingroup$
OK, that makes sense. By the way, does every sigma compact metrizable space have a Heine-Borel metric which induces its topology?
$endgroup$
– Keshav Srinivasan
Dec 19 '18 at 6:29
$begingroup$
It must also be locally compact. $mathbb Q$ is $sigma$-compact but not locally compact. Maybe a locally compact separable metrizable space has a Heine–Borel metric?
$endgroup$
– bof
Dec 19 '18 at 9:17
$begingroup$
It must also be locally compact. $mathbb Q$ is $sigma$-compact but not locally compact. Maybe a locally compact separable metrizable space has a Heine–Borel metric?
$endgroup$
– bof
Dec 19 '18 at 9:17
|
show 4 more comments
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