Is every separable locally compact metrizable topology induced by a Heine-Borel metric?
$begingroup$
This is a follow-up to my question here. A metric has the Heine-Borel property if a set is closed and bounded with respect to the metric if and only if it is compact. Now if a metrizable topology is induced by a metric with the Heine-Borel property, then it is locally compact and separable.
My question is, is the converse true? That is, if a topologicy is separable, locally compact, and metrizable, then is it induced by some metric with the Heine-Borel property?
If not, what is an example of such a topology all of whose metrics fail to have the Heine-Borel property?
general-topology metric-spaces compactness examples-counterexamples separable-spaces
asked Dec 20 '18 at 6:39
Keshav SrinivasanKeshav Srinivasan
2,33021445
$endgroup$
add a comment |
$begingroup$
This is a follow-up to my question here. A metric has the Heine-Borel property if a set is closed and bounded with respect to the metric if and only if it is compact. Now if a metrizable topology is induced by a metric with the Heine-Borel property, then it is locally compact and separable.
My question is, is the converse true? That is, if a topologicy is separable, locally compact, and metrizable, then is it induced by some metric with the Heine-Borel property?
If not, what is an example of such a topology all of whose metrics fail to have the Heine-Borel property?
general-topology metric-spaces compactness examples-counterexamples separable-spaces
asked Dec 20 '18 at 6:39
Keshav SrinivasanKeshav Srinivasan
2,33021445
$endgroup$
add a comment |
$begingroup$
This is a follow-up to my question here. A metric has the Heine-Borel property if a set is closed and bounded with respect to the metric if and only if it is compact. Now if a metrizable topology is induced by a metric with the Heine-Borel property, then it is locally compact and separable.
My question is, is the converse true? That is, if a topologicy is separable, locally compact, and metrizable, then is it induced by some metric with the Heine-Borel property?
If not, what is an example of such a topology all of whose metrics fail to have the Heine-Borel property?
general-topology metric-spaces compactness examples-counterexamples separable-spaces
asked Dec 20 '18 at 6:39
Keshav SrinivasanKeshav Srinivasan
2,33021445
$endgroup$
This is a follow-up to my question here. A metric has the Heine-Borel property if a set is closed and bounded with respect to the metric if and only if it is compact. Now if a metrizable topology is induced by a metric with the Heine-Borel property, then it is locally compact and separable.
My question is, is the converse true? That is, if a topologicy is separable, locally compact, and metrizable, then is it induced by some metric with the Heine-Borel property?
If not, what is an example of such a topology all of whose metrics fail to have the Heine-Borel property?
general-topology metric-spaces compactness examples-counterexamples separable-spaces
general-topology metric-spaces compactness examples-counterexamples separable-spaces
asked Dec 20 '18 at 6:39
Keshav SrinivasanKeshav Srinivasan
2,33021445
asked Dec 20 '18 at 6:39
Keshav SrinivasanKeshav Srinivasan
2,33021445
asked Dec 20 '18 at 6:39
Keshav SrinivasanKeshav Srinivasan
2,33021445
asked Dec 20 '18 at 6:39
Keshav SrinivasanKeshav Srinivasan
2,33021445
asked Dec 20 '18 at 6:39
Keshav SrinivasanKeshav Srinivasan
2,33021445
2,33021445
add a comment |
add a comment |
1 Answer
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$begingroup$
Such a locally compact separable metric space has a one-point compactification $Y$ that is metrisable. This is classical and well-known. Let $d$ be a metric for $Y$, consider $X$ to be a subset of $Y$ with $Ysetminus X={infty}$ and define $d'(x_1, x_2) = d(x_1, x_2) + |frac{1}{d(x_1,infty)} - frac{1}{d(x_2,infty)}|$ for $x_1,x_2 in X$.
One can check that $d$ is a compatible metric on $X$ such that $(X,d)$ has the Heine-Borel property. Boundedness implies that we "stay away from $infty$", essentially.
answered Dec 20 '18 at 7:07
Henno BrandsmaHenno Brandsma
111k348119
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1 Answer
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1 Answer
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$begingroup$
Such a locally compact separable metric space has a one-point compactification $Y$ that is metrisable. This is classical and well-known. Let $d$ be a metric for $Y$, consider $X$ to be a subset of $Y$ with $Ysetminus X={infty}$ and define $d'(x_1, x_2) = d(x_1, x_2) + |frac{1}{d(x_1,infty)} - frac{1}{d(x_2,infty)}|$ for $x_1,x_2 in X$.
One can check that $d$ is a compatible metric on $X$ such that $(X,d)$ has the Heine-Borel property. Boundedness implies that we "stay away from $infty$", essentially.
answered Dec 20 '18 at 7:07
Henno BrandsmaHenno Brandsma
111k348119
$endgroup$
add a comment |
$begingroup$
Such a locally compact separable metric space has a one-point compactification $Y$ that is metrisable. This is classical and well-known. Let $d$ be a metric for $Y$, consider $X$ to be a subset of $Y$ with $Ysetminus X={infty}$ and define $d'(x_1, x_2) = d(x_1, x_2) + |frac{1}{d(x_1,infty)} - frac{1}{d(x_2,infty)}|$ for $x_1,x_2 in X$.
One can check that $d$ is a compatible metric on $X$ such that $(X,d)$ has the Heine-Borel property. Boundedness implies that we "stay away from $infty$", essentially.
answered Dec 20 '18 at 7:07
Henno BrandsmaHenno Brandsma
111k348119
$endgroup$
add a comment |
$begingroup$
Such a locally compact separable metric space has a one-point compactification $Y$ that is metrisable. This is classical and well-known. Let $d$ be a metric for $Y$, consider $X$ to be a subset of $Y$ with $Ysetminus X={infty}$ and define $d'(x_1, x_2) = d(x_1, x_2) + |frac{1}{d(x_1,infty)} - frac{1}{d(x_2,infty)}|$ for $x_1,x_2 in X$.
One can check that $d$ is a compatible metric on $X$ such that $(X,d)$ has the Heine-Borel property. Boundedness implies that we "stay away from $infty$", essentially.
answered Dec 20 '18 at 7:07
Henno BrandsmaHenno Brandsma
111k348119
$endgroup$
Such a locally compact separable metric space has a one-point compactification $Y$ that is metrisable. This is classical and well-known. Let $d$ be a metric for $Y$, consider $X$ to be a subset of $Y$ with $Ysetminus X={infty}$ and define $d'(x_1, x_2) = d(x_1, x_2) + |frac{1}{d(x_1,infty)} - frac{1}{d(x_2,infty)}|$ for $x_1,x_2 in X$.
One can check that $d$ is a compatible metric on $X$ such that $(X,d)$ has the Heine-Borel property. Boundedness implies that we "stay away from $infty$", essentially.
answered Dec 20 '18 at 7:07
Henno BrandsmaHenno Brandsma
111k348119
answered Dec 20 '18 at 7:07
Henno BrandsmaHenno Brandsma
111k348119
answered Dec 20 '18 at 7:07
Henno BrandsmaHenno Brandsma
111k348119
answered Dec 20 '18 at 7:07
Henno BrandsmaHenno Brandsma
111k348119
111k348119
add a comment |
add a comment |
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