Mixing
Is every uniformizable topology induced by a Heine-Borel uniformity?
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This is a follow-up to my question here. A subset $A$ of a uniform space is said to be bounded if for each entourage $V$, $A$ is a subset of $V^n[F]$ for some natural number $n$ and some finite set $F$. Just like for metric spaces, a compact subset of a uniform space is always closed and bounded, but the converse need not be true. A uniformity has the Heine-Borel property if a set is compact if and only if it is closed and bounded.
Suppose $X$ is a uniformizable topological space, AKA a completely regular topological space. My question is, is it necessarily the case that the topology on $X$ is induced by some Heine-Borel uniformity?
Now this answer shows that a metrizable topology is induced a metric with the Heine-Borel property if and only if it is locally compact and separable. But that doesn’t answer this question, because even if we take a topology which isn’t locally compact or isn’t separable, it’s possible that such a topology is induced by a non-metrizable uniformity with the Heine-Borel property.
general-topology metric-spaces compactness examples-counterexamples uniform-spaces
asked Dec 22 '18 at 7:42
Keshav SrinivasanKeshav Srinivasan
2,37221445
$endgroup$
add a comment |
$begingroup$
This is a follow-up to my question here. A subset $A$ of a uniform space is said to be bounded if for each entourage $V$, $A$ is a subset of $V^n[F]$ for some natural number $n$ and some finite set $F$. Just like for metric spaces, a compact subset of a uniform space is always closed and bounded, but the converse need not be true. A uniformity has the Heine-Borel property if a set is compact if and only if it is closed and bounded.
Suppose $X$ is a uniformizable topological space, AKA a completely regular topological space. My question is, is it necessarily the case that the topology on $X$ is induced by some Heine-Borel uniformity?
Now this answer shows that a metrizable topology is induced a metric with the Heine-Borel property if and only if it is locally compact and separable. But that doesn’t answer this question, because even if we take a topology which isn’t locally compact or isn’t separable, it’s possible that such a topology is induced by a non-metrizable uniformity with the Heine-Borel property.
general-topology metric-spaces compactness examples-counterexamples uniform-spaces
asked Dec 22 '18 at 7:42
Keshav SrinivasanKeshav Srinivasan
2,37221445
$endgroup$
$begingroup$
It's probably false, but one would need some more theory on the classs of uniform spaces with the HB property to disprove it, I think. Is this class closed under products, uniformly continuous images etc. ?
$endgroup$
– Henno Brandsma
Dec 22 '18 at 7:59
$begingroup$
@HennoBrandsma It may be even easier; the proofs that topologies induced by Heine-Borel metrics are locally compact and separable may be able to be modified to show that topologies induced by Heine-Borel uniformities are locally compact and separable.
$endgroup$
– Keshav Srinivasan
Dec 22 '18 at 8:19
$begingroup$
I don't think that would hold (separability and local compactness), btu I'm not sure. Just a hunch.
$endgroup$
– Henno Brandsma
Dec 22 '18 at 10:25
add a comment |
$begingroup$
This is a follow-up to my question here. A subset $A$ of a uniform space is said to be bounded if for each entourage $V$, $A$ is a subset of $V^n[F]$ for some natural number $n$ and some finite set $F$. Just like for metric spaces, a compact subset of a uniform space is always closed and bounded, but the converse need not be true. A uniformity has the Heine-Borel property if a set is compact if and only if it is closed and bounded.
Suppose $X$ is a uniformizable topological space, AKA a completely regular topological space. My question is, is it necessarily the case that the topology on $X$ is induced by some Heine-Borel uniformity?
Now this answer shows that a metrizable topology is induced a metric with the Heine-Borel property if and only if it is locally compact and separable. But that doesn’t answer this question, because even if we take a topology which isn’t locally compact or isn’t separable, it’s possible that such a topology is induced by a non-metrizable uniformity with the Heine-Borel property.
general-topology metric-spaces compactness examples-counterexamples uniform-spaces
asked Dec 22 '18 at 7:42
Keshav SrinivasanKeshav Srinivasan
2,37221445
$endgroup$
This is a follow-up to my question here. A subset $A$ of a uniform space is said to be bounded if for each entourage $V$, $A$ is a subset of $V^n[F]$ for some natural number $n$ and some finite set $F$. Just like for metric spaces, a compact subset of a uniform space is always closed and bounded, but the converse need not be true. A uniformity has the Heine-Borel property if a set is compact if and only if it is closed and bounded.
Suppose $X$ is a uniformizable topological space, AKA a completely regular topological space. My question is, is it necessarily the case that the topology on $X$ is induced by some Heine-Borel uniformity?
Now this answer shows that a metrizable topology is induced a metric with the Heine-Borel property if and only if it is locally compact and separable. But that doesn’t answer this question, because even if we take a topology which isn’t locally compact or isn’t separable, it’s possible that such a topology is induced by a non-metrizable uniformity with the Heine-Borel property.
general-topology metric-spaces compactness examples-counterexamples uniform-spaces
general-topology metric-spaces compactness examples-counterexamples uniform-spaces
asked Dec 22 '18 at 7:42
Keshav SrinivasanKeshav Srinivasan
2,37221445
asked Dec 22 '18 at 7:42
Keshav SrinivasanKeshav Srinivasan
2,37221445
asked Dec 22 '18 at 7:42
Keshav SrinivasanKeshav Srinivasan
2,37221445
asked Dec 22 '18 at 7:42
Keshav SrinivasanKeshav Srinivasan
2,37221445
asked Dec 22 '18 at 7:42
Keshav SrinivasanKeshav Srinivasan
2,37221445
2,37221445
$begingroup$
It's probably false, but one would need some more theory on the classs of uniform spaces with the HB property to disprove it, I think. Is this class closed under products, uniformly continuous images etc. ?
$endgroup$
– Henno Brandsma
Dec 22 '18 at 7:59
$begingroup$
@HennoBrandsma It may be even easier; the proofs that topologies induced by Heine-Borel metrics are locally compact and separable may be able to be modified to show that topologies induced by Heine-Borel uniformities are locally compact and separable.
$endgroup$
– Keshav Srinivasan
Dec 22 '18 at 8:19
$begingroup$
I don't think that would hold (separability and local compactness), btu I'm not sure. Just a hunch.
$endgroup$
– Henno Brandsma
Dec 22 '18 at 10:25
add a comment |
$begingroup$
It's probably false, but one would need some more theory on the classs of uniform spaces with the HB property to disprove it, I think. Is this class closed under products, uniformly continuous images etc. ?
$endgroup$
– Henno Brandsma
Dec 22 '18 at 7:59
$begingroup$
@HennoBrandsma It may be even easier; the proofs that topologies induced by Heine-Borel metrics are locally compact and separable may be able to be modified to show that topologies induced by Heine-Borel uniformities are locally compact and separable.
$endgroup$
– Keshav Srinivasan
Dec 22 '18 at 8:19
$begingroup$
I don't think that would hold (separability and local compactness), btu I'm not sure. Just a hunch.
$endgroup$
– Henno Brandsma
Dec 22 '18 at 10:25
$begingroup$
It's probably false, but one would need some more theory on the classs of uniform spaces with the HB property to disprove it, I think. Is this class closed under products, uniformly continuous images etc. ?
$endgroup$
– Henno Brandsma
Dec 22 '18 at 7:59
$begingroup$
It's probably false, but one would need some more theory on the classs of uniform spaces with the HB property to disprove it, I think. Is this class closed under products, uniformly continuous images etc. ?
$endgroup$
– Henno Brandsma
Dec 22 '18 at 7:59
$begingroup$
@HennoBrandsma It may be even easier; the proofs that topologies induced by Heine-Borel metrics are locally compact and separable may be able to be modified to show that topologies induced by Heine-Borel uniformities are locally compact and separable.
$endgroup$
– Keshav Srinivasan
Dec 22 '18 at 8:19
$begingroup$
@HennoBrandsma It may be even easier; the proofs that topologies induced by Heine-Borel metrics are locally compact and separable may be able to be modified to show that topologies induced by Heine-Borel uniformities are locally compact and separable.
$endgroup$
– Keshav Srinivasan
Dec 22 '18 at 8:19
$begingroup$
I don't think that would hold (separability and local compactness), btu I'm not sure. Just a hunch.
$endgroup$
– Henno Brandsma
Dec 22 '18 at 10:25
$begingroup$
I don't think that would hold (separability and local compactness), btu I'm not sure. Just a hunch.
$endgroup$
– Henno Brandsma
Dec 22 '18 at 10:25
add a comment |
1 Answer
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A space $X$ is pseudocompact if any continuous $Xtomathbb R$ is bounded.
There exist completely regular non-compact pseudocompact spaces $X$ such as the long line (there are other examples at π-Base). By pseudocompactness, $X$ is bounded in any continuous pseudometric. By the positive answer to your previous question Is a set bounded in every metric for a uniformity bounded in the uniformity?, noting that the answers also work for pseudometrics in non-metrisable spaces, $X$ is bounded in every uniformity. So $X$ does not admit any Heine-Borel uniformity.
answered Jan 26 at 20:29
DapDap
19k842
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$begingroup$
A space $X$ is pseudocompact if any continuous $Xtomathbb R$ is bounded.
There exist completely regular non-compact pseudocompact spaces $X$ such as the long line (there are other examples at π-Base). By pseudocompactness, $X$ is bounded in any continuous pseudometric. By the positive answer to your previous question Is a set bounded in every metric for a uniformity bounded in the uniformity?, noting that the answers also work for pseudometrics in non-metrisable spaces, $X$ is bounded in every uniformity. So $X$ does not admit any Heine-Borel uniformity.
answered Jan 26 at 20:29
DapDap
19k842
$endgroup$
add a comment |
$begingroup$
A space $X$ is pseudocompact if any continuous $Xtomathbb R$ is bounded.
There exist completely regular non-compact pseudocompact spaces $X$ such as the long line (there are other examples at π-Base). By pseudocompactness, $X$ is bounded in any continuous pseudometric. By the positive answer to your previous question Is a set bounded in every metric for a uniformity bounded in the uniformity?, noting that the answers also work for pseudometrics in non-metrisable spaces, $X$ is bounded in every uniformity. So $X$ does not admit any Heine-Borel uniformity.
answered Jan 26 at 20:29
DapDap
19k842
$endgroup$
add a comment |
$begingroup$
A space $X$ is pseudocompact if any continuous $Xtomathbb R$ is bounded.
There exist completely regular non-compact pseudocompact spaces $X$ such as the long line (there are other examples at π-Base). By pseudocompactness, $X$ is bounded in any continuous pseudometric. By the positive answer to your previous question Is a set bounded in every metric for a uniformity bounded in the uniformity?, noting that the answers also work for pseudometrics in non-metrisable spaces, $X$ is bounded in every uniformity. So $X$ does not admit any Heine-Borel uniformity.
answered Jan 26 at 20:29
DapDap
19k842
$endgroup$
A space $X$ is pseudocompact if any continuous $Xtomathbb R$ is bounded.
There exist completely regular non-compact pseudocompact spaces $X$ such as the long line (there are other examples at π-Base). By pseudocompactness, $X$ is bounded in any continuous pseudometric. By the positive answer to your previous question Is a set bounded in every metric for a uniformity bounded in the uniformity?, noting that the answers also work for pseudometrics in non-metrisable spaces, $X$ is bounded in every uniformity. So $X$ does not admit any Heine-Borel uniformity.
answered Jan 26 at 20:29
DapDap
19k842
answered Jan 26 at 20:29
DapDap
19k842
answered Jan 26 at 20:29
DapDap
19k842
answered Jan 26 at 20:29
DapDap
19k842
19k842
add a comment |
add a comment |
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$begingroup$
It's probably false, but one would need some more theory on the classs of uniform spaces with the HB property to disprove it, I think. Is this class closed under products, uniformly continuous images etc. ?
$endgroup$
– Henno Brandsma
Dec 22 '18 at 7:59
$begingroup$
@HennoBrandsma It may be even easier; the proofs that topologies induced by Heine-Borel metrics are locally compact and separable may be able to be modified to show that topologies induced by Heine-Borel uniformities are locally compact and separable.
$endgroup$
– Keshav Srinivasan
Dec 22 '18 at 8:19
$begingroup$
I don't think that would hold (separability and local compactness), btu I'm not sure. Just a hunch.
$endgroup$
– Henno Brandsma
Dec 22 '18 at 10:25