Complete+bounded homeomorphic to incomplete+unbounded
I'm aware that completeness and (total) boundedness are not preserved under homeomorphism, with $(0,1) cong mathbb R$ being a counterexample to both simultaneously. I'm curious if there exists a "double counterexample" in another way:
Are there homeomorphic metric spaces $M$ and $N$ such that
$M$ is both complete and bounded,
but $N$ is neither complete nor bounded?
Of course in that case, $M$ could not be totally bounded, else it would be compact, which is certainly a topological property!
general-topology metric-spaces
asked Dec 25 at 0:23
Evan Chen
1,160516
add a comment |
I'm aware that completeness and (total) boundedness are not preserved under homeomorphism, with $(0,1) cong mathbb R$ being a counterexample to both simultaneously. I'm curious if there exists a "double counterexample" in another way:
Are there homeomorphic metric spaces $M$ and $N$ such that
$M$ is both complete and bounded,
but $N$ is neither complete nor bounded?
Of course in that case, $M$ could not be totally bounded, else it would be compact, which is certainly a topological property!
general-topology metric-spaces
asked Dec 25 at 0:23
Evan Chen
1,160516
add a comment |
I'm aware that completeness and (total) boundedness are not preserved under homeomorphism, with $(0,1) cong mathbb R$ being a counterexample to both simultaneously. I'm curious if there exists a "double counterexample" in another way:
Are there homeomorphic metric spaces $M$ and $N$ such that
$M$ is both complete and bounded,
but $N$ is neither complete nor bounded?
Of course in that case, $M$ could not be totally bounded, else it would be compact, which is certainly a topological property!
general-topology metric-spaces
asked Dec 25 at 0:23
Evan Chen
1,160516
I'm aware that completeness and (total) boundedness are not preserved under homeomorphism, with $(0,1) cong mathbb R$ being a counterexample to both simultaneously. I'm curious if there exists a "double counterexample" in another way:
Are there homeomorphic metric spaces $M$ and $N$ such that
$M$ is both complete and bounded,
but $N$ is neither complete nor bounded?
Of course in that case, $M$ could not be totally bounded, else it would be compact, which is certainly a topological property!
general-topology metric-spaces
general-topology metric-spaces
asked Dec 25 at 0:23
Evan Chen
1,160516
asked Dec 25 at 0:23
Evan Chen
1,160516
asked Dec 25 at 0:23
Evan Chen
1,160516
asked Dec 25 at 0:23
Evan Chen
1,160516
asked Dec 25 at 0:23
Evan Chen
1,160516
1,160516
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
You could let $M$ be $mathbb R$ with the metric $d(x,y)=min(1,|x-y|)$ and $N$ be $(0,infty)$ with the usual metric.
answered Dec 25 at 0:35
Henning Makholm
237k16302537
Thanks! Do you mind if I include this example in my metric topology notes? I would credit you and link here.
– Evan Chen
Dec 25 at 0:46
@EvanChen: Sure, go ahead.
– Henning Makholm
Dec 25 at 0:49
add a comment |
Yes. Let $M$ be $mathbb N$ with the discrete metric, and let $N= {1/n:nin mathbb N}cup {2,3,4,dots}$ with the usual $mathbb R$ metric.
answered Dec 25 at 0:48
zhw.
71.6k43075
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
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active
oldest
votes
You could let $M$ be $mathbb R$ with the metric $d(x,y)=min(1,|x-y|)$ and $N$ be $(0,infty)$ with the usual metric.
answered Dec 25 at 0:35
Henning Makholm
237k16302537
Thanks! Do you mind if I include this example in my metric topology notes? I would credit you and link here.
– Evan Chen
Dec 25 at 0:46
@EvanChen: Sure, go ahead.
– Henning Makholm
Dec 25 at 0:49
add a comment |
You could let $M$ be $mathbb R$ with the metric $d(x,y)=min(1,|x-y|)$ and $N$ be $(0,infty)$ with the usual metric.
answered Dec 25 at 0:35
Henning Makholm
237k16302537
Thanks! Do you mind if I include this example in my metric topology notes? I would credit you and link here.
– Evan Chen
Dec 25 at 0:46
@EvanChen: Sure, go ahead.
– Henning Makholm
Dec 25 at 0:49
add a comment |
You could let $M$ be $mathbb R$ with the metric $d(x,y)=min(1,|x-y|)$ and $N$ be $(0,infty)$ with the usual metric.
answered Dec 25 at 0:35
Henning Makholm
237k16302537
You could let $M$ be $mathbb R$ with the metric $d(x,y)=min(1,|x-y|)$ and $N$ be $(0,infty)$ with the usual metric.
answered Dec 25 at 0:35
Henning Makholm
237k16302537
answered Dec 25 at 0:35
Henning Makholm
237k16302537
answered Dec 25 at 0:35
Henning Makholm
237k16302537
answered Dec 25 at 0:35
Henning Makholm
237k16302537
237k16302537
Thanks! Do you mind if I include this example in my metric topology notes? I would credit you and link here.
– Evan Chen
Dec 25 at 0:46
@EvanChen: Sure, go ahead.
– Henning Makholm
Dec 25 at 0:49
add a comment |
Thanks! Do you mind if I include this example in my metric topology notes? I would credit you and link here.
– Evan Chen
Dec 25 at 0:46
@EvanChen: Sure, go ahead.
– Henning Makholm
Dec 25 at 0:49
Thanks! Do you mind if I include this example in my metric topology notes? I would credit you and link here.
– Evan Chen
Dec 25 at 0:46
Thanks! Do you mind if I include this example in my metric topology notes? I would credit you and link here.
– Evan Chen
Dec 25 at 0:46
@EvanChen: Sure, go ahead.
– Henning Makholm
Dec 25 at 0:49
@EvanChen: Sure, go ahead.
– Henning Makholm
Dec 25 at 0:49
add a comment |
Yes. Let $M$ be $mathbb N$ with the discrete metric, and let $N= {1/n:nin mathbb N}cup {2,3,4,dots}$ with the usual $mathbb R$ metric.
answered Dec 25 at 0:48
zhw.
71.6k43075
add a comment |
Yes. Let $M$ be $mathbb N$ with the discrete metric, and let $N= {1/n:nin mathbb N}cup {2,3,4,dots}$ with the usual $mathbb R$ metric.
answered Dec 25 at 0:48
zhw.
71.6k43075
add a comment |
Yes. Let $M$ be $mathbb N$ with the discrete metric, and let $N= {1/n:nin mathbb N}cup {2,3,4,dots}$ with the usual $mathbb R$ metric.
answered Dec 25 at 0:48
zhw.
71.6k43075
Yes. Let $M$ be $mathbb N$ with the discrete metric, and let $N= {1/n:nin mathbb N}cup {2,3,4,dots}$ with the usual $mathbb R$ metric.
answered Dec 25 at 0:48
zhw.
71.6k43075
answered Dec 25 at 0:48
zhw.
71.6k43075
answered Dec 25 at 0:48
zhw.
71.6k43075
answered Dec 25 at 0:48
zhw.
71.6k43075
71.6k43075
add a comment |
add a comment |
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