Example of a Baire metric space which is not completely metrizable
I know that some Baire metric spaces are not complete metric spaces but all examples, that I know, are completely metrizable. Help me to find an example of Baire metric space which is not completely metrizable. $[$Please give some short proofs or references$]$
general-topology metric-spaces
asked Nov 18 at 15:05
Offlaw
2649
add a comment |
I know that some Baire metric spaces are not complete metric spaces but all examples, that I know, are completely metrizable. Help me to find an example of Baire metric space which is not completely metrizable. $[$Please give some short proofs or references$]$
general-topology metric-spaces
asked Nov 18 at 15:05
Offlaw
2649
1
See Completely Metrizable Space and Baire Theorem.
– Dave L. Renfro
Nov 18 at 15:15
add a comment |
I know that some Baire metric spaces are not complete metric spaces but all examples, that I know, are completely metrizable. Help me to find an example of Baire metric space which is not completely metrizable. $[$Please give some short proofs or references$]$
general-topology metric-spaces
asked Nov 18 at 15:05
Offlaw
2649
I know that some Baire metric spaces are not complete metric spaces but all examples, that I know, are completely metrizable. Help me to find an example of Baire metric space which is not completely metrizable. $[$Please give some short proofs or references$]$
general-topology metric-spaces
general-topology metric-spaces
asked Nov 18 at 15:05
Offlaw
2649
asked Nov 18 at 15:05
Offlaw
2649
asked Nov 18 at 15:05
Offlaw
2649
asked Nov 18 at 15:05
Offlaw
2649
asked Nov 18 at 15:05
Offlaw
2649
2649
1
See Completely Metrizable Space and Baire Theorem.
– Dave L. Renfro
Nov 18 at 15:15
add a comment |
1
See Completely Metrizable Space and Baire Theorem.
– Dave L. Renfro
Nov 18 at 15:15
1
1
See Completely Metrizable Space and Baire Theorem.
– Dave L. Renfro
Nov 18 at 15:15
See Completely Metrizable Space and Baire Theorem.
– Dave L. Renfro
Nov 18 at 15:15
add a comment |
1 Answer
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A classic example is the open upper half plane with the rationals on the $x$-axis:
$X = {(x,y) in mathbb{R}^2: y >0 text{ or } y=0, x in mathbb{Q}}$ in the Euclidean metric.
This is Baire as it has an open dense Baire subspace $mathbb{R} times (0,infty)$ (which is completely metrisable) and not completely metrisable as it has a closed homeomorphic copy of $mathbb{Q}$.
answered Nov 18 at 17:30
Henno Brandsma
105k346114
If I'm not wrong, $X=mathbb{Q}times (0,infty)$, then considering $U_n=Xsetminus {r_n}times (0, infty)$ we got $cap U_n = phi$
– Offlaw
Nov 19 at 13:44
@Offlaw $mathbb{Q}times (0,infty)$ is irrelevant.
– Henno Brandsma
Nov 19 at 21:06
I got it. $X=mathbb{R} times (0,infty) cup mathbb{Q} times {0}$.
– Offlaw
Nov 20 at 1:29
@Offlaw indeed. That’s the space
– Henno Brandsma
Nov 20 at 4:24
add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
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votes
A classic example is the open upper half plane with the rationals on the $x$-axis:
$X = {(x,y) in mathbb{R}^2: y >0 text{ or } y=0, x in mathbb{Q}}$ in the Euclidean metric.
This is Baire as it has an open dense Baire subspace $mathbb{R} times (0,infty)$ (which is completely metrisable) and not completely metrisable as it has a closed homeomorphic copy of $mathbb{Q}$.
answered Nov 18 at 17:30
Henno Brandsma
105k346114
If I'm not wrong, $X=mathbb{Q}times (0,infty)$, then considering $U_n=Xsetminus {r_n}times (0, infty)$ we got $cap U_n = phi$
– Offlaw
Nov 19 at 13:44
@Offlaw $mathbb{Q}times (0,infty)$ is irrelevant.
– Henno Brandsma
Nov 19 at 21:06
I got it. $X=mathbb{R} times (0,infty) cup mathbb{Q} times {0}$.
– Offlaw
Nov 20 at 1:29
@Offlaw indeed. That’s the space
– Henno Brandsma
Nov 20 at 4:24
add a comment |
A classic example is the open upper half plane with the rationals on the $x$-axis:
$X = {(x,y) in mathbb{R}^2: y >0 text{ or } y=0, x in mathbb{Q}}$ in the Euclidean metric.
This is Baire as it has an open dense Baire subspace $mathbb{R} times (0,infty)$ (which is completely metrisable) and not completely metrisable as it has a closed homeomorphic copy of $mathbb{Q}$.
answered Nov 18 at 17:30
Henno Brandsma
105k346114
If I'm not wrong, $X=mathbb{Q}times (0,infty)$, then considering $U_n=Xsetminus {r_n}times (0, infty)$ we got $cap U_n = phi$
– Offlaw
Nov 19 at 13:44
@Offlaw $mathbb{Q}times (0,infty)$ is irrelevant.
– Henno Brandsma
Nov 19 at 21:06
I got it. $X=mathbb{R} times (0,infty) cup mathbb{Q} times {0}$.
– Offlaw
Nov 20 at 1:29
@Offlaw indeed. That’s the space
– Henno Brandsma
Nov 20 at 4:24
add a comment |
A classic example is the open upper half plane with the rationals on the $x$-axis:
$X = {(x,y) in mathbb{R}^2: y >0 text{ or } y=0, x in mathbb{Q}}$ in the Euclidean metric.
This is Baire as it has an open dense Baire subspace $mathbb{R} times (0,infty)$ (which is completely metrisable) and not completely metrisable as it has a closed homeomorphic copy of $mathbb{Q}$.
answered Nov 18 at 17:30
Henno Brandsma
105k346114
A classic example is the open upper half plane with the rationals on the $x$-axis:
$X = {(x,y) in mathbb{R}^2: y >0 text{ or } y=0, x in mathbb{Q}}$ in the Euclidean metric.
This is Baire as it has an open dense Baire subspace $mathbb{R} times (0,infty)$ (which is completely metrisable) and not completely metrisable as it has a closed homeomorphic copy of $mathbb{Q}$.
answered Nov 18 at 17:30
Henno Brandsma
105k346114
answered Nov 18 at 17:30
Henno Brandsma
105k346114
answered Nov 18 at 17:30
Henno Brandsma
105k346114
answered Nov 18 at 17:30
Henno Brandsma
105k346114
105k346114
If I'm not wrong, $X=mathbb{Q}times (0,infty)$, then considering $U_n=Xsetminus {r_n}times (0, infty)$ we got $cap U_n = phi$
– Offlaw
Nov 19 at 13:44
@Offlaw $mathbb{Q}times (0,infty)$ is irrelevant.
– Henno Brandsma
Nov 19 at 21:06
I got it. $X=mathbb{R} times (0,infty) cup mathbb{Q} times {0}$.
– Offlaw
Nov 20 at 1:29
@Offlaw indeed. That’s the space
– Henno Brandsma
Nov 20 at 4:24
add a comment |
If I'm not wrong, $X=mathbb{Q}times (0,infty)$, then considering $U_n=Xsetminus {r_n}times (0, infty)$ we got $cap U_n = phi$
– Offlaw
Nov 19 at 13:44
@Offlaw $mathbb{Q}times (0,infty)$ is irrelevant.
– Henno Brandsma
Nov 19 at 21:06
I got it. $X=mathbb{R} times (0,infty) cup mathbb{Q} times {0}$.
– Offlaw
Nov 20 at 1:29
@Offlaw indeed. That’s the space
– Henno Brandsma
Nov 20 at 4:24
If I'm not wrong, $X=mathbb{Q}times (0,infty)$, then considering $U_n=Xsetminus {r_n}times (0, infty)$ we got $cap U_n = phi$
– Offlaw
Nov 19 at 13:44
If I'm not wrong, $X=mathbb{Q}times (0,infty)$, then considering $U_n=Xsetminus {r_n}times (0, infty)$ we got $cap U_n = phi$
– Offlaw
Nov 19 at 13:44
@Offlaw $mathbb{Q}times (0,infty)$ is irrelevant.
– Henno Brandsma
Nov 19 at 21:06
@Offlaw $mathbb{Q}times (0,infty)$ is irrelevant.
– Henno Brandsma
Nov 19 at 21:06
I got it. $X=mathbb{R} times (0,infty) cup mathbb{Q} times {0}$.
– Offlaw
Nov 20 at 1:29
I got it. $X=mathbb{R} times (0,infty) cup mathbb{Q} times {0}$.
– Offlaw
Nov 20 at 1:29
@Offlaw indeed. That’s the space
– Henno Brandsma
Nov 20 at 4:24
@Offlaw indeed. That’s the space
– Henno Brandsma
Nov 20 at 4:24
add a comment |
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1
See Completely Metrizable Space and Baire Theorem.
– Dave L. Renfro
Nov 18 at 15:15