construct two non-homeomorphic topological spaces
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Construct two topological spaces $X,Y$ satisfying that $X$ is not homeomorphic to $Y$, but $Xtimes[0,1]$ is homeomorphic to $Ytimes[0,1]$
i just solve the condition with $[0,1)$. Have no good idea of this construction.
Any idea is helpful. thanks
general-topology
edited Nov 17 at 13:25
Paul Frost
7,8041527
asked Nov 17 at 13:08
yufeng lu
111
|
show 2 more comments
up vote
2
down vote
favorite
Construct two topological spaces $X,Y$ satisfying that $X$ is not homeomorphic to $Y$, but $Xtimes[0,1]$ is homeomorphic to $Ytimes[0,1]$
i just solve the condition with $[0,1)$. Have no good idea of this construction.
Any idea is helpful. thanks
general-topology
edited Nov 17 at 13:25
Paul Frost
7,8041527
asked Nov 17 at 13:08
yufeng lu
111
Yes, for $[0,1)$ as the second factor, $X=[0,1)$ and $Y=[0,1]$ work, as a well known example. There are metric examples for $[0,1]$ too, though I don’t remember them now.
– Henno Brandsma
Nov 17 at 14:31
@HennoBrandsma I hope I'm not missing something, but $[0,1)times [0,1]$ is not compact unlike $[0,1]times [0,1]$, so they're not homeomorphic.
– Scientifica
Nov 17 at 15:51
@Scientifica That's right, but $[0,1] times [0,1)$ and $[0,1) times [0,1)$ are. The second factor $[0,1)$ case.
– Henno Brandsma
Nov 17 at 15:53
@HennoBrandsma Oh ok sry I missunderstood. Thank you.
– Scientifica
Nov 17 at 15:53
@Scientifica Ok, I was just referring to the OP mentioning he already did the $[0,1)$ case, which is classical.
– Henno Brandsma
Nov 17 at 15:55
|
show 2 more comments
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Construct two topological spaces $X,Y$ satisfying that $X$ is not homeomorphic to $Y$, but $Xtimes[0,1]$ is homeomorphic to $Ytimes[0,1]$
i just solve the condition with $[0,1)$. Have no good idea of this construction.
Any idea is helpful. thanks
general-topology
edited Nov 17 at 13:25
Paul Frost
7,8041527
asked Nov 17 at 13:08
yufeng lu
111
Construct two topological spaces $X,Y$ satisfying that $X$ is not homeomorphic to $Y$, but $Xtimes[0,1]$ is homeomorphic to $Ytimes[0,1]$
i just solve the condition with $[0,1)$. Have no good idea of this construction.
Any idea is helpful. thanks
general-topology
general-topology
edited Nov 17 at 13:25
Paul Frost
7,8041527
asked Nov 17 at 13:08
yufeng lu
111
edited Nov 17 at 13:25
Paul Frost
7,8041527
asked Nov 17 at 13:08
yufeng lu
111
edited Nov 17 at 13:25
Paul Frost
7,8041527
edited Nov 17 at 13:25
Paul Frost
7,8041527
edited Nov 17 at 13:25
Paul Frost
7,8041527
7,8041527
asked Nov 17 at 13:08
yufeng lu
111
asked Nov 17 at 13:08
yufeng lu
111
asked Nov 17 at 13:08
yufeng lu
111
111
Yes, for $[0,1)$ as the second factor, $X=[0,1)$ and $Y=[0,1]$ work, as a well known example. There are metric examples for $[0,1]$ too, though I don’t remember them now.
– Henno Brandsma
Nov 17 at 14:31
@HennoBrandsma I hope I'm not missing something, but $[0,1)times [0,1]$ is not compact unlike $[0,1]times [0,1]$, so they're not homeomorphic.
– Scientifica
Nov 17 at 15:51
@Scientifica That's right, but $[0,1] times [0,1)$ and $[0,1) times [0,1)$ are. The second factor $[0,1)$ case.
– Henno Brandsma
Nov 17 at 15:53
@HennoBrandsma Oh ok sry I missunderstood. Thank you.
– Scientifica
Nov 17 at 15:53
@Scientifica Ok, I was just referring to the OP mentioning he already did the $[0,1)$ case, which is classical.
– Henno Brandsma
Nov 17 at 15:55
|
show 2 more comments
Yes, for $[0,1)$ as the second factor, $X=[0,1)$ and $Y=[0,1]$ work, as a well known example. There are metric examples for $[0,1]$ too, though I don’t remember them now.
– Henno Brandsma
Nov 17 at 14:31
@HennoBrandsma I hope I'm not missing something, but $[0,1)times [0,1]$ is not compact unlike $[0,1]times [0,1]$, so they're not homeomorphic.
– Scientifica
Nov 17 at 15:51
@Scientifica That's right, but $[0,1] times [0,1)$ and $[0,1) times [0,1)$ are. The second factor $[0,1)$ case.
– Henno Brandsma
Nov 17 at 15:53
@HennoBrandsma Oh ok sry I missunderstood. Thank you.
– Scientifica
Nov 17 at 15:53
@Scientifica Ok, I was just referring to the OP mentioning he already did the $[0,1)$ case, which is classical.
– Henno Brandsma
Nov 17 at 15:55
Yes, for $[0,1)$ as the second factor, $X=[0,1)$ and $Y=[0,1]$ work, as a well known example. There are metric examples for $[0,1]$ too, though I don’t remember them now.
– Henno Brandsma
Nov 17 at 14:31
Yes, for $[0,1)$ as the second factor, $X=[0,1)$ and $Y=[0,1]$ work, as a well known example. There are metric examples for $[0,1]$ too, though I don’t remember them now.
– Henno Brandsma
Nov 17 at 14:31
@HennoBrandsma I hope I'm not missing something, but $[0,1)times [0,1]$ is not compact unlike $[0,1]times [0,1]$, so they're not homeomorphic.
– Scientifica
Nov 17 at 15:51
@HennoBrandsma I hope I'm not missing something, but $[0,1)times [0,1]$ is not compact unlike $[0,1]times [0,1]$, so they're not homeomorphic.
– Scientifica
Nov 17 at 15:51
@Scientifica That's right, but $[0,1] times [0,1)$ and $[0,1) times [0,1)$ are. The second factor $[0,1)$ case.
– Henno Brandsma
Nov 17 at 15:53
@Scientifica That's right, but $[0,1] times [0,1)$ and $[0,1) times [0,1)$ are. The second factor $[0,1)$ case.
– Henno Brandsma
Nov 17 at 15:53
@HennoBrandsma Oh ok sry I missunderstood. Thank you.
– Scientifica
Nov 17 at 15:53
@HennoBrandsma Oh ok sry I missunderstood. Thank you.
– Scientifica
Nov 17 at 15:53
@Scientifica Ok, I was just referring to the OP mentioning he already did the $[0,1)$ case, which is classical.
– Henno Brandsma
Nov 17 at 15:55
@Scientifica Ok, I was just referring to the OP mentioning he already did the $[0,1)$ case, which is classical.
– Henno Brandsma
Nov 17 at 15:55
|
show 2 more comments
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Yes, for $[0,1)$ as the second factor, $X=[0,1)$ and $Y=[0,1]$ work, as a well known example. There are metric examples for $[0,1]$ too, though I don’t remember them now.
– Henno Brandsma
Nov 17 at 14:31
@HennoBrandsma I hope I'm not missing something, but $[0,1)times [0,1]$ is not compact unlike $[0,1]times [0,1]$, so they're not homeomorphic.
– Scientifica
Nov 17 at 15:51
@Scientifica That's right, but $[0,1] times [0,1)$ and $[0,1) times [0,1)$ are. The second factor $[0,1)$ case.
– Henno Brandsma
Nov 17 at 15:53
@HennoBrandsma Oh ok sry I missunderstood. Thank you.
– Scientifica
Nov 17 at 15:53
@Scientifica Ok, I was just referring to the OP mentioning he already did the $[0,1)$ case, which is classical.
– Henno Brandsma
Nov 17 at 15:55