Prove: Euclidean and Cofinite topologies on $mathbb{R}$ are not Homeomorphic
Let $mathbb{R}$ denote the real numbers with the standard topology and $mathbb{R}_f$ denote the real numbers with the cofinite topology.
Explicitly show that the identity map between these two spaces is not a homeomorphism.
How do we know there is no homeomorphism between them? (think about a property that must get preserved under a homeomorphism)
Since I am a new with topology. Please give me some advice.
Thanks
general-topology
edited Nov 24 at 18:19
Cassius12
10611
asked Oct 16 '16 at 23:22
Vui Tinh
658
add a comment |
Let $mathbb{R}$ denote the real numbers with the standard topology and $mathbb{R}_f$ denote the real numbers with the cofinite topology.
Explicitly show that the identity map between these two spaces is not a homeomorphism.
How do we know there is no homeomorphism between them? (think about a property that must get preserved under a homeomorphism)
Since I am a new with topology. Please give me some advice.
Thanks
general-topology
edited Nov 24 at 18:19
Cassius12
10611
asked Oct 16 '16 at 23:22
Vui Tinh
658
Well, what properties does a map need to satisfy to be a homeomorphism?
– Mike Pierce
Oct 16 '16 at 23:29
If $tau_1$ and $tau_2$ are two topologies on $X$, the definition yields that the identity map $id_X:(X,tau_1)to (X,tau_2)$ is a homeomorphism if and only if $tau_1=tau_2$.
– user228113
Oct 16 '16 at 23:40
For the second part: how many infinite closed sets are there are in $R$ and in $R_f$?
– Rob Arthan
Oct 16 '16 at 23:44
For the second part you could show that $Bbb R_f$ is compact, while $Bbb R$ is not.
– Brian M. Scott
Oct 17 '16 at 10:20
add a comment |
Let $mathbb{R}$ denote the real numbers with the standard topology and $mathbb{R}_f$ denote the real numbers with the cofinite topology.
Explicitly show that the identity map between these two spaces is not a homeomorphism.
How do we know there is no homeomorphism between them? (think about a property that must get preserved under a homeomorphism)
Since I am a new with topology. Please give me some advice.
Thanks
general-topology
edited Nov 24 at 18:19
Cassius12
10611
asked Oct 16 '16 at 23:22
Vui Tinh
658
Let $mathbb{R}$ denote the real numbers with the standard topology and $mathbb{R}_f$ denote the real numbers with the cofinite topology.
Explicitly show that the identity map between these two spaces is not a homeomorphism.
How do we know there is no homeomorphism between them? (think about a property that must get preserved under a homeomorphism)
Since I am a new with topology. Please give me some advice.
Thanks
general-topology
general-topology
edited Nov 24 at 18:19
Cassius12
10611
asked Oct 16 '16 at 23:22
Vui Tinh
658
edited Nov 24 at 18:19
Cassius12
10611
asked Oct 16 '16 at 23:22
Vui Tinh
658
edited Nov 24 at 18:19
Cassius12
10611
edited Nov 24 at 18:19
Cassius12
10611
edited Nov 24 at 18:19
Cassius12
10611
10611
asked Oct 16 '16 at 23:22
Vui Tinh
658
asked Oct 16 '16 at 23:22
Vui Tinh
658
asked Oct 16 '16 at 23:22
Vui Tinh
658
658
Well, what properties does a map need to satisfy to be a homeomorphism?
– Mike Pierce
Oct 16 '16 at 23:29
If $tau_1$ and $tau_2$ are two topologies on $X$, the definition yields that the identity map $id_X:(X,tau_1)to (X,tau_2)$ is a homeomorphism if and only if $tau_1=tau_2$.
– user228113
Oct 16 '16 at 23:40
For the second part: how many infinite closed sets are there are in $R$ and in $R_f$?
– Rob Arthan
Oct 16 '16 at 23:44
For the second part you could show that $Bbb R_f$ is compact, while $Bbb R$ is not.
– Brian M. Scott
Oct 17 '16 at 10:20
add a comment |
Well, what properties does a map need to satisfy to be a homeomorphism?
– Mike Pierce
Oct 16 '16 at 23:29
If $tau_1$ and $tau_2$ are two topologies on $X$, the definition yields that the identity map $id_X:(X,tau_1)to (X,tau_2)$ is a homeomorphism if and only if $tau_1=tau_2$.
– user228113
Oct 16 '16 at 23:40
For the second part: how many infinite closed sets are there are in $R$ and in $R_f$?
– Rob Arthan
Oct 16 '16 at 23:44
For the second part you could show that $Bbb R_f$ is compact, while $Bbb R$ is not.
– Brian M. Scott
Oct 17 '16 at 10:20
Well, what properties does a map need to satisfy to be a homeomorphism?
– Mike Pierce
Oct 16 '16 at 23:29
Well, what properties does a map need to satisfy to be a homeomorphism?
– Mike Pierce
Oct 16 '16 at 23:29
If $tau_1$ and $tau_2$ are two topologies on $X$, the definition yields that the identity map $id_X:(X,tau_1)to (X,tau_2)$ is a homeomorphism if and only if $tau_1=tau_2$.
– user228113
Oct 16 '16 at 23:40
If $tau_1$ and $tau_2$ are two topologies on $X$, the definition yields that the identity map $id_X:(X,tau_1)to (X,tau_2)$ is a homeomorphism if and only if $tau_1=tau_2$.
– user228113
Oct 16 '16 at 23:40
For the second part: how many infinite closed sets are there are in $R$ and in $R_f$?
– Rob Arthan
Oct 16 '16 at 23:44
For the second part: how many infinite closed sets are there are in $R$ and in $R_f$?
– Rob Arthan
Oct 16 '16 at 23:44
For the second part you could show that $Bbb R_f$ is compact, while $Bbb R$ is not.
– Brian M. Scott
Oct 17 '16 at 10:20
For the second part you could show that $Bbb R_f$ is compact, while $Bbb R$ is not.
– Brian M. Scott
Oct 17 '16 at 10:20
add a comment |
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Well, what properties does a map need to satisfy to be a homeomorphism?
– Mike Pierce
Oct 16 '16 at 23:29
If $tau_1$ and $tau_2$ are two topologies on $X$, the definition yields that the identity map $id_X:(X,tau_1)to (X,tau_2)$ is a homeomorphism if and only if $tau_1=tau_2$.
– user228113
Oct 16 '16 at 23:40
For the second part: how many infinite closed sets are there are in $R$ and in $R_f$?
– Rob Arthan
Oct 16 '16 at 23:44
For the second part you could show that $Bbb R_f$ is compact, while $Bbb R$ is not.
– Brian M. Scott
Oct 17 '16 at 10:20