Show continuous functions need not be open maps and open maps need not be continuous.
up vote
1
down vote
favorite
A function from one metric space to another is said to be an open map
if it maps open sets to open sets. Similarly one can define a closed
map.
1-Provide a continuous function which does not map an open set to another open set?
2- Provide a function which maps every open set to another one but it is not a continuous function?
real-analysis general-topology metric-spaces
asked Nov 13 at 14:04
Saeed
407110
add a comment |
up vote
1
down vote
favorite
A function from one metric space to another is said to be an open map
if it maps open sets to open sets. Similarly one can define a closed
map.
1-Provide a continuous function which does not map an open set to another open set?
2- Provide a function which maps every open set to another one but it is not a continuous function?
real-analysis general-topology metric-spaces
asked Nov 13 at 14:04
Saeed
407110
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
A function from one metric space to another is said to be an open map
if it maps open sets to open sets. Similarly one can define a closed
map.
1-Provide a continuous function which does not map an open set to another open set?
2- Provide a function which maps every open set to another one but it is not a continuous function?
real-analysis general-topology metric-spaces
asked Nov 13 at 14:04
Saeed
407110
A function from one metric space to another is said to be an open map
if it maps open sets to open sets. Similarly one can define a closed
map.
1-Provide a continuous function which does not map an open set to another open set?
2- Provide a function which maps every open set to another one but it is not a continuous function?
real-analysis general-topology metric-spaces
real-analysis general-topology metric-spaces
asked Nov 13 at 14:04
Saeed
407110
asked Nov 13 at 14:04
Saeed
407110
asked Nov 13 at 14:04
Saeed
407110
asked Nov 13 at 14:04
Saeed
407110
asked Nov 13 at 14:04
Saeed
407110
407110
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
up vote
3
down vote
accepted
For 1) take a constant function and let it be that singletons in the codomain are not open.
For 2) take the identity $mathbb Rtomathbb R$ where codomain is discrete (so $d(x,y)=0$ if $x=y$ and $d(x,y)=1$ otherwise) and domain is not discrete.
answered Nov 13 at 14:17
drhab
94.5k543125
For 1) You mean the range of the constant function be singletons? Then how it can be constant? Could you explain it or represent is using $f: X rightarrow Y$?
– Saeed
Nov 13 at 14:26
Every constant function is continuous. If $U$ is an open subset of $Y$ and $f$ is prescribed by $xmapsto c$ where $c$ is an element of $Y$ then $f^{-1}(U)=varnothing$ (this if $cnotin U$) or $f^{-1}(U)=X$ (this if $cin U$). The sets $varnothing$ and $X$ are always open. Further open set $U$ is sent to the not open set ${c}$ (provided that singletons in $Y$ are not open).
– drhab
Nov 13 at 14:31
add a comment |
up vote
2
down vote
A) To violate openness you can give any map which is not bijective.
B) Open maps which are not continuous
edited Nov 13 at 14:50
Mefitico
881117
answered Nov 13 at 14:31
Shubham
1,2931518
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
For 1) take a constant function and let it be that singletons in the codomain are not open.
For 2) take the identity $mathbb Rtomathbb R$ where codomain is discrete (so $d(x,y)=0$ if $x=y$ and $d(x,y)=1$ otherwise) and domain is not discrete.
answered Nov 13 at 14:17
drhab
94.5k543125
For 1) You mean the range of the constant function be singletons? Then how it can be constant? Could you explain it or represent is using $f: X rightarrow Y$?
– Saeed
Nov 13 at 14:26
Every constant function is continuous. If $U$ is an open subset of $Y$ and $f$ is prescribed by $xmapsto c$ where $c$ is an element of $Y$ then $f^{-1}(U)=varnothing$ (this if $cnotin U$) or $f^{-1}(U)=X$ (this if $cin U$). The sets $varnothing$ and $X$ are always open. Further open set $U$ is sent to the not open set ${c}$ (provided that singletons in $Y$ are not open).
– drhab
Nov 13 at 14:31
add a comment |
up vote
3
down vote
accepted
For 1) take a constant function and let it be that singletons in the codomain are not open.
For 2) take the identity $mathbb Rtomathbb R$ where codomain is discrete (so $d(x,y)=0$ if $x=y$ and $d(x,y)=1$ otherwise) and domain is not discrete.
answered Nov 13 at 14:17
drhab
94.5k543125
For 1) You mean the range of the constant function be singletons? Then how it can be constant? Could you explain it or represent is using $f: X rightarrow Y$?
– Saeed
Nov 13 at 14:26
Every constant function is continuous. If $U$ is an open subset of $Y$ and $f$ is prescribed by $xmapsto c$ where $c$ is an element of $Y$ then $f^{-1}(U)=varnothing$ (this if $cnotin U$) or $f^{-1}(U)=X$ (this if $cin U$). The sets $varnothing$ and $X$ are always open. Further open set $U$ is sent to the not open set ${c}$ (provided that singletons in $Y$ are not open).
– drhab
Nov 13 at 14:31
add a comment |
up vote
3
down vote
accepted
up vote
3
down vote
accepted
For 1) take a constant function and let it be that singletons in the codomain are not open.
For 2) take the identity $mathbb Rtomathbb R$ where codomain is discrete (so $d(x,y)=0$ if $x=y$ and $d(x,y)=1$ otherwise) and domain is not discrete.
answered Nov 13 at 14:17
drhab
94.5k543125
For 1) take a constant function and let it be that singletons in the codomain are not open.
For 2) take the identity $mathbb Rtomathbb R$ where codomain is discrete (so $d(x,y)=0$ if $x=y$ and $d(x,y)=1$ otherwise) and domain is not discrete.
answered Nov 13 at 14:17
drhab
94.5k543125
answered Nov 13 at 14:17
drhab
94.5k543125
answered Nov 13 at 14:17
drhab
94.5k543125
answered Nov 13 at 14:17
drhab
94.5k543125
94.5k543125
For 1) You mean the range of the constant function be singletons? Then how it can be constant? Could you explain it or represent is using $f: X rightarrow Y$?
– Saeed
Nov 13 at 14:26
Every constant function is continuous. If $U$ is an open subset of $Y$ and $f$ is prescribed by $xmapsto c$ where $c$ is an element of $Y$ then $f^{-1}(U)=varnothing$ (this if $cnotin U$) or $f^{-1}(U)=X$ (this if $cin U$). The sets $varnothing$ and $X$ are always open. Further open set $U$ is sent to the not open set ${c}$ (provided that singletons in $Y$ are not open).
– drhab
Nov 13 at 14:31
add a comment |
For 1) You mean the range of the constant function be singletons? Then how it can be constant? Could you explain it or represent is using $f: X rightarrow Y$?
– Saeed
Nov 13 at 14:26
Every constant function is continuous. If $U$ is an open subset of $Y$ and $f$ is prescribed by $xmapsto c$ where $c$ is an element of $Y$ then $f^{-1}(U)=varnothing$ (this if $cnotin U$) or $f^{-1}(U)=X$ (this if $cin U$). The sets $varnothing$ and $X$ are always open. Further open set $U$ is sent to the not open set ${c}$ (provided that singletons in $Y$ are not open).
– drhab
Nov 13 at 14:31
For 1) You mean the range of the constant function be singletons? Then how it can be constant? Could you explain it or represent is using $f: X rightarrow Y$?
– Saeed
Nov 13 at 14:26
For 1) You mean the range of the constant function be singletons? Then how it can be constant? Could you explain it or represent is using $f: X rightarrow Y$?
– Saeed
Nov 13 at 14:26
Every constant function is continuous. If $U$ is an open subset of $Y$ and $f$ is prescribed by $xmapsto c$ where $c$ is an element of $Y$ then $f^{-1}(U)=varnothing$ (this if $cnotin U$) or $f^{-1}(U)=X$ (this if $cin U$). The sets $varnothing$ and $X$ are always open. Further open set $U$ is sent to the not open set ${c}$ (provided that singletons in $Y$ are not open).
– drhab
Nov 13 at 14:31
Every constant function is continuous. If $U$ is an open subset of $Y$ and $f$ is prescribed by $xmapsto c$ where $c$ is an element of $Y$ then $f^{-1}(U)=varnothing$ (this if $cnotin U$) or $f^{-1}(U)=X$ (this if $cin U$). The sets $varnothing$ and $X$ are always open. Further open set $U$ is sent to the not open set ${c}$ (provided that singletons in $Y$ are not open).
– drhab
Nov 13 at 14:31
add a comment |
up vote
2
down vote
A) To violate openness you can give any map which is not bijective.
B) Open maps which are not continuous
edited Nov 13 at 14:50
Mefitico
881117
answered Nov 13 at 14:31
Shubham
1,2931518
add a comment |
up vote
2
down vote
A) To violate openness you can give any map which is not bijective.
B) Open maps which are not continuous
edited Nov 13 at 14:50
Mefitico
881117
answered Nov 13 at 14:31
Shubham
1,2931518
add a comment |
up vote
2
down vote
up vote
2
down vote
A) To violate openness you can give any map which is not bijective.
B) Open maps which are not continuous
edited Nov 13 at 14:50
Mefitico
881117
answered Nov 13 at 14:31
Shubham
1,2931518
A) To violate openness you can give any map which is not bijective.
B) Open maps which are not continuous
edited Nov 13 at 14:50
Mefitico
881117
answered Nov 13 at 14:31
Shubham
1,2931518
edited Nov 13 at 14:50
Mefitico
881117
edited Nov 13 at 14:50
Mefitico
881117
edited Nov 13 at 14:50
Mefitico
881117
881117
answered Nov 13 at 14:31
Shubham
1,2931518
answered Nov 13 at 14:31
Shubham
1,2931518
answered Nov 13 at 14:31
Shubham
1,2931518
1,2931518
add a comment |
add a comment |
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