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Identity function continuous function between usual and discrete metric space
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Let $X = (mathbb R, d)$ be the usual real line and $Y =(mathbb R, d')$ be the set R with discrete metric.
Show that identity map from $X$ to $Y$ is not continuous but open as well as closed.
On the other hand, the identity map from $Y$ to $X$ is continuous which is neither open nor closed.
My attempt :
we know that for continuity we want to show that inverse image of the open set is open.
Any singleton set $x$ is open in discrete metric space and hence its inverse image under identity map is also $x$, which is not an open set in usual metric.
Hence, identity map is not continuous.
Again if identity map have domain with discrete metric then it is always continuous.
Am I correct??
Now for open map (closed), we want to show that image of open (closed) set is open (closed).
How to use this definition to prove open and closed.
Please help. Thank you.
metric-spaces continuity
edited Jul 15 at 9:18
José Carlos Santos
114k1698177
asked Jul 15 at 9:09
Golam biswas
399
add a comment |Â
up vote
0
down vote
favorite
Let $X = (mathbb R, d)$ be the usual real line and $Y =(mathbb R, d')$ be the set R with discrete metric.
Show that identity map from $X$ to $Y$ is not continuous but open as well as closed.
On the other hand, the identity map from $Y$ to $X$ is continuous which is neither open nor closed.
My attempt :
we know that for continuity we want to show that inverse image of the open set is open.
Any singleton set $x$ is open in discrete metric space and hence its inverse image under identity map is also $x$, which is not an open set in usual metric.
Hence, identity map is not continuous.
Again if identity map have domain with discrete metric then it is always continuous.
Am I correct??
Now for open map (closed), we want to show that image of open (closed) set is open (closed).
How to use this definition to prove open and closed.
Please help. Thank you.
metric-spaces continuity
edited Jul 15 at 9:18
José Carlos Santos
114k1698177
asked Jul 15 at 9:09
Golam biswas
399
Your answer is correct. Note that every map out of a discrete space is continuous. Note also that every subset of a discrete space is both open and closed.
– Tyrone
Aug 5 at 10:53
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $X = (mathbb R, d)$ be the usual real line and $Y =(mathbb R, d')$ be the set R with discrete metric.
Show that identity map from $X$ to $Y$ is not continuous but open as well as closed.
On the other hand, the identity map from $Y$ to $X$ is continuous which is neither open nor closed.
My attempt :
we know that for continuity we want to show that inverse image of the open set is open.
Any singleton set $x$ is open in discrete metric space and hence its inverse image under identity map is also $x$, which is not an open set in usual metric.
Hence, identity map is not continuous.
Again if identity map have domain with discrete metric then it is always continuous.
Am I correct??
Now for open map (closed), we want to show that image of open (closed) set is open (closed).
How to use this definition to prove open and closed.
Please help. Thank you.
metric-spaces continuity
edited Jul 15 at 9:18
José Carlos Santos
114k1698177
asked Jul 15 at 9:09
Golam biswas
399
Let $X = (mathbb R, d)$ be the usual real line and $Y =(mathbb R, d')$ be the set R with discrete metric.
Show that identity map from $X$ to $Y$ is not continuous but open as well as closed.
On the other hand, the identity map from $Y$ to $X$ is continuous which is neither open nor closed.
My attempt :
we know that for continuity we want to show that inverse image of the open set is open.
Any singleton set $x$ is open in discrete metric space and hence its inverse image under identity map is also $x$, which is not an open set in usual metric.
Hence, identity map is not continuous.
Again if identity map have domain with discrete metric then it is always continuous.
Am I correct??
Now for open map (closed), we want to show that image of open (closed) set is open (closed).
How to use this definition to prove open and closed.
Please help. Thank you.
metric-spaces continuity
edited Jul 15 at 9:18
José Carlos Santos
114k1698177
asked Jul 15 at 9:09
Golam biswas
399
edited Jul 15 at 9:18
José Carlos Santos
114k1698177
edited Jul 15 at 9:18
José Carlos Santos
114k1698177
edited Jul 15 at 9:18
José Carlos Santos
114k1698177
114k1698177
asked Jul 15 at 9:09
Golam biswas
399
asked Jul 15 at 9:09
Golam biswas
399
asked Jul 15 at 9:09
Golam biswas
399
399
Your answer is correct. Note that every map out of a discrete space is continuous. Note also that every subset of a discrete space is both open and closed.
– Tyrone
Aug 5 at 10:53
add a comment |Â
Your answer is correct. Note that every map out of a discrete space is continuous. Note also that every subset of a discrete space is both open and closed.
– Tyrone
Aug 5 at 10:53
Your answer is correct. Note that every map out of a discrete space is continuous. Note also that every subset of a discrete space is both open and closed.
– Tyrone
Aug 5 at 10:53
Your answer is correct. Note that every map out of a discrete space is continuous. Note also that every subset of a discrete space is both open and closed.
– Tyrone
Aug 5 at 10:53
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
What you did is correct.
Now, you have to keep in mind that, with respect to the discrete metric every set is open and every set is closed. In fact, given a set $S$, $S=bigcup_xin Sx$ and, since each singleton is open, $S$ is open. And since every set is open, every set is closed too. Therefore, every map from an arbitrary metric space into a discrete one is both open and closed.
In order to prove the the identity from $Y$ to $X$ is neither open nor closed, you can take, for instance, the set $A=[0,1)$. It is neither open nor closed in $X$ but it is both open and closed in $Y$. Since $operatornameid(A)$, $operatornameid$ is neither open nor closed.
answered Jul 15 at 9:14
José Carlos Santos
114k1698177
I ask my procedure is correct or not for continuous part?
– Golam biswas
Jul 15 at 10:20
It is. That's the first sentence of my answer.
– José Carlos Santos
Jul 15 at 10:26
In the second part, identity map Y to X is neither open nor closed. How to proceed?
– Golam biswas
Jul 15 at 10:28
1
I've added the answer to that question.
– José Carlos Santos
Jul 15 at 10:32
1
Yes, you are right.
– José Carlos Santos
Jul 15 at 10:44
 |Â
show 4 more comments
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
What you did is correct.
Now, you have to keep in mind that, with respect to the discrete metric every set is open and every set is closed. In fact, given a set $S$, $S=bigcup_xin Sx$ and, since each singleton is open, $S$ is open. And since every set is open, every set is closed too. Therefore, every map from an arbitrary metric space into a discrete one is both open and closed.
In order to prove the the identity from $Y$ to $X$ is neither open nor closed, you can take, for instance, the set $A=[0,1)$. It is neither open nor closed in $X$ but it is both open and closed in $Y$. Since $operatornameid(A)$, $operatornameid$ is neither open nor closed.
answered Jul 15 at 9:14
José Carlos Santos
114k1698177
I ask my procedure is correct or not for continuous part?
– Golam biswas
Jul 15 at 10:20
It is. That's the first sentence of my answer.
– José Carlos Santos
Jul 15 at 10:26
In the second part, identity map Y to X is neither open nor closed. How to proceed?
– Golam biswas
Jul 15 at 10:28
1
I've added the answer to that question.
– José Carlos Santos
Jul 15 at 10:32
1
Yes, you are right.
– José Carlos Santos
Jul 15 at 10:44
 |Â
show 4 more comments
up vote
1
down vote
accepted
What you did is correct.
Now, you have to keep in mind that, with respect to the discrete metric every set is open and every set is closed. In fact, given a set $S$, $S=bigcup_xin Sx$ and, since each singleton is open, $S$ is open. And since every set is open, every set is closed too. Therefore, every map from an arbitrary metric space into a discrete one is both open and closed.
In order to prove the the identity from $Y$ to $X$ is neither open nor closed, you can take, for instance, the set $A=[0,1)$. It is neither open nor closed in $X$ but it is both open and closed in $Y$. Since $operatornameid(A)$, $operatornameid$ is neither open nor closed.
answered Jul 15 at 9:14
José Carlos Santos
114k1698177
I ask my procedure is correct or not for continuous part?
– Golam biswas
Jul 15 at 10:20
It is. That's the first sentence of my answer.
– José Carlos Santos
Jul 15 at 10:26
In the second part, identity map Y to X is neither open nor closed. How to proceed?
– Golam biswas
Jul 15 at 10:28
1
I've added the answer to that question.
– José Carlos Santos
Jul 15 at 10:32
1
Yes, you are right.
– José Carlos Santos
Jul 15 at 10:44
 |Â
show 4 more comments
up vote
1
down vote
accepted
up vote
1
down vote
accepted
What you did is correct.
Now, you have to keep in mind that, with respect to the discrete metric every set is open and every set is closed. In fact, given a set $S$, $S=bigcup_xin Sx$ and, since each singleton is open, $S$ is open. And since every set is open, every set is closed too. Therefore, every map from an arbitrary metric space into a discrete one is both open and closed.
In order to prove the the identity from $Y$ to $X$ is neither open nor closed, you can take, for instance, the set $A=[0,1)$. It is neither open nor closed in $X$ but it is both open and closed in $Y$. Since $operatornameid(A)$, $operatornameid$ is neither open nor closed.
answered Jul 15 at 9:14
José Carlos Santos
114k1698177
What you did is correct.
Now, you have to keep in mind that, with respect to the discrete metric every set is open and every set is closed. In fact, given a set $S$, $S=bigcup_xin Sx$ and, since each singleton is open, $S$ is open. And since every set is open, every set is closed too. Therefore, every map from an arbitrary metric space into a discrete one is both open and closed.
In order to prove the the identity from $Y$ to $X$ is neither open nor closed, you can take, for instance, the set $A=[0,1)$. It is neither open nor closed in $X$ but it is both open and closed in $Y$. Since $operatornameid(A)$, $operatornameid$ is neither open nor closed.
answered Jul 15 at 9:14
José Carlos Santos
114k1698177
edited Jul 15 at 10:31
answered Jul 15 at 9:14
José Carlos Santos
114k1698177
answered Jul 15 at 9:14
José Carlos Santos
114k1698177
answered Jul 15 at 9:14
José Carlos Santos
114k1698177
114k1698177
I ask my procedure is correct or not for continuous part?
– Golam biswas
Jul 15 at 10:20
It is. That's the first sentence of my answer.
– José Carlos Santos
Jul 15 at 10:26
In the second part, identity map Y to X is neither open nor closed. How to proceed?
– Golam biswas
Jul 15 at 10:28
1
I've added the answer to that question.
– José Carlos Santos
Jul 15 at 10:32
1
Yes, you are right.
– José Carlos Santos
Jul 15 at 10:44
 |Â
show 4 more comments
I ask my procedure is correct or not for continuous part?
– Golam biswas
Jul 15 at 10:20
It is. That's the first sentence of my answer.
– José Carlos Santos
Jul 15 at 10:26
In the second part, identity map Y to X is neither open nor closed. How to proceed?
– Golam biswas
Jul 15 at 10:28
1
I've added the answer to that question.
– José Carlos Santos
Jul 15 at 10:32
1
Yes, you are right.
– José Carlos Santos
Jul 15 at 10:44
I ask my procedure is correct or not for continuous part?
– Golam biswas
Jul 15 at 10:20
I ask my procedure is correct or not for continuous part?
– Golam biswas
Jul 15 at 10:20
It is. That's the first sentence of my answer.
– José Carlos Santos
Jul 15 at 10:26
It is. That's the first sentence of my answer.
– José Carlos Santos
Jul 15 at 10:26
In the second part, identity map Y to X is neither open nor closed. How to proceed?
– Golam biswas
Jul 15 at 10:28
In the second part, identity map Y to X is neither open nor closed. How to proceed?
– Golam biswas
Jul 15 at 10:28
1
1
I've added the answer to that question.
– José Carlos Santos
Jul 15 at 10:32
I've added the answer to that question.
– José Carlos Santos
Jul 15 at 10:32
1
1
Yes, you are right.
– José Carlos Santos
Jul 15 at 10:44
Yes, you are right.
– José Carlos Santos
Jul 15 at 10:44
 |Â
show 4 more comments
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Your answer is correct. Note that every map out of a discrete space is continuous. Note also that every subset of a discrete space is both open and closed.
– Tyrone
Aug 5 at 10:53