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What do you call this property involving a function between two complete metric spaces?
Clash Royale CLAN TAG#URR8PPP
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I have a notion, for which I am not able to find any reference name, as I am not that familiar with these concepts. Please help me by pointing to a definition for the below scenario.
Is there a name for the following property of the setup?
There is a a continuous and onto function $e : A to B$, $A$ and $B$ being two different complete metric spaces.
For any element $bin B$, and for any element $a in e^-1(b)$,
(where $e^-1(b)$ is the pre-image of the element $b$ in the function $e$),
For every punctured neighbourhood of $b$ denoted as $P_epsilon(b)$, the pre-image $e^-1(P_epsilon(b))$ contains a sequence $a_n$, such that $a_n to a$
real-analysis general-topology analysis functions metric-spaces
asked Jul 28 at 15:43
Rajesh Dachiraju
1,39922564
 |Â
show 3 more comments
up vote
9
down vote
favorite
I have a notion, for which I am not able to find any reference name, as I am not that familiar with these concepts. Please help me by pointing to a definition for the below scenario.
Is there a name for the following property of the setup?
There is a a continuous and onto function $e : A to B$, $A$ and $B$ being two different complete metric spaces.
For any element $bin B$, and for any element $a in e^-1(b)$,
(where $e^-1(b)$ is the pre-image of the element $b$ in the function $e$),
For every punctured neighbourhood of $b$ denoted as $P_epsilon(b)$, the pre-image $e^-1(P_epsilon(b))$ contains a sequence $a_n$, such that $a_n to a$
real-analysis general-topology analysis functions metric-spaces
asked Jul 28 at 15:43
Rajesh Dachiraju
1,39922564
By $e(b_n)$ did you mean something like "an element of $e^-1(b_n)$?
– Mark S.
Jul 28 at 15:47
If the $b_n$ are elements of $B$, it does not make sense to look at $e(b_n)$.
– Keenan Kidwell
Jul 28 at 15:47
@KeenanKidwell : Sorry I messed up. Let me correct it.
– Rajesh Dachiraju
Jul 28 at 15:49
@MarkS. : I have corrected.
– Rajesh Dachiraju
Jul 28 at 16:03
@KeenanKidwell : I have corrected.
– Rajesh Dachiraju
Jul 28 at 16:03
 |Â
show 3 more comments
up vote
9
down vote
favorite
up vote
9
down vote
favorite
I have a notion, for which I am not able to find any reference name, as I am not that familiar with these concepts. Please help me by pointing to a definition for the below scenario.
Is there a name for the following property of the setup?
There is a a continuous and onto function $e : A to B$, $A$ and $B$ being two different complete metric spaces.
For any element $bin B$, and for any element $a in e^-1(b)$,
(where $e^-1(b)$ is the pre-image of the element $b$ in the function $e$),
For every punctured neighbourhood of $b$ denoted as $P_epsilon(b)$, the pre-image $e^-1(P_epsilon(b))$ contains a sequence $a_n$, such that $a_n to a$
real-analysis general-topology analysis functions metric-spaces
asked Jul 28 at 15:43
Rajesh Dachiraju
1,39922564
I have a notion, for which I am not able to find any reference name, as I am not that familiar with these concepts. Please help me by pointing to a definition for the below scenario.
Is there a name for the following property of the setup?
There is a a continuous and onto function $e : A to B$, $A$ and $B$ being two different complete metric spaces.
For any element $bin B$, and for any element $a in e^-1(b)$,
(where $e^-1(b)$ is the pre-image of the element $b$ in the function $e$),
For every punctured neighbourhood of $b$ denoted as $P_epsilon(b)$, the pre-image $e^-1(P_epsilon(b))$ contains a sequence $a_n$, such that $a_n to a$
real-analysis general-topology analysis functions metric-spaces
asked Jul 28 at 15:43
Rajesh Dachiraju
1,39922564
edited Jul 28 at 21:14
asked Jul 28 at 15:43
Rajesh Dachiraju
1,39922564
asked Jul 28 at 15:43
Rajesh Dachiraju
1,39922564
asked Jul 28 at 15:43
Rajesh Dachiraju
1,39922564
1,39922564
By $e(b_n)$ did you mean something like "an element of $e^-1(b_n)$?
– Mark S.
Jul 28 at 15:47
If the $b_n$ are elements of $B$, it does not make sense to look at $e(b_n)$.
– Keenan Kidwell
Jul 28 at 15:47
@KeenanKidwell : Sorry I messed up. Let me correct it.
– Rajesh Dachiraju
Jul 28 at 15:49
@MarkS. : I have corrected.
– Rajesh Dachiraju
Jul 28 at 16:03
@KeenanKidwell : I have corrected.
– Rajesh Dachiraju
Jul 28 at 16:03
 |Â
show 3 more comments
By $e(b_n)$ did you mean something like "an element of $e^-1(b_n)$?
– Mark S.
Jul 28 at 15:47
If the $b_n$ are elements of $B$, it does not make sense to look at $e(b_n)$.
– Keenan Kidwell
Jul 28 at 15:47
@KeenanKidwell : Sorry I messed up. Let me correct it.
– Rajesh Dachiraju
Jul 28 at 15:49
@MarkS. : I have corrected.
– Rajesh Dachiraju
Jul 28 at 16:03
@KeenanKidwell : I have corrected.
– Rajesh Dachiraju
Jul 28 at 16:03
By $e(b_n)$ did you mean something like "an element of $e^-1(b_n)$?
– Mark S.
Jul 28 at 15:47
By $e(b_n)$ did you mean something like "an element of $e^-1(b_n)$?
– Mark S.
Jul 28 at 15:47
If the $b_n$ are elements of $B$, it does not make sense to look at $e(b_n)$.
– Keenan Kidwell
Jul 28 at 15:47
If the $b_n$ are elements of $B$, it does not make sense to look at $e(b_n)$.
– Keenan Kidwell
Jul 28 at 15:47
@KeenanKidwell : Sorry I messed up. Let me correct it.
– Rajesh Dachiraju
Jul 28 at 15:49
@KeenanKidwell : Sorry I messed up. Let me correct it.
– Rajesh Dachiraju
Jul 28 at 15:49
@MarkS. : I have corrected.
– Rajesh Dachiraju
Jul 28 at 16:03
@MarkS. : I have corrected.
– Rajesh Dachiraju
Jul 28 at 16:03
@KeenanKidwell : I have corrected.
– Rajesh Dachiraju
Jul 28 at 16:03
@KeenanKidwell : I have corrected.
– Rajesh Dachiraju
Jul 28 at 16:03
 |Â
show 3 more comments
2 Answers
2
active
oldest
votes
up vote
3
down vote
accepted
Let us call the property described in question as Property P. Continuing the observations made in Stefan Böttner's answer we get the following.
Observation. Let $A$ and $B$ be metric spaces and $ecolon Ato B$ be a continuous function. Then $e$ has the Property P if and only if $A$ has no isolated points and $e$ is nowhere constant. (I am not sure to which extent this is a standard therm, but it seems to ba a natural name for this. It also appears in some books.)
By nowhere constant I mean that there is no non-empty open subset $Usubseteq A$ such that $e|_U$ is constant.
Proof. $boxedRightarrow$ If $a$ is any point of $A$ then property P implies existence of a sequence converging to $a$, hence $a$ is not isolated.
Let $Uneemptyset$ and $ain U$. Let $b=e(a)$. Let $varepsilon>0$. The set $e^-1[P_varepsilon(b)]$ contains sequence $(a_n)$ converging to $a$. Starting with some $n_0$, terms of these sequence belong to $U$ and we also have $e(a_n)ne e(b)$. Therefore $e|_U$
$boxedLeftarrow$ Let $B(b,varepsilon)$ be the open ball around $b$. By continuity we get that there is a $delta$ such that $B(a,delta)subseteq e^-1[B(b,varepsilon)]$. Let us choose $n_0$ with $1/n_0<delta$. Then each ball $B(a,frac1n_0+k)$ lies inside $e^-1[B(b,varepsilon)]$. And since the function $e$ is not constant on this ball, we can choose $a_kin B(a,frac1n_0+k)$ such that $e(a_k)ne e(a)$, i.e., $a_kin e^-1[P_varepsilon(b)]$.
answered yesterday
Martin Sleziak
43.5k6113259
add a comment |Â
up vote
3
down vote
By continuity, the pre-image of a neighborhood of $b$ should be a neighborhood of $a$. Now if $a$ is a discrete point in $A$ (i.e. the set $asubset A$ is open), that neighborhood may be trivial such that the pre-image of the punctured neighborhood of $b$ does not contain any other points in the vicinity of $a$. Otherwise you should be able to find such a sequence. Hence that might be the concept you are looking for.
answered Aug 3 at 14:03
Stefan Böttner
33916
1
This is not exactly what he desires. Note that continuity does not have the aforementioned property for punctured neighborhoods in $B$
– Rushabh Mehta
Aug 6 at 19:12
An example: Suppose $A=mathbbR$ and $$forall c>0inmathbbRquad e(c) = b_0$$$$forall -1leq cleq 0inmathbbRquad e(c)=b_1$$$$forall -c<-1inmathbbRquad e(c)=b_2$$ Note that if $exists P_epsilon(b_0)$ that contains $b_2$ but not $b_1$, OP's property doesn't hold, despite there being no discrete points in $A$.
– Rushabh Mehta
Aug 6 at 19:18
@RushabhMehta Your function $e$ is neither continuous nor surjective (onto).
– Stefan Böttner
Aug 7 at 11:39
Well that's fair, I'll have to think about that
– Rushabh Mehta
Aug 7 at 11:39
Let $U_b$ be open such that $b in U_b subset B$. Then $U_a=e^-1(U_b)$ is open and $a in U_a subset A$. Unless $a$ was a discrete point, you should be able to find points $a_1,a_2,ldots in U_asetminusa$ such that $a_nrightarrow a$.
– Stefan Böttner
Aug 7 at 12:01
 |Â
show 8 more comments
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
Let us call the property described in question as Property P. Continuing the observations made in Stefan Böttner's answer we get the following.
Observation. Let $A$ and $B$ be metric spaces and $ecolon Ato B$ be a continuous function. Then $e$ has the Property P if and only if $A$ has no isolated points and $e$ is nowhere constant. (I am not sure to which extent this is a standard therm, but it seems to ba a natural name for this. It also appears in some books.)
By nowhere constant I mean that there is no non-empty open subset $Usubseteq A$ such that $e|_U$ is constant.
Proof. $boxedRightarrow$ If $a$ is any point of $A$ then property P implies existence of a sequence converging to $a$, hence $a$ is not isolated.
Let $Uneemptyset$ and $ain U$. Let $b=e(a)$. Let $varepsilon>0$. The set $e^-1[P_varepsilon(b)]$ contains sequence $(a_n)$ converging to $a$. Starting with some $n_0$, terms of these sequence belong to $U$ and we also have $e(a_n)ne e(b)$. Therefore $e|_U$
$boxedLeftarrow$ Let $B(b,varepsilon)$ be the open ball around $b$. By continuity we get that there is a $delta$ such that $B(a,delta)subseteq e^-1[B(b,varepsilon)]$. Let us choose $n_0$ with $1/n_0<delta$. Then each ball $B(a,frac1n_0+k)$ lies inside $e^-1[B(b,varepsilon)]$. And since the function $e$ is not constant on this ball, we can choose $a_kin B(a,frac1n_0+k)$ such that $e(a_k)ne e(a)$, i.e., $a_kin e^-1[P_varepsilon(b)]$.
answered yesterday
Martin Sleziak
43.5k6113259
add a comment |Â
up vote
3
down vote
accepted
Let us call the property described in question as Property P. Continuing the observations made in Stefan Böttner's answer we get the following.
Observation. Let $A$ and $B$ be metric spaces and $ecolon Ato B$ be a continuous function. Then $e$ has the Property P if and only if $A$ has no isolated points and $e$ is nowhere constant. (I am not sure to which extent this is a standard therm, but it seems to ba a natural name for this. It also appears in some books.)
By nowhere constant I mean that there is no non-empty open subset $Usubseteq A$ such that $e|_U$ is constant.
Proof. $boxedRightarrow$ If $a$ is any point of $A$ then property P implies existence of a sequence converging to $a$, hence $a$ is not isolated.
Let $Uneemptyset$ and $ain U$. Let $b=e(a)$. Let $varepsilon>0$. The set $e^-1[P_varepsilon(b)]$ contains sequence $(a_n)$ converging to $a$. Starting with some $n_0$, terms of these sequence belong to $U$ and we also have $e(a_n)ne e(b)$. Therefore $e|_U$
$boxedLeftarrow$ Let $B(b,varepsilon)$ be the open ball around $b$. By continuity we get that there is a $delta$ such that $B(a,delta)subseteq e^-1[B(b,varepsilon)]$. Let us choose $n_0$ with $1/n_0<delta$. Then each ball $B(a,frac1n_0+k)$ lies inside $e^-1[B(b,varepsilon)]$. And since the function $e$ is not constant on this ball, we can choose $a_kin B(a,frac1n_0+k)$ such that $e(a_k)ne e(a)$, i.e., $a_kin e^-1[P_varepsilon(b)]$.
answered yesterday
Martin Sleziak
43.5k6113259
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
Let us call the property described in question as Property P. Continuing the observations made in Stefan Böttner's answer we get the following.
Observation. Let $A$ and $B$ be metric spaces and $ecolon Ato B$ be a continuous function. Then $e$ has the Property P if and only if $A$ has no isolated points and $e$ is nowhere constant. (I am not sure to which extent this is a standard therm, but it seems to ba a natural name for this. It also appears in some books.)
By nowhere constant I mean that there is no non-empty open subset $Usubseteq A$ such that $e|_U$ is constant.
Proof. $boxedRightarrow$ If $a$ is any point of $A$ then property P implies existence of a sequence converging to $a$, hence $a$ is not isolated.
Let $Uneemptyset$ and $ain U$. Let $b=e(a)$. Let $varepsilon>0$. The set $e^-1[P_varepsilon(b)]$ contains sequence $(a_n)$ converging to $a$. Starting with some $n_0$, terms of these sequence belong to $U$ and we also have $e(a_n)ne e(b)$. Therefore $e|_U$
$boxedLeftarrow$ Let $B(b,varepsilon)$ be the open ball around $b$. By continuity we get that there is a $delta$ such that $B(a,delta)subseteq e^-1[B(b,varepsilon)]$. Let us choose $n_0$ with $1/n_0<delta$. Then each ball $B(a,frac1n_0+k)$ lies inside $e^-1[B(b,varepsilon)]$. And since the function $e$ is not constant on this ball, we can choose $a_kin B(a,frac1n_0+k)$ such that $e(a_k)ne e(a)$, i.e., $a_kin e^-1[P_varepsilon(b)]$.
answered yesterday
Martin Sleziak
43.5k6113259
Let us call the property described in question as Property P. Continuing the observations made in Stefan Böttner's answer we get the following.
Observation. Let $A$ and $B$ be metric spaces and $ecolon Ato B$ be a continuous function. Then $e$ has the Property P if and only if $A$ has no isolated points and $e$ is nowhere constant. (I am not sure to which extent this is a standard therm, but it seems to ba a natural name for this. It also appears in some books.)
By nowhere constant I mean that there is no non-empty open subset $Usubseteq A$ such that $e|_U$ is constant.
Proof. $boxedRightarrow$ If $a$ is any point of $A$ then property P implies existence of a sequence converging to $a$, hence $a$ is not isolated.
Let $Uneemptyset$ and $ain U$. Let $b=e(a)$. Let $varepsilon>0$. The set $e^-1[P_varepsilon(b)]$ contains sequence $(a_n)$ converging to $a$. Starting with some $n_0$, terms of these sequence belong to $U$ and we also have $e(a_n)ne e(b)$. Therefore $e|_U$
$boxedLeftarrow$ Let $B(b,varepsilon)$ be the open ball around $b$. By continuity we get that there is a $delta$ such that $B(a,delta)subseteq e^-1[B(b,varepsilon)]$. Let us choose $n_0$ with $1/n_0<delta$. Then each ball $B(a,frac1n_0+k)$ lies inside $e^-1[B(b,varepsilon)]$. And since the function $e$ is not constant on this ball, we can choose $a_kin B(a,frac1n_0+k)$ such that $e(a_k)ne e(a)$, i.e., $a_kin e^-1[P_varepsilon(b)]$.
answered yesterday
Martin Sleziak
43.5k6113259
answered yesterday
Martin Sleziak
43.5k6113259
answered yesterday
Martin Sleziak
43.5k6113259
answered yesterday
Martin Sleziak
43.5k6113259
43.5k6113259
add a comment |Â
add a comment |Â
up vote
3
down vote
By continuity, the pre-image of a neighborhood of $b$ should be a neighborhood of $a$. Now if $a$ is a discrete point in $A$ (i.e. the set $asubset A$ is open), that neighborhood may be trivial such that the pre-image of the punctured neighborhood of $b$ does not contain any other points in the vicinity of $a$. Otherwise you should be able to find such a sequence. Hence that might be the concept you are looking for.
answered Aug 3 at 14:03
Stefan Böttner
33916
1
This is not exactly what he desires. Note that continuity does not have the aforementioned property for punctured neighborhoods in $B$
– Rushabh Mehta
Aug 6 at 19:12
An example: Suppose $A=mathbbR$ and $$forall c>0inmathbbRquad e(c) = b_0$$$$forall -1leq cleq 0inmathbbRquad e(c)=b_1$$$$forall -c<-1inmathbbRquad e(c)=b_2$$ Note that if $exists P_epsilon(b_0)$ that contains $b_2$ but not $b_1$, OP's property doesn't hold, despite there being no discrete points in $A$.
– Rushabh Mehta
Aug 6 at 19:18
@RushabhMehta Your function $e$ is neither continuous nor surjective (onto).
– Stefan Böttner
Aug 7 at 11:39
Well that's fair, I'll have to think about that
– Rushabh Mehta
Aug 7 at 11:39
Let $U_b$ be open such that $b in U_b subset B$. Then $U_a=e^-1(U_b)$ is open and $a in U_a subset A$. Unless $a$ was a discrete point, you should be able to find points $a_1,a_2,ldots in U_asetminusa$ such that $a_nrightarrow a$.
– Stefan Böttner
Aug 7 at 12:01
 |Â
show 8 more comments
up vote
3
down vote
By continuity, the pre-image of a neighborhood of $b$ should be a neighborhood of $a$. Now if $a$ is a discrete point in $A$ (i.e. the set $asubset A$ is open), that neighborhood may be trivial such that the pre-image of the punctured neighborhood of $b$ does not contain any other points in the vicinity of $a$. Otherwise you should be able to find such a sequence. Hence that might be the concept you are looking for.
answered Aug 3 at 14:03
Stefan Böttner
33916
1
This is not exactly what he desires. Note that continuity does not have the aforementioned property for punctured neighborhoods in $B$
– Rushabh Mehta
Aug 6 at 19:12
An example: Suppose $A=mathbbR$ and $$forall c>0inmathbbRquad e(c) = b_0$$$$forall -1leq cleq 0inmathbbRquad e(c)=b_1$$$$forall -c<-1inmathbbRquad e(c)=b_2$$ Note that if $exists P_epsilon(b_0)$ that contains $b_2$ but not $b_1$, OP's property doesn't hold, despite there being no discrete points in $A$.
– Rushabh Mehta
Aug 6 at 19:18
@RushabhMehta Your function $e$ is neither continuous nor surjective (onto).
– Stefan Böttner
Aug 7 at 11:39
Well that's fair, I'll have to think about that
– Rushabh Mehta
Aug 7 at 11:39
Let $U_b$ be open such that $b in U_b subset B$. Then $U_a=e^-1(U_b)$ is open and $a in U_a subset A$. Unless $a$ was a discrete point, you should be able to find points $a_1,a_2,ldots in U_asetminusa$ such that $a_nrightarrow a$.
– Stefan Böttner
Aug 7 at 12:01
 |Â
show 8 more comments
up vote
3
down vote
up vote
3
down vote
By continuity, the pre-image of a neighborhood of $b$ should be a neighborhood of $a$. Now if $a$ is a discrete point in $A$ (i.e. the set $asubset A$ is open), that neighborhood may be trivial such that the pre-image of the punctured neighborhood of $b$ does not contain any other points in the vicinity of $a$. Otherwise you should be able to find such a sequence. Hence that might be the concept you are looking for.
answered Aug 3 at 14:03
Stefan Böttner
33916
By continuity, the pre-image of a neighborhood of $b$ should be a neighborhood of $a$. Now if $a$ is a discrete point in $A$ (i.e. the set $asubset A$ is open), that neighborhood may be trivial such that the pre-image of the punctured neighborhood of $b$ does not contain any other points in the vicinity of $a$. Otherwise you should be able to find such a sequence. Hence that might be the concept you are looking for.
answered Aug 3 at 14:03
Stefan Böttner
33916
answered Aug 3 at 14:03
Stefan Böttner
33916
answered Aug 3 at 14:03
Stefan Böttner
33916
answered Aug 3 at 14:03
Stefan Böttner
33916
33916
1
This is not exactly what he desires. Note that continuity does not have the aforementioned property for punctured neighborhoods in $B$
– Rushabh Mehta
Aug 6 at 19:12
An example: Suppose $A=mathbbR$ and $$forall c>0inmathbbRquad e(c) = b_0$$$$forall -1leq cleq 0inmathbbRquad e(c)=b_1$$$$forall -c<-1inmathbbRquad e(c)=b_2$$ Note that if $exists P_epsilon(b_0)$ that contains $b_2$ but not $b_1$, OP's property doesn't hold, despite there being no discrete points in $A$.
– Rushabh Mehta
Aug 6 at 19:18
@RushabhMehta Your function $e$ is neither continuous nor surjective (onto).
– Stefan Böttner
Aug 7 at 11:39
Well that's fair, I'll have to think about that
– Rushabh Mehta
Aug 7 at 11:39
Let $U_b$ be open such that $b in U_b subset B$. Then $U_a=e^-1(U_b)$ is open and $a in U_a subset A$. Unless $a$ was a discrete point, you should be able to find points $a_1,a_2,ldots in U_asetminusa$ such that $a_nrightarrow a$.
– Stefan Böttner
Aug 7 at 12:01
 |Â
show 8 more comments
1
This is not exactly what he desires. Note that continuity does not have the aforementioned property for punctured neighborhoods in $B$
– Rushabh Mehta
Aug 6 at 19:12
An example: Suppose $A=mathbbR$ and $$forall c>0inmathbbRquad e(c) = b_0$$$$forall -1leq cleq 0inmathbbRquad e(c)=b_1$$$$forall -c<-1inmathbbRquad e(c)=b_2$$ Note that if $exists P_epsilon(b_0)$ that contains $b_2$ but not $b_1$, OP's property doesn't hold, despite there being no discrete points in $A$.
– Rushabh Mehta
Aug 6 at 19:18
@RushabhMehta Your function $e$ is neither continuous nor surjective (onto).
– Stefan Böttner
Aug 7 at 11:39
Well that's fair, I'll have to think about that
– Rushabh Mehta
Aug 7 at 11:39
Let $U_b$ be open such that $b in U_b subset B$. Then $U_a=e^-1(U_b)$ is open and $a in U_a subset A$. Unless $a$ was a discrete point, you should be able to find points $a_1,a_2,ldots in U_asetminusa$ such that $a_nrightarrow a$.
– Stefan Böttner
Aug 7 at 12:01
1
1
This is not exactly what he desires. Note that continuity does not have the aforementioned property for punctured neighborhoods in $B$
– Rushabh Mehta
Aug 6 at 19:12
This is not exactly what he desires. Note that continuity does not have the aforementioned property for punctured neighborhoods in $B$
– Rushabh Mehta
Aug 6 at 19:12
An example: Suppose $A=mathbbR$ and $$forall c>0inmathbbRquad e(c) = b_0$$$$forall -1leq cleq 0inmathbbRquad e(c)=b_1$$$$forall -c<-1inmathbbRquad e(c)=b_2$$ Note that if $exists P_epsilon(b_0)$ that contains $b_2$ but not $b_1$, OP's property doesn't hold, despite there being no discrete points in $A$.
– Rushabh Mehta
Aug 6 at 19:18
An example: Suppose $A=mathbbR$ and $$forall c>0inmathbbRquad e(c) = b_0$$$$forall -1leq cleq 0inmathbbRquad e(c)=b_1$$$$forall -c<-1inmathbbRquad e(c)=b_2$$ Note that if $exists P_epsilon(b_0)$ that contains $b_2$ but not $b_1$, OP's property doesn't hold, despite there being no discrete points in $A$.
– Rushabh Mehta
Aug 6 at 19:18
@RushabhMehta Your function $e$ is neither continuous nor surjective (onto).
– Stefan Böttner
Aug 7 at 11:39
@RushabhMehta Your function $e$ is neither continuous nor surjective (onto).
– Stefan Böttner
Aug 7 at 11:39
Well that's fair, I'll have to think about that
– Rushabh Mehta
Aug 7 at 11:39
Well that's fair, I'll have to think about that
– Rushabh Mehta
Aug 7 at 11:39
Let $U_b$ be open such that $b in U_b subset B$. Then $U_a=e^-1(U_b)$ is open and $a in U_a subset A$. Unless $a$ was a discrete point, you should be able to find points $a_1,a_2,ldots in U_asetminusa$ such that $a_nrightarrow a$.
– Stefan Böttner
Aug 7 at 12:01
Let $U_b$ be open such that $b in U_b subset B$. Then $U_a=e^-1(U_b)$ is open and $a in U_a subset A$. Unless $a$ was a discrete point, you should be able to find points $a_1,a_2,ldots in U_asetminusa$ such that $a_nrightarrow a$.
– Stefan Böttner
Aug 7 at 12:01
 |Â
show 8 more comments
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By $e(b_n)$ did you mean something like "an element of $e^-1(b_n)$?
– Mark S.
Jul 28 at 15:47
If the $b_n$ are elements of $B$, it does not make sense to look at $e(b_n)$.
– Keenan Kidwell
Jul 28 at 15:47
@KeenanKidwell : Sorry I messed up. Let me correct it.
– Rajesh Dachiraju
Jul 28 at 15:49
@MarkS. : I have corrected.
– Rajesh Dachiraju
Jul 28 at 16:03
@KeenanKidwell : I have corrected.
– Rajesh Dachiraju
Jul 28 at 16:03