Continuous function of $R_0$-space into non-$R_0$
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Let $f:Xrightarrow Y$ be a continuous function of an $R_0$-space into a topological space (non-necessary $R_0$-space) (see Wiki to recall what an $R_0$-space is).
I wish to find out the conditions to ensure that if $X$ is $R_0$ then $f(X)$ also is.
For the momment, I have found that if $f$ is open and injective then $f(X)$ is $R_0$. Moreover, I know that if $f$ is not open or is not injective, then $f(X)$ may be no $R_0$:
For if $f$ is not open, given two separated points $x,yin X$ and two (open) neighbourhoods $V_x$ and $V_y$ such that $ynotin V_x$ and $xnotin V_y$, then $f(y)notin f(V_x)$ and $f(x)notin f(V_y)$, but $f(V_x)$ and $f(V_y)$ needn't to be neigbhbourhoods of $f(x)$ and $f(y)$ resp. An example illustrating that is the identity map of ${0,1}$ (discrete) onto ${0,1}$ (Sierpinski space).
Similarly, if $f$ is open but not injective, despite that $f(V_x)$ and $f(V_y)$ are neighbourhoods of $f(x)$ and $f(y)$ and that $ynotin V_x$ and $xnotin V_y$, it is poosible that $f(x)in f(V_y)$ and that $f(y)in f(V_x)$ (for example if $f(x)=f(y)$).
However, I can't find an example illustrating the second case. My problem is that I can't find continuous open functions of an $R_0$-space into a non-$R_0$-space. My first try was to consider a map from the Sorgrenfey line onto the Sierpinski space (because it is disconnected) but that map was not open. Then I considered the identity from $[-1,1]$ with the usual subspace topology onto $[-1,1]$ endowed with the overlapping interval topology, but it was open neither.
Can you provide an example for the second case, please?
EDIT: The result is trivial if $f$ is constant. I'm interested when $f$ is non-constant, i.e. $f(X)$ has at least two points, so in the above, asuume that $f$ is non-constant.
general-topology examples-counterexamples
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Let $f:Xrightarrow Y$ be a continuous function of an $R_0$-space into a topological space (non-necessary $R_0$-space) (see Wiki to recall what an $R_0$-space is).
I wish to find out the conditions to ensure that if $X$ is $R_0$ then $f(X)$ also is.
For the momment, I have found that if $f$ is open and injective then $f(X)$ is $R_0$. Moreover, I know that if $f$ is not open or is not injective, then $f(X)$ may be no $R_0$:
For if $f$ is not open, given two separated points $x,yin X$ and two (open) neighbourhoods $V_x$ and $V_y$ such that $ynotin V_x$ and $xnotin V_y$, then $f(y)notin f(V_x)$ and $f(x)notin f(V_y)$, but $f(V_x)$ and $f(V_y)$ needn't to be neigbhbourhoods of $f(x)$ and $f(y)$ resp. An example illustrating that is the identity map of ${0,1}$ (discrete) onto ${0,1}$ (Sierpinski space).
Similarly, if $f$ is open but not injective, despite that $f(V_x)$ and $f(V_y)$ are neighbourhoods of $f(x)$ and $f(y)$ and that $ynotin V_x$ and $xnotin V_y$, it is poosible that $f(x)in f(V_y)$ and that $f(y)in f(V_x)$ (for example if $f(x)=f(y)$).
However, I can't find an example illustrating the second case. My problem is that I can't find continuous open functions of an $R_0$-space into a non-$R_0$-space. My first try was to consider a map from the Sorgrenfey line onto the Sierpinski space (because it is disconnected) but that map was not open. Then I considered the identity from $[-1,1]$ with the usual subspace topology onto $[-1,1]$ endowed with the overlapping interval topology, but it was open neither.
Can you provide an example for the second case, please?
EDIT: The result is trivial if $f$ is constant. I'm interested when $f$ is non-constant, i.e. $f(X)$ has at least two points, so in the above, asuume that $f$ is non-constant.
general-topology examples-counterexamples
I'm fairly sure neither of your conditions is necessary: Take $X$ to be the disjoint union of an $R_0$ space $A$ with a space $B$ that is not $R_0$, with the obvious topology. This is not $R_0$ because of the $B$ part. Take $f$ to be open and injective (and whatever other conditions are needed to ensure that $f(B)$ is $R_0$) on $B$, and neither open nor injective on $A$. Then $f$ is neither open nor injective, but $f(X) = f(A)cup f(B)$ is $R_0$, since $A$ is $R_0$ and $f|_B$ is open and injective. More explictly, the constant map is neither open nor injective, but $f(X)$ is $R_0$.
– user3482749
Nov 16 at 10:45
I can't follows your reasoning. If $B$ is not $R_0$, I can't imagine how $f(B)$ (that for you is homeomorphict to $B$) can be $R_0$. On the other hand, if $f$ is neither open nor injective, then $f(A)$ needn't to be $R_0$ (see my example). Regarding your final sentence, I'l edit your question to exclude a single-point spaces.
– Dog_69
Nov 16 at 11:00
@Dog_69 The statement "i'm interested in spaces with at least two points" is unrelated to "$f$ being constant is trivial". Unless you additionally assume that $f$ is surjective?
– freakish
Nov 16 at 11:08
Also, consider $f:Xto Y$ continuous, surjective, non-open, non-injective with both $X,Y$ being $R_0$, e.g. the double wrapping map $[0,2)to S^1$. These conditions are clearly not necessary.
– freakish
Nov 16 at 11:16
@Dog_69 It's nothing to do with 1-point spaces. Take any $R_0$ space $X$ with more than one point and any space $Y$. Choose some $y_0 in Y$. Then $f: X to Y$ such that $f(x) = y_0$ for all $x$ has $f(X) = {y_0}$ is both trivial and discrete, so is $R_0$, but $f$ is neither open nor injective. You being able to find one map that is not open/injective but does not have your property does not mean that all non-open/injective maps do not have your property.
– user3482749
Nov 16 at 11:43
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show 3 more comments
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up vote
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Let $f:Xrightarrow Y$ be a continuous function of an $R_0$-space into a topological space (non-necessary $R_0$-space) (see Wiki to recall what an $R_0$-space is).
I wish to find out the conditions to ensure that if $X$ is $R_0$ then $f(X)$ also is.
For the momment, I have found that if $f$ is open and injective then $f(X)$ is $R_0$. Moreover, I know that if $f$ is not open or is not injective, then $f(X)$ may be no $R_0$:
For if $f$ is not open, given two separated points $x,yin X$ and two (open) neighbourhoods $V_x$ and $V_y$ such that $ynotin V_x$ and $xnotin V_y$, then $f(y)notin f(V_x)$ and $f(x)notin f(V_y)$, but $f(V_x)$ and $f(V_y)$ needn't to be neigbhbourhoods of $f(x)$ and $f(y)$ resp. An example illustrating that is the identity map of ${0,1}$ (discrete) onto ${0,1}$ (Sierpinski space).
Similarly, if $f$ is open but not injective, despite that $f(V_x)$ and $f(V_y)$ are neighbourhoods of $f(x)$ and $f(y)$ and that $ynotin V_x$ and $xnotin V_y$, it is poosible that $f(x)in f(V_y)$ and that $f(y)in f(V_x)$ (for example if $f(x)=f(y)$).
However, I can't find an example illustrating the second case. My problem is that I can't find continuous open functions of an $R_0$-space into a non-$R_0$-space. My first try was to consider a map from the Sorgrenfey line onto the Sierpinski space (because it is disconnected) but that map was not open. Then I considered the identity from $[-1,1]$ with the usual subspace topology onto $[-1,1]$ endowed with the overlapping interval topology, but it was open neither.
Can you provide an example for the second case, please?
EDIT: The result is trivial if $f$ is constant. I'm interested when $f$ is non-constant, i.e. $f(X)$ has at least two points, so in the above, asuume that $f$ is non-constant.
general-topology examples-counterexamples
Let $f:Xrightarrow Y$ be a continuous function of an $R_0$-space into a topological space (non-necessary $R_0$-space) (see Wiki to recall what an $R_0$-space is).
I wish to find out the conditions to ensure that if $X$ is $R_0$ then $f(X)$ also is.
For the momment, I have found that if $f$ is open and injective then $f(X)$ is $R_0$. Moreover, I know that if $f$ is not open or is not injective, then $f(X)$ may be no $R_0$:
For if $f$ is not open, given two separated points $x,yin X$ and two (open) neighbourhoods $V_x$ and $V_y$ such that $ynotin V_x$ and $xnotin V_y$, then $f(y)notin f(V_x)$ and $f(x)notin f(V_y)$, but $f(V_x)$ and $f(V_y)$ needn't to be neigbhbourhoods of $f(x)$ and $f(y)$ resp. An example illustrating that is the identity map of ${0,1}$ (discrete) onto ${0,1}$ (Sierpinski space).
Similarly, if $f$ is open but not injective, despite that $f(V_x)$ and $f(V_y)$ are neighbourhoods of $f(x)$ and $f(y)$ and that $ynotin V_x$ and $xnotin V_y$, it is poosible that $f(x)in f(V_y)$ and that $f(y)in f(V_x)$ (for example if $f(x)=f(y)$).
However, I can't find an example illustrating the second case. My problem is that I can't find continuous open functions of an $R_0$-space into a non-$R_0$-space. My first try was to consider a map from the Sorgrenfey line onto the Sierpinski space (because it is disconnected) but that map was not open. Then I considered the identity from $[-1,1]$ with the usual subspace topology onto $[-1,1]$ endowed with the overlapping interval topology, but it was open neither.
Can you provide an example for the second case, please?
EDIT: The result is trivial if $f$ is constant. I'm interested when $f$ is non-constant, i.e. $f(X)$ has at least two points, so in the above, asuume that $f$ is non-constant.
general-topology examples-counterexamples
general-topology examples-counterexamples
edited Nov 16 at 15:44
asked Nov 16 at 10:33
Dog_69
5211422
5211422
I'm fairly sure neither of your conditions is necessary: Take $X$ to be the disjoint union of an $R_0$ space $A$ with a space $B$ that is not $R_0$, with the obvious topology. This is not $R_0$ because of the $B$ part. Take $f$ to be open and injective (and whatever other conditions are needed to ensure that $f(B)$ is $R_0$) on $B$, and neither open nor injective on $A$. Then $f$ is neither open nor injective, but $f(X) = f(A)cup f(B)$ is $R_0$, since $A$ is $R_0$ and $f|_B$ is open and injective. More explictly, the constant map is neither open nor injective, but $f(X)$ is $R_0$.
– user3482749
Nov 16 at 10:45
I can't follows your reasoning. If $B$ is not $R_0$, I can't imagine how $f(B)$ (that for you is homeomorphict to $B$) can be $R_0$. On the other hand, if $f$ is neither open nor injective, then $f(A)$ needn't to be $R_0$ (see my example). Regarding your final sentence, I'l edit your question to exclude a single-point spaces.
– Dog_69
Nov 16 at 11:00
@Dog_69 The statement "i'm interested in spaces with at least two points" is unrelated to "$f$ being constant is trivial". Unless you additionally assume that $f$ is surjective?
– freakish
Nov 16 at 11:08
Also, consider $f:Xto Y$ continuous, surjective, non-open, non-injective with both $X,Y$ being $R_0$, e.g. the double wrapping map $[0,2)to S^1$. These conditions are clearly not necessary.
– freakish
Nov 16 at 11:16
@Dog_69 It's nothing to do with 1-point spaces. Take any $R_0$ space $X$ with more than one point and any space $Y$. Choose some $y_0 in Y$. Then $f: X to Y$ such that $f(x) = y_0$ for all $x$ has $f(X) = {y_0}$ is both trivial and discrete, so is $R_0$, but $f$ is neither open nor injective. You being able to find one map that is not open/injective but does not have your property does not mean that all non-open/injective maps do not have your property.
– user3482749
Nov 16 at 11:43
|
show 3 more comments
I'm fairly sure neither of your conditions is necessary: Take $X$ to be the disjoint union of an $R_0$ space $A$ with a space $B$ that is not $R_0$, with the obvious topology. This is not $R_0$ because of the $B$ part. Take $f$ to be open and injective (and whatever other conditions are needed to ensure that $f(B)$ is $R_0$) on $B$, and neither open nor injective on $A$. Then $f$ is neither open nor injective, but $f(X) = f(A)cup f(B)$ is $R_0$, since $A$ is $R_0$ and $f|_B$ is open and injective. More explictly, the constant map is neither open nor injective, but $f(X)$ is $R_0$.
– user3482749
Nov 16 at 10:45
I can't follows your reasoning. If $B$ is not $R_0$, I can't imagine how $f(B)$ (that for you is homeomorphict to $B$) can be $R_0$. On the other hand, if $f$ is neither open nor injective, then $f(A)$ needn't to be $R_0$ (see my example). Regarding your final sentence, I'l edit your question to exclude a single-point spaces.
– Dog_69
Nov 16 at 11:00
@Dog_69 The statement "i'm interested in spaces with at least two points" is unrelated to "$f$ being constant is trivial". Unless you additionally assume that $f$ is surjective?
– freakish
Nov 16 at 11:08
Also, consider $f:Xto Y$ continuous, surjective, non-open, non-injective with both $X,Y$ being $R_0$, e.g. the double wrapping map $[0,2)to S^1$. These conditions are clearly not necessary.
– freakish
Nov 16 at 11:16
@Dog_69 It's nothing to do with 1-point spaces. Take any $R_0$ space $X$ with more than one point and any space $Y$. Choose some $y_0 in Y$. Then $f: X to Y$ such that $f(x) = y_0$ for all $x$ has $f(X) = {y_0}$ is both trivial and discrete, so is $R_0$, but $f$ is neither open nor injective. You being able to find one map that is not open/injective but does not have your property does not mean that all non-open/injective maps do not have your property.
– user3482749
Nov 16 at 11:43
I'm fairly sure neither of your conditions is necessary: Take $X$ to be the disjoint union of an $R_0$ space $A$ with a space $B$ that is not $R_0$, with the obvious topology. This is not $R_0$ because of the $B$ part. Take $f$ to be open and injective (and whatever other conditions are needed to ensure that $f(B)$ is $R_0$) on $B$, and neither open nor injective on $A$. Then $f$ is neither open nor injective, but $f(X) = f(A)cup f(B)$ is $R_0$, since $A$ is $R_0$ and $f|_B$ is open and injective. More explictly, the constant map is neither open nor injective, but $f(X)$ is $R_0$.
– user3482749
Nov 16 at 10:45
I'm fairly sure neither of your conditions is necessary: Take $X$ to be the disjoint union of an $R_0$ space $A$ with a space $B$ that is not $R_0$, with the obvious topology. This is not $R_0$ because of the $B$ part. Take $f$ to be open and injective (and whatever other conditions are needed to ensure that $f(B)$ is $R_0$) on $B$, and neither open nor injective on $A$. Then $f$ is neither open nor injective, but $f(X) = f(A)cup f(B)$ is $R_0$, since $A$ is $R_0$ and $f|_B$ is open and injective. More explictly, the constant map is neither open nor injective, but $f(X)$ is $R_0$.
– user3482749
Nov 16 at 10:45
I can't follows your reasoning. If $B$ is not $R_0$, I can't imagine how $f(B)$ (that for you is homeomorphict to $B$) can be $R_0$. On the other hand, if $f$ is neither open nor injective, then $f(A)$ needn't to be $R_0$ (see my example). Regarding your final sentence, I'l edit your question to exclude a single-point spaces.
– Dog_69
Nov 16 at 11:00
I can't follows your reasoning. If $B$ is not $R_0$, I can't imagine how $f(B)$ (that for you is homeomorphict to $B$) can be $R_0$. On the other hand, if $f$ is neither open nor injective, then $f(A)$ needn't to be $R_0$ (see my example). Regarding your final sentence, I'l edit your question to exclude a single-point spaces.
– Dog_69
Nov 16 at 11:00
@Dog_69 The statement "i'm interested in spaces with at least two points" is unrelated to "$f$ being constant is trivial". Unless you additionally assume that $f$ is surjective?
– freakish
Nov 16 at 11:08
@Dog_69 The statement "i'm interested in spaces with at least two points" is unrelated to "$f$ being constant is trivial". Unless you additionally assume that $f$ is surjective?
– freakish
Nov 16 at 11:08
Also, consider $f:Xto Y$ continuous, surjective, non-open, non-injective with both $X,Y$ being $R_0$, e.g. the double wrapping map $[0,2)to S^1$. These conditions are clearly not necessary.
– freakish
Nov 16 at 11:16
Also, consider $f:Xto Y$ continuous, surjective, non-open, non-injective with both $X,Y$ being $R_0$, e.g. the double wrapping map $[0,2)to S^1$. These conditions are clearly not necessary.
– freakish
Nov 16 at 11:16
@Dog_69 It's nothing to do with 1-point spaces. Take any $R_0$ space $X$ with more than one point and any space $Y$. Choose some $y_0 in Y$. Then $f: X to Y$ such that $f(x) = y_0$ for all $x$ has $f(X) = {y_0}$ is both trivial and discrete, so is $R_0$, but $f$ is neither open nor injective. You being able to find one map that is not open/injective but does not have your property does not mean that all non-open/injective maps do not have your property.
– user3482749
Nov 16 at 11:43
@Dog_69 It's nothing to do with 1-point spaces. Take any $R_0$ space $X$ with more than one point and any space $Y$. Choose some $y_0 in Y$. Then $f: X to Y$ such that $f(x) = y_0$ for all $x$ has $f(X) = {y_0}$ is both trivial and discrete, so is $R_0$, but $f$ is neither open nor injective. You being able to find one map that is not open/injective but does not have your property does not mean that all non-open/injective maps do not have your property.
– user3482749
Nov 16 at 11:43
|
show 3 more comments
1 Answer
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It seems there was a simple example:
Let $X=mathbb R$ with the usual topology and $Y={0,1}$ be the Sierpinski space (with ${1}$ open). Consider $f:Xrightarrow Y$ defined as $f(0)=0$ and $1$ otherwise. Then:
$f$ is continuous because $f^{-1}({1})=(-infty,0)cup(0,+infty)$ is open and $f^{-1}({0,1})=mathbb R$ is also open.
$f$ is open because $f((a,b))={1}$ if $0notin (a,b)$ and ${0,1}$ if $0in(a,b)$; in both cases they are open.
The Sierpinski space is not symmetric because $0$ and $1$ are topologically distinguishable but $1$ belongs to each neighbourhood of $0$.
add a comment |
1 Answer
1
active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
It seems there was a simple example:
Let $X=mathbb R$ with the usual topology and $Y={0,1}$ be the Sierpinski space (with ${1}$ open). Consider $f:Xrightarrow Y$ defined as $f(0)=0$ and $1$ otherwise. Then:
$f$ is continuous because $f^{-1}({1})=(-infty,0)cup(0,+infty)$ is open and $f^{-1}({0,1})=mathbb R$ is also open.
$f$ is open because $f((a,b))={1}$ if $0notin (a,b)$ and ${0,1}$ if $0in(a,b)$; in both cases they are open.
The Sierpinski space is not symmetric because $0$ and $1$ are topologically distinguishable but $1$ belongs to each neighbourhood of $0$.
add a comment |
up vote
0
down vote
It seems there was a simple example:
Let $X=mathbb R$ with the usual topology and $Y={0,1}$ be the Sierpinski space (with ${1}$ open). Consider $f:Xrightarrow Y$ defined as $f(0)=0$ and $1$ otherwise. Then:
$f$ is continuous because $f^{-1}({1})=(-infty,0)cup(0,+infty)$ is open and $f^{-1}({0,1})=mathbb R$ is also open.
$f$ is open because $f((a,b))={1}$ if $0notin (a,b)$ and ${0,1}$ if $0in(a,b)$; in both cases they are open.
The Sierpinski space is not symmetric because $0$ and $1$ are topologically distinguishable but $1$ belongs to each neighbourhood of $0$.
add a comment |
up vote
0
down vote
up vote
0
down vote
It seems there was a simple example:
Let $X=mathbb R$ with the usual topology and $Y={0,1}$ be the Sierpinski space (with ${1}$ open). Consider $f:Xrightarrow Y$ defined as $f(0)=0$ and $1$ otherwise. Then:
$f$ is continuous because $f^{-1}({1})=(-infty,0)cup(0,+infty)$ is open and $f^{-1}({0,1})=mathbb R$ is also open.
$f$ is open because $f((a,b))={1}$ if $0notin (a,b)$ and ${0,1}$ if $0in(a,b)$; in both cases they are open.
The Sierpinski space is not symmetric because $0$ and $1$ are topologically distinguishable but $1$ belongs to each neighbourhood of $0$.
It seems there was a simple example:
Let $X=mathbb R$ with the usual topology and $Y={0,1}$ be the Sierpinski space (with ${1}$ open). Consider $f:Xrightarrow Y$ defined as $f(0)=0$ and $1$ otherwise. Then:
$f$ is continuous because $f^{-1}({1})=(-infty,0)cup(0,+infty)$ is open and $f^{-1}({0,1})=mathbb R$ is also open.
$f$ is open because $f((a,b))={1}$ if $0notin (a,b)$ and ${0,1}$ if $0in(a,b)$; in both cases they are open.
The Sierpinski space is not symmetric because $0$ and $1$ are topologically distinguishable but $1$ belongs to each neighbourhood of $0$.
answered Nov 19 at 16:07
Dog_69
5211422
5211422
add a comment |
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I'm fairly sure neither of your conditions is necessary: Take $X$ to be the disjoint union of an $R_0$ space $A$ with a space $B$ that is not $R_0$, with the obvious topology. This is not $R_0$ because of the $B$ part. Take $f$ to be open and injective (and whatever other conditions are needed to ensure that $f(B)$ is $R_0$) on $B$, and neither open nor injective on $A$. Then $f$ is neither open nor injective, but $f(X) = f(A)cup f(B)$ is $R_0$, since $A$ is $R_0$ and $f|_B$ is open and injective. More explictly, the constant map is neither open nor injective, but $f(X)$ is $R_0$.
– user3482749
Nov 16 at 10:45
I can't follows your reasoning. If $B$ is not $R_0$, I can't imagine how $f(B)$ (that for you is homeomorphict to $B$) can be $R_0$. On the other hand, if $f$ is neither open nor injective, then $f(A)$ needn't to be $R_0$ (see my example). Regarding your final sentence, I'l edit your question to exclude a single-point spaces.
– Dog_69
Nov 16 at 11:00
@Dog_69 The statement "i'm interested in spaces with at least two points" is unrelated to "$f$ being constant is trivial". Unless you additionally assume that $f$ is surjective?
– freakish
Nov 16 at 11:08
Also, consider $f:Xto Y$ continuous, surjective, non-open, non-injective with both $X,Y$ being $R_0$, e.g. the double wrapping map $[0,2)to S^1$. These conditions are clearly not necessary.
– freakish
Nov 16 at 11:16
@Dog_69 It's nothing to do with 1-point spaces. Take any $R_0$ space $X$ with more than one point and any space $Y$. Choose some $y_0 in Y$. Then $f: X to Y$ such that $f(x) = y_0$ for all $x$ has $f(X) = {y_0}$ is both trivial and discrete, so is $R_0$, but $f$ is neither open nor injective. You being able to find one map that is not open/injective but does not have your property does not mean that all non-open/injective maps do not have your property.
– user3482749
Nov 16 at 11:43