Mixing
If the image of the interior equals the interior of the image, the map is continuous
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Let $f$ be a map between topological spaces $f : X to Y$ such that for every $A subset X$ we have that $f(operatorname{int} A) = operatorname{int} f(A)$. Prove $f$ is continuous.
I think I can prove the converse isn't true, any continuous non-open map should do, take $f : mathbb{R} to mathbb{R} : x mapsto x^2$ with the standard topology, and $A = mathbb{R} = operatorname{int}A$. Then $f(operatorname{int}A) = [0, +infty) neq (0, +infty) = operatorname{int}f(A)$
I have no clue how to prove this. I have tried using the fact that $f$ is continuous if $f(overline{B}) subset overline{f(B)}$ for all $B$, but I can't seem to construct the right $A$ that implies this.
general-topology
asked Jan 27 at 0:45
kmmkmm
1637
$endgroup$
add a comment |
$begingroup$
Let $f$ be a map between topological spaces $f : X to Y$ such that for every $A subset X$ we have that $f(operatorname{int} A) = operatorname{int} f(A)$. Prove $f$ is continuous.
I think I can prove the converse isn't true, any continuous non-open map should do, take $f : mathbb{R} to mathbb{R} : x mapsto x^2$ with the standard topology, and $A = mathbb{R} = operatorname{int}A$. Then $f(operatorname{int}A) = [0, +infty) neq (0, +infty) = operatorname{int}f(A)$
I have no clue how to prove this. I have tried using the fact that $f$ is continuous if $f(overline{B}) subset overline{f(B)}$ for all $B$, but I can't seem to construct the right $A$ that implies this.
general-topology
asked Jan 27 at 0:45
kmmkmm
1637
$endgroup$
1
$begingroup$
Concerning the converse: The projection $p : [0,1] times [0,1] to [0,1]$ onto the first coordinate is continuous, open, closed and surjective but does not satisfy $p(operatorname{int} A) = operatorname{int} p(A)$ (consider $A$ = diagonal).
$endgroup$
– Paul Frost
Jan 27 at 17:17
$begingroup$
What is the origin of your question? Do you have any indication that $f$ is continuous or should one look for counterexamples?
$endgroup$
– Paul Frost
Jan 27 at 23:24
$begingroup$
I got it from a list of exercises of a university topology course. It asks to prove $f$ is continuous, so the statement must be true. There is no further information given that what I have written down here, the other exercises are very clear about the properties of the functions and topological spaces, so I don't think are more conditions. I think I will contact the TA associated to the course to ask if something is missing.
$endgroup$
– kmm
Jan 28 at 0:07
$begingroup$
Do not believe that something must be true simply because it appears in a list of exercises ;-) However, I do not claim it is wrong, perhaps I just do not see how to prove it.
$endgroup$
– Paul Frost
Jan 28 at 16:01
$begingroup$
I got a reply back with a rough proof sketch that I think I was able to work out. I added my own answer, sorry for the mess, thank you for your remarks.
$endgroup$
– kmm
Jan 29 at 1:03
add a comment |
$begingroup$
Let $f$ be a map between topological spaces $f : X to Y$ such that for every $A subset X$ we have that $f(operatorname{int} A) = operatorname{int} f(A)$. Prove $f$ is continuous.
I think I can prove the converse isn't true, any continuous non-open map should do, take $f : mathbb{R} to mathbb{R} : x mapsto x^2$ with the standard topology, and $A = mathbb{R} = operatorname{int}A$. Then $f(operatorname{int}A) = [0, +infty) neq (0, +infty) = operatorname{int}f(A)$
I have no clue how to prove this. I have tried using the fact that $f$ is continuous if $f(overline{B}) subset overline{f(B)}$ for all $B$, but I can't seem to construct the right $A$ that implies this.
general-topology
asked Jan 27 at 0:45
kmmkmm
1637
$endgroup$
Let $f$ be a map between topological spaces $f : X to Y$ such that for every $A subset X$ we have that $f(operatorname{int} A) = operatorname{int} f(A)$. Prove $f$ is continuous.
I think I can prove the converse isn't true, any continuous non-open map should do, take $f : mathbb{R} to mathbb{R} : x mapsto x^2$ with the standard topology, and $A = mathbb{R} = operatorname{int}A$. Then $f(operatorname{int}A) = [0, +infty) neq (0, +infty) = operatorname{int}f(A)$
I have no clue how to prove this. I have tried using the fact that $f$ is continuous if $f(overline{B}) subset overline{f(B)}$ for all $B$, but I can't seem to construct the right $A$ that implies this.
general-topology
general-topology
asked Jan 27 at 0:45
kmmkmm
1637
asked Jan 27 at 0:45
kmmkmm
1637
asked Jan 27 at 0:45
kmmkmm
1637
asked Jan 27 at 0:45
kmmkmm
1637
asked Jan 27 at 0:45
kmmkmm
1637
1637
1
$begingroup$
Concerning the converse: The projection $p : [0,1] times [0,1] to [0,1]$ onto the first coordinate is continuous, open, closed and surjective but does not satisfy $p(operatorname{int} A) = operatorname{int} p(A)$ (consider $A$ = diagonal).
$endgroup$
– Paul Frost
Jan 27 at 17:17
$begingroup$
What is the origin of your question? Do you have any indication that $f$ is continuous or should one look for counterexamples?
$endgroup$
– Paul Frost
Jan 27 at 23:24
$begingroup$
I got it from a list of exercises of a university topology course. It asks to prove $f$ is continuous, so the statement must be true. There is no further information given that what I have written down here, the other exercises are very clear about the properties of the functions and topological spaces, so I don't think are more conditions. I think I will contact the TA associated to the course to ask if something is missing.
$endgroup$
– kmm
Jan 28 at 0:07
$begingroup$
Do not believe that something must be true simply because it appears in a list of exercises ;-) However, I do not claim it is wrong, perhaps I just do not see how to prove it.
$endgroup$
– Paul Frost
Jan 28 at 16:01
$begingroup$
I got a reply back with a rough proof sketch that I think I was able to work out. I added my own answer, sorry for the mess, thank you for your remarks.
$endgroup$
– kmm
Jan 29 at 1:03
add a comment |
1
$begingroup$
Concerning the converse: The projection $p : [0,1] times [0,1] to [0,1]$ onto the first coordinate is continuous, open, closed and surjective but does not satisfy $p(operatorname{int} A) = operatorname{int} p(A)$ (consider $A$ = diagonal).
$endgroup$
– Paul Frost
Jan 27 at 17:17
$begingroup$
What is the origin of your question? Do you have any indication that $f$ is continuous or should one look for counterexamples?
$endgroup$
– Paul Frost
Jan 27 at 23:24
$begingroup$
I got it from a list of exercises of a university topology course. It asks to prove $f$ is continuous, so the statement must be true. There is no further information given that what I have written down here, the other exercises are very clear about the properties of the functions and topological spaces, so I don't think are more conditions. I think I will contact the TA associated to the course to ask if something is missing.
$endgroup$
– kmm
Jan 28 at 0:07
$begingroup$
Do not believe that something must be true simply because it appears in a list of exercises ;-) However, I do not claim it is wrong, perhaps I just do not see how to prove it.
$endgroup$
– Paul Frost
Jan 28 at 16:01
$begingroup$
I got a reply back with a rough proof sketch that I think I was able to work out. I added my own answer, sorry for the mess, thank you for your remarks.
$endgroup$
– kmm
Jan 29 at 1:03
1
1
$begingroup$
Concerning the converse: The projection $p : [0,1] times [0,1] to [0,1]$ onto the first coordinate is continuous, open, closed and surjective but does not satisfy $p(operatorname{int} A) = operatorname{int} p(A)$ (consider $A$ = diagonal).
$endgroup$
– Paul Frost
Jan 27 at 17:17
$begingroup$
Concerning the converse: The projection $p : [0,1] times [0,1] to [0,1]$ onto the first coordinate is continuous, open, closed and surjective but does not satisfy $p(operatorname{int} A) = operatorname{int} p(A)$ (consider $A$ = diagonal).
$endgroup$
– Paul Frost
Jan 27 at 17:17
$begingroup$
What is the origin of your question? Do you have any indication that $f$ is continuous or should one look for counterexamples?
$endgroup$
– Paul Frost
Jan 27 at 23:24
$begingroup$
What is the origin of your question? Do you have any indication that $f$ is continuous or should one look for counterexamples?
$endgroup$
– Paul Frost
Jan 27 at 23:24
$begingroup$
I got it from a list of exercises of a university topology course. It asks to prove $f$ is continuous, so the statement must be true. There is no further information given that what I have written down here, the other exercises are very clear about the properties of the functions and topological spaces, so I don't think are more conditions. I think I will contact the TA associated to the course to ask if something is missing.
$endgroup$
– kmm
Jan 28 at 0:07
$begingroup$
I got it from a list of exercises of a university topology course. It asks to prove $f$ is continuous, so the statement must be true. There is no further information given that what I have written down here, the other exercises are very clear about the properties of the functions and topological spaces, so I don't think are more conditions. I think I will contact the TA associated to the course to ask if something is missing.
$endgroup$
– kmm
Jan 28 at 0:07
$begingroup$
Do not believe that something must be true simply because it appears in a list of exercises ;-) However, I do not claim it is wrong, perhaps I just do not see how to prove it.
$endgroup$
– Paul Frost
Jan 28 at 16:01
$begingroup$
Do not believe that something must be true simply because it appears in a list of exercises ;-) However, I do not claim it is wrong, perhaps I just do not see how to prove it.
$endgroup$
– Paul Frost
Jan 28 at 16:01
$begingroup$
I got a reply back with a rough proof sketch that I think I was able to work out. I added my own answer, sorry for the mess, thank you for your remarks.
$endgroup$
– kmm
Jan 29 at 1:03
$begingroup$
I got a reply back with a rough proof sketch that I think I was able to work out. I added my own answer, sorry for the mess, thank you for your remarks.
$endgroup$
– kmm
Jan 29 at 1:03
add a comment |
2 Answers
2
active
oldest
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$begingroup$
From a rough sketch given by the TA of the course, I was able to formulate a proof.
Consider an open $B subset Y$. Let $A = f^{-1}(B)$. Let $C = A setminus operatorname{int} A$. The goal is to prove $C$ is empty.
Consider now $G = C cup (A setminus f^{-1}(f(C)))$. Then $f(G) = f(A) = B$. By the assumption, $f(operatorname{int} G) = operatorname{int}(f(G)) = operatorname{int} B = B$
Now $operatorname{int}G =operatorname{int} C cup operatorname{int}(A setminus f^{-1}(f(C))) = operatorname{int}(A setminus f^{-1}(f(C)))$, as $C$ has an empty interior. Therefore $f(operatorname{int} G) = f(operatorname{int} (A setminus f^{-1}(f(C))) = B$ so it follows that $ B = f(A setminus f^{-1}(f(C))) = f(A) setminus f(C) = B setminus f(C)$
So $B = B setminus f(C)$, which implies that $f(C) = varnothing$ and thus $C = varnothing$, so $A$ is open, and we can finally conclude that $f$ is continuous.
answered Jan 28 at 20:33
kmmkmm
1637
$endgroup$
$begingroup$
Here are two problems in the proof. (1) In general $f(A) subsetneqq B$. This can be resolved as follows. We know that $f$ is an open map so that $f(X)$ is open in $Y$. Hence $B' = B cap f(X)$ is open and $f^{-1}(B') = f^{-1}(B)$. Thus we may assume w.l.o.g. $B subset f(X)$. In that case $f(A) = B$. (2) We only have $operatorname{int} (M cup N) supset operatorname{int} M cup operatorname{int} N$ and in general both sides are not equal. I do not see how to resolve this since we cannot conclude $operatorname{int}G = operatorname{int}(A setminus f^{-1}(f(C)))$.
$endgroup$
– Paul Frost
Jan 29 at 10:52
$begingroup$
Good points, thank you. I did forget to clarify that we can assume that $f$ is subjective. With regard to the second problem, the interior of a union equals the union of the interiors in the case that the sets are separated. I think that is true here, $C = A setminus operatorname{int} A subset overline{A} setminus operatorname{int} A = partial A$, and $A setminus f^{-1}(f(C)) subset A setminus C = A setminus (A setminus operatorname{int}A) = operatorname{int} A$. And the boundary and interior of a set are always separate.
$endgroup$
– kmm
Jan 29 at 20:18
$begingroup$
The usual definition that sets $M,N$ are separated is that $overline{M} cap N = M cap overline{N} = emptyset$. But $partial A$ and $operatorname{int} A$ are in general not separated. So perhaps the better argument is this. We have $G subset A$. Hence $operatorname{int} G subset operatorname{int} A$ so that $operatorname{int} G cap partial A = emptyset$. Since $C subset partial A$, we get $operatorname{int} G subset A setminus f^{-1}(f(C))$ whence $operatorname{int} G subset operatorname{int} (A setminus f^{-1}(f(C))) subset operatorname{int} G$.
$endgroup$
– Paul Frost
Jan 30 at 0:50
add a comment |
$begingroup$
This is only a partial answer. Obviously $f$ is open. Concerning continuity we may
w.l.o.g. assume that $f$ is surjective. In fact, consider the map $f' : X stackrel{f}{rightarrow} f(X)$. Then $f$ is continuous iff $f'$ is continuous. But $f'$ also satisfies $f'(operatorname{int} A) = operatorname{int} f'(A)$ because $f(X)$ is open in $Y$ (that is, for $B subset f(X)$ we have $operatorname{int}_Y B = operatorname{int}_{f(X)} B$).
Now it is clear that if $f$ is injective, then it is continuous. Our above argument shows that it suffices to consider a bijective $f$. In that case let $V subset Y$ be open. Then $f(operatorname{int} ( f^{-1}(V)) = operatorname{int} f(f^{-1}(V)) = operatorname{int} V = V = f(f^{-1}(V))$, hence $operatorname{int} f^{-1}(V) = f^{-1}(V)$, i.e. $f^{-1}(V)$ is open.
answered Jan 27 at 17:42
Paul FrostPaul Frost
11.6k3935
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2 Answers
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$begingroup$
From a rough sketch given by the TA of the course, I was able to formulate a proof.
Consider an open $B subset Y$. Let $A = f^{-1}(B)$. Let $C = A setminus operatorname{int} A$. The goal is to prove $C$ is empty.
Consider now $G = C cup (A setminus f^{-1}(f(C)))$. Then $f(G) = f(A) = B$. By the assumption, $f(operatorname{int} G) = operatorname{int}(f(G)) = operatorname{int} B = B$
Now $operatorname{int}G =operatorname{int} C cup operatorname{int}(A setminus f^{-1}(f(C))) = operatorname{int}(A setminus f^{-1}(f(C)))$, as $C$ has an empty interior. Therefore $f(operatorname{int} G) = f(operatorname{int} (A setminus f^{-1}(f(C))) = B$ so it follows that $ B = f(A setminus f^{-1}(f(C))) = f(A) setminus f(C) = B setminus f(C)$
So $B = B setminus f(C)$, which implies that $f(C) = varnothing$ and thus $C = varnothing$, so $A$ is open, and we can finally conclude that $f$ is continuous.
answered Jan 28 at 20:33
kmmkmm
1637
$endgroup$
$begingroup$
Here are two problems in the proof. (1) In general $f(A) subsetneqq B$. This can be resolved as follows. We know that $f$ is an open map so that $f(X)$ is open in $Y$. Hence $B' = B cap f(X)$ is open and $f^{-1}(B') = f^{-1}(B)$. Thus we may assume w.l.o.g. $B subset f(X)$. In that case $f(A) = B$. (2) We only have $operatorname{int} (M cup N) supset operatorname{int} M cup operatorname{int} N$ and in general both sides are not equal. I do not see how to resolve this since we cannot conclude $operatorname{int}G = operatorname{int}(A setminus f^{-1}(f(C)))$.
$endgroup$
– Paul Frost
Jan 29 at 10:52
$begingroup$
Good points, thank you. I did forget to clarify that we can assume that $f$ is subjective. With regard to the second problem, the interior of a union equals the union of the interiors in the case that the sets are separated. I think that is true here, $C = A setminus operatorname{int} A subset overline{A} setminus operatorname{int} A = partial A$, and $A setminus f^{-1}(f(C)) subset A setminus C = A setminus (A setminus operatorname{int}A) = operatorname{int} A$. And the boundary and interior of a set are always separate.
$endgroup$
– kmm
Jan 29 at 20:18
$begingroup$
The usual definition that sets $M,N$ are separated is that $overline{M} cap N = M cap overline{N} = emptyset$. But $partial A$ and $operatorname{int} A$ are in general not separated. So perhaps the better argument is this. We have $G subset A$. Hence $operatorname{int} G subset operatorname{int} A$ so that $operatorname{int} G cap partial A = emptyset$. Since $C subset partial A$, we get $operatorname{int} G subset A setminus f^{-1}(f(C))$ whence $operatorname{int} G subset operatorname{int} (A setminus f^{-1}(f(C))) subset operatorname{int} G$.
$endgroup$
– Paul Frost
Jan 30 at 0:50
add a comment |
$begingroup$
From a rough sketch given by the TA of the course, I was able to formulate a proof.
Consider an open $B subset Y$. Let $A = f^{-1}(B)$. Let $C = A setminus operatorname{int} A$. The goal is to prove $C$ is empty.
Consider now $G = C cup (A setminus f^{-1}(f(C)))$. Then $f(G) = f(A) = B$. By the assumption, $f(operatorname{int} G) = operatorname{int}(f(G)) = operatorname{int} B = B$
Now $operatorname{int}G =operatorname{int} C cup operatorname{int}(A setminus f^{-1}(f(C))) = operatorname{int}(A setminus f^{-1}(f(C)))$, as $C$ has an empty interior. Therefore $f(operatorname{int} G) = f(operatorname{int} (A setminus f^{-1}(f(C))) = B$ so it follows that $ B = f(A setminus f^{-1}(f(C))) = f(A) setminus f(C) = B setminus f(C)$
So $B = B setminus f(C)$, which implies that $f(C) = varnothing$ and thus $C = varnothing$, so $A$ is open, and we can finally conclude that $f$ is continuous.
answered Jan 28 at 20:33
kmmkmm
1637
$endgroup$
$begingroup$
Here are two problems in the proof. (1) In general $f(A) subsetneqq B$. This can be resolved as follows. We know that $f$ is an open map so that $f(X)$ is open in $Y$. Hence $B' = B cap f(X)$ is open and $f^{-1}(B') = f^{-1}(B)$. Thus we may assume w.l.o.g. $B subset f(X)$. In that case $f(A) = B$. (2) We only have $operatorname{int} (M cup N) supset operatorname{int} M cup operatorname{int} N$ and in general both sides are not equal. I do not see how to resolve this since we cannot conclude $operatorname{int}G = operatorname{int}(A setminus f^{-1}(f(C)))$.
$endgroup$
– Paul Frost
Jan 29 at 10:52
$begingroup$
Good points, thank you. I did forget to clarify that we can assume that $f$ is subjective. With regard to the second problem, the interior of a union equals the union of the interiors in the case that the sets are separated. I think that is true here, $C = A setminus operatorname{int} A subset overline{A} setminus operatorname{int} A = partial A$, and $A setminus f^{-1}(f(C)) subset A setminus C = A setminus (A setminus operatorname{int}A) = operatorname{int} A$. And the boundary and interior of a set are always separate.
$endgroup$
– kmm
Jan 29 at 20:18
$begingroup$
The usual definition that sets $M,N$ are separated is that $overline{M} cap N = M cap overline{N} = emptyset$. But $partial A$ and $operatorname{int} A$ are in general not separated. So perhaps the better argument is this. We have $G subset A$. Hence $operatorname{int} G subset operatorname{int} A$ so that $operatorname{int} G cap partial A = emptyset$. Since $C subset partial A$, we get $operatorname{int} G subset A setminus f^{-1}(f(C))$ whence $operatorname{int} G subset operatorname{int} (A setminus f^{-1}(f(C))) subset operatorname{int} G$.
$endgroup$
– Paul Frost
Jan 30 at 0:50
add a comment |
$begingroup$
From a rough sketch given by the TA of the course, I was able to formulate a proof.
Consider an open $B subset Y$. Let $A = f^{-1}(B)$. Let $C = A setminus operatorname{int} A$. The goal is to prove $C$ is empty.
Consider now $G = C cup (A setminus f^{-1}(f(C)))$. Then $f(G) = f(A) = B$. By the assumption, $f(operatorname{int} G) = operatorname{int}(f(G)) = operatorname{int} B = B$
Now $operatorname{int}G =operatorname{int} C cup operatorname{int}(A setminus f^{-1}(f(C))) = operatorname{int}(A setminus f^{-1}(f(C)))$, as $C$ has an empty interior. Therefore $f(operatorname{int} G) = f(operatorname{int} (A setminus f^{-1}(f(C))) = B$ so it follows that $ B = f(A setminus f^{-1}(f(C))) = f(A) setminus f(C) = B setminus f(C)$
So $B = B setminus f(C)$, which implies that $f(C) = varnothing$ and thus $C = varnothing$, so $A$ is open, and we can finally conclude that $f$ is continuous.
answered Jan 28 at 20:33
kmmkmm
1637
$endgroup$
From a rough sketch given by the TA of the course, I was able to formulate a proof.
Consider an open $B subset Y$. Let $A = f^{-1}(B)$. Let $C = A setminus operatorname{int} A$. The goal is to prove $C$ is empty.
Consider now $G = C cup (A setminus f^{-1}(f(C)))$. Then $f(G) = f(A) = B$. By the assumption, $f(operatorname{int} G) = operatorname{int}(f(G)) = operatorname{int} B = B$
Now $operatorname{int}G =operatorname{int} C cup operatorname{int}(A setminus f^{-1}(f(C))) = operatorname{int}(A setminus f^{-1}(f(C)))$, as $C$ has an empty interior. Therefore $f(operatorname{int} G) = f(operatorname{int} (A setminus f^{-1}(f(C))) = B$ so it follows that $ B = f(A setminus f^{-1}(f(C))) = f(A) setminus f(C) = B setminus f(C)$
So $B = B setminus f(C)$, which implies that $f(C) = varnothing$ and thus $C = varnothing$, so $A$ is open, and we can finally conclude that $f$ is continuous.
answered Jan 28 at 20:33
kmmkmm
1637
answered Jan 28 at 20:33
kmmkmm
1637
answered Jan 28 at 20:33
kmmkmm
1637
answered Jan 28 at 20:33
kmmkmm
1637
1637
$begingroup$
Here are two problems in the proof. (1) In general $f(A) subsetneqq B$. This can be resolved as follows. We know that $f$ is an open map so that $f(X)$ is open in $Y$. Hence $B' = B cap f(X)$ is open and $f^{-1}(B') = f^{-1}(B)$. Thus we may assume w.l.o.g. $B subset f(X)$. In that case $f(A) = B$. (2) We only have $operatorname{int} (M cup N) supset operatorname{int} M cup operatorname{int} N$ and in general both sides are not equal. I do not see how to resolve this since we cannot conclude $operatorname{int}G = operatorname{int}(A setminus f^{-1}(f(C)))$.
$endgroup$
– Paul Frost
Jan 29 at 10:52
$begingroup$
Good points, thank you. I did forget to clarify that we can assume that $f$ is subjective. With regard to the second problem, the interior of a union equals the union of the interiors in the case that the sets are separated. I think that is true here, $C = A setminus operatorname{int} A subset overline{A} setminus operatorname{int} A = partial A$, and $A setminus f^{-1}(f(C)) subset A setminus C = A setminus (A setminus operatorname{int}A) = operatorname{int} A$. And the boundary and interior of a set are always separate.
$endgroup$
– kmm
Jan 29 at 20:18
$begingroup$
The usual definition that sets $M,N$ are separated is that $overline{M} cap N = M cap overline{N} = emptyset$. But $partial A$ and $operatorname{int} A$ are in general not separated. So perhaps the better argument is this. We have $G subset A$. Hence $operatorname{int} G subset operatorname{int} A$ so that $operatorname{int} G cap partial A = emptyset$. Since $C subset partial A$, we get $operatorname{int} G subset A setminus f^{-1}(f(C))$ whence $operatorname{int} G subset operatorname{int} (A setminus f^{-1}(f(C))) subset operatorname{int} G$.
$endgroup$
– Paul Frost
Jan 30 at 0:50
add a comment |
$begingroup$
Here are two problems in the proof. (1) In general $f(A) subsetneqq B$. This can be resolved as follows. We know that $f$ is an open map so that $f(X)$ is open in $Y$. Hence $B' = B cap f(X)$ is open and $f^{-1}(B') = f^{-1}(B)$. Thus we may assume w.l.o.g. $B subset f(X)$. In that case $f(A) = B$. (2) We only have $operatorname{int} (M cup N) supset operatorname{int} M cup operatorname{int} N$ and in general both sides are not equal. I do not see how to resolve this since we cannot conclude $operatorname{int}G = operatorname{int}(A setminus f^{-1}(f(C)))$.
$endgroup$
– Paul Frost
Jan 29 at 10:52
$begingroup$
Good points, thank you. I did forget to clarify that we can assume that $f$ is subjective. With regard to the second problem, the interior of a union equals the union of the interiors in the case that the sets are separated. I think that is true here, $C = A setminus operatorname{int} A subset overline{A} setminus operatorname{int} A = partial A$, and $A setminus f^{-1}(f(C)) subset A setminus C = A setminus (A setminus operatorname{int}A) = operatorname{int} A$. And the boundary and interior of a set are always separate.
$endgroup$
– kmm
Jan 29 at 20:18
$begingroup$
The usual definition that sets $M,N$ are separated is that $overline{M} cap N = M cap overline{N} = emptyset$. But $partial A$ and $operatorname{int} A$ are in general not separated. So perhaps the better argument is this. We have $G subset A$. Hence $operatorname{int} G subset operatorname{int} A$ so that $operatorname{int} G cap partial A = emptyset$. Since $C subset partial A$, we get $operatorname{int} G subset A setminus f^{-1}(f(C))$ whence $operatorname{int} G subset operatorname{int} (A setminus f^{-1}(f(C))) subset operatorname{int} G$.
$endgroup$
– Paul Frost
Jan 30 at 0:50
$begingroup$
Here are two problems in the proof. (1) In general $f(A) subsetneqq B$. This can be resolved as follows. We know that $f$ is an open map so that $f(X)$ is open in $Y$. Hence $B' = B cap f(X)$ is open and $f^{-1}(B') = f^{-1}(B)$. Thus we may assume w.l.o.g. $B subset f(X)$. In that case $f(A) = B$. (2) We only have $operatorname{int} (M cup N) supset operatorname{int} M cup operatorname{int} N$ and in general both sides are not equal. I do not see how to resolve this since we cannot conclude $operatorname{int}G = operatorname{int}(A setminus f^{-1}(f(C)))$.
$endgroup$
– Paul Frost
Jan 29 at 10:52
$begingroup$
Here are two problems in the proof. (1) In general $f(A) subsetneqq B$. This can be resolved as follows. We know that $f$ is an open map so that $f(X)$ is open in $Y$. Hence $B' = B cap f(X)$ is open and $f^{-1}(B') = f^{-1}(B)$. Thus we may assume w.l.o.g. $B subset f(X)$. In that case $f(A) = B$. (2) We only have $operatorname{int} (M cup N) supset operatorname{int} M cup operatorname{int} N$ and in general both sides are not equal. I do not see how to resolve this since we cannot conclude $operatorname{int}G = operatorname{int}(A setminus f^{-1}(f(C)))$.
$endgroup$
– Paul Frost
Jan 29 at 10:52
$begingroup$
Good points, thank you. I did forget to clarify that we can assume that $f$ is subjective. With regard to the second problem, the interior of a union equals the union of the interiors in the case that the sets are separated. I think that is true here, $C = A setminus operatorname{int} A subset overline{A} setminus operatorname{int} A = partial A$, and $A setminus f^{-1}(f(C)) subset A setminus C = A setminus (A setminus operatorname{int}A) = operatorname{int} A$. And the boundary and interior of a set are always separate.
$endgroup$
– kmm
Jan 29 at 20:18
$begingroup$
Good points, thank you. I did forget to clarify that we can assume that $f$ is subjective. With regard to the second problem, the interior of a union equals the union of the interiors in the case that the sets are separated. I think that is true here, $C = A setminus operatorname{int} A subset overline{A} setminus operatorname{int} A = partial A$, and $A setminus f^{-1}(f(C)) subset A setminus C = A setminus (A setminus operatorname{int}A) = operatorname{int} A$. And the boundary and interior of a set are always separate.
$endgroup$
– kmm
Jan 29 at 20:18
$begingroup$
The usual definition that sets $M,N$ are separated is that $overline{M} cap N = M cap overline{N} = emptyset$. But $partial A$ and $operatorname{int} A$ are in general not separated. So perhaps the better argument is this. We have $G subset A$. Hence $operatorname{int} G subset operatorname{int} A$ so that $operatorname{int} G cap partial A = emptyset$. Since $C subset partial A$, we get $operatorname{int} G subset A setminus f^{-1}(f(C))$ whence $operatorname{int} G subset operatorname{int} (A setminus f^{-1}(f(C))) subset operatorname{int} G$.
$endgroup$
– Paul Frost
Jan 30 at 0:50
$begingroup$
The usual definition that sets $M,N$ are separated is that $overline{M} cap N = M cap overline{N} = emptyset$. But $partial A$ and $operatorname{int} A$ are in general not separated. So perhaps the better argument is this. We have $G subset A$. Hence $operatorname{int} G subset operatorname{int} A$ so that $operatorname{int} G cap partial A = emptyset$. Since $C subset partial A$, we get $operatorname{int} G subset A setminus f^{-1}(f(C))$ whence $operatorname{int} G subset operatorname{int} (A setminus f^{-1}(f(C))) subset operatorname{int} G$.
$endgroup$
– Paul Frost
Jan 30 at 0:50
add a comment |
$begingroup$
This is only a partial answer. Obviously $f$ is open. Concerning continuity we may
w.l.o.g. assume that $f$ is surjective. In fact, consider the map $f' : X stackrel{f}{rightarrow} f(X)$. Then $f$ is continuous iff $f'$ is continuous. But $f'$ also satisfies $f'(operatorname{int} A) = operatorname{int} f'(A)$ because $f(X)$ is open in $Y$ (that is, for $B subset f(X)$ we have $operatorname{int}_Y B = operatorname{int}_{f(X)} B$).
Now it is clear that if $f$ is injective, then it is continuous. Our above argument shows that it suffices to consider a bijective $f$. In that case let $V subset Y$ be open. Then $f(operatorname{int} ( f^{-1}(V)) = operatorname{int} f(f^{-1}(V)) = operatorname{int} V = V = f(f^{-1}(V))$, hence $operatorname{int} f^{-1}(V) = f^{-1}(V)$, i.e. $f^{-1}(V)$ is open.
answered Jan 27 at 17:42
Paul FrostPaul Frost
11.6k3935
$endgroup$
add a comment |
$begingroup$
This is only a partial answer. Obviously $f$ is open. Concerning continuity we may
w.l.o.g. assume that $f$ is surjective. In fact, consider the map $f' : X stackrel{f}{rightarrow} f(X)$. Then $f$ is continuous iff $f'$ is continuous. But $f'$ also satisfies $f'(operatorname{int} A) = operatorname{int} f'(A)$ because $f(X)$ is open in $Y$ (that is, for $B subset f(X)$ we have $operatorname{int}_Y B = operatorname{int}_{f(X)} B$).
Now it is clear that if $f$ is injective, then it is continuous. Our above argument shows that it suffices to consider a bijective $f$. In that case let $V subset Y$ be open. Then $f(operatorname{int} ( f^{-1}(V)) = operatorname{int} f(f^{-1}(V)) = operatorname{int} V = V = f(f^{-1}(V))$, hence $operatorname{int} f^{-1}(V) = f^{-1}(V)$, i.e. $f^{-1}(V)$ is open.
answered Jan 27 at 17:42
Paul FrostPaul Frost
11.6k3935
$endgroup$
add a comment |
$begingroup$
This is only a partial answer. Obviously $f$ is open. Concerning continuity we may
w.l.o.g. assume that $f$ is surjective. In fact, consider the map $f' : X stackrel{f}{rightarrow} f(X)$. Then $f$ is continuous iff $f'$ is continuous. But $f'$ also satisfies $f'(operatorname{int} A) = operatorname{int} f'(A)$ because $f(X)$ is open in $Y$ (that is, for $B subset f(X)$ we have $operatorname{int}_Y B = operatorname{int}_{f(X)} B$).
Now it is clear that if $f$ is injective, then it is continuous. Our above argument shows that it suffices to consider a bijective $f$. In that case let $V subset Y$ be open. Then $f(operatorname{int} ( f^{-1}(V)) = operatorname{int} f(f^{-1}(V)) = operatorname{int} V = V = f(f^{-1}(V))$, hence $operatorname{int} f^{-1}(V) = f^{-1}(V)$, i.e. $f^{-1}(V)$ is open.
answered Jan 27 at 17:42
Paul FrostPaul Frost
11.6k3935
$endgroup$
This is only a partial answer. Obviously $f$ is open. Concerning continuity we may
w.l.o.g. assume that $f$ is surjective. In fact, consider the map $f' : X stackrel{f}{rightarrow} f(X)$. Then $f$ is continuous iff $f'$ is continuous. But $f'$ also satisfies $f'(operatorname{int} A) = operatorname{int} f'(A)$ because $f(X)$ is open in $Y$ (that is, for $B subset f(X)$ we have $operatorname{int}_Y B = operatorname{int}_{f(X)} B$).
Now it is clear that if $f$ is injective, then it is continuous. Our above argument shows that it suffices to consider a bijective $f$. In that case let $V subset Y$ be open. Then $f(operatorname{int} ( f^{-1}(V)) = operatorname{int} f(f^{-1}(V)) = operatorname{int} V = V = f(f^{-1}(V))$, hence $operatorname{int} f^{-1}(V) = f^{-1}(V)$, i.e. $f^{-1}(V)$ is open.
answered Jan 27 at 17:42
Paul FrostPaul Frost
11.6k3935
answered Jan 27 at 17:42
Paul FrostPaul Frost
11.6k3935
answered Jan 27 at 17:42
Paul FrostPaul Frost
11.6k3935
answered Jan 27 at 17:42
Paul FrostPaul Frost
11.6k3935
11.6k3935
add a comment |
add a comment |
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$begingroup$
Concerning the converse: The projection $p : [0,1] times [0,1] to [0,1]$ onto the first coordinate is continuous, open, closed and surjective but does not satisfy $p(operatorname{int} A) = operatorname{int} p(A)$ (consider $A$ = diagonal).
$endgroup$
– Paul Frost
Jan 27 at 17:17
$begingroup$
What is the origin of your question? Do you have any indication that $f$ is continuous or should one look for counterexamples?
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– Paul Frost
Jan 27 at 23:24
$begingroup$
I got it from a list of exercises of a university topology course. It asks to prove $f$ is continuous, so the statement must be true. There is no further information given that what I have written down here, the other exercises are very clear about the properties of the functions and topological spaces, so I don't think are more conditions. I think I will contact the TA associated to the course to ask if something is missing.
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– kmm
Jan 28 at 0:07
$begingroup$
Do not believe that something must be true simply because it appears in a list of exercises ;-) However, I do not claim it is wrong, perhaps I just do not see how to prove it.
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– Paul Frost
Jan 28 at 16:01
$begingroup$
I got a reply back with a rough proof sketch that I think I was able to work out. I added my own answer, sorry for the mess, thank you for your remarks.
$endgroup$
– kmm
Jan 29 at 1:03