Does there exist a bijection of $mathbb{R}^n$ with itself such that the forward map is connected but the...
$begingroup$
Let $(X,tau), (Y,sigma)$ be two topological spaces. We say that a map $f: mathcal{P}(X)to mathcal{P}(Y)$ between their power sets is connected if for every $Ssubset X$ connected, $f(S)subset Y$ is connected.
Question: Assume $f:mathbb{R}^ntomathbb{R}^n$ is a bijection, where $mathbb{R}^n$ is equipped with the standard topology. Does the connectedness of (the induced power set map) $f$ imply that of $f^{-1}$?
Full disclosure: I've now crossposted the question on MO.
Various remarks
If we remove the bijection requirement then the answer is clearly "no". For example, $f(x) = sin(x)$ when $n = 1$ is a map whose forward map preserves connectedness but the inverse map does not.
With the bijection, it holds true in $n = 1$. But this is using the order structure of $mathbb{R}$: a bijection that preserves connectedness on $mathbb{R}$ must be monotone.
As a result of the invariance of domain theorem if either $f$ or $f^{-1}$ is continuous, we must have that $f$ is a homeomorphism, which would imply that both $f$ and $f^{-1}$ must be connected. (See Is bijection mapping connected sets to connected homeomorphism? which inspired this question for more about this.)
Invariance of domain, in fact, asserts a positive answer to the following question which is very similar in shape and spirit to the one I asked above:
Assume $f:mathbb{R}^ntomathbb{R}^n$ is a bijection, where $mathbb{R}^n$ is equipped with the standard topology. Does the fact that $f$ is an open map imply that $f^{-1}$ is open?
Some properties of $mathbb{R}^n$ must factor in heavily in the answer. If we replace the question and consider, instead of self-maps of $mathbb{R}^n$ with the standard topology to itself, by self-maps of some arbitrary topological space, it is easy to make the answer go either way.
For example, if the topological space $(X,tau)$ is such that there are only finitely many connected subsets of $X$ (which would be the case if $X$ itself is a finite set), then by cardinality argument we have that the answer is yes, $f^{-1}$ must also be connected.
On the other hand, one can easily cook up examples where the answer is no; a large number of examples can be constructed as variants of the following: let $X = mathbb{Z}$ equipped with the topology generated by
$$ {mathbb{N}} cup { {k} mid k in mathbb{Z} setminus mathbb{N} } $$
then the map $k mapsto k+ell$ for any $ell > 0$ maps connected sets to connected sets, but its inverse $kmapsto k-ell$ can map connected sets to disconnected ones.
Working a bit harder one can construct in similar vein examples which are Hausdorff.
general-topology metric-spaces examples-counterexamples connectedness
edited Apr 13 '17 at 12:58
Community♦
1
asked Sep 30 '14 at 12:18
Willie WongWillie Wong
55.3k10108209
$endgroup$
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show 28 more comments
$begingroup$
Let $(X,tau), (Y,sigma)$ be two topological spaces. We say that a map $f: mathcal{P}(X)to mathcal{P}(Y)$ between their power sets is connected if for every $Ssubset X$ connected, $f(S)subset Y$ is connected.
Question: Assume $f:mathbb{R}^ntomathbb{R}^n$ is a bijection, where $mathbb{R}^n$ is equipped with the standard topology. Does the connectedness of (the induced power set map) $f$ imply that of $f^{-1}$?
Full disclosure: I've now crossposted the question on MO.
Various remarks
If we remove the bijection requirement then the answer is clearly "no". For example, $f(x) = sin(x)$ when $n = 1$ is a map whose forward map preserves connectedness but the inverse map does not.
With the bijection, it holds true in $n = 1$. But this is using the order structure of $mathbb{R}$: a bijection that preserves connectedness on $mathbb{R}$ must be monotone.
As a result of the invariance of domain theorem if either $f$ or $f^{-1}$ is continuous, we must have that $f$ is a homeomorphism, which would imply that both $f$ and $f^{-1}$ must be connected. (See Is bijection mapping connected sets to connected homeomorphism? which inspired this question for more about this.)
Invariance of domain, in fact, asserts a positive answer to the following question which is very similar in shape and spirit to the one I asked above:
Assume $f:mathbb{R}^ntomathbb{R}^n$ is a bijection, where $mathbb{R}^n$ is equipped with the standard topology. Does the fact that $f$ is an open map imply that $f^{-1}$ is open?
Some properties of $mathbb{R}^n$ must factor in heavily in the answer. If we replace the question and consider, instead of self-maps of $mathbb{R}^n$ with the standard topology to itself, by self-maps of some arbitrary topological space, it is easy to make the answer go either way.
For example, if the topological space $(X,tau)$ is such that there are only finitely many connected subsets of $X$ (which would be the case if $X$ itself is a finite set), then by cardinality argument we have that the answer is yes, $f^{-1}$ must also be connected.
On the other hand, one can easily cook up examples where the answer is no; a large number of examples can be constructed as variants of the following: let $X = mathbb{Z}$ equipped with the topology generated by
$$ {mathbb{N}} cup { {k} mid k in mathbb{Z} setminus mathbb{N} } $$
then the map $k mapsto k+ell$ for any $ell > 0$ maps connected sets to connected sets, but its inverse $kmapsto k-ell$ can map connected sets to disconnected ones.
Working a bit harder one can construct in similar vein examples which are Hausdorff.
general-topology metric-spaces examples-counterexamples connectedness
edited Apr 13 '17 at 12:58
Community♦
1
asked Sep 30 '14 at 12:18
Willie WongWillie Wong
55.3k10108209
$endgroup$
23
$begingroup$
To discourage other broken attempts by users unaware of the subtleties would it be a good idea to tell that as a consequence of invariance of domain an eventual counterexample cannot be continuous everywhere? The problem is easy to understand, so an unsuspecting noob familiar with the terms may think they can say something helpful out of ignorance.
$endgroup$
– Jyrki Lahtonen
Jun 20 '15 at 18:00
6
$begingroup$
@JyrkiLahtonen: Great idea! Let me edit that and some other remarks in. (I also protected this question.)
$endgroup$
– Willie Wong
Jun 22 '15 at 8:03
5
$begingroup$
@user2520938: the quoted theorem from the other post says that if both $f$ and $f^{-1}$ are connected then $f$ is continuous. It does not, as you seem to think, say that $f^{-1}$ connected $implies$ $f$ is continuous. (The connectedness of the forward map is used implicitly in the contradiction step.)
$endgroup$
– Willie Wong
Aug 21 '15 at 2:24
5
$begingroup$
@Greg Martin: Consider $f:mathbb{R}^2tomathbb{R}^2$ with $f(0,t)=(0,t)$ and $f(s,t)=t+sin(pi/s)$ for $sneq0$. It seems to me that $f$ is a bijection and $f$ and $f^{-1}$ are both connected and discontinuous. I had hoped that $f$ could be proved continuous, but obviously not! A positive theorem, however, is that if $f^{-1}$ maps separated pairs of sets to separated pairs, and $Y$ is $T_3$, then $f$ is continuous. My example, however, does not disprove the conjecture that $f^{-1}$ is connected.
$endgroup$
– Alan U. Kennington
Aug 29 '15 at 21:03
5
$begingroup$
For a concrete example of a connected set $S$ such that $R:=f(S)$ is not connected, consider $R:= {(0,1)}cup (0,1]times{0}$ and $S:=f^{-1}(R )$.
$endgroup$
– Pietro Majer
Jan 26 '17 at 17:38
|
show 28 more comments
$begingroup$
Let $(X,tau), (Y,sigma)$ be two topological spaces. We say that a map $f: mathcal{P}(X)to mathcal{P}(Y)$ between their power sets is connected if for every $Ssubset X$ connected, $f(S)subset Y$ is connected.
Question: Assume $f:mathbb{R}^ntomathbb{R}^n$ is a bijection, where $mathbb{R}^n$ is equipped with the standard topology. Does the connectedness of (the induced power set map) $f$ imply that of $f^{-1}$?
Full disclosure: I've now crossposted the question on MO.
Various remarks
If we remove the bijection requirement then the answer is clearly "no". For example, $f(x) = sin(x)$ when $n = 1$ is a map whose forward map preserves connectedness but the inverse map does not.
With the bijection, it holds true in $n = 1$. But this is using the order structure of $mathbb{R}$: a bijection that preserves connectedness on $mathbb{R}$ must be monotone.
As a result of the invariance of domain theorem if either $f$ or $f^{-1}$ is continuous, we must have that $f$ is a homeomorphism, which would imply that both $f$ and $f^{-1}$ must be connected. (See Is bijection mapping connected sets to connected homeomorphism? which inspired this question for more about this.)
Invariance of domain, in fact, asserts a positive answer to the following question which is very similar in shape and spirit to the one I asked above:
Assume $f:mathbb{R}^ntomathbb{R}^n$ is a bijection, where $mathbb{R}^n$ is equipped with the standard topology. Does the fact that $f$ is an open map imply that $f^{-1}$ is open?
Some properties of $mathbb{R}^n$ must factor in heavily in the answer. If we replace the question and consider, instead of self-maps of $mathbb{R}^n$ with the standard topology to itself, by self-maps of some arbitrary topological space, it is easy to make the answer go either way.
For example, if the topological space $(X,tau)$ is such that there are only finitely many connected subsets of $X$ (which would be the case if $X$ itself is a finite set), then by cardinality argument we have that the answer is yes, $f^{-1}$ must also be connected.
On the other hand, one can easily cook up examples where the answer is no; a large number of examples can be constructed as variants of the following: let $X = mathbb{Z}$ equipped with the topology generated by
$$ {mathbb{N}} cup { {k} mid k in mathbb{Z} setminus mathbb{N} } $$
then the map $k mapsto k+ell$ for any $ell > 0$ maps connected sets to connected sets, but its inverse $kmapsto k-ell$ can map connected sets to disconnected ones.
Working a bit harder one can construct in similar vein examples which are Hausdorff.
general-topology metric-spaces examples-counterexamples connectedness
edited Apr 13 '17 at 12:58
Community♦
1
asked Sep 30 '14 at 12:18
Willie WongWillie Wong
55.3k10108209
$endgroup$
Let $(X,tau), (Y,sigma)$ be two topological spaces. We say that a map $f: mathcal{P}(X)to mathcal{P}(Y)$ between their power sets is connected if for every $Ssubset X$ connected, $f(S)subset Y$ is connected.
Question: Assume $f:mathbb{R}^ntomathbb{R}^n$ is a bijection, where $mathbb{R}^n$ is equipped with the standard topology. Does the connectedness of (the induced power set map) $f$ imply that of $f^{-1}$?
Full disclosure: I've now crossposted the question on MO.
Various remarks
If we remove the bijection requirement then the answer is clearly "no". For example, $f(x) = sin(x)$ when $n = 1$ is a map whose forward map preserves connectedness but the inverse map does not.
With the bijection, it holds true in $n = 1$. But this is using the order structure of $mathbb{R}$: a bijection that preserves connectedness on $mathbb{R}$ must be monotone.
As a result of the invariance of domain theorem if either $f$ or $f^{-1}$ is continuous, we must have that $f$ is a homeomorphism, which would imply that both $f$ and $f^{-1}$ must be connected. (See Is bijection mapping connected sets to connected homeomorphism? which inspired this question for more about this.)
Invariance of domain, in fact, asserts a positive answer to the following question which is very similar in shape and spirit to the one I asked above:
Assume $f:mathbb{R}^ntomathbb{R}^n$ is a bijection, where $mathbb{R}^n$ is equipped with the standard topology. Does the fact that $f$ is an open map imply that $f^{-1}$ is open?
Some properties of $mathbb{R}^n$ must factor in heavily in the answer. If we replace the question and consider, instead of self-maps of $mathbb{R}^n$ with the standard topology to itself, by self-maps of some arbitrary topological space, it is easy to make the answer go either way.
For example, if the topological space $(X,tau)$ is such that there are only finitely many connected subsets of $X$ (which would be the case if $X$ itself is a finite set), then by cardinality argument we have that the answer is yes, $f^{-1}$ must also be connected.
On the other hand, one can easily cook up examples where the answer is no; a large number of examples can be constructed as variants of the following: let $X = mathbb{Z}$ equipped with the topology generated by
$$ {mathbb{N}} cup { {k} mid k in mathbb{Z} setminus mathbb{N} } $$
then the map $k mapsto k+ell$ for any $ell > 0$ maps connected sets to connected sets, but its inverse $kmapsto k-ell$ can map connected sets to disconnected ones.
Working a bit harder one can construct in similar vein examples which are Hausdorff.
general-topology metric-spaces examples-counterexamples connectedness
general-topology metric-spaces examples-counterexamples connectedness
edited Apr 13 '17 at 12:58
Community♦
1
asked Sep 30 '14 at 12:18
Willie WongWillie Wong
55.3k10108209
edited Apr 13 '17 at 12:58
Community♦
1
asked Sep 30 '14 at 12:18
Willie WongWillie Wong
55.3k10108209
edited Apr 13 '17 at 12:58
Community♦
1
edited Apr 13 '17 at 12:58
Community♦
1
edited Apr 13 '17 at 12:58
Community♦
1
1
asked Sep 30 '14 at 12:18
Willie WongWillie Wong
55.3k10108209
asked Sep 30 '14 at 12:18
Willie WongWillie Wong
55.3k10108209
asked Sep 30 '14 at 12:18
Willie WongWillie Wong
55.3k10108209
55.3k10108209
23
$begingroup$
To discourage other broken attempts by users unaware of the subtleties would it be a good idea to tell that as a consequence of invariance of domain an eventual counterexample cannot be continuous everywhere? The problem is easy to understand, so an unsuspecting noob familiar with the terms may think they can say something helpful out of ignorance.
$endgroup$
– Jyrki Lahtonen
Jun 20 '15 at 18:00
6
$begingroup$
@JyrkiLahtonen: Great idea! Let me edit that and some other remarks in. (I also protected this question.)
$endgroup$
– Willie Wong
Jun 22 '15 at 8:03
5
$begingroup$
@user2520938: the quoted theorem from the other post says that if both $f$ and $f^{-1}$ are connected then $f$ is continuous. It does not, as you seem to think, say that $f^{-1}$ connected $implies$ $f$ is continuous. (The connectedness of the forward map is used implicitly in the contradiction step.)
$endgroup$
– Willie Wong
Aug 21 '15 at 2:24
5
$begingroup$
@Greg Martin: Consider $f:mathbb{R}^2tomathbb{R}^2$ with $f(0,t)=(0,t)$ and $f(s,t)=t+sin(pi/s)$ for $sneq0$. It seems to me that $f$ is a bijection and $f$ and $f^{-1}$ are both connected and discontinuous. I had hoped that $f$ could be proved continuous, but obviously not! A positive theorem, however, is that if $f^{-1}$ maps separated pairs of sets to separated pairs, and $Y$ is $T_3$, then $f$ is continuous. My example, however, does not disprove the conjecture that $f^{-1}$ is connected.
$endgroup$
– Alan U. Kennington
Aug 29 '15 at 21:03
5
$begingroup$
For a concrete example of a connected set $S$ such that $R:=f(S)$ is not connected, consider $R:= {(0,1)}cup (0,1]times{0}$ and $S:=f^{-1}(R )$.
$endgroup$
– Pietro Majer
Jan 26 '17 at 17:38
|
show 28 more comments
23
$begingroup$
To discourage other broken attempts by users unaware of the subtleties would it be a good idea to tell that as a consequence of invariance of domain an eventual counterexample cannot be continuous everywhere? The problem is easy to understand, so an unsuspecting noob familiar with the terms may think they can say something helpful out of ignorance.
$endgroup$
– Jyrki Lahtonen
Jun 20 '15 at 18:00
6
$begingroup$
@JyrkiLahtonen: Great idea! Let me edit that and some other remarks in. (I also protected this question.)
$endgroup$
– Willie Wong
Jun 22 '15 at 8:03
5
$begingroup$
@user2520938: the quoted theorem from the other post says that if both $f$ and $f^{-1}$ are connected then $f$ is continuous. It does not, as you seem to think, say that $f^{-1}$ connected $implies$ $f$ is continuous. (The connectedness of the forward map is used implicitly in the contradiction step.)
$endgroup$
– Willie Wong
Aug 21 '15 at 2:24
5
$begingroup$
@Greg Martin: Consider $f:mathbb{R}^2tomathbb{R}^2$ with $f(0,t)=(0,t)$ and $f(s,t)=t+sin(pi/s)$ for $sneq0$. It seems to me that $f$ is a bijection and $f$ and $f^{-1}$ are both connected and discontinuous. I had hoped that $f$ could be proved continuous, but obviously not! A positive theorem, however, is that if $f^{-1}$ maps separated pairs of sets to separated pairs, and $Y$ is $T_3$, then $f$ is continuous. My example, however, does not disprove the conjecture that $f^{-1}$ is connected.
$endgroup$
– Alan U. Kennington
Aug 29 '15 at 21:03
5
$begingroup$
For a concrete example of a connected set $S$ such that $R:=f(S)$ is not connected, consider $R:= {(0,1)}cup (0,1]times{0}$ and $S:=f^{-1}(R )$.
$endgroup$
– Pietro Majer
Jan 26 '17 at 17:38
23
23
$begingroup$
To discourage other broken attempts by users unaware of the subtleties would it be a good idea to tell that as a consequence of invariance of domain an eventual counterexample cannot be continuous everywhere? The problem is easy to understand, so an unsuspecting noob familiar with the terms may think they can say something helpful out of ignorance.
$endgroup$
– Jyrki Lahtonen
Jun 20 '15 at 18:00
$begingroup$
To discourage other broken attempts by users unaware of the subtleties would it be a good idea to tell that as a consequence of invariance of domain an eventual counterexample cannot be continuous everywhere? The problem is easy to understand, so an unsuspecting noob familiar with the terms may think they can say something helpful out of ignorance.
$endgroup$
– Jyrki Lahtonen
Jun 20 '15 at 18:00
6
6
$begingroup$
@JyrkiLahtonen: Great idea! Let me edit that and some other remarks in. (I also protected this question.)
$endgroup$
– Willie Wong
Jun 22 '15 at 8:03
$begingroup$
@JyrkiLahtonen: Great idea! Let me edit that and some other remarks in. (I also protected this question.)
$endgroup$
– Willie Wong
Jun 22 '15 at 8:03
5
5
$begingroup$
@user2520938: the quoted theorem from the other post says that if both $f$ and $f^{-1}$ are connected then $f$ is continuous. It does not, as you seem to think, say that $f^{-1}$ connected $implies$ $f$ is continuous. (The connectedness of the forward map is used implicitly in the contradiction step.)
$endgroup$
– Willie Wong
Aug 21 '15 at 2:24
$begingroup$
@user2520938: the quoted theorem from the other post says that if both $f$ and $f^{-1}$ are connected then $f$ is continuous. It does not, as you seem to think, say that $f^{-1}$ connected $implies$ $f$ is continuous. (The connectedness of the forward map is used implicitly in the contradiction step.)
$endgroup$
– Willie Wong
Aug 21 '15 at 2:24
5
5
$begingroup$
@Greg Martin: Consider $f:mathbb{R}^2tomathbb{R}^2$ with $f(0,t)=(0,t)$ and $f(s,t)=t+sin(pi/s)$ for $sneq0$. It seems to me that $f$ is a bijection and $f$ and $f^{-1}$ are both connected and discontinuous. I had hoped that $f$ could be proved continuous, but obviously not! A positive theorem, however, is that if $f^{-1}$ maps separated pairs of sets to separated pairs, and $Y$ is $T_3$, then $f$ is continuous. My example, however, does not disprove the conjecture that $f^{-1}$ is connected.
$endgroup$
– Alan U. Kennington
Aug 29 '15 at 21:03
$begingroup$
@Greg Martin: Consider $f:mathbb{R}^2tomathbb{R}^2$ with $f(0,t)=(0,t)$ and $f(s,t)=t+sin(pi/s)$ for $sneq0$. It seems to me that $f$ is a bijection and $f$ and $f^{-1}$ are both connected and discontinuous. I had hoped that $f$ could be proved continuous, but obviously not! A positive theorem, however, is that if $f^{-1}$ maps separated pairs of sets to separated pairs, and $Y$ is $T_3$, then $f$ is continuous. My example, however, does not disprove the conjecture that $f^{-1}$ is connected.
$endgroup$
– Alan U. Kennington
Aug 29 '15 at 21:03
5
5
$begingroup$
For a concrete example of a connected set $S$ such that $R:=f(S)$ is not connected, consider $R:= {(0,1)}cup (0,1]times{0}$ and $S:=f^{-1}(R )$.
$endgroup$
– Pietro Majer
Jan 26 '17 at 17:38
$begingroup$
For a concrete example of a connected set $S$ such that $R:=f(S)$ is not connected, consider $R:= {(0,1)}cup (0,1]times{0}$ and $S:=f^{-1}(R )$.
$endgroup$
– Pietro Majer
Jan 26 '17 at 17:38
|
show 28 more comments
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protected by Willie Wong Jun 22 '15 at 8:02
Thank you for your interest in this question.
Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).
Would you like to answer one of these unanswered questions instead?
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protected by Willie Wong Jun 22 '15 at 8:02
Thank you for your interest in this question.
Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).
Would you like to answer one of these unanswered questions instead?
23
$begingroup$
To discourage other broken attempts by users unaware of the subtleties would it be a good idea to tell that as a consequence of invariance of domain an eventual counterexample cannot be continuous everywhere? The problem is easy to understand, so an unsuspecting noob familiar with the terms may think they can say something helpful out of ignorance.
$endgroup$
– Jyrki Lahtonen
Jun 20 '15 at 18:00
6
$begingroup$
@JyrkiLahtonen: Great idea! Let me edit that and some other remarks in. (I also protected this question.)
$endgroup$
– Willie Wong
Jun 22 '15 at 8:03
5
$begingroup$
@user2520938: the quoted theorem from the other post says that if both $f$ and $f^{-1}$ are connected then $f$ is continuous. It does not, as you seem to think, say that $f^{-1}$ connected $implies$ $f$ is continuous. (The connectedness of the forward map is used implicitly in the contradiction step.)
$endgroup$
– Willie Wong
Aug 21 '15 at 2:24
5
$begingroup$
@Greg Martin: Consider $f:mathbb{R}^2tomathbb{R}^2$ with $f(0,t)=(0,t)$ and $f(s,t)=t+sin(pi/s)$ for $sneq0$. It seems to me that $f$ is a bijection and $f$ and $f^{-1}$ are both connected and discontinuous. I had hoped that $f$ could be proved continuous, but obviously not! A positive theorem, however, is that if $f^{-1}$ maps separated pairs of sets to separated pairs, and $Y$ is $T_3$, then $f$ is continuous. My example, however, does not disprove the conjecture that $f^{-1}$ is connected.
$endgroup$
– Alan U. Kennington
Aug 29 '15 at 21:03
5
$begingroup$
For a concrete example of a connected set $S$ such that $R:=f(S)$ is not connected, consider $R:= {(0,1)}cup (0,1]times{0}$ and $S:=f^{-1}(R )$.
$endgroup$
– Pietro Majer
Jan 26 '17 at 17:38