$f : M → M$ be a transitive continuous transformation, if $phi circ f=phi$, then $phi$ is constant. The 2019 Stack Overflow Developer Survey Results Are InProof based on distance function continuityDefinition of ContinuityProblem 13 chapter 4 from baby RudinProof based on convergence arguments that, if $phi in mathbbR^X$ is continuous, then $ x $ is closedIs an expanding map on a compact metric space continuous?If $f$ pulls back closed sets to closed sets then $f$ is continuousLimit at set boundary for uniform function$f$ continuous at point if and only if $limlimits_ktoinfty(sup(f)-inf(f))=0$ on diminishing set $N_k$Show that there is a continuous real function $h$ on $[0,1]$ such that $lim sup |frach(x+t)-h(x)t| = infty$ for all $x in [0,1)$Show that $f(x,y)=begincases fracphi(x) - phi(y)x-y & x not = y\ phi'(x) & x=y endcases$ is continuous
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$f : M → M$ be a transitive continuous transformation, if $phi circ f=phi$, then $phi$ is constant.
The 2019 Stack Overflow Developer Survey Results Are InProof based on distance function continuityDefinition of ContinuityProblem 13 chapter 4 from baby RudinProof based on convergence arguments that, if $phi in mathbbR^X$ is continuous, then $ x $ is closedIs an expanding map on a compact metric space continuous?If $f$ pulls back closed sets to closed sets then $f$ is continuousLimit at set boundary for uniform function$f$ continuous at point if and only if $limlimits_ktoinfty(sup(f)-inf(f))=0$ on diminishing set $N_k$Show that there is a continuous real function $h$ on $[0,1]$ such that $lim sup |frach(x+t)-h(x)t| = infty$ for all $x in [0,1)$Show that $f(x,y)=begincases fracphi(x) - phi(y)x-y & x not = y\ phi'(x) & x=y endcases$ is continuous
$begingroup$
We say that $f$ is transitive if there exists some $x ∈ M$ such that $ f^n(x) :
n ∈ N $ is dense in $M$.
Let $f : M → M$ be a transitive continuous transformation in a compact metric
space, and $phi:M rightarrow mathbbR$ continuous. If $phi circ f=phi$, then $phi$ is constant.
Attempt:I have tried to show that $ phi (x) = phi (y)$ for any $y$ in $M$. Using the orbit of $x$ is dense in $M$, I find a sequence $f^n_s(x)$ converging to $y$. Then using that $phi circ f=phi$, by recursion we obtain $phi circ f^n=phi$. Then $limlimits_s to infty ( phi circ f^n_s)(x)=phi(x) $, but by continuity $phi (limlimits_s to infty f^n_s(x))=phi(x)$ implying $phi (y)=phi(x)$.
I imagine the proof is something of that kind, because the fact of being $ f^n(x) :n ∈ N $ dense, and of equality $phi circ f^n=phi$ says that $ phi(x)$ coincides with almost every point in $M$. Does anyone have any better suggestions?
proof-verification continuity metric-spaces
edited Apr 8 at 1:41
ZeroXLR
1,441519
asked Apr 8 at 1:28
Ricardo FreireRicardo Freire
611211
$endgroup$
add a comment |
$begingroup$
We say that $f$ is transitive if there exists some $x ∈ M$ such that $ f^n(x) :
n ∈ N $ is dense in $M$.
Let $f : M → M$ be a transitive continuous transformation in a compact metric
space, and $phi:M rightarrow mathbbR$ continuous. If $phi circ f=phi$, then $phi$ is constant.
Attempt:I have tried to show that $ phi (x) = phi (y)$ for any $y$ in $M$. Using the orbit of $x$ is dense in $M$, I find a sequence $f^n_s(x)$ converging to $y$. Then using that $phi circ f=phi$, by recursion we obtain $phi circ f^n=phi$. Then $limlimits_s to infty ( phi circ f^n_s)(x)=phi(x) $, but by continuity $phi (limlimits_s to infty f^n_s(x))=phi(x)$ implying $phi (y)=phi(x)$.
I imagine the proof is something of that kind, because the fact of being $ f^n(x) :n ∈ N $ dense, and of equality $phi circ f^n=phi$ says that $ phi(x)$ coincides with almost every point in $M$. Does anyone have any better suggestions?
proof-verification continuity metric-spaces
edited Apr 8 at 1:41
ZeroXLR
1,441519
asked Apr 8 at 1:28
Ricardo FreireRicardo Freire
611211
$endgroup$
1
$begingroup$
To me, this is already better.
$endgroup$
– HK Lee
Apr 8 at 1:59
add a comment |
$begingroup$
We say that $f$ is transitive if there exists some $x ∈ M$ such that $ f^n(x) :
n ∈ N $ is dense in $M$.
Let $f : M → M$ be a transitive continuous transformation in a compact metric
space, and $phi:M rightarrow mathbbR$ continuous. If $phi circ f=phi$, then $phi$ is constant.
Attempt:I have tried to show that $ phi (x) = phi (y)$ for any $y$ in $M$. Using the orbit of $x$ is dense in $M$, I find a sequence $f^n_s(x)$ converging to $y$. Then using that $phi circ f=phi$, by recursion we obtain $phi circ f^n=phi$. Then $limlimits_s to infty ( phi circ f^n_s)(x)=phi(x) $, but by continuity $phi (limlimits_s to infty f^n_s(x))=phi(x)$ implying $phi (y)=phi(x)$.
I imagine the proof is something of that kind, because the fact of being $ f^n(x) :n ∈ N $ dense, and of equality $phi circ f^n=phi$ says that $ phi(x)$ coincides with almost every point in $M$. Does anyone have any better suggestions?
proof-verification continuity metric-spaces
edited Apr 8 at 1:41
ZeroXLR
1,441519
asked Apr 8 at 1:28
Ricardo FreireRicardo Freire
611211
$endgroup$
We say that $f$ is transitive if there exists some $x ∈ M$ such that $ f^n(x) :
n ∈ N $ is dense in $M$.
Let $f : M → M$ be a transitive continuous transformation in a compact metric
space, and $phi:M rightarrow mathbbR$ continuous. If $phi circ f=phi$, then $phi$ is constant.
Attempt:I have tried to show that $ phi (x) = phi (y)$ for any $y$ in $M$. Using the orbit of $x$ is dense in $M$, I find a sequence $f^n_s(x)$ converging to $y$. Then using that $phi circ f=phi$, by recursion we obtain $phi circ f^n=phi$. Then $limlimits_s to infty ( phi circ f^n_s)(x)=phi(x) $, but by continuity $phi (limlimits_s to infty f^n_s(x))=phi(x)$ implying $phi (y)=phi(x)$.
I imagine the proof is something of that kind, because the fact of being $ f^n(x) :n ∈ N $ dense, and of equality $phi circ f^n=phi$ says that $ phi(x)$ coincides with almost every point in $M$. Does anyone have any better suggestions?
proof-verification continuity metric-spaces
proof-verification continuity metric-spaces
edited Apr 8 at 1:41
ZeroXLR
1,441519
asked Apr 8 at 1:28
Ricardo FreireRicardo Freire
611211
edited Apr 8 at 1:41
ZeroXLR
1,441519
asked Apr 8 at 1:28
Ricardo FreireRicardo Freire
611211
edited Apr 8 at 1:41
ZeroXLR
1,441519
edited Apr 8 at 1:41
ZeroXLR
1,441519
edited Apr 8 at 1:41
ZeroXLR
1,441519
1,441519
asked Apr 8 at 1:28
Ricardo FreireRicardo Freire
611211
asked Apr 8 at 1:28
Ricardo FreireRicardo Freire
611211
asked Apr 8 at 1:28
Ricardo FreireRicardo Freire
611211
611211
1
$begingroup$
To me, this is already better.
$endgroup$
– HK Lee
Apr 8 at 1:59
add a comment |
1
$begingroup$
To me, this is already better.
$endgroup$
– HK Lee
Apr 8 at 1:59
1
1
$begingroup$
To me, this is already better.
$endgroup$
– HK Lee
Apr 8 at 1:59
$begingroup$
To me, this is already better.
$endgroup$
– HK Lee
Apr 8 at 1:59
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Since $phicirc f=phi$, for every integer $n$, $phicirc f^n=phi$. This implies that $phi(f^n(x))=phi(x)$, we deduce that the restriction of $phi$ on $Y=f^n(x)$ is constant, and $phi$ is constant on the adherence $overlineY=X$ of $Y$.
answered Apr 8 at 2:42
Tsemo AristideTsemo Aristide
60.4k11446
$endgroup$
$begingroup$
It suffices that $M$ is a Hausdorff space. If $X$ is $any$ space and if $M$ is Hausdorff and if $f:Xto M,, g:Xto M$ are continuous then $yin X: f(y)=g(y) $ is closed in $X$... With $X=M$ and with $g(y)=phi(x)$ for all $yin M.$...+1
$endgroup$
– DanielWainfleet
Apr 8 at 8:12
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Since $phicirc f=phi$, for every integer $n$, $phicirc f^n=phi$. This implies that $phi(f^n(x))=phi(x)$, we deduce that the restriction of $phi$ on $Y=f^n(x)$ is constant, and $phi$ is constant on the adherence $overlineY=X$ of $Y$.
answered Apr 8 at 2:42
Tsemo AristideTsemo Aristide
60.4k11446
$endgroup$
$begingroup$
It suffices that $M$ is a Hausdorff space. If $X$ is $any$ space and if $M$ is Hausdorff and if $f:Xto M,, g:Xto M$ are continuous then $yin X: f(y)=g(y) $ is closed in $X$... With $X=M$ and with $g(y)=phi(x)$ for all $yin M.$...+1
$endgroup$
– DanielWainfleet
Apr 8 at 8:12
add a comment |
$begingroup$
Since $phicirc f=phi$, for every integer $n$, $phicirc f^n=phi$. This implies that $phi(f^n(x))=phi(x)$, we deduce that the restriction of $phi$ on $Y=f^n(x)$ is constant, and $phi$ is constant on the adherence $overlineY=X$ of $Y$.
answered Apr 8 at 2:42
Tsemo AristideTsemo Aristide
60.4k11446
$endgroup$
$begingroup$
It suffices that $M$ is a Hausdorff space. If $X$ is $any$ space and if $M$ is Hausdorff and if $f:Xto M,, g:Xto M$ are continuous then $yin X: f(y)=g(y) $ is closed in $X$... With $X=M$ and with $g(y)=phi(x)$ for all $yin M.$...+1
$endgroup$
– DanielWainfleet
Apr 8 at 8:12
add a comment |
$begingroup$
Since $phicirc f=phi$, for every integer $n$, $phicirc f^n=phi$. This implies that $phi(f^n(x))=phi(x)$, we deduce that the restriction of $phi$ on $Y=f^n(x)$ is constant, and $phi$ is constant on the adherence $overlineY=X$ of $Y$.
answered Apr 8 at 2:42
Tsemo AristideTsemo Aristide
60.4k11446
$endgroup$
Since $phicirc f=phi$, for every integer $n$, $phicirc f^n=phi$. This implies that $phi(f^n(x))=phi(x)$, we deduce that the restriction of $phi$ on $Y=f^n(x)$ is constant, and $phi$ is constant on the adherence $overlineY=X$ of $Y$.
answered Apr 8 at 2:42
Tsemo AristideTsemo Aristide
60.4k11446
answered Apr 8 at 2:42
Tsemo AristideTsemo Aristide
60.4k11446
answered Apr 8 at 2:42
Tsemo AristideTsemo Aristide
60.4k11446
answered Apr 8 at 2:42
Tsemo AristideTsemo Aristide
60.4k11446
60.4k11446
$begingroup$
It suffices that $M$ is a Hausdorff space. If $X$ is $any$ space and if $M$ is Hausdorff and if $f:Xto M,, g:Xto M$ are continuous then $yin X: f(y)=g(y) $ is closed in $X$... With $X=M$ and with $g(y)=phi(x)$ for all $yin M.$...+1
$endgroup$
– DanielWainfleet
Apr 8 at 8:12
add a comment |
$begingroup$
It suffices that $M$ is a Hausdorff space. If $X$ is $any$ space and if $M$ is Hausdorff and if $f:Xto M,, g:Xto M$ are continuous then $yin X: f(y)=g(y) $ is closed in $X$... With $X=M$ and with $g(y)=phi(x)$ for all $yin M.$...+1
$endgroup$
– DanielWainfleet
Apr 8 at 8:12
$begingroup$
It suffices that $M$ is a Hausdorff space. If $X$ is $any$ space and if $M$ is Hausdorff and if $f:Xto M,, g:Xto M$ are continuous then $yin X: f(y)=g(y) $ is closed in $X$... With $X=M$ and with $g(y)=phi(x)$ for all $yin M.$...+1
$endgroup$
– DanielWainfleet
Apr 8 at 8:12
$begingroup$
It suffices that $M$ is a Hausdorff space. If $X$ is $any$ space and if $M$ is Hausdorff and if $f:Xto M,, g:Xto M$ are continuous then $yin X: f(y)=g(y) $ is closed in $X$... With $X=M$ and with $g(y)=phi(x)$ for all $yin M.$...+1
$endgroup$
– DanielWainfleet
Apr 8 at 8:12
add a comment |
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$begingroup$
To me, this is already better.
$endgroup$
– HK Lee
Apr 8 at 1:59