Continuous function on a compact metric space with no fixed point
$begingroup$
Let $(X,d)$ be a compact metric space, and suppose $f:X to X$ is a continuous map with no fixed point. Show that there is an $epsilon > 0$ so that $d(f(x),x) ge epsilon forall x in X$.
I considered a function $g:X to mathbb{R}$ such that $g(x) = d(f(x),x)$ so that there is a fixed point only when $g(x)=0$. Is there a way to proceed using this function?
general-topology metric-spaces
edited Dec 5 '18 at 4:41
gt6989b
34.6k22456
asked Dec 5 '18 at 4:30
general1597general1597
313
$endgroup$
add a comment |
$begingroup$
Let $(X,d)$ be a compact metric space, and suppose $f:X to X$ is a continuous map with no fixed point. Show that there is an $epsilon > 0$ so that $d(f(x),x) ge epsilon forall x in X$.
I considered a function $g:X to mathbb{R}$ such that $g(x) = d(f(x),x)$ so that there is a fixed point only when $g(x)=0$. Is there a way to proceed using this function?
general-topology metric-spaces
edited Dec 5 '18 at 4:41
gt6989b
34.6k22456
asked Dec 5 '18 at 4:30
general1597general1597
313
$endgroup$
$begingroup$
$f: X to X$for example.
$endgroup$
– Randall
Dec 5 '18 at 4:30
1
$begingroup$
If $forall epsilon > 0$, $exists x in X$ such that $dleft( fleft( x right), x right) < epsilon$, then $x$ will be a fixed point of $f$. Hence, the result.
$endgroup$
– Aniruddha Deshmukh
Dec 5 '18 at 4:35
add a comment |
$begingroup$
Let $(X,d)$ be a compact metric space, and suppose $f:X to X$ is a continuous map with no fixed point. Show that there is an $epsilon > 0$ so that $d(f(x),x) ge epsilon forall x in X$.
I considered a function $g:X to mathbb{R}$ such that $g(x) = d(f(x),x)$ so that there is a fixed point only when $g(x)=0$. Is there a way to proceed using this function?
general-topology metric-spaces
edited Dec 5 '18 at 4:41
gt6989b
34.6k22456
asked Dec 5 '18 at 4:30
general1597general1597
313
$endgroup$
Let $(X,d)$ be a compact metric space, and suppose $f:X to X$ is a continuous map with no fixed point. Show that there is an $epsilon > 0$ so that $d(f(x),x) ge epsilon forall x in X$.
I considered a function $g:X to mathbb{R}$ such that $g(x) = d(f(x),x)$ so that there is a fixed point only when $g(x)=0$. Is there a way to proceed using this function?
general-topology metric-spaces
general-topology metric-spaces
edited Dec 5 '18 at 4:41
gt6989b
34.6k22456
asked Dec 5 '18 at 4:30
general1597general1597
313
edited Dec 5 '18 at 4:41
gt6989b
34.6k22456
asked Dec 5 '18 at 4:30
general1597general1597
313
edited Dec 5 '18 at 4:41
gt6989b
34.6k22456
edited Dec 5 '18 at 4:41
gt6989b
34.6k22456
edited Dec 5 '18 at 4:41
gt6989b
34.6k22456
34.6k22456
asked Dec 5 '18 at 4:30
general1597general1597
313
asked Dec 5 '18 at 4:30
general1597general1597
313
asked Dec 5 '18 at 4:30
general1597general1597
313
313
$begingroup$
$f: X to X$for example.
$endgroup$
– Randall
Dec 5 '18 at 4:30
1
$begingroup$
If $forall epsilon > 0$, $exists x in X$ such that $dleft( fleft( x right), x right) < epsilon$, then $x$ will be a fixed point of $f$. Hence, the result.
$endgroup$
– Aniruddha Deshmukh
Dec 5 '18 at 4:35
add a comment |
$begingroup$
$f: X to X$for example.
$endgroup$
– Randall
Dec 5 '18 at 4:30
1
$begingroup$
If $forall epsilon > 0$, $exists x in X$ such that $dleft( fleft( x right), x right) < epsilon$, then $x$ will be a fixed point of $f$. Hence, the result.
$endgroup$
– Aniruddha Deshmukh
Dec 5 '18 at 4:35
$begingroup$
$f: X to X$ for example.$endgroup$
– Randall
Dec 5 '18 at 4:30
$begingroup$
$f: X to X$ for example.$endgroup$
– Randall
Dec 5 '18 at 4:30
1
1
$begingroup$
If $forall epsilon > 0$, $exists x in X$ such that $dleft( fleft( x right), x right) < epsilon$, then $x$ will be a fixed point of $f$. Hence, the result.
$endgroup$
– Aniruddha Deshmukh
Dec 5 '18 at 4:35
$begingroup$
If $forall epsilon > 0$, $exists x in X$ such that $dleft( fleft( x right), x right) < epsilon$, then $x$ will be a fixed point of $f$. Hence, the result.
$endgroup$
– Aniruddha Deshmukh
Dec 5 '18 at 4:35
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The function $g : X to mathbb{R}$ defined $g(x) = d(f(x),x)$ is continuous because it is a composition of continuous functions. For every $epsilon > 0$ let $I_{varepsilon} = {rin mathbb{R} mid r > varepsilon}$, which is an open set in $mathbb{R}$. Then $g^{-1}(I_{varepsilon})$ is open in $X$, and the family of sets ${g^{-1}(I_{varepsilon}) mid varepsilon > 0 }$ covers $X$ because $X$ has no fixed point. Since $X$ is compact, any open cover of it will have a finite subcover: $X = bigcup_{i=1}^{n} g^{-1}(I_{varepsilon_i})$. Let $varepsilon_{min} = min{varepsilon_i,dots,varepsilon_n}$. Then for every $xin X$ we have $x in g^{-1}(I_{varepsilon_i})$ for some $i$, hence $d(f(x),x) = g(x) in I_{varepsilon_i}$, so $d(f(x),x) > varepsilon_i geq varepsilon_{min}$, and we are done.
answered Dec 5 '18 at 4:58
mlerma54mlerma54
1,177148
$endgroup$
1
$begingroup$
The continuous image of a compact space is compact. So if $X$ is not empty then $g(X)$ is a compact ( and hence closed ) non-empty subset of $[0,infty)$ and therefore $g(X)$ has a minimum member
$endgroup$
– DanielWainfleet
Dec 5 '18 at 10:08
add a comment |
$begingroup$
For metric spaces, compactness and sequential compactness are the same. Assume that $forall epsilon > 0$, $exists x in X$ such that $d(f(x),x)< epsilon$. Then we get a sequence $x_i$ of points such that $d(f(x_i),x_i)$ converges to 0. Sequential compactness gives us the desired limit point $x^*$ that is a fixed point. This is a contradiction.
answered Dec 5 '18 at 5:06
zoidbergzoidberg
1,080113
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
The function $g : X to mathbb{R}$ defined $g(x) = d(f(x),x)$ is continuous because it is a composition of continuous functions. For every $epsilon > 0$ let $I_{varepsilon} = {rin mathbb{R} mid r > varepsilon}$, which is an open set in $mathbb{R}$. Then $g^{-1}(I_{varepsilon})$ is open in $X$, and the family of sets ${g^{-1}(I_{varepsilon}) mid varepsilon > 0 }$ covers $X$ because $X$ has no fixed point. Since $X$ is compact, any open cover of it will have a finite subcover: $X = bigcup_{i=1}^{n} g^{-1}(I_{varepsilon_i})$. Let $varepsilon_{min} = min{varepsilon_i,dots,varepsilon_n}$. Then for every $xin X$ we have $x in g^{-1}(I_{varepsilon_i})$ for some $i$, hence $d(f(x),x) = g(x) in I_{varepsilon_i}$, so $d(f(x),x) > varepsilon_i geq varepsilon_{min}$, and we are done.
answered Dec 5 '18 at 4:58
mlerma54mlerma54
1,177148
$endgroup$
1
$begingroup$
The continuous image of a compact space is compact. So if $X$ is not empty then $g(X)$ is a compact ( and hence closed ) non-empty subset of $[0,infty)$ and therefore $g(X)$ has a minimum member
$endgroup$
– DanielWainfleet
Dec 5 '18 at 10:08
add a comment |
$begingroup$
The function $g : X to mathbb{R}$ defined $g(x) = d(f(x),x)$ is continuous because it is a composition of continuous functions. For every $epsilon > 0$ let $I_{varepsilon} = {rin mathbb{R} mid r > varepsilon}$, which is an open set in $mathbb{R}$. Then $g^{-1}(I_{varepsilon})$ is open in $X$, and the family of sets ${g^{-1}(I_{varepsilon}) mid varepsilon > 0 }$ covers $X$ because $X$ has no fixed point. Since $X$ is compact, any open cover of it will have a finite subcover: $X = bigcup_{i=1}^{n} g^{-1}(I_{varepsilon_i})$. Let $varepsilon_{min} = min{varepsilon_i,dots,varepsilon_n}$. Then for every $xin X$ we have $x in g^{-1}(I_{varepsilon_i})$ for some $i$, hence $d(f(x),x) = g(x) in I_{varepsilon_i}$, so $d(f(x),x) > varepsilon_i geq varepsilon_{min}$, and we are done.
answered Dec 5 '18 at 4:58
mlerma54mlerma54
1,177148
$endgroup$
1
$begingroup$
The continuous image of a compact space is compact. So if $X$ is not empty then $g(X)$ is a compact ( and hence closed ) non-empty subset of $[0,infty)$ and therefore $g(X)$ has a minimum member
$endgroup$
– DanielWainfleet
Dec 5 '18 at 10:08
add a comment |
$begingroup$
The function $g : X to mathbb{R}$ defined $g(x) = d(f(x),x)$ is continuous because it is a composition of continuous functions. For every $epsilon > 0$ let $I_{varepsilon} = {rin mathbb{R} mid r > varepsilon}$, which is an open set in $mathbb{R}$. Then $g^{-1}(I_{varepsilon})$ is open in $X$, and the family of sets ${g^{-1}(I_{varepsilon}) mid varepsilon > 0 }$ covers $X$ because $X$ has no fixed point. Since $X$ is compact, any open cover of it will have a finite subcover: $X = bigcup_{i=1}^{n} g^{-1}(I_{varepsilon_i})$. Let $varepsilon_{min} = min{varepsilon_i,dots,varepsilon_n}$. Then for every $xin X$ we have $x in g^{-1}(I_{varepsilon_i})$ for some $i$, hence $d(f(x),x) = g(x) in I_{varepsilon_i}$, so $d(f(x),x) > varepsilon_i geq varepsilon_{min}$, and we are done.
answered Dec 5 '18 at 4:58
mlerma54mlerma54
1,177148
$endgroup$
The function $g : X to mathbb{R}$ defined $g(x) = d(f(x),x)$ is continuous because it is a composition of continuous functions. For every $epsilon > 0$ let $I_{varepsilon} = {rin mathbb{R} mid r > varepsilon}$, which is an open set in $mathbb{R}$. Then $g^{-1}(I_{varepsilon})$ is open in $X$, and the family of sets ${g^{-1}(I_{varepsilon}) mid varepsilon > 0 }$ covers $X$ because $X$ has no fixed point. Since $X$ is compact, any open cover of it will have a finite subcover: $X = bigcup_{i=1}^{n} g^{-1}(I_{varepsilon_i})$. Let $varepsilon_{min} = min{varepsilon_i,dots,varepsilon_n}$. Then for every $xin X$ we have $x in g^{-1}(I_{varepsilon_i})$ for some $i$, hence $d(f(x),x) = g(x) in I_{varepsilon_i}$, so $d(f(x),x) > varepsilon_i geq varepsilon_{min}$, and we are done.
answered Dec 5 '18 at 4:58
mlerma54mlerma54
1,177148
edited Dec 5 '18 at 5:26
answered Dec 5 '18 at 4:58
mlerma54mlerma54
1,177148
answered Dec 5 '18 at 4:58
mlerma54mlerma54
1,177148
answered Dec 5 '18 at 4:58
mlerma54mlerma54
1,177148
1,177148
1
$begingroup$
The continuous image of a compact space is compact. So if $X$ is not empty then $g(X)$ is a compact ( and hence closed ) non-empty subset of $[0,infty)$ and therefore $g(X)$ has a minimum member
$endgroup$
– DanielWainfleet
Dec 5 '18 at 10:08
add a comment |
1
$begingroup$
The continuous image of a compact space is compact. So if $X$ is not empty then $g(X)$ is a compact ( and hence closed ) non-empty subset of $[0,infty)$ and therefore $g(X)$ has a minimum member
$endgroup$
– DanielWainfleet
Dec 5 '18 at 10:08
1
1
$begingroup$
The continuous image of a compact space is compact. So if $X$ is not empty then $g(X)$ is a compact ( and hence closed ) non-empty subset of $[0,infty)$ and therefore $g(X)$ has a minimum member
$endgroup$
– DanielWainfleet
Dec 5 '18 at 10:08
$begingroup$
The continuous image of a compact space is compact. So if $X$ is not empty then $g(X)$ is a compact ( and hence closed ) non-empty subset of $[0,infty)$ and therefore $g(X)$ has a minimum member
$endgroup$
– DanielWainfleet
Dec 5 '18 at 10:08
add a comment |
$begingroup$
For metric spaces, compactness and sequential compactness are the same. Assume that $forall epsilon > 0$, $exists x in X$ such that $d(f(x),x)< epsilon$. Then we get a sequence $x_i$ of points such that $d(f(x_i),x_i)$ converges to 0. Sequential compactness gives us the desired limit point $x^*$ that is a fixed point. This is a contradiction.
answered Dec 5 '18 at 5:06
zoidbergzoidberg
1,080113
$endgroup$
add a comment |
$begingroup$
For metric spaces, compactness and sequential compactness are the same. Assume that $forall epsilon > 0$, $exists x in X$ such that $d(f(x),x)< epsilon$. Then we get a sequence $x_i$ of points such that $d(f(x_i),x_i)$ converges to 0. Sequential compactness gives us the desired limit point $x^*$ that is a fixed point. This is a contradiction.
answered Dec 5 '18 at 5:06
zoidbergzoidberg
1,080113
$endgroup$
add a comment |
$begingroup$
For metric spaces, compactness and sequential compactness are the same. Assume that $forall epsilon > 0$, $exists x in X$ such that $d(f(x),x)< epsilon$. Then we get a sequence $x_i$ of points such that $d(f(x_i),x_i)$ converges to 0. Sequential compactness gives us the desired limit point $x^*$ that is a fixed point. This is a contradiction.
answered Dec 5 '18 at 5:06
zoidbergzoidberg
1,080113
$endgroup$
For metric spaces, compactness and sequential compactness are the same. Assume that $forall epsilon > 0$, $exists x in X$ such that $d(f(x),x)< epsilon$. Then we get a sequence $x_i$ of points such that $d(f(x_i),x_i)$ converges to 0. Sequential compactness gives us the desired limit point $x^*$ that is a fixed point. This is a contradiction.
answered Dec 5 '18 at 5:06
zoidbergzoidberg
1,080113
answered Dec 5 '18 at 5:06
zoidbergzoidberg
1,080113
answered Dec 5 '18 at 5:06
zoidbergzoidberg
1,080113
answered Dec 5 '18 at 5:06
zoidbergzoidberg
1,080113
1,080113
add a comment |
add a comment |
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$begingroup$
$f: X to X$for example.$endgroup$
– Randall
Dec 5 '18 at 4:30
1
$begingroup$
If $forall epsilon > 0$, $exists x in X$ such that $dleft( fleft( x right), x right) < epsilon$, then $x$ will be a fixed point of $f$. Hence, the result.
$endgroup$
– Aniruddha Deshmukh
Dec 5 '18 at 4:35