If $f:Xrightarrow X$ is continuous and $(X,d)$ is compact, show there is $epsilon>0$ such that...
$begingroup$
If $f:Xrightarrow X$ is a continuous function with no fixed points and $(X,d)$ is compact, show there is some $epsilon>0$ such that $d(x,f(x))>epsilon$ for all $xin X$.
My hope is to argue as follows, but I seem to be ending up with a slightly stronger statement which is what worries me a bit. Since $X$ is compact we know that $Xtimes f(X)$ is compact and therefore $d: Xtimes f(X)rightarrow (0,infty)$ has compact image. So $d$ attains a minimum, thus there is some $epsilon>0$ for which $d(x,f(x))geq epsilon$ for all $xin X$. Does this seem ok?
general-topology proof-verification metric-spaces compactness
asked Jan 7 at 14:26
ArbutusArbutus
723715
$endgroup$
add a comment |
$begingroup$
If $f:Xrightarrow X$ is a continuous function with no fixed points and $(X,d)$ is compact, show there is some $epsilon>0$ such that $d(x,f(x))>epsilon$ for all $xin X$.
My hope is to argue as follows, but I seem to be ending up with a slightly stronger statement which is what worries me a bit. Since $X$ is compact we know that $Xtimes f(X)$ is compact and therefore $d: Xtimes f(X)rightarrow (0,infty)$ has compact image. So $d$ attains a minimum, thus there is some $epsilon>0$ for which $d(x,f(x))geq epsilon$ for all $xin X$. Does this seem ok?
general-topology proof-verification metric-spaces compactness
asked Jan 7 at 14:26
ArbutusArbutus
723715
$endgroup$
$begingroup$
You need to explain why the minimum that $d$ attains is positive (i.e., not zero). It's only one short line to do this.
$endgroup$
– Michael Burr
Jan 7 at 14:30
$begingroup$
That's just because $f$ has no fixed points. I left it out as I figured it was obvious, but you're right.
$endgroup$
– Arbutus
Jan 7 at 14:31
$begingroup$
Your thought seems on the right track; I would just rephrase it a bit. Let $g: X to mathbb{R}_{geq 0}$ be given by $g(x) = d(x, f(x))$. Then $g$ is continuous, so its image in $mathbb{R}_{geq 0}$ is a compact connected set; in particular, it cannot be an interval with an open endpoint at $0$. The no-fixed-points condition also rules out a closed endpoint at $0$, and you're done. Note that $X times f(X)$ is not the same thing as the set ${(x, f(x)): x in X}$, which is what I think you want.
$endgroup$
– Connor Harris
Jan 7 at 14:33
add a comment |
$begingroup$
If $f:Xrightarrow X$ is a continuous function with no fixed points and $(X,d)$ is compact, show there is some $epsilon>0$ such that $d(x,f(x))>epsilon$ for all $xin X$.
My hope is to argue as follows, but I seem to be ending up with a slightly stronger statement which is what worries me a bit. Since $X$ is compact we know that $Xtimes f(X)$ is compact and therefore $d: Xtimes f(X)rightarrow (0,infty)$ has compact image. So $d$ attains a minimum, thus there is some $epsilon>0$ for which $d(x,f(x))geq epsilon$ for all $xin X$. Does this seem ok?
general-topology proof-verification metric-spaces compactness
asked Jan 7 at 14:26
ArbutusArbutus
723715
$endgroup$
If $f:Xrightarrow X$ is a continuous function with no fixed points and $(X,d)$ is compact, show there is some $epsilon>0$ such that $d(x,f(x))>epsilon$ for all $xin X$.
My hope is to argue as follows, but I seem to be ending up with a slightly stronger statement which is what worries me a bit. Since $X$ is compact we know that $Xtimes f(X)$ is compact and therefore $d: Xtimes f(X)rightarrow (0,infty)$ has compact image. So $d$ attains a minimum, thus there is some $epsilon>0$ for which $d(x,f(x))geq epsilon$ for all $xin X$. Does this seem ok?
general-topology proof-verification metric-spaces compactness
general-topology proof-verification metric-spaces compactness
asked Jan 7 at 14:26
ArbutusArbutus
723715
asked Jan 7 at 14:26
ArbutusArbutus
723715
asked Jan 7 at 14:26
ArbutusArbutus
723715
asked Jan 7 at 14:26
ArbutusArbutus
723715
asked Jan 7 at 14:26
ArbutusArbutus
723715
723715
$begingroup$
You need to explain why the minimum that $d$ attains is positive (i.e., not zero). It's only one short line to do this.
$endgroup$
– Michael Burr
Jan 7 at 14:30
$begingroup$
That's just because $f$ has no fixed points. I left it out as I figured it was obvious, but you're right.
$endgroup$
– Arbutus
Jan 7 at 14:31
$begingroup$
Your thought seems on the right track; I would just rephrase it a bit. Let $g: X to mathbb{R}_{geq 0}$ be given by $g(x) = d(x, f(x))$. Then $g$ is continuous, so its image in $mathbb{R}_{geq 0}$ is a compact connected set; in particular, it cannot be an interval with an open endpoint at $0$. The no-fixed-points condition also rules out a closed endpoint at $0$, and you're done. Note that $X times f(X)$ is not the same thing as the set ${(x, f(x)): x in X}$, which is what I think you want.
$endgroup$
– Connor Harris
Jan 7 at 14:33
add a comment |
$begingroup$
You need to explain why the minimum that $d$ attains is positive (i.e., not zero). It's only one short line to do this.
$endgroup$
– Michael Burr
Jan 7 at 14:30
$begingroup$
That's just because $f$ has no fixed points. I left it out as I figured it was obvious, but you're right.
$endgroup$
– Arbutus
Jan 7 at 14:31
$begingroup$
Your thought seems on the right track; I would just rephrase it a bit. Let $g: X to mathbb{R}_{geq 0}$ be given by $g(x) = d(x, f(x))$. Then $g$ is continuous, so its image in $mathbb{R}_{geq 0}$ is a compact connected set; in particular, it cannot be an interval with an open endpoint at $0$. The no-fixed-points condition also rules out a closed endpoint at $0$, and you're done. Note that $X times f(X)$ is not the same thing as the set ${(x, f(x)): x in X}$, which is what I think you want.
$endgroup$
– Connor Harris
Jan 7 at 14:33
$begingroup$
You need to explain why the minimum that $d$ attains is positive (i.e., not zero). It's only one short line to do this.
$endgroup$
– Michael Burr
Jan 7 at 14:30
$begingroup$
You need to explain why the minimum that $d$ attains is positive (i.e., not zero). It's only one short line to do this.
$endgroup$
– Michael Burr
Jan 7 at 14:30
$begingroup$
That's just because $f$ has no fixed points. I left it out as I figured it was obvious, but you're right.
$endgroup$
– Arbutus
Jan 7 at 14:31
$begingroup$
That's just because $f$ has no fixed points. I left it out as I figured it was obvious, but you're right.
$endgroup$
– Arbutus
Jan 7 at 14:31
$begingroup$
Your thought seems on the right track; I would just rephrase it a bit. Let $g: X to mathbb{R}_{geq 0}$ be given by $g(x) = d(x, f(x))$. Then $g$ is continuous, so its image in $mathbb{R}_{geq 0}$ is a compact connected set; in particular, it cannot be an interval with an open endpoint at $0$. The no-fixed-points condition also rules out a closed endpoint at $0$, and you're done. Note that $X times f(X)$ is not the same thing as the set ${(x, f(x)): x in X}$, which is what I think you want.
$endgroup$
– Connor Harris
Jan 7 at 14:33
$begingroup$
Your thought seems on the right track; I would just rephrase it a bit. Let $g: X to mathbb{R}_{geq 0}$ be given by $g(x) = d(x, f(x))$. Then $g$ is continuous, so its image in $mathbb{R}_{geq 0}$ is a compact connected set; in particular, it cannot be an interval with an open endpoint at $0$. The no-fixed-points condition also rules out a closed endpoint at $0$, and you're done. Note that $X times f(X)$ is not the same thing as the set ${(x, f(x)): x in X}$, which is what I think you want.
$endgroup$
– Connor Harris
Jan 7 at 14:33
add a comment |
1 Answer
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No, this is not correct. It doesn't follow from your assumptions that the range of $d$ restricted to $Xtimes f(X)$ is a subset of $(0,infty)$. You are only assuming that, for each individual $xin X$, $xneq f(x)$.
Use the fact that the map$$begin{array}{ccc}X&longrightarrow&[0,infty)\x&mapsto&dbigl(x,f(x)bigr)end{array}$$is continuous with compact domain and such that $0$ doesn't belong to its range. So, the range is contained in some set of the type $[varepsilon,M]$ for some $varepsilon,Min(0,infty)$ (and $varepsilonleqslant M$).
answered Jan 7 at 14:32
José Carlos SantosJosé Carlos Santos
170k23132239
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add a comment |
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$begingroup$
No, this is not correct. It doesn't follow from your assumptions that the range of $d$ restricted to $Xtimes f(X)$ is a subset of $(0,infty)$. You are only assuming that, for each individual $xin X$, $xneq f(x)$.
Use the fact that the map$$begin{array}{ccc}X&longrightarrow&[0,infty)\x&mapsto&dbigl(x,f(x)bigr)end{array}$$is continuous with compact domain and such that $0$ doesn't belong to its range. So, the range is contained in some set of the type $[varepsilon,M]$ for some $varepsilon,Min(0,infty)$ (and $varepsilonleqslant M$).
answered Jan 7 at 14:32
José Carlos SantosJosé Carlos Santos
170k23132239
$endgroup$
add a comment |
$begingroup$
No, this is not correct. It doesn't follow from your assumptions that the range of $d$ restricted to $Xtimes f(X)$ is a subset of $(0,infty)$. You are only assuming that, for each individual $xin X$, $xneq f(x)$.
Use the fact that the map$$begin{array}{ccc}X&longrightarrow&[0,infty)\x&mapsto&dbigl(x,f(x)bigr)end{array}$$is continuous with compact domain and such that $0$ doesn't belong to its range. So, the range is contained in some set of the type $[varepsilon,M]$ for some $varepsilon,Min(0,infty)$ (and $varepsilonleqslant M$).
answered Jan 7 at 14:32
José Carlos SantosJosé Carlos Santos
170k23132239
$endgroup$
add a comment |
$begingroup$
No, this is not correct. It doesn't follow from your assumptions that the range of $d$ restricted to $Xtimes f(X)$ is a subset of $(0,infty)$. You are only assuming that, for each individual $xin X$, $xneq f(x)$.
Use the fact that the map$$begin{array}{ccc}X&longrightarrow&[0,infty)\x&mapsto&dbigl(x,f(x)bigr)end{array}$$is continuous with compact domain and such that $0$ doesn't belong to its range. So, the range is contained in some set of the type $[varepsilon,M]$ for some $varepsilon,Min(0,infty)$ (and $varepsilonleqslant M$).
answered Jan 7 at 14:32
José Carlos SantosJosé Carlos Santos
170k23132239
$endgroup$
No, this is not correct. It doesn't follow from your assumptions that the range of $d$ restricted to $Xtimes f(X)$ is a subset of $(0,infty)$. You are only assuming that, for each individual $xin X$, $xneq f(x)$.
Use the fact that the map$$begin{array}{ccc}X&longrightarrow&[0,infty)\x&mapsto&dbigl(x,f(x)bigr)end{array}$$is continuous with compact domain and such that $0$ doesn't belong to its range. So, the range is contained in some set of the type $[varepsilon,M]$ for some $varepsilon,Min(0,infty)$ (and $varepsilonleqslant M$).
answered Jan 7 at 14:32
José Carlos SantosJosé Carlos Santos
170k23132239
edited Jan 7 at 14:44
answered Jan 7 at 14:32
José Carlos SantosJosé Carlos Santos
170k23132239
answered Jan 7 at 14:32
José Carlos SantosJosé Carlos Santos
170k23132239
answered Jan 7 at 14:32
José Carlos SantosJosé Carlos Santos
170k23132239
170k23132239
add a comment |
add a comment |
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$begingroup$
You need to explain why the minimum that $d$ attains is positive (i.e., not zero). It's only one short line to do this.
$endgroup$
– Michael Burr
Jan 7 at 14:30
$begingroup$
That's just because $f$ has no fixed points. I left it out as I figured it was obvious, but you're right.
$endgroup$
– Arbutus
Jan 7 at 14:31
$begingroup$
Your thought seems on the right track; I would just rephrase it a bit. Let $g: X to mathbb{R}_{geq 0}$ be given by $g(x) = d(x, f(x))$. Then $g$ is continuous, so its image in $mathbb{R}_{geq 0}$ is a compact connected set; in particular, it cannot be an interval with an open endpoint at $0$. The no-fixed-points condition also rules out a closed endpoint at $0$, and you're done. Note that $X times f(X)$ is not the same thing as the set ${(x, f(x)): x in X}$, which is what I think you want.
$endgroup$
– Connor Harris
Jan 7 at 14:33