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Do locally path connected locally compact metric spaces have an intrinsic metric?
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Given a metric space $(X,d)$ we define the path metric $rho(x,y)$ as the infimum of all paths lengths from $x$ to $y$. The length of a path $gamma$ means the supremum of all $sum_i d big (gamma(x_i),gamma(x_i+1) big )$ for all choices of points $0 le x_1 < x_2 <ldots < x_n le 1$.
It is known $rho$ induces a finer topology than $d$. Sometimes strictly finer. For example consider the harmonic fan: the union of all line segments from points $(1,1/n)$ and $(1,0)$ to $(0,0)$. Then $(X,d)$ is compact but $(X,rho)$ is noncompact.
The harmonic fan is not locally arc-connected. I wonder what happens if we assume the space is arc-connected.
Suppose the metric space $X$ is arc-connected, locally compact and locally arc-connected. Is the metric on $X$ equivalent to the path metric?
I believe this is equivalent to the following (which I cannot prove either).
Suppose $X$ is above and $x in X$. Let $U_1 supset U_2 supset ldots $ be a chain of open neighborhoods of $x$ with diameter decreasing to zero. Then the path-diameter of $U_i$ also decreases to zero.
Any ideas?
If it helps we can also assume each point has a basis of closed arc-connected neighborhoods.
general-topology metric-spaces connectedness path-connected
edited Aug 2 at 17:01
Paul Frost
3,483420
asked Aug 2 at 8:11
Daron
4,2731923
 |Â
show 2 more comments
up vote
1
down vote
favorite
Given a metric space $(X,d)$ we define the path metric $rho(x,y)$ as the infimum of all paths lengths from $x$ to $y$. The length of a path $gamma$ means the supremum of all $sum_i d big (gamma(x_i),gamma(x_i+1) big )$ for all choices of points $0 le x_1 < x_2 <ldots < x_n le 1$.
It is known $rho$ induces a finer topology than $d$. Sometimes strictly finer. For example consider the harmonic fan: the union of all line segments from points $(1,1/n)$ and $(1,0)$ to $(0,0)$. Then $(X,d)$ is compact but $(X,rho)$ is noncompact.
The harmonic fan is not locally arc-connected. I wonder what happens if we assume the space is arc-connected.
Suppose the metric space $X$ is arc-connected, locally compact and locally arc-connected. Is the metric on $X$ equivalent to the path metric?
I believe this is equivalent to the following (which I cannot prove either).
Suppose $X$ is above and $x in X$. Let $U_1 supset U_2 supset ldots $ be a chain of open neighborhoods of $x$ with diameter decreasing to zero. Then the path-diameter of $U_i$ also decreases to zero.
Any ideas?
If it helps we can also assume each point has a basis of closed arc-connected neighborhoods.
general-topology metric-spaces connectedness path-connected
edited Aug 2 at 17:01
Paul Frost
3,483420
asked Aug 2 at 8:11
Daron
4,2731923
You definition of the path length is not correct. It is the supremum of such sums where $x_i = gamma(t_i)$ and $t_1,...,t_n$ is a partition of the interval on which $gamma$ is defined. See for example encyclopediaofmath.org/index.php/Rectifiable_curve.
– Paul Frost
Aug 2 at 11:49
How do you treat the case that there exist paths with infinite length?
– Paul Frost
Aug 2 at 11:57
Correct. There was an error in the definition of path length.
– Daron
Aug 2 at 12:17
We're assuming local arc-connectedness. Arcs in $X$ have finite length. For let $Gamma subset X$ be an arc. By compactness the homeomorphism $Gamma to [0,1]$ is uniformly continuous and the modulus lets us put a bound on the length of $Gamma$ in $X$.
– Daron
Aug 2 at 12:21
Error not yet corrected. If you have a path $gamma : [0,1] to X$ and allow all choices $x_i in gamma([0,1])$, the supremum will always be infinite. You have to take the supremum over all sequences $x_1,..., x_n$ having the form $x_i = gamma(t_i)$ with $0 = t_1 < t_2 < ... < t_n = 1$.
– Paul Frost
Aug 2 at 12:34
 |Â
show 2 more comments
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Given a metric space $(X,d)$ we define the path metric $rho(x,y)$ as the infimum of all paths lengths from $x$ to $y$. The length of a path $gamma$ means the supremum of all $sum_i d big (gamma(x_i),gamma(x_i+1) big )$ for all choices of points $0 le x_1 < x_2 <ldots < x_n le 1$.
It is known $rho$ induces a finer topology than $d$. Sometimes strictly finer. For example consider the harmonic fan: the union of all line segments from points $(1,1/n)$ and $(1,0)$ to $(0,0)$. Then $(X,d)$ is compact but $(X,rho)$ is noncompact.
The harmonic fan is not locally arc-connected. I wonder what happens if we assume the space is arc-connected.
Suppose the metric space $X$ is arc-connected, locally compact and locally arc-connected. Is the metric on $X$ equivalent to the path metric?
I believe this is equivalent to the following (which I cannot prove either).
Suppose $X$ is above and $x in X$. Let $U_1 supset U_2 supset ldots $ be a chain of open neighborhoods of $x$ with diameter decreasing to zero. Then the path-diameter of $U_i$ also decreases to zero.
Any ideas?
If it helps we can also assume each point has a basis of closed arc-connected neighborhoods.
general-topology metric-spaces connectedness path-connected
edited Aug 2 at 17:01
Paul Frost
3,483420
asked Aug 2 at 8:11
Daron
4,2731923
Given a metric space $(X,d)$ we define the path metric $rho(x,y)$ as the infimum of all paths lengths from $x$ to $y$. The length of a path $gamma$ means the supremum of all $sum_i d big (gamma(x_i),gamma(x_i+1) big )$ for all choices of points $0 le x_1 < x_2 <ldots < x_n le 1$.
It is known $rho$ induces a finer topology than $d$. Sometimes strictly finer. For example consider the harmonic fan: the union of all line segments from points $(1,1/n)$ and $(1,0)$ to $(0,0)$. Then $(X,d)$ is compact but $(X,rho)$ is noncompact.
The harmonic fan is not locally arc-connected. I wonder what happens if we assume the space is arc-connected.
Suppose the metric space $X$ is arc-connected, locally compact and locally arc-connected. Is the metric on $X$ equivalent to the path metric?
I believe this is equivalent to the following (which I cannot prove either).
Suppose $X$ is above and $x in X$. Let $U_1 supset U_2 supset ldots $ be a chain of open neighborhoods of $x$ with diameter decreasing to zero. Then the path-diameter of $U_i$ also decreases to zero.
Any ideas?
If it helps we can also assume each point has a basis of closed arc-connected neighborhoods.
general-topology metric-spaces connectedness path-connected
edited Aug 2 at 17:01
Paul Frost
3,483420
asked Aug 2 at 8:11
Daron
4,2731923
edited Aug 2 at 17:01
Paul Frost
3,483420
edited Aug 2 at 17:01
Paul Frost
3,483420
edited Aug 2 at 17:01
Paul Frost
3,483420
3,483420
asked Aug 2 at 8:11
Daron
4,2731923
asked Aug 2 at 8:11
Daron
4,2731923
asked Aug 2 at 8:11
Daron
4,2731923
4,2731923
You definition of the path length is not correct. It is the supremum of such sums where $x_i = gamma(t_i)$ and $t_1,...,t_n$ is a partition of the interval on which $gamma$ is defined. See for example encyclopediaofmath.org/index.php/Rectifiable_curve.
– Paul Frost
Aug 2 at 11:49
How do you treat the case that there exist paths with infinite length?
– Paul Frost
Aug 2 at 11:57
Correct. There was an error in the definition of path length.
– Daron
Aug 2 at 12:17
We're assuming local arc-connectedness. Arcs in $X$ have finite length. For let $Gamma subset X$ be an arc. By compactness the homeomorphism $Gamma to [0,1]$ is uniformly continuous and the modulus lets us put a bound on the length of $Gamma$ in $X$.
– Daron
Aug 2 at 12:21
Error not yet corrected. If you have a path $gamma : [0,1] to X$ and allow all choices $x_i in gamma([0,1])$, the supremum will always be infinite. You have to take the supremum over all sequences $x_1,..., x_n$ having the form $x_i = gamma(t_i)$ with $0 = t_1 < t_2 < ... < t_n = 1$.
– Paul Frost
Aug 2 at 12:34
 |Â
show 2 more comments
You definition of the path length is not correct. It is the supremum of such sums where $x_i = gamma(t_i)$ and $t_1,...,t_n$ is a partition of the interval on which $gamma$ is defined. See for example encyclopediaofmath.org/index.php/Rectifiable_curve.
– Paul Frost
Aug 2 at 11:49
How do you treat the case that there exist paths with infinite length?
– Paul Frost
Aug 2 at 11:57
Correct. There was an error in the definition of path length.
– Daron
Aug 2 at 12:17
We're assuming local arc-connectedness. Arcs in $X$ have finite length. For let $Gamma subset X$ be an arc. By compactness the homeomorphism $Gamma to [0,1]$ is uniformly continuous and the modulus lets us put a bound on the length of $Gamma$ in $X$.
– Daron
Aug 2 at 12:21
Error not yet corrected. If you have a path $gamma : [0,1] to X$ and allow all choices $x_i in gamma([0,1])$, the supremum will always be infinite. You have to take the supremum over all sequences $x_1,..., x_n$ having the form $x_i = gamma(t_i)$ with $0 = t_1 < t_2 < ... < t_n = 1$.
– Paul Frost
Aug 2 at 12:34
You definition of the path length is not correct. It is the supremum of such sums where $x_i = gamma(t_i)$ and $t_1,...,t_n$ is a partition of the interval on which $gamma$ is defined. See for example encyclopediaofmath.org/index.php/Rectifiable_curve.
– Paul Frost
Aug 2 at 11:49
You definition of the path length is not correct. It is the supremum of such sums where $x_i = gamma(t_i)$ and $t_1,...,t_n$ is a partition of the interval on which $gamma$ is defined. See for example encyclopediaofmath.org/index.php/Rectifiable_curve.
– Paul Frost
Aug 2 at 11:49
How do you treat the case that there exist paths with infinite length?
– Paul Frost
Aug 2 at 11:57
How do you treat the case that there exist paths with infinite length?
– Paul Frost
Aug 2 at 11:57
Correct. There was an error in the definition of path length.
– Daron
Aug 2 at 12:17
Correct. There was an error in the definition of path length.
– Daron
Aug 2 at 12:17
We're assuming local arc-connectedness. Arcs in $X$ have finite length. For let $Gamma subset X$ be an arc. By compactness the homeomorphism $Gamma to [0,1]$ is uniformly continuous and the modulus lets us put a bound on the length of $Gamma$ in $X$.
– Daron
Aug 2 at 12:21
We're assuming local arc-connectedness. Arcs in $X$ have finite length. For let $Gamma subset X$ be an arc. By compactness the homeomorphism $Gamma to [0,1]$ is uniformly continuous and the modulus lets us put a bound on the length of $Gamma$ in $X$.
– Daron
Aug 2 at 12:21
Error not yet corrected. If you have a path $gamma : [0,1] to X$ and allow all choices $x_i in gamma([0,1])$, the supremum will always be infinite. You have to take the supremum over all sequences $x_1,..., x_n$ having the form $x_i = gamma(t_i)$ with $0 = t_1 < t_2 < ... < t_n = 1$.
– Paul Frost
Aug 2 at 12:34
Error not yet corrected. If you have a path $gamma : [0,1] to X$ and allow all choices $x_i in gamma([0,1])$, the supremum will always be infinite. You have to take the supremum over all sequences $x_1,..., x_n$ having the form $x_i = gamma(t_i)$ with $0 = t_1 < t_2 < ... < t_n = 1$.
– Paul Frost
Aug 2 at 12:34
 |Â
show 2 more comments
1 Answer
1
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up vote
1
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Here are two old results. The first:
Theorem (Bing: A convex metric for a locally-connected continuum, 1928)
Each locally connected metric continuum admits an equivalent convex metrix.
The second:
Theorem (Goebel & Kirk: Topics in Metric Fixed Point Theory p 24)
Each complete and convex metric space has the path metric equivalent to the original metric.
Now suppose $(X,d)$ is as described in your question. The path metric topology is always finer than the original topology. We need to show it is coarser too. So suppose $x_n to x$ with respect to $d$ and let $K$ be a locally-connected continuum neighborhood of $x$.
Some tail $y_m$ of $x_n$ is contained in the metric subspace $(K,d_K)$. The above theorems say $d_K$ is equivalent to the $d_K$ path metric $rho_K$. Let $epsilon>0$ be arbitrary. Some tail $y_N,y_N+1, ldots$ has all $rho(y_i,x)<epsilon$. That means there are paths $gamma_i subset K$ from $y_i$ to $x$ of $d_K$ length less than $epsilon$.
But since $d_K$ is the restriction of $K$ these paths $gamma_i$ are also paths of $d$-length length less than $epsilon$. Since $epsilon$ is arbitrary we get $rho(y_m,x) to 0$ as required.
answered Aug 3 at 14:18
Daron
4,2731923
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
Here are two old results. The first:
Theorem (Bing: A convex metric for a locally-connected continuum, 1928)
Each locally connected metric continuum admits an equivalent convex metrix.
The second:
Theorem (Goebel & Kirk: Topics in Metric Fixed Point Theory p 24)
Each complete and convex metric space has the path metric equivalent to the original metric.
Now suppose $(X,d)$ is as described in your question. The path metric topology is always finer than the original topology. We need to show it is coarser too. So suppose $x_n to x$ with respect to $d$ and let $K$ be a locally-connected continuum neighborhood of $x$.
Some tail $y_m$ of $x_n$ is contained in the metric subspace $(K,d_K)$. The above theorems say $d_K$ is equivalent to the $d_K$ path metric $rho_K$. Let $epsilon>0$ be arbitrary. Some tail $y_N,y_N+1, ldots$ has all $rho(y_i,x)<epsilon$. That means there are paths $gamma_i subset K$ from $y_i$ to $x$ of $d_K$ length less than $epsilon$.
But since $d_K$ is the restriction of $K$ these paths $gamma_i$ are also paths of $d$-length length less than $epsilon$. Since $epsilon$ is arbitrary we get $rho(y_m,x) to 0$ as required.
answered Aug 3 at 14:18
Daron
4,2731923
add a comment |Â
up vote
1
down vote
Here are two old results. The first:
Theorem (Bing: A convex metric for a locally-connected continuum, 1928)
Each locally connected metric continuum admits an equivalent convex metrix.
The second:
Theorem (Goebel & Kirk: Topics in Metric Fixed Point Theory p 24)
Each complete and convex metric space has the path metric equivalent to the original metric.
Now suppose $(X,d)$ is as described in your question. The path metric topology is always finer than the original topology. We need to show it is coarser too. So suppose $x_n to x$ with respect to $d$ and let $K$ be a locally-connected continuum neighborhood of $x$.
Some tail $y_m$ of $x_n$ is contained in the metric subspace $(K,d_K)$. The above theorems say $d_K$ is equivalent to the $d_K$ path metric $rho_K$. Let $epsilon>0$ be arbitrary. Some tail $y_N,y_N+1, ldots$ has all $rho(y_i,x)<epsilon$. That means there are paths $gamma_i subset K$ from $y_i$ to $x$ of $d_K$ length less than $epsilon$.
But since $d_K$ is the restriction of $K$ these paths $gamma_i$ are also paths of $d$-length length less than $epsilon$. Since $epsilon$ is arbitrary we get $rho(y_m,x) to 0$ as required.
answered Aug 3 at 14:18
Daron
4,2731923
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Here are two old results. The first:
Theorem (Bing: A convex metric for a locally-connected continuum, 1928)
Each locally connected metric continuum admits an equivalent convex metrix.
The second:
Theorem (Goebel & Kirk: Topics in Metric Fixed Point Theory p 24)
Each complete and convex metric space has the path metric equivalent to the original metric.
Now suppose $(X,d)$ is as described in your question. The path metric topology is always finer than the original topology. We need to show it is coarser too. So suppose $x_n to x$ with respect to $d$ and let $K$ be a locally-connected continuum neighborhood of $x$.
Some tail $y_m$ of $x_n$ is contained in the metric subspace $(K,d_K)$. The above theorems say $d_K$ is equivalent to the $d_K$ path metric $rho_K$. Let $epsilon>0$ be arbitrary. Some tail $y_N,y_N+1, ldots$ has all $rho(y_i,x)<epsilon$. That means there are paths $gamma_i subset K$ from $y_i$ to $x$ of $d_K$ length less than $epsilon$.
But since $d_K$ is the restriction of $K$ these paths $gamma_i$ are also paths of $d$-length length less than $epsilon$. Since $epsilon$ is arbitrary we get $rho(y_m,x) to 0$ as required.
answered Aug 3 at 14:18
Daron
4,2731923
Here are two old results. The first:
Theorem (Bing: A convex metric for a locally-connected continuum, 1928)
Each locally connected metric continuum admits an equivalent convex metrix.
The second:
Theorem (Goebel & Kirk: Topics in Metric Fixed Point Theory p 24)
Each complete and convex metric space has the path metric equivalent to the original metric.
Now suppose $(X,d)$ is as described in your question. The path metric topology is always finer than the original topology. We need to show it is coarser too. So suppose $x_n to x$ with respect to $d$ and let $K$ be a locally-connected continuum neighborhood of $x$.
Some tail $y_m$ of $x_n$ is contained in the metric subspace $(K,d_K)$. The above theorems say $d_K$ is equivalent to the $d_K$ path metric $rho_K$. Let $epsilon>0$ be arbitrary. Some tail $y_N,y_N+1, ldots$ has all $rho(y_i,x)<epsilon$. That means there are paths $gamma_i subset K$ from $y_i$ to $x$ of $d_K$ length less than $epsilon$.
But since $d_K$ is the restriction of $K$ these paths $gamma_i$ are also paths of $d$-length length less than $epsilon$. Since $epsilon$ is arbitrary we get $rho(y_m,x) to 0$ as required.
answered Aug 3 at 14:18
Daron
4,2731923
answered Aug 3 at 14:18
Daron
4,2731923
answered Aug 3 at 14:18
Daron
4,2731923
answered Aug 3 at 14:18
Daron
4,2731923
4,2731923
add a comment |Â
add a comment |Â
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You definition of the path length is not correct. It is the supremum of such sums where $x_i = gamma(t_i)$ and $t_1,...,t_n$ is a partition of the interval on which $gamma$ is defined. See for example encyclopediaofmath.org/index.php/Rectifiable_curve.
– Paul Frost
Aug 2 at 11:49
How do you treat the case that there exist paths with infinite length?
– Paul Frost
Aug 2 at 11:57
Correct. There was an error in the definition of path length.
– Daron
Aug 2 at 12:17
We're assuming local arc-connectedness. Arcs in $X$ have finite length. For let $Gamma subset X$ be an arc. By compactness the homeomorphism $Gamma to [0,1]$ is uniformly continuous and the modulus lets us put a bound on the length of $Gamma$ in $X$.
– Daron
Aug 2 at 12:21
Error not yet corrected. If you have a path $gamma : [0,1] to X$ and allow all choices $x_i in gamma([0,1])$, the supremum will always be infinite. You have to take the supremum over all sequences $x_1,..., x_n$ having the form $x_i = gamma(t_i)$ with $0 = t_1 < t_2 < ... < t_n = 1$.
– Paul Frost
Aug 2 at 12:34