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Is there a standard construction of a metric on one-point compactification of a proper metric space?

Comments:

• A metric space is proper if all bounded closed sets are compact.
• Standard means found in literature.

From the answers and comments:

Here is a simplification of the construction given here (thanks to Jonas for ref). Let $d$ be the original metric. Fix a point $p$ and set $h(x)=1/(1+d(p,x))$. Then take the metric $$\hat d(x,y)=\min\{d(x,y),\,h(x)+h(y)\},\ \ \ \ \hat d(\infty,x)=h(x).$$

A more complicated construction is given here (thanks to LK for ref), some call it "sphericalization". One takes $$\bar d(x,y)=d(x,y)\cdot h(x)\cdot h(y),\ \ \ \ \bar d(\infty,x)=h(x).$$ The function $\bar d$ does not satisfies triangle inequality, but one can show that there is a metric $\rho$ such that $\tfrac14\cdot \bar d\le \rho\le \bar d$.

Is there a standard construction of a metric on one-point compactification of a proper metric space?

Comments:

• A metric space is proper if all bounded closed sets are compact.
• Standard means found in literature.

From the answers and comments:

Here is a simplification of the construction given here¹ (thanks to Jonas for ref). Let $d$ be the original metric. Fix a point $p$ and set $h(x)=1/(1+d(p,x))$. Then take the metric $$\hat d(x,y)=\min\{d(x,y),\,h(x)+h(y)\},\ \ \ \ \hat d(\infty,x)=h(x).$$

A more complicated construction is given here² (thanks to LK for ref), some call it "sphericalization". One takes $$\bar d(x,y)=d(x,y)\cdot h(x)\cdot h(y),\ \ \ \ \bar d(\infty,x)=h(x).$$ The function $\bar d$ does not satisfies triangle inequality, but one can show that there is a metric $\rho$ such that $\tfrac14\cdot \bar d\le \rho\le \bar d$.

¹Mandelkern, Mark. “Metrization of the One-Point Compactification.” Proceedings of the American Mathematical Society, vol. 107, no. 4, American Mathematical Society, 1989, pp. 1111–15, https://doi.org/10.2307/2047675.

²Mario Bonk, Bruce Kleiner: Rigidity for Quasi-Mobius group actions

Is there a standard construction of a metric on one-point compactification of a proper metric space?

Comments:

• A metric space is proper if all bounded closed sets are compact.
• Standard means found in literature.

From the answers and comments:

Here is a simplification of the construction given here (thanks to Jonas for ref). Let $d$ be the original metric. Fix a point $p$ and set $h(x)=1/(1+d(p,x))$. Then take the metric $$\hat d(x,y)=\min\{d(x,y),\\,h(x)+h(y)\},\ \ \ \ \hat d(\infty,x)=h(x).$$

A more complicated construction is given here (thanks to LK for ref), some call it "sphericalization". One takes $$\bar d(x,y)=d(x,y)\cdot h(x)\cdot h(y),\ \ \ \ \bar d(\infty,x)=h(x).$$ The function $\bar d$ does not satisfies triangle inequality, but one can show that there is a metric $\rho$ such that $\tfrac14\cdot \bar d\le \rho\le \bar d$.

Is there a standard construction of a metric on one-point compactification of a proper metric space?

Comments:

• A metric space is proper if all bounded closed sets are compact.
• Standard means found in literature.

From the answers and comments:

Here is a simplification of the construction given here (thanks to Jonas for ref). Let $d$ be the original metric. Fix a point $p$ and set $h(x)=1/(1+d(p,x))$. Then take the metric $$\hat d(x,y)=\min\{d(x,y),\,h(x)+h(y)\},\ \ \ \ \hat d(\infty,x)=h(x).$$

A more complicated construction is given here (thanks to LK for ref), some call it "sphericalization". One takes $$\bar d(x,y)=d(x,y)\cdot h(x)\cdot h(y),\ \ \ \ \bar d(\infty,x)=h(x).$$ The function $\bar d$ does not satisfies triangle inequality, but one can show that there is a metric $\rho$ such that $\tfrac14\cdot \bar d\le \rho\le \bar d$.

Is there a standard construction of a metric on one-point compactification of a proper metric space?

Comments:

• A metric space is proper if all bounded closed sets are compact.
• Standard means found in literature.

From the answers and comments:

Here is a simplification of the construction given here (thanks to Jonas for ref). Let $d$ be the original metric. Fix a point $p$ and set $h(x)=1/(1+d(p,x))$. Then take the metric $$\hat d(x,y)=\min\{d(x,y),\\,h(x)+h(y)\},\ \ \ \ \hat d(\infty,x)=h(x).$$

A more complicated construction is given here (thanks to LK for ref), some call it "sphericalization". One takes $$\bar d(x,y)=d(x,y)\cdot h(x)\cdot h(y),\ \ \ \ \hat d(\infty,x)=h(x).$$ The function $\bar d$ does not satisfies triangle inequality, but one can show that there is a metric $\rho$ such that $\tfrac14\cdot \bar d\le \rho\le \bar d$.

Is there a standard construction of a metric on one-point compactification of a proper metric space?

Comments:

• A metric space is proper if all bounded closed sets are compact.
• Standard means found in literature.

From the answers and comments:

Here is a simplification of the construction given here (thanks to Jonas for ref). Let $d$ be the original metric. Fix a point $p$ and set $h(x)=1/(1+d(p,x))$. Then take the metric $$\hat d(x,y)=\min\{d(x,y),\\,h(x)+h(y)\},\ \ \ \ \hat d(\infty,x)=h(x).$$

A more complicated construction is given here (thanks to LK for ref), some call it "sphericalization". One takes $$\bar d(x,y)=d(x,y)\cdot h(x)\cdot h(y),\ \ \ \ \bar d(\infty,x)=h(x).$$ The function $\bar d$ does not satisfies triangle inequality, but one can show that there is a metric $\rho$ such that $\tfrac14\cdot \bar d\le \rho\le \bar d$.