Return to Answer

There exists a theorem--by Hausdorff--about extending subspace metrics, according to which:

Given a metric space $\ (X\ d),\ $ a closed subset $\ A\subseteq X,\ $ and a metrics $\ \rho_0\ $ in $\ A\ $ such that metrics $\ \rho_0\ $ and $\ d\,|\,A\times A\ $ are topologically equivalent (in $\ A$),  there exists a metrics $\ \rho\ $ in $\ X\ $ which is topologically equivalent to $\ d,\ $ and such that $\ \rho\,|\,A\times A = \rho_0$.

An especially elegant proof was obtained by Henryk Toruńczyk, "A short proof of Hausdorff ’s theorem on extending metrics", Fund. Math. 77 (1972), no. 2, 191–193. MR 47:9559.

(A search will provide you with more information).

There exists a theorem--by Hausdorff--about extending subspace metrics, according to which:

Given a metric space $\ (X\ d),\ $ a closed subset $\ A\subseteq X,\ $ and a metrics $\ \rho_0\ $ in $\ A\ $ such that metrics $\ \rho_0\ $ and $\ d\,|\,A\times A\ $ are topologically equivalent (in $\ A$),  there exists a metrics $\ \rho\ $ in $\ X\ $ which is topologically equivalent to $\ d,\ $ and such that $\ \rho\,|\,A\times A = \rho_0$.

An especially elegant proof was obtained by Henryk Toruńczyk, "A short proof of Hausdorff’s theorem on extending metrics", Fund. Math. 77 (1972), no. 2, 191–193. MR 47:9559.

(A search will provide you with more information).

There exists a theorem--by Hausdorff--about extending subspace metrics, according to which:

Given a metric space $\ (X\ d),\ $ a closed subset $\ A\subseteq X,\ $ and a metrics $\ \rho_0\ $ in $\ A\ $ such that metrics $\ \rho_0\ $ and $\ d\,|\,A\times A\ $ are topologically equivalent (in $\ A$),  there exists a metrics $\ \rho\ $ in $\ X\ $ which is topologically equivalent to $\ d,\ $ and such that $\ \rho\,|\,A\times A = \rho_0$.

An especially elegant proof was obtained by Henryk Toruńczyk.

(A search will provide you with more information).

There exists a theorem--by Hausdorff--about extending subspace metrics, according to which:

Given a metric space $\ (X\ d),\ $ a closed subset $\ A\subseteq X,\ $ and a metrics $\ \rho_0\ $ in $\ A\ $ such that metrics $\ \rho_0\ $ and $\ d\,|\,A\times A\ $ are topologically equivalent (in $\ A$),  there exists a metrics $\ \rho\ $ in $\ X\ $ which is topologically equivalent to $\ d,\ $ and such that $\ \rho\,|\,A\times A = \rho_0$.

An especially elegant proof was obtained by Henryk Toruńczyk, "A short proof of Hausdorff ’s theorem on extending metrics", Fund. Math. 77 (1972), no. 2, 191–193. MR 47:9559.

(A search will provide you with more information).

There exists a theorem (I think that it is Hausdorff theorem?) about extending subspace metrics, according to which:

Given a metric space $\ (X\ d),\ $ a closed subset $\ A\subseteq X,\ $ and a metrics $\ \rho_0\ $ in $\ A\ $ such that metrics $\ \rho_0\ $ and $\ d\,|\,A\times A\ $ are topologically equivalent (in $\ A$),  there exists a metrics $\ \rho\ $ in $\ X\ $ which is topologically equivalent to $\ d,\ $ and such that $\ \rho\,|\,A\times A = \rho_0$.

An especially elegant proof was obtained by Henryk Toruńczyk.

There exists a theorem--by Hausdorff--about extending subspace metrics, according to which:

Given a metric space $\ (X\ d),\ $ a closed subset $\ A\subseteq X,\ $ and a metrics $\ \rho_0\ $ in $\ A\ $ such that metrics $\ \rho_0\ $ and $\ d\,|\,A\times A\ $ are topologically equivalent (in $\ A$),  there exists a metrics $\ \rho\ $ in $\ X\ $ which is topologically equivalent to $\ d,\ $ and such that $\ \rho\,|\,A\times A = \rho_0$.

An especially elegant proof was obtained by Henryk Toruńczyk.

(A search will provide you with more information).