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If two spaces are homeomorphic and one is a metric space must the other be as well?
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Suppose $X$, and $Y$ are topological spaces, and that they are homeomorphic. If $X$ is a metric space must $Y$ be as well?
Here is my thought process.
If $X$ is a metric space then there exists some distance function on $X$ given by $d:Xtimes XrightarrowBbb R$. Let $varphi:Xrightarrow Y$ be a homeomorphism between $X$ and $Y$.
Then for any distinct points $y_1,y_2in Y$ we can determine a "distance" between them by $d(varphi^-1(y_1),varphi^-1(y_2))inBbb R$
To be a metric space however, I need some way of combining this idea of passing the points back to $X$ from $Y$ in a nice way, and then applying the distance function on $X$. Would it be as simple as defining $d'(y_1,y_2) = d(varphi^-1(y_1),varphi^-1(y_2))$?
If not is there a way to do that? Is this enough to say that $Y$ is a metric space? Or am I just out of my mind?
general-topology metric-spaces
asked yesterday
Logan Toll
797417
add a comment |Â
up vote
8
down vote
favorite
Suppose $X$, and $Y$ are topological spaces, and that they are homeomorphic. If $X$ is a metric space must $Y$ be as well?
Here is my thought process.
If $X$ is a metric space then there exists some distance function on $X$ given by $d:Xtimes XrightarrowBbb R$. Let $varphi:Xrightarrow Y$ be a homeomorphism between $X$ and $Y$.
Then for any distinct points $y_1,y_2in Y$ we can determine a "distance" between them by $d(varphi^-1(y_1),varphi^-1(y_2))inBbb R$
To be a metric space however, I need some way of combining this idea of passing the points back to $X$ from $Y$ in a nice way, and then applying the distance function on $X$. Would it be as simple as defining $d'(y_1,y_2) = d(varphi^-1(y_1),varphi^-1(y_2))$?
If not is there a way to do that? Is this enough to say that $Y$ is a metric space? Or am I just out of my mind?
general-topology metric-spaces
asked yesterday
Logan Toll
797417
3
I don't really think this question makes sense. However, a space that is homeomorphic to a metric space is called a metrisable space.
– Lord Shark the Unknown
yesterday
add a comment |Â
up vote
8
down vote
favorite
up vote
8
down vote
favorite
Suppose $X$, and $Y$ are topological spaces, and that they are homeomorphic. If $X$ is a metric space must $Y$ be as well?
Here is my thought process.
If $X$ is a metric space then there exists some distance function on $X$ given by $d:Xtimes XrightarrowBbb R$. Let $varphi:Xrightarrow Y$ be a homeomorphism between $X$ and $Y$.
Then for any distinct points $y_1,y_2in Y$ we can determine a "distance" between them by $d(varphi^-1(y_1),varphi^-1(y_2))inBbb R$
To be a metric space however, I need some way of combining this idea of passing the points back to $X$ from $Y$ in a nice way, and then applying the distance function on $X$. Would it be as simple as defining $d'(y_1,y_2) = d(varphi^-1(y_1),varphi^-1(y_2))$?
If not is there a way to do that? Is this enough to say that $Y$ is a metric space? Or am I just out of my mind?
general-topology metric-spaces
asked yesterday
Logan Toll
797417
Suppose $X$, and $Y$ are topological spaces, and that they are homeomorphic. If $X$ is a metric space must $Y$ be as well?
Here is my thought process.
If $X$ is a metric space then there exists some distance function on $X$ given by $d:Xtimes XrightarrowBbb R$. Let $varphi:Xrightarrow Y$ be a homeomorphism between $X$ and $Y$.
Then for any distinct points $y_1,y_2in Y$ we can determine a "distance" between them by $d(varphi^-1(y_1),varphi^-1(y_2))inBbb R$
To be a metric space however, I need some way of combining this idea of passing the points back to $X$ from $Y$ in a nice way, and then applying the distance function on $X$. Would it be as simple as defining $d'(y_1,y_2) = d(varphi^-1(y_1),varphi^-1(y_2))$?
If not is there a way to do that? Is this enough to say that $Y$ is a metric space? Or am I just out of my mind?
general-topology metric-spaces
general-topology metric-spaces
asked yesterday
Logan Toll
797417
asked yesterday
Logan Toll
797417
asked yesterday
Logan Toll
797417
asked yesterday
Logan Toll
797417
asked yesterday
Logan Toll
797417
797417
3
I don't really think this question makes sense. However, a space that is homeomorphic to a metric space is called a metrisable space.
– Lord Shark the Unknown
yesterday
add a comment |Â
3
I don't really think this question makes sense. However, a space that is homeomorphic to a metric space is called a metrisable space.
– Lord Shark the Unknown
yesterday
3
3
I don't really think this question makes sense. However, a space that is homeomorphic to a metric space is called a metrisable space.
– Lord Shark the Unknown
yesterday
I don't really think this question makes sense. However, a space that is homeomorphic to a metric space is called a metrisable space.
– Lord Shark the Unknown
yesterday
add a comment |Â
4 Answers
4
active
oldest
votes
up vote
11
down vote
accepted
If $(X, d_X)$ is a metric space then the homeomorphic $Y$ is a so-called metrisable space: a space that can be given a metric that induces its topology. A metric space comes with a pre-given metric (and has an induced topology by this metric), and is a different type of structure.
The metric is indeed a transportation of the given one on $X$: $phi: X to Y$ (the homeomorphism) has a well defined continuous inverse (I'll call it $psi: Y to X$ so that $psi circ phi = 1_X$ and $phi circ psi = 1_Y$), and we can define $$d_Y: Y times Y to mathbbR text by: d_Y(y_1,y_2) = d_X(psi(y_1), psi(y_2))$$
One easily checks the axioms for a metric for $d_Y$ and by bijectiveness and the definitions we get that $phi[B_d_X(x, r)] = B_d_Y(phi(x), r)$ for all $x in X, r>0$ etc. so that balls in the $d_X$ metric get mapped to balls in the $d_Y$-metric showing (with a little thought) that indeed open balls under $d_Y$ are open and form a base for the topology of $Y$, so that $Y$ is metrisable.
answered yesterday
Henno Brandsma
93.1k342101
add a comment |Â
up vote
6
down vote
You construction can be done regardless of $varphi$ being a homeomorphism: for any bijection you will have a metric structure on $Y$. What might be interesting to verify is if the topology induced by the metric $d' = d(varphi^-1(-),varphi^-1(-))$ coincides with the original topology on $Y$. Let's describe the open balls of $d'$: let $y in Y$ and $varepsilon > 0$ . Now,
$$
B'_varepsilon(y) := x:d'(x,y) < varepsilon = x : d(varphi^-1(x), varphi^-1(y)) < varepsilon = varphi(B_varepsilon(varphi^-1(y))).
$$
Since $B_varepsilon(varphi^-1(y))$ is a ball in $X$ and $varphi$ is homeo, $B'_varepsilon(y)$ will be open on $Y$ with the original topology. In the same fashion, if $U subseteq Y$ is open for the original topology,
$$
U = varphi(varphi^-1(U)) = varphi(bigcup_i in IB_r_i(x_i)) = bigcup_i in Ivarphi(B_r_i(x_i)) = bigcup_i in IB'_r_i(phi(x_i))
$$
and so $U$ is open for the metric topology. Here we use that $varphi^-1(U)$ is an open set of $X$ and so it is a union of open balls.
answered yesterday
Guido A.
4,644726
add a comment |Â
up vote
1
down vote
I would say is as simple as that, since $varphi$ is a homeomorphism it seems straightforward to check the metric space axioms, you may want to be carefull with the triangular inequality though, but at a glimpse, it is straightforward. Regards.
answered yesterday
Astro
264
add a comment |Â
up vote
0
down vote
Given any sets A and B and a bijection $varphi$ between them, any structure in one set induces an isomorhpic structure in the other; if A has a metric, or a topology, or an additive structure, etc., then a corresponding structure can be defined on B. For instance, if we define $b_1+b_2$ as being $varphi(varphi^-1(b_1)+varphi^-1(b_2))$, then $varphi$ will be a group isomorphism with respect to addition in A and addition in B.
Strictly speaking, a metric space is not a set; it's a tuple $(S,d)$ where $S$ is a set and $d$ is a distance function. So neither $X$ nor $Y$ are metric spaces, but $X$ along with its distance function is a metric space, and $Y$ along with the metric induced by $varphi$ is a metric space. The only question I see that is remotely non-trival is whether the topology in $Y$ induced by the metric in $Y$ induced by the metric in $X$ is the same as the original topology in $Y$. Since the topology in $Y$ is isomorphic to the topology in $X$, and the topology in $X$ is presumably the topology induced by its metric, the two topologies in $Y$ would indeed be the same.
answered yesterday
Acccumulation
5,7742616
add a comment |Â
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
11
down vote
accepted
If $(X, d_X)$ is a metric space then the homeomorphic $Y$ is a so-called metrisable space: a space that can be given a metric that induces its topology. A metric space comes with a pre-given metric (and has an induced topology by this metric), and is a different type of structure.
The metric is indeed a transportation of the given one on $X$: $phi: X to Y$ (the homeomorphism) has a well defined continuous inverse (I'll call it $psi: Y to X$ so that $psi circ phi = 1_X$ and $phi circ psi = 1_Y$), and we can define $$d_Y: Y times Y to mathbbR text by: d_Y(y_1,y_2) = d_X(psi(y_1), psi(y_2))$$
One easily checks the axioms for a metric for $d_Y$ and by bijectiveness and the definitions we get that $phi[B_d_X(x, r)] = B_d_Y(phi(x), r)$ for all $x in X, r>0$ etc. so that balls in the $d_X$ metric get mapped to balls in the $d_Y$-metric showing (with a little thought) that indeed open balls under $d_Y$ are open and form a base for the topology of $Y$, so that $Y$ is metrisable.
answered yesterday
Henno Brandsma
93.1k342101
add a comment |Â
up vote
11
down vote
accepted
If $(X, d_X)$ is a metric space then the homeomorphic $Y$ is a so-called metrisable space: a space that can be given a metric that induces its topology. A metric space comes with a pre-given metric (and has an induced topology by this metric), and is a different type of structure.
The metric is indeed a transportation of the given one on $X$: $phi: X to Y$ (the homeomorphism) has a well defined continuous inverse (I'll call it $psi: Y to X$ so that $psi circ phi = 1_X$ and $phi circ psi = 1_Y$), and we can define $$d_Y: Y times Y to mathbbR text by: d_Y(y_1,y_2) = d_X(psi(y_1), psi(y_2))$$
One easily checks the axioms for a metric for $d_Y$ and by bijectiveness and the definitions we get that $phi[B_d_X(x, r)] = B_d_Y(phi(x), r)$ for all $x in X, r>0$ etc. so that balls in the $d_X$ metric get mapped to balls in the $d_Y$-metric showing (with a little thought) that indeed open balls under $d_Y$ are open and form a base for the topology of $Y$, so that $Y$ is metrisable.
answered yesterday
Henno Brandsma
93.1k342101
add a comment |Â
up vote
11
down vote
accepted
up vote
11
down vote
accepted
If $(X, d_X)$ is a metric space then the homeomorphic $Y$ is a so-called metrisable space: a space that can be given a metric that induces its topology. A metric space comes with a pre-given metric (and has an induced topology by this metric), and is a different type of structure.
The metric is indeed a transportation of the given one on $X$: $phi: X to Y$ (the homeomorphism) has a well defined continuous inverse (I'll call it $psi: Y to X$ so that $psi circ phi = 1_X$ and $phi circ psi = 1_Y$), and we can define $$d_Y: Y times Y to mathbbR text by: d_Y(y_1,y_2) = d_X(psi(y_1), psi(y_2))$$
One easily checks the axioms for a metric for $d_Y$ and by bijectiveness and the definitions we get that $phi[B_d_X(x, r)] = B_d_Y(phi(x), r)$ for all $x in X, r>0$ etc. so that balls in the $d_X$ metric get mapped to balls in the $d_Y$-metric showing (with a little thought) that indeed open balls under $d_Y$ are open and form a base for the topology of $Y$, so that $Y$ is metrisable.
answered yesterday
Henno Brandsma
93.1k342101
If $(X, d_X)$ is a metric space then the homeomorphic $Y$ is a so-called metrisable space: a space that can be given a metric that induces its topology. A metric space comes with a pre-given metric (and has an induced topology by this metric), and is a different type of structure.
The metric is indeed a transportation of the given one on $X$: $phi: X to Y$ (the homeomorphism) has a well defined continuous inverse (I'll call it $psi: Y to X$ so that $psi circ phi = 1_X$ and $phi circ psi = 1_Y$), and we can define $$d_Y: Y times Y to mathbbR text by: d_Y(y_1,y_2) = d_X(psi(y_1), psi(y_2))$$
One easily checks the axioms for a metric for $d_Y$ and by bijectiveness and the definitions we get that $phi[B_d_X(x, r)] = B_d_Y(phi(x), r)$ for all $x in X, r>0$ etc. so that balls in the $d_X$ metric get mapped to balls in the $d_Y$-metric showing (with a little thought) that indeed open balls under $d_Y$ are open and form a base for the topology of $Y$, so that $Y$ is metrisable.
answered yesterday
Henno Brandsma
93.1k342101
answered yesterday
Henno Brandsma
93.1k342101
answered yesterday
Henno Brandsma
93.1k342101
answered yesterday
Henno Brandsma
93.1k342101
93.1k342101
add a comment |Â
add a comment |Â
up vote
6
down vote
You construction can be done regardless of $varphi$ being a homeomorphism: for any bijection you will have a metric structure on $Y$. What might be interesting to verify is if the topology induced by the metric $d' = d(varphi^-1(-),varphi^-1(-))$ coincides with the original topology on $Y$. Let's describe the open balls of $d'$: let $y in Y$ and $varepsilon > 0$ . Now,
$$
B'_varepsilon(y) := x:d'(x,y) < varepsilon = x : d(varphi^-1(x), varphi^-1(y)) < varepsilon = varphi(B_varepsilon(varphi^-1(y))).
$$
Since $B_varepsilon(varphi^-1(y))$ is a ball in $X$ and $varphi$ is homeo, $B'_varepsilon(y)$ will be open on $Y$ with the original topology. In the same fashion, if $U subseteq Y$ is open for the original topology,
$$
U = varphi(varphi^-1(U)) = varphi(bigcup_i in IB_r_i(x_i)) = bigcup_i in Ivarphi(B_r_i(x_i)) = bigcup_i in IB'_r_i(phi(x_i))
$$
and so $U$ is open for the metric topology. Here we use that $varphi^-1(U)$ is an open set of $X$ and so it is a union of open balls.
answered yesterday
Guido A.
4,644726
add a comment |Â
up vote
6
down vote
You construction can be done regardless of $varphi$ being a homeomorphism: for any bijection you will have a metric structure on $Y$. What might be interesting to verify is if the topology induced by the metric $d' = d(varphi^-1(-),varphi^-1(-))$ coincides with the original topology on $Y$. Let's describe the open balls of $d'$: let $y in Y$ and $varepsilon > 0$ . Now,
$$
B'_varepsilon(y) := x:d'(x,y) < varepsilon = x : d(varphi^-1(x), varphi^-1(y)) < varepsilon = varphi(B_varepsilon(varphi^-1(y))).
$$
Since $B_varepsilon(varphi^-1(y))$ is a ball in $X$ and $varphi$ is homeo, $B'_varepsilon(y)$ will be open on $Y$ with the original topology. In the same fashion, if $U subseteq Y$ is open for the original topology,
$$
U = varphi(varphi^-1(U)) = varphi(bigcup_i in IB_r_i(x_i)) = bigcup_i in Ivarphi(B_r_i(x_i)) = bigcup_i in IB'_r_i(phi(x_i))
$$
and so $U$ is open for the metric topology. Here we use that $varphi^-1(U)$ is an open set of $X$ and so it is a union of open balls.
answered yesterday
Guido A.
4,644726
add a comment |Â
up vote
6
down vote
up vote
6
down vote
You construction can be done regardless of $varphi$ being a homeomorphism: for any bijection you will have a metric structure on $Y$. What might be interesting to verify is if the topology induced by the metric $d' = d(varphi^-1(-),varphi^-1(-))$ coincides with the original topology on $Y$. Let's describe the open balls of $d'$: let $y in Y$ and $varepsilon > 0$ . Now,
$$
B'_varepsilon(y) := x:d'(x,y) < varepsilon = x : d(varphi^-1(x), varphi^-1(y)) < varepsilon = varphi(B_varepsilon(varphi^-1(y))).
$$
Since $B_varepsilon(varphi^-1(y))$ is a ball in $X$ and $varphi$ is homeo, $B'_varepsilon(y)$ will be open on $Y$ with the original topology. In the same fashion, if $U subseteq Y$ is open for the original topology,
$$
U = varphi(varphi^-1(U)) = varphi(bigcup_i in IB_r_i(x_i)) = bigcup_i in Ivarphi(B_r_i(x_i)) = bigcup_i in IB'_r_i(phi(x_i))
$$
and so $U$ is open for the metric topology. Here we use that $varphi^-1(U)$ is an open set of $X$ and so it is a union of open balls.
answered yesterday
Guido A.
4,644726
You construction can be done regardless of $varphi$ being a homeomorphism: for any bijection you will have a metric structure on $Y$. What might be interesting to verify is if the topology induced by the metric $d' = d(varphi^-1(-),varphi^-1(-))$ coincides with the original topology on $Y$. Let's describe the open balls of $d'$: let $y in Y$ and $varepsilon > 0$ . Now,
$$
B'_varepsilon(y) := x:d'(x,y) < varepsilon = x : d(varphi^-1(x), varphi^-1(y)) < varepsilon = varphi(B_varepsilon(varphi^-1(y))).
$$
Since $B_varepsilon(varphi^-1(y))$ is a ball in $X$ and $varphi$ is homeo, $B'_varepsilon(y)$ will be open on $Y$ with the original topology. In the same fashion, if $U subseteq Y$ is open for the original topology,
$$
U = varphi(varphi^-1(U)) = varphi(bigcup_i in IB_r_i(x_i)) = bigcup_i in Ivarphi(B_r_i(x_i)) = bigcup_i in IB'_r_i(phi(x_i))
$$
and so $U$ is open for the metric topology. Here we use that $varphi^-1(U)$ is an open set of $X$ and so it is a union of open balls.
answered yesterday
Guido A.
4,644726
answered yesterday
Guido A.
4,644726
answered yesterday
Guido A.
4,644726
answered yesterday
Guido A.
4,644726
4,644726
add a comment |Â
add a comment |Â
up vote
1
down vote
I would say is as simple as that, since $varphi$ is a homeomorphism it seems straightforward to check the metric space axioms, you may want to be carefull with the triangular inequality though, but at a glimpse, it is straightforward. Regards.
answered yesterday
Astro
264
add a comment |Â
up vote
1
down vote
I would say is as simple as that, since $varphi$ is a homeomorphism it seems straightforward to check the metric space axioms, you may want to be carefull with the triangular inequality though, but at a glimpse, it is straightforward. Regards.
answered yesterday
Astro
264
add a comment |Â
up vote
1
down vote
up vote
1
down vote
I would say is as simple as that, since $varphi$ is a homeomorphism it seems straightforward to check the metric space axioms, you may want to be carefull with the triangular inequality though, but at a glimpse, it is straightforward. Regards.
answered yesterday
Astro
264
I would say is as simple as that, since $varphi$ is a homeomorphism it seems straightforward to check the metric space axioms, you may want to be carefull with the triangular inequality though, but at a glimpse, it is straightforward. Regards.
answered yesterday
Astro
264
answered yesterday
Astro
264
answered yesterday
Astro
264
answered yesterday
Astro
264
264
add a comment |Â
add a comment |Â
up vote
0
down vote
Given any sets A and B and a bijection $varphi$ between them, any structure in one set induces an isomorhpic structure in the other; if A has a metric, or a topology, or an additive structure, etc., then a corresponding structure can be defined on B. For instance, if we define $b_1+b_2$ as being $varphi(varphi^-1(b_1)+varphi^-1(b_2))$, then $varphi$ will be a group isomorphism with respect to addition in A and addition in B.
Strictly speaking, a metric space is not a set; it's a tuple $(S,d)$ where $S$ is a set and $d$ is a distance function. So neither $X$ nor $Y$ are metric spaces, but $X$ along with its distance function is a metric space, and $Y$ along with the metric induced by $varphi$ is a metric space. The only question I see that is remotely non-trival is whether the topology in $Y$ induced by the metric in $Y$ induced by the metric in $X$ is the same as the original topology in $Y$. Since the topology in $Y$ is isomorphic to the topology in $X$, and the topology in $X$ is presumably the topology induced by its metric, the two topologies in $Y$ would indeed be the same.
answered yesterday
Acccumulation
5,7742616
add a comment |Â
up vote
0
down vote
Given any sets A and B and a bijection $varphi$ between them, any structure in one set induces an isomorhpic structure in the other; if A has a metric, or a topology, or an additive structure, etc., then a corresponding structure can be defined on B. For instance, if we define $b_1+b_2$ as being $varphi(varphi^-1(b_1)+varphi^-1(b_2))$, then $varphi$ will be a group isomorphism with respect to addition in A and addition in B.
Strictly speaking, a metric space is not a set; it's a tuple $(S,d)$ where $S$ is a set and $d$ is a distance function. So neither $X$ nor $Y$ are metric spaces, but $X$ along with its distance function is a metric space, and $Y$ along with the metric induced by $varphi$ is a metric space. The only question I see that is remotely non-trival is whether the topology in $Y$ induced by the metric in $Y$ induced by the metric in $X$ is the same as the original topology in $Y$. Since the topology in $Y$ is isomorphic to the topology in $X$, and the topology in $X$ is presumably the topology induced by its metric, the two topologies in $Y$ would indeed be the same.
answered yesterday
Acccumulation
5,7742616
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Given any sets A and B and a bijection $varphi$ between them, any structure in one set induces an isomorhpic structure in the other; if A has a metric, or a topology, or an additive structure, etc., then a corresponding structure can be defined on B. For instance, if we define $b_1+b_2$ as being $varphi(varphi^-1(b_1)+varphi^-1(b_2))$, then $varphi$ will be a group isomorphism with respect to addition in A and addition in B.
Strictly speaking, a metric space is not a set; it's a tuple $(S,d)$ where $S$ is a set and $d$ is a distance function. So neither $X$ nor $Y$ are metric spaces, but $X$ along with its distance function is a metric space, and $Y$ along with the metric induced by $varphi$ is a metric space. The only question I see that is remotely non-trival is whether the topology in $Y$ induced by the metric in $Y$ induced by the metric in $X$ is the same as the original topology in $Y$. Since the topology in $Y$ is isomorphic to the topology in $X$, and the topology in $X$ is presumably the topology induced by its metric, the two topologies in $Y$ would indeed be the same.
answered yesterday
Acccumulation
5,7742616
Given any sets A and B and a bijection $varphi$ between them, any structure in one set induces an isomorhpic structure in the other; if A has a metric, or a topology, or an additive structure, etc., then a corresponding structure can be defined on B. For instance, if we define $b_1+b_2$ as being $varphi(varphi^-1(b_1)+varphi^-1(b_2))$, then $varphi$ will be a group isomorphism with respect to addition in A and addition in B.
Strictly speaking, a metric space is not a set; it's a tuple $(S,d)$ where $S$ is a set and $d$ is a distance function. So neither $X$ nor $Y$ are metric spaces, but $X$ along with its distance function is a metric space, and $Y$ along with the metric induced by $varphi$ is a metric space. The only question I see that is remotely non-trival is whether the topology in $Y$ induced by the metric in $Y$ induced by the metric in $X$ is the same as the original topology in $Y$. Since the topology in $Y$ is isomorphic to the topology in $X$, and the topology in $X$ is presumably the topology induced by its metric, the two topologies in $Y$ would indeed be the same.
answered yesterday
Acccumulation
5,7742616
answered yesterday
Acccumulation
5,7742616
answered yesterday
Acccumulation
5,7742616
answered yesterday
Acccumulation
5,7742616
5,7742616
add a comment |Â
add a comment |Â
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3
I don't really think this question makes sense. However, a space that is homeomorphic to a metric space is called a metrisable space.
– Lord Shark the Unknown
yesterday