Metric Homeomorphism: Isometry Equivalence?

[facenian]

Is every homeomorphism between a metric space X and a topological Y equivalent to an isometry?
I think it is, but I need to confirm.

 

[MathematicalPhysicist]

In what sense do you mean "equivalent" there are different notions of isomorphisms?

Anyway if Y doesn't have some norm defined on it how do you want this homeomorphism to be an isometry?

An Isometry conserves the norm from one normed space to another normed space.

 

[facenian]

There is no norm neither in X nor in Y. X does have a metric, Y is a topological space with no metric. The question is whether the topology in Y is generated by isometry.
The question is a gnerenal one, normed spaces are particular cases of metric spaces defined on vector spaces.

 

[member 587159]

Is every homeomorphism between a metric space X and a topological Y equivalent to an isometry?
I think it is, but I need to confirm.

Define isometry where the codomain is an arbitrary topological space? Isn't an isometry a map that preserves distances?

 

[facenian]

Define isometry where the codomain is an arbitrary topological space? Isn't an isometry a map that preserves distances?

Yes, that is the correct concept of isometry. The codomain Y has a topolgy (NO METRIC) which is homeomorfic with X(whic has a metric topology). Now you define a metric in Y by isometry. The question is whether the NEW topology generated in Y by isometry coincides wih the previous existing one.

 

[MathematicalPhysicist]

If Y doesn't have a distance defined on it, then how can the homeomorphism conserve the distances from X to Y?

 

[MathematicalPhysicist]

I mean an isometry between two metric spaces, X and Y is one such that $d_Y(f(x),f(y))=d_X(x,y)$ where $d_Y$ is the metric on $Y$ and $d_X$ the metric on $X$.If there isn't a metric on $Y$ then how can there be an isometry between the two spaces?

 

[fresh_42]

Let me rephrase the question.

Given a homeomorphism $\varphi\, : \,X \longrightarrow Y$ where $X$ carries a metric induced topology, and $Y$ an arbitrary one.
a) Can we equip $Y$ with a metric induced by $\varphi$?
b) Will this automatically result in an isometry $\hat{\varphi}$?
c) Is there an example where $(X,d) \cong (Y,\tau)$ and $\tau$ is not given by a metric?

 

[MathematicalPhysicist]

I believe you indeed can define a metric on Y by defining: $d_Y(x,y)=d_X(\varphi^{-1}(x),\varphi^{-1}y)$.

 

[facenian]

Let me rephrase the question.

Given a homeomorphism $\varphi\, : \,X \longrightarrow Y$ where $X$ carries a metric induced topology, and $Y$ an arbitrary one.
a) Can we equip $Y$ with a metric induced by $\varphi$?
b) Will this automatically result in an isometry $\hat{\varphi}$?
c) Is there an example where $(X,d) \cong (Y,\tau)$ and $\tau$ is not given by a metric?

The answer to a) is afirmative and it is the isometry. Is this isometry the same topology as the one previously given in Y?. I believe the answer is yes but I'm not sure. In this case c) is not posible

 

[fresh_42]

The answer to a) is afirmative and it is the isometry.

Proof? I don't see how the triangle inequality follows from @MathematicalPhysicist 's definition? If we have an isometry, then $d$ and $\varphi$ commute, but not the other way around. I guess one has to define an appropriate topology on $Y$ first.

In this case c) is not posible

... which exactly has been your question. Is there an example or not? Or how does a metric and a homeomorphism relate? As mentioned above, as long as metric and homeomorphism do not commute, and I can't see why they have to, as long there is a counterexample. One could even have two different metrics on $Y$, one which extends the homeomorphism, and one which is a completely different one as e.g. the discrete metric.

 

[MathematicalPhysicist]

Proof? I don't see how the triangle inequality follows from @MathematicalPhysicist 's definition? If we have an isometry, then $d$ and $\varphi$ commute, but not the other way around. I guess one has to define an appropriate topology on $Y$ first.

... which exactly has been your question. Is there an example or not? Or how does a metric and a homeomorphism relate? As mentioned above, as long as metric and homeomorphism do not commute, and I can't see why they have to, as long there is a counterexample. One could even have two different metrics on $Y$, one which extends the homeomorphism, and one which is a completely different one as e.g. the discrete metric.

Why shouldn't the triangle inequality from my definition $d_Y(x,y)=d_X(\varphi^{-1} x , \varphi^{-1} y) \le d_X(\varphi^{-1}x,\varphi^{-1}z)+d_X(\varphi^{-1}z , \varphi^{-1}y)= d_Y(x,z)+d_Y(z,y)$, I used the triangle inequality on the metric of the space $X$.

 

[fresh_42]

Why shouldn't the triangle inequality from my definition $d_Y(x,y)=d_X(\varphi^{-1} x , \varphi^{-1} y) \le d_X(\varphi^{-1}x,\varphi^{-1}z)+d_X(\varphi^{-1}z , \varphi^{-1}y)= d_Y(x,z)+d_Y(z,y)$, I used the triangle inequality on the metric of the space $X$.

So the answers to a) and b) is yes, we can copy the topology isometrically from $X$ on $Y$ by a given bijection.

Thus we have sufficiency. Is it also necessary? I think not, that is it remains to give a counterexample for c), i.e. homeomorphic spaces with incompatible metrics, or one without a metric at all.

 

[wrobel]

I believe you indeed can define a metric on Y by defining: $d_Y(x,y)=d_X(\varphi^{-1}(x),\varphi^{-1}y)$.

I believe that this prompt closes all the questions. This metric generates a topology in $Y$that is equivalent to $\tau$

 

[WWGD]

Is every homeomorphism between a metric space X and a topological Y equivalent to an isometry?
I think it is, but I need to confirm.

I think the case of Arctan between an interval and the Real line is a standard counter. There is no mention/need of choice of isometry between the two, so you can use any metric in the interval, and not all metrics are equivalent. But, yes, if there is an isomorphism, you may pull-back the metric as mentioned in other posts, to give you an isometry. EDIT: Basically, we use in e.g., (-1,1) , the subspace metric, under which (-1,1) is bounded, and similar for the Real line with the standard metric is unbounded.EDIT2: Notice too, that with the subspace metric d(x,y):=|x-y|, that (-1,1) is not complete, e.g., take $a_n =1+1/n$ , while the Reals are. Notice too, that unless you include the condition of a bijection, an isometry is not necessarily a homeomorphism.

 

[WWGD]

I just thought of a much simpler example: take f(x)=2x ( f(x)= kx , $k \neq 1$ will do ).; $f: \mathbb R \rightarrow \mathbb R$, both with standard metric) Then (x,y) with d(x,y)=|x-y| is sent to (2x,2y) with d(2x,2y)= 2|x-y|.