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facenian
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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 it is, but I need to confirm.
Last edited:
facenian said: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.
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.Math_QED said:Define isometry where the codomain is an arbitrary topological space? Isn't an isometry a map that preserves distances?
fresh_42 said: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?
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.facenian said:The answer to a) is afirmative and it is the isometry.
... 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.facenian said:In this case c) is not posible
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 said: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.
So the answers to a) and b) is yes, we can copy the topology isometrically from ##X## on ##Y## by a given bijection.MathematicalPhysicist said: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##.
I believe that this prompt closes all the questions. This metric generates a topology in ##Y##that is equivalent to ##\tau##MathematicalPhysicist said: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 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.facenian said: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.
A metric homeomorphism is a mathematical concept that refers to the relationship between two metric spaces. It is a type of function that maps points from one metric space to another, while preserving the distance between the points. In other words, a metric homeomorphism is a function that maintains the same metric structure between two spaces.
Metric homeomorphism and isometry are often used interchangeably, but there is a slight difference between the two. While both concepts refer to preserving the distance between points in a metric space, isometry implies that the function also preserves other geometric properties such as angles and shapes.
Metric homeomorphism has various applications in real life, particularly in fields such as physics, engineering, and computer science. For example, it can be used to model physical systems, analyze data, and design algorithms.
Yes, metric homeomorphism can exist between different types of metric spaces as long as they have the same underlying metric structure. This means that the distance between points can be preserved even if the spaces have different shapes or dimensions.
Metric homeomorphism and topological equivalence are similar concepts that both deal with preserving the structure of mathematical spaces. However, metric homeomorphism focuses on preserving the distance between points, while topological equivalence is concerned with preserving the open sets and continuity of functions between spaces.