Equivalent Metrics From Clopen Sets

[jamilmalik]

Homework Statement

Prove that if $(X,d)$ is a metric space and $C$ and $X \setminus C$ are nonempty clopen sets, then there is an equivalent metric $\rho$ on $X$ such that $\forall a \in C, \quad \forall b \in X \setminus C, \quad \rho(a,b) \geq 1$.

I know the term "clopen" is not a very formal definition, at least not to my knowledge, but it does describe the two properties of the given sets: they are both open and closed.

Homework Equations

The Attempt at a Solution

Would I have to show that the metric $\rho$ satisfies the properties of a metric or would I need to show that the metric $d$ and the metric $\rho$ generate the same topology to show they are equivalent?

If I define a function $\rho: X \times X \to \mathbb{R}_+$ by the following:

if both $x,y$ lie in either $C$ or $X \setminus C$, then $\rho(x,y) = d(x,y)$. Otherwise, let $\rho(x,y) = d(x,y) +100$, say. How do I proceed to show that this function satisfies the properties of a metric, that it is equivalent to $d$, and that it satisfies the desired property of $\rho(x,y) \geq 1$?

As a side thought, would using sequences be useful here, or would this be a completely different approach?

Any help with this problem is greatly appreciated as I do not know where to begin or how to proceed in writing up a correct proof.

Thanks in advance for your time and patience.

 

[WWGD]

Do you know what property follows , i.e., is equivalent to having a clopen set other than X and the empty set?

 

[jamilmalik]

Would it have something to do with being connected? My textbook states that a space $X$ is connected if there do not exist open subsets $A, B$ of $X$ such that $A \neq \emptyset, \quad B \neq \emptyset, \quad A \cap B = \emptyset, \quad A \cup B = X$. Is this equivalent? Thank you for your prompt response.

 

[WWGD]

Precisely, good going.

 

[jamilmalik]

Ok, so if $C \cup X \setminus C = X$, then this creates a separation, according to what I am reading from Fred H. Croom's Principles of Topology. How do I tie this together with metric spaces to show equivalence? Again, many thanks for your feedback.

 

[Dick]

Ok, so if $C \cup X \setminus C = X$, then this creates a separation, according to what I am reading from Fred H. Croom's Principles of Topology. How do I tie this together with metric spaces to show equivalence? Again, many thanks for your feedback.

There are two notions of equivalent metrics. http://en.wikipedia.org/wiki/Equivalence_of_metrics Which one are you using? I'm guessing you are just looking for topological equivalence.

 

[jamilmalik]

Yes, we are using topological equivalence. To be honest, I do not think I have seen strong equivalence before. Thank you for this clarification.

 

[Dick]

Yes, we are using topological equivalence. To be honest, I do not think I have seen strong equivalence before. Thank you for this clarification.

Well then, just go ahead with your idea. Show $\rho$ is a metric. That should be easy. Then show they both generate the same topology. That's where you use the 'clopen'.

 

[WWGD]

The basic properties of \rho as a metric follow from those of d(x,y). Just consider the cases for x,y,z are not all in the same component.

 

[jamilmalik]

So by construction, $\rho(x,y) = d(x,y)$ if both $x,y \in C$ or $x,y \in X \setminus C$. As for considering the cases where $x,y,z$ are not all in the same component, my book defines component of a topological space as a connected subset $C$ of $X$ which is not a proper subset of any connected subset of $X$. There is also a list of properties of the components which follow the definition, but I am unsure how to properly use this.

I get the feeling that showing the triangle inequality would look something like this:

if $\rho(x,z) \leq \rho(x,y) + \rho(y,z)$, then let $\rho(x,y) = d(x,y)+50$ and let $\rho(y,z) = d(x,y)+50$ so that we get $\rho(x,z) \leq d(x,y) +100$. Does this seem correct? Many thanks for your time and assistance.

 

[Dick]

So by construction, $\rho(x,y) = d(x,y)$ if both $x,y \in C$ or $x,y \in X \setminus C$. As for considering the cases where $x,y,z$ are not all in the same component, my book defines component of a topological space as a connected subset $C$ of $X$ which is not a proper subset of any connected subset of $X$. There is also a list of properties of the components which follow the definition, but I am unsure how to properly use this.

I get the feeling that showing the triangle inequality would look something like this:

if $\rho(x,z) \leq \rho(x,y) + \rho(y,z)$, then let $\rho(x,y) = d(x,y)+50$ and let $\rho(y,z) = d(x,y)+50$ so that we get $\rho(x,z) \leq d(x,y) +100$. Does this seem correct? Many thanks for your time and assistance.

WWGD was just using the word 'component' to refer to either $C$ or $X \setminus C$. You know $d$ is a metric so use that to show the $\rho$ you defined in first post is also a metric.

So you can start by assuming $d(x,y) \le d(x,z) + d(z,y)$ for any $x, y, z$. You want to show $\rho(x,y) \le \rho(x,z) + \rho(z,y)$. Just consider the cases. It's pretty obvious if $x$, $y$ and $z$ lie in the same set, yes? Now just consider other cases, like $x$ and $y$ lie in the same set and $z$ lies in the other. Or $x$ and $z$ are in the same set and $y$ is in the other. Is the $\rho$ inequality also true then? Just substitute for what $\rho$ is in terms of $d$. Do you still get true inequalities?