Definition of a function by a metric.

In summary, the author is trying to prove that a specific homeomorphism between two sets is possible, but is stuck on the second step.
  • #1
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Let (X,p) be a metric space with p metric on X, define for each subset A of X
[tex]p(x,A)=inf_{y\in A}p(x,y)[/tex]
prove that: f_A(x)=p(x,A) f:X->R is continuous.
basically, if U is open in R, we need to show that f^-1_A(U) is open in X, i.e that
{x in X|p(x,A) in U} is open in X, now because U is open in R, then there's an open interval or half open interval contained in U that it contains p(x,A), i.e p(x,A) is bounded, but from this to show that f^-1(U) is open I'm kind of in a stuck.

any hints?

thanks in advance.
 
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  • #2
It's probably easier to use the good old epsilon-delta definition of continuity here. The following observation will be helpful: p(x,A) <= p(x,z) + p(z,A), for any z in X.
 
  • #3
Morphism is right. You have to do what he suggested twice, in order to get the absolute value on the left side. This is assuming that you've already learned that the epsilon-delta definition of continuity applies to metric spaces in general (and it does, calculus clases specialize in R^n, but it can be proven that it applies to metric spaces in general).
 
  • #4
ok, I have another question, let [tex]S^n[/tex]={[tex]x\in R^n|||x||=1[/tex]}
let x_0=(1,0,...,0) (n entries), prove that S^n-{x_0} is homeomorphic to R^(n+1).
well I thought of dividing by the norm, i.e defining a homeomorphism such as this:
x=(x1,...,x_n+1) in R^n+1 f(x)=(x2,...,x_n-1,1)/||(x2,...,1)|| when (x1,...,xn)=(1,0,...,0)
when (x1,...,x_n)!=(1,0,...,0) then f(x)=(1,x2,...,x_n-1)/||(1,x2,...,x_n-1)||
I know that somewhere I need to divide by the norm in order to keep it that the norm of the vector is 1, but don'y know how, anyone has some hints on this?

thanks in advance.
 
  • #5
hint: http://en.wikipedia.org/wiki/Stereographic_projection" .
btw, the set you defined as S^n is more usually written as S^{n-1}, and you need to show that removing a point makes it homeomorphic to R^{n-1}. It can't be homeomorphic to R^{n+1}, as they have different dimensions.
 
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  • #6
so basically in order to prove it, all you need is to use g(x)=x/||x|| and the fact that S^n-1 is bijective to S^n-1\{x0}, so there's a bijection f between them then gof is the appropiate function.
 

Related to Definition of a function by a metric.

1. What is a function?

A function is a mathematical concept that describes the relationship between two sets of numbers or variables. It maps each element in one set (the input) to a unique element in another set (the output).

2. What is a metric?

A metric is a mathematical tool used to measure distance or similarity between objects or points in a set. It assigns a numerical value to the distance between two points, which can then be used to compare different objects or points.

3. How is a function defined by a metric?

A function can be defined by a metric by using the metric to measure the distance between the input and output values of the function. This means that the function will produce output values that are in some way related to the input values based on the metric used.

4. What is the importance of defining a function by a metric?

Defining a function by a metric allows us to quantitatively measure the relationship between the input and output values of the function. This can help us understand the behavior of the function and make predictions about its output based on its input.

5. Can different metrics be used to define the same function?

Yes, different metrics can be used to define the same function. This is because a function is defined by its input-output relationship, not the specific metric used to measure that relationship. However, different metrics may produce different results and have different implications for the behavior of the function.

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