MHB Contraction and contractive mapping

ozkan12
Messages
145
Reaction score
0
Let $(X,d)$ be a complete metric space, and suppose that $f:X \to X$ satisfies the condition: for each $\epsilon >0$, there exists $\delta > 0$ such that for all $x,y \in X$

$$ \epsilon \le d(x,y) < \epsilon+\delta \implies d(f(x),f(y)) < \epsilon.$$Clearly, this condition implies that the mapping $f$ is contractive... also $f$ map is contraction...but I don't understand ? how this contraction, how this happened ? please help me :) thanks a lot :)
 
Last edited by a moderator:
Physics news on Phys.org
Here's the definition of a contraction mapping on a metric space:

A function $f:M \to M$ is a contraction mapping iff there exists a real number $k\in [0,1)$ such that for all $x,y \in M$, we have $d(f(x),f(y)) \le k \, d(x,y)$.

It seems to me that we need to massage the condition you were given to look like the definition of a contraction mapping. How do you think we could do that?
 
Let $(X,d)$ be a complete metric space, and suppose that $f:X\to X$ satisfies the condition: for each $\epsilon >0$, there exists $\delta > 0$ such that for all $x,y \in X$,

$$\epsilon \le d(x,y) \implies d(f(x),f(y)) < \epsilon.$$this write in theorem...and this case has a relation with contraction mapping but I don't understand...I wrote this as in the theorem
 
Last edited by a moderator:
ozkan12 said:
Let (X,d) be a complete metric space, and suppose that f:X>>>>X satisfies the condition: for each epsilon >0, there exists delta > 0 such that for all x,y in X

epsilon <=d(x,y) d(f(x),f(y))< epsilonthis write in theorem...and this case has a relation with contraction mapping but I don't understand...I wrote this as in the theorem

Hi ozkan12,

It's not clear what the statement is. Are supposing that $f : X \to X$ satisfies the property that for every $\epsilon$, there is a $\delta > 0$ such that for all $x,y \in X$, $d(x,y) \ge \mathbf{\delta}$ implies $d(f(x),f(y)) < \epsilon$?
 
no, $f$ satisfies this condition

$$\epsilon \le d(x,y) <\epsilon+\delta \implies d(f(x),f(y))< \epsilon$$

then this concept has a relation contraction map...but it how happen ? I don't understand
 
Last edited by a moderator:
We all know the definition of n-dimensional topological manifold uses open sets and homeomorphisms onto the image as open set in ##\mathbb R^n##. It should be possible to reformulate the definition of n-dimensional topological manifold using closed sets on the manifold's topology and on ##\mathbb R^n## ? I'm positive for this. Perhaps the definition of smooth manifold would be problematic, though.

Similar threads

  • · Replies 0 ·
Replies
0
Views
1K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 5 ·
Replies
5
Views
799
  • · Replies 1 ·
Replies
1
Views
2K
Replies
8
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 16 ·
Replies
16
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 11 ·
Replies
11
Views
2K
Replies
2
Views
2K