MHB Contraction and contractive mapping

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In a complete metric space (X,d), a function f:X to X is discussed under the condition that for every ε > 0, there exists a δ > 0 such that if ε ≤ d(x,y) < ε + δ, then d(f(x),f(y)) < ε. This condition implies that f behaves like a contraction mapping, which is defined as a function where a constant k in [0,1) exists such that d(f(x),f(y)) ≤ k * d(x,y) for all x,y in M. The discussion revolves around clarifying how the given condition relates to the standard definition of contraction mappings. Understanding this relationship is crucial for grasping the concept of contraction in metric spaces. The participants seek to reconcile the provided condition with the formal definition of a contraction mapping.
ozkan12
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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 :)
 
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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
 
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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
 
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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.

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