MHB Contraction and contractive mapping

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SUMMARY

The discussion centers on the concept of contraction mappings within complete metric spaces, specifically addressing 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 the mapping f is contractive, aligning with the formal definition of a contraction mapping, which requires a constant k in [0,1) such that d(f(x),f(y)) ≤ k * d(x,y) for all x, y in the space. The participants seek clarification on how the given condition relates to the established definition of contraction mappings.

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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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